If the acid in question is not on this list, you may assume it is weak for this class.
If the base in question is not on this list, you may assume it is weak for this class.
Ions with more than one atom. (You should know the name, formula, and charge!)
Charge | Name | Formula |
---|---|---|
-1 | Acetate | C_{2}H_{3}O_{2}^{-} |
-1 | Chlorate | ClO_{3}^{-} |
-1 | Chlorite | ClO_{2}^{-} |
-1 | Nitrate | NO_{3}^{-} |
-1 | Nitrite | NO_{2}^{-} |
-1 | Cyanide | CN^{-} |
-1 | Hydroxide | OH^{-} |
+1 | Ammonium | NH_{4}^{+} |
-2 | Chromate | CrO_{4}^{2-} |
-2 | Carbonate | CO_{3}^{2-} |
-2 | Sulfate | SO_{4}^{2-} |
-2 | Sulfite | SO_{3}^{2-} |
-3 | phosphate | PO_{4}^{3-} |
-3 | phosphite | PO_{3}^{3-} |
Contains these Ions | Soluble | Exceptions |
---|---|---|
Li^{+}, Na^{+}, K^{+}, Rb^{+}, Cs^{+}, NH_{4}^{+} | Most salts (ionic compounds) containing alkali metals (Group 1) and ammonium are soluble. | none |
NO_{3}^{-} | Most salts containing nitrate are soluble | |
Cl^{-}, Br^{-}, I^{-} | Most salts containing halogen anions (Main Group 7) are soluble. (Chloride, bromide, and iodide salts) | Salts containing Cu^{+}, Ag^{+}, Pb^{2+}, and Hg_{2}^{2+} |
SO_{4}^{2-} | Many salts containing sulfate are soluble. | Salts containing Sr^{2+}, Ba^{2+}, Pb^{2+}, and Hg_{2}^{2+} |
Contains these Ions | Insoluble | Exceptions |
---|---|---|
OH^{-} and S^{2-} | Most salts containing hydroxide or sulfide anions are insoluble. | Salts containing alkali metals (Li^{+}, Na^{+}, K^{+}, Rb^{+}, and Cs^{+}), and the “heavy” alkaline earth metals (Ca^{2+}, Sr^{2+}, and Ba^{2+}) |
CO_{3}^{2-}, PO_{4}^{3-} | Most salts containing carbonate or phosphate anions are insoluble. | Salts containing alkali metals (Group 1) and ammonium are soluble. (Li^{+}, Na^{+}, K^{+}, Rb^{+}, Cs^{+}, NH_{4}^{+}) |
SO_{4}^{2-} | Many salts containing sulfate and cations with charges greater than or equal to 2 are insoluble, including salts containing Sr^{2+}, Ba^{2+}, Pb^{2+}, and Hg_{2}^{2+}. | Salts containing Cu^{+}, Ag^{+}, Pb^{2+}, and Hg_{2}^{2+} |
SO_{4}^{2-} | Many salts containing sulfate are soluble. | Salts containing alkali metals (Group 1) and ammonium are soluble. (Li^{+}, Na^{+}, K^{+}, Rb^{+}, Cs^{+}, NH_{4}^{+}), and Mg^{2+} |
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]]>Table 1. Elemental Composition of Earth | ||||||
---|---|---|---|---|---|---|
Element | Symbol | Percent Mass | Element | Symbol | Percent Mass | |
oxygen | O | 49.20 | chlorine | Cl | 0.19 | |
silicon | Si | 25.67 | phosphorus | P | 0.11 | |
aluminum | Al | 7.50 | manganese | Mn | 0.09 | |
iron | Fe | 4.71 | carbon | C | 0.08 | |
calcium | Ca | 3.39 | sulfur | S | 0.06 | |
sodium | Na | 2.63 | barium | Ba | 0.04 | |
potassium | K | 2.40 | nitrogen | N | 0.03 | |
magnesium | Mg | 1.93 | fluorine | F | 0.03 | |
hydrogen | H | 0.87 | strontium | Sr | 0.02 | |
titanium | Ti | 0.58 | all others | - | 0.47 |
The National Fire Protection Agency (NFPA) 704 Hazard Identification System was developed by NFPA to provide safety information about certain substances. The system details flammability, reactivity, health, and other hazards. Within the overall diamond symbol, the top (red) diamond specifies the level of fire hazard (temperature range for flash point). The blue (left) diamond indicates the level of health hazard. The yellow (right) diamond describes reactivity hazards, such as how readily the substance will undergo detonation or a violent chemical change. The white (bottom) diamond points out special hazards, such as if it is an oxidizer (which allows the substance to burn in the absence of air/oxygen), undergoes an unusual or dangerous reaction with water, is corrosive, acidic, alkaline, a biological hazard, radioactive, and so on. Each hazard is rated on a scale from 0 to 4, with 0 being no hazard and 4 being extremely hazardous.
While many elements differ dramatically in their chemical and physical properties, some elements have similar properties. We can identify sets of elements that exhibit common behaviors. For example, many elements conduct heat and electricity well, whereas others are poor conductors. These properties can be used to sort the elements into three classes: metals (elements that conduct well), nonmetals (elements that conduct poorly), and metalloids (elements that have properties of both metals and nonmetals).Table 1. Base Units of the SI System | ||
---|---|---|
Property Measured | Name of Unit | Symbol of Unit |
length | meter | m |
mass | kilogram | kg |
time | second | s |
temperature | kelvin | K |
electric current | ampere | A |
amount of substance | mole | mol |
luminous intensity | candela | cd |
Table 2. Common Unit Prefixes | |||
---|---|---|---|
Prefix | Symbol | Factor | Example |
femto | f | 10^{−15} | 1 femtosecond (fs) = 1 × 10^{−15} m (0.000000000001 m) |
pico | p | 10^{−12} | 1 picometer (pm) = 1 × 10^{−12} m (0.000000000001 m) |
nano | n | 10^{−9} | 4 nanograms (ng) = 4 × 10^{−9} g (0.000000004 g) |
micro | µ | 10^{−6} | 1 microliter (μL) = 1 × 10^{−6} L (0.000001 L) |
milli | m | 10^{−3} | 2 millimoles (mmol) = 2 × 10^{−3} mol (0.002 mol) |
centi | c | 10^{−2} | 7 centimeters (cm) = 7 × 10^{−2} m (0.07 m) |
deci | d | 10^{−1} | 1 deciliter (dL) = 1 × 10^{−1} L (0.1 L ) |
kilo | k | 10^{3} | 1 kilometer (km) = 1 × 10^{3} m (1000 m) |
mega | M | 10^{6} | 3 megahertz (MHz) = 3 × 10^{6} Hz (3,000,000 Hz) |
giga | G | 10^{9} | 8 gigayears (Gyr) = 8 × 10^{9} yr (8,000,000,000 Gyr) |
tera | T | 10^{12} | 5 terawatts (TW) = 5 × 10^{12} W (5,000,000,000,000 W) |
Table 3. Densities of Common Substances | ||
---|---|---|
Solids | Liquids | Gases (at 25 °C and 1 atm) |
ice (at 0 °C) 0.92 g/cm^{3} | water 1.0 g/cm^{3} | dry air 1.20 g/L |
oak (wood) 0.60–0.90 g/cm^{3} | ethanol 0.79 g/cm^{3} | oxygen 1.31 g/L |
iron 7.9 g/cm^{3} | acetone 0.79 g/cm^{3} | nitrogen 1.14 g/L |
copper 9.0 g/cm^{3} | glycerin 1.26 g/cm^{3} | carbon dioxide 1.80 g/L |
lead 11.3 g/cm^{3} | olive oil 0.92 g/cm^{3} | helium 0.16 g/L |
silver 10.5 g/cm^{3} | gasoline 0.70–0.77 g/cm^{3} | neon 0.83 g/L |
gold 19.3 g/cm^{3} | mercury 13.6 g/cm^{3} | radon 9.1 g/L |
[latex]\text{density}=\frac{\text{mass}}{\text{volume}}[/latex]
[latex]\text{density}=\frac{\text{mass}}{\text{volume}}=\frac{\text{5.00 kg}}{\text{1.25 L}}=4.00 kg/L[/latex]
Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:[latex]\text{density}=\frac{\text{mass}}{\text{volume}}=\frac{\text{5.00 kg}}{10.00 L}=\text{0.500 kg/L}[/latex]
[/hidden-answer][reveal-answer q="998865"]Show Answer[/reveal-answer] [hidden-answer a="998865"]
[latex]\text{volume}=\text{22.4 mL}-\text{13.5 mL}=\text{8.9 mL}={\text{8.9 cm}}^{3}[/latex]
(rounded to the nearest 0.1 mL, per the rule for addition and subtraction) The density is the mass-to-volume ratio:[latex]\text{density}=\frac{\text{mass}}{\text{volume}}=\frac{\text{69.658 g}}{{\text{8.9 cm}}^{3}}={\text{7.8 g/cm}}^{3}[/latex]
(rounded to two significant figures, per the rule for multiplication and division) The density of iron is 7.9 g/cm^{3}, very close to that of rebar, which lends some support to the fact that rebar is mostly iron. [/hidden-answer]Table 1. Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers | ||
---|---|---|
Dispenser #1 | Dispenser #2 | Dispenser #3 |
283.3 | 298.3 | 296.1 |
284.1 | 294.2 | 295.9 |
283.9 | 296.0 | 296.1 |
284.0 | 297.8 | 296.0 |
284.1 | 293.9 | 296.1 |
[latex]\text{speed}=\frac{\text{distance}}{\text{time}}[/latex]
An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of [latex]\frac{\text{100 m}}{\text{10 s}}=\text{10 m/s}[/latex]. Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:[latex]\text{time}=\frac{\text{distance}}{\text{speed}}[/latex]
The time can then be computed as [latex]\frac{\text{25 m}}{\text{10 m/s}}=\text{2.5 s}[/latex]. Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.” These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.[latex]\frac{\text{2.54 cm}}{\text{1 in.}}\text{(2.54 cm}=\text{1 in.) or 2.54}\frac{\text{cm}}{\text{in.}}[/latex]
Several other commonly used conversion factors are given in Table 1.Table 1. Common Conversion Factors | ||
---|---|---|
Length | Volume | Mass |
1 m = 1.0936 yd | 1 L = 1.0567 qt | 1 kg = 2.2046 lb |
1 in. = 2.54 cm (exact) | 1 qt = 0.94635 L | 1 lb = 453.59 g |
1 km = 0.62137 mi | 1 ft^{3} = 28.317 L | 1 (avoirdupois) oz = 28.349 g |
1 mi = 1609.3 m | 1 tbsp = 14.787 mL | 1 (troy) oz = 31.103 g |
[latex]\text{34 in.}\times \frac{\text{2.54 cm}}{1\cancel{\text{in.}}}=\text{86 cm}[/latex]
Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield [latex]\frac{\text{in.}\times \text{cm}}{\text{in.}}[/latex] . Just as for numbers, a ratio of identical units is also numerically equal to one, [latex]\frac{\text{in.}}{\text{in.}}=\text{1,}[/latex] and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.[latex]x\text{ oz}=\text{125 g}\times \text{unit conversion factor}[/latex]
We write the unit conversion factor in its two forms:[latex]\frac{\text{1 oz}}{\text{28.35 g}}\text{ and }\frac{\text{28.349 g}}{\text{1 oz}}[/latex]
The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.[latex]\begin{array}{ccc}x\text{ oz}\hfill & \text{=}\hfill & 125\cancel{\text{g}}\times \frac{\text{1 oz}}{\text{28.349}\cancel{\text{g}}}\hfill \\ \hfill & =\hfill & \left(\frac{125}{\text{28.349}}\right)\text{oz}\hfill \\ \hfill & =\hfill & \text{4.41 oz (three significant figures)}\hfill \end{array}[/latex]
[/hidden-answer][latex]\text{9.26}\cancel{\text{lb}}\times \frac{\text{453.59 g}}{1\cancel{\text{lb}}}=4.20\times {10}^{3}\text{g}[/latex]
We need to use two steps to convert volume from quarts to milliliters.[latex]\frac{\text{9.26}\cancel{\text{lb}}}{\text{4.00}\cancel{\text{qt}}}\times \frac{\text{453.59 g}}{1\cancel{\text{lb}}}\times \frac{\text{1.0567}\cancel{\text{qt}}}{1\cancel{\text{L}}}\times \frac{1\cancel{\text{L}}}{\text{1000 mL}}=\text{1.11 g/mL}[/latex]
[/hidden-answer][latex]\text{(average) mileage}=\frac{\text{777 mi}}{\text{56.3 gal}}=\text{13.8 miles/gallon}=\text{13.8 mpg}[/latex]
Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:[latex]\frac{1250\cancel{\text{km}}}{213\cancel{\text{L}}}\times \frac{\text{0.62137 mi}}{1\cancel{\text{km}}}\times \frac{1\cancel{\text{L}}}{\text{1.0567}\cancel{\text{qt}}}\times \frac{4\cancel{\text{qt}}}{\text{1 gal}}=\text{13.8 mpg}[/latex]
[latex]\text{length in feet}=\left(\frac{\text{1 ft}}{\text{12 in.}}\right)\times \text{length in inches}[/latex]
where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (b). The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as x and the Fahrenheit temperature as y, the slope, m, is computed to be:[latex]\displaystyle{m}=\frac{\Delta y}{\Delta x}=\frac{212^{\circ}\text{ F}-32^{\circ}\text{ F}}{100^{\circ}\text{ C}-0^{\circ}\text{ C}}=\frac{180^{\circ}\text{ F}}{100^{\circ}\text{ C}}=\frac{9^{\circ}\text{ F}}{5^{\circ}\text{ C}}[/latex]
The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:[latex]b=y-mx=32^{\circ}\text{ F}-\frac{9 ^{\circ}\text{ F}}{5 ^{\circ}\text{ C}}\times{0}^{\circ}\text{ C}=32^{\circ}\text{ F}[/latex]
The equation relating the temperature scales is then:[latex]{T}_{\text{ }^{\circ}\text{F}}=\left(\frac{9^{\circ}\text{ F}}{5^{\circ}\text{ C}}\times{T}_{\text{ }^{\circ}\text{C}}\right)+32 ^{\circ}\text{ C}[/latex]
An abbreviated form of this equation that omits the measurement units is:[latex]{T}_{\text{ }^{\circ}\text{F}}=\frac{9}{5}\times{T}_{^{\circ}\text{ C}}+32[/latex]
Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:[latex]{T}_{\text{ }^{\circ}\text{C}}=\frac{5}{9}\left({T}_{^{\circ}\text{ F}}-32\right)[/latex]
As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases). The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of [latex]1\frac{\text{K}}{^{\circ}\text{C}}[/latex] . Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:[latex]{T}_{\text{K}}={T}_{^{\circ}\text{C}}+\text{273.15}[/latex]
[latex]{T}_{\text{ }^{\circ}\text{C}}={T}_{\text{K}}-\text{273.15}[/latex]
The 273.15 in these equations has been determined experimentally, so it is not exact. Figure 1 shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale. Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.[latex]\text{K}=\text{ }^{\circ}\text{C}+273.15=37.0+273.2=\text{310.2 K}[/latex]
[latex]\text{ }^{\circ}\text{F}=\frac{9}{5}^{\circ}\text{ C}+32.0=\left(\frac{9}{5}\times 37.0\right)+32.0=66.6+32.0=98.6^{\circ}\text{ F}[/latex]
[/hidden-answer][latex]\text{ }^{\circ}\text{C}=\frac{5}{9}(^{\circ}\text{ F}-\text{32)}=\frac{5}{9}\left(450-32\right)=\frac{5}{9}\times 418=232^{\circ}\text{ C}\rightarrow\text{set oven to 230}^{\circ}\text{ C}\left(\text{two significant figures}\right)[/latex]
[latex]\text{K}=\text{ }^{\circ}\text{ C}+273.15=230+273=\text{503 K}\rightarrow 5.0\times {10}^{2}\text{K}\left(\text{two significant figures}\right)[/latex]
[/hidden-answer]Trial | Mass |
1 | 1.201 ± 0.001 |
2 | 1.202 ± 0.001 |
3 | 1.200 ± 0.001 |
[reveal-answer q="777641"]Show Answer[/reveal-answer] [hidden-answer a="777641"]The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)[/hidden-answer]
Table 1. Constant Composition of Isooctane | |||
---|---|---|---|
Sample | Carbon | Hydrogen | Mass Ratio |
A | 14.82 g | 2.78 g | [latex]\displaystyle\frac{\text{14.82 g carbon}}{\text{2.78 g hydrogen}}=\frac{\text{5.33 g carbon}}{\text{1.00 g hydrogen}}[/latex] |
B | 22.33 g | 4.19 g | [latex]\displaystyle\frac{\text{22.33 g carbon}}{\text{4.19 g hydrogen}}=\frac{\text{5.33 g carbon}}{\text{1.00 g hydrogen}}[/latex] |
C | 19.40 g | 3.64 g | [latex]\displaystyle\frac{\text{19.40 g carbon}}{\text{3.63 g hydrogen}}=\frac{\text{5.33 g carbon}}{\text{1.00 g hydrogen}}[/latex] |
[latex]\displaystyle\frac{\frac{\text{1.116 g Cl}}{\text{1 g Cu}}}{\frac{\text{0.558 g Cl}}{\text{1 g Cu}}}=\frac{2}{1}[/latex]
This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound. This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure 4).[latex]\displaystyle\frac{\frac{\text{14.13 g C}}{\text{2.96 g H}}}{\frac{\text{19.91 g C}}{\text{3.34 g H}}}=\frac{\text{4.77 g C/g H}}{\text{5.96 g C/g H}}=0.800=\frac{4}{5}[/latex]
This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds. [/hidden-answer]Compound | Description | Mass of Carbon | Mass of Hydrogen |
---|---|---|---|
X | clear, colorless, liquid with strong odor | 1.776 g | 0.148 g |
Y | clear, colorless, liquid with strong odor | 1.974 g | 0.329 g |
Z | clear, colorless, liquid with strong odor | 7.812 g | 0.651 g |
[latex]\text{Mass of electron}=1.602\times {10}^{-19}\text{C}\times\frac{1\text{kg}}{1.759\times {10}^{11}\text{C}}=9.107\times {10}^{-31}\text{kg}[/latex]
Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure 3). [caption id="" align="aligncenter" width="878"] Figure 3. (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with embedded raisins (“plums”). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of electrons surrounding a positive “planet.” (credit a: modification of work by “Man vyi”/Wikimedia Commons; credit b: modification of work by “NASA”/Wikimedia Commons)[/caption] The next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle. What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure 4). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” [caption id="" align="aligncenter" width="880"] Figure 4. Geiger and Rutherford fired α particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected.[/caption] Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:Table 1. Properties of Subatomic Particles | |||||
---|---|---|---|---|---|
Name | Location | Charge (C) | Unit Charge | Mass (amu) | Mass (g) |
electron | outside nucleus | −1.602 × 10^{−19} | 1− | 0.00055 | 0.00091 × 10^{−24} |
proton | nucleus | 1.602 × 10^{−19} | 1+ | 1.00727 | 1.67262 × 10^{−24} |
neutron | nucleus | 0 | 0 | 1.00866 | 1.67493 × 10^{−24} |
[latex]\begin{array}{ccc}\hfill \text{atomic number}\left(\text{Z}\right)& =& \text{number of protons}\hfill \\ \hfill \text{atomic mass}\left(\text{A}\right)& =& \text{number of protons}+\text{number of neutrons}\hfill \\ \hfill \text{A}-\text{Z}& =& \text{number of neutrons}\hfill \end{array}[/latex]
Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:Atomic charge = number of protons − number of electrons
As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+). A neutral oxygen atom (Z = 8) has eight electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).Table 2. Some Common Elements and Their Symbols | |||
---|---|---|---|
Element | Symbol | Element | Symbol |
aluminum | Al | iron | Fe (from ferrum) |
bromine | Br | lead | Pb (from plumbum) |
calcium | Ca | magnesium | Mg |
carbon | C | mercury | Hg (from hydrargyrum) |
chlorine | Cl | nitrogen | N |
chromium | Cr | oxygen | O |
cobalt | Co | potassium | K (from kalium) |
copper | Cu (from cuprum) | silicon | Si |
fluorine | F | silver | Ag (from argentum) |
gold | Au (from aurum) | sodium | Na (from natrium) |
helium | He | sulfur | S |
hydrogen | H | tin | Sn (from stannum) |
iodine | I | zinc | Zn |
Table 3. Nuclear Compositions of Atoms of the Very Light Elements | ||||||
---|---|---|---|---|---|---|
Element | Symbol | Atomic Number | Number of Protons | Number of Neutrons | Mass (amu) | % Natural Abundance |
hydrogen | [latex]{}_{1}^{1}\text{H}[/latex] (protium) | 1 | 1 | 0 | 1.0078 | 99.989 |
[latex]{}_{1}^{2}\text{H}[/latex] (deuterium) | 1 | 1 | 1 | 2.0141 | 0.0115 | |
[latex]{}_{1}^{3}\text{H}[/latex] (tritium) | 1 | 1 | 2 | 3.01605 | — (trace) | |
helium | [latex]{}_{2}^{3}\text{He}[/latex] | 2 | 2 | 1 | 3.01603 | 0.00013 |
[latex]{}_{2}^{4}\text{He}[/latex] | 2 | 2 | 2 | 4.0026 | 100 | |
lithium | [latex]{}_{3}^{6}\text{Li}[/latex] | 3 | 3 | 3 | 6.0151 | 7.59 |
[latex]{}_{3}^{7}\text{Li}[/latex] | 3 | 3 | 4 | 7.0160 | 92.41 | |
beryllium | [latex]{}_{4}^{9}\text{Be}[/latex] | 4 | 4 | 5 | 9.0122 | 100 |
boron | [latex]{}_{5}^{10}\text{B}[/latex] | 5 | 5 | 5 | 10.0129 | 19.9 |
[latex]{}_{5}^{11}\text{B}[/latex] | 5 | 5 | 6 | 11.0093 | 80.1 | |
carbon | [latex]{}_{6}^{12}\text{C}[/latex] | 6 | 6 | 6 | 12.0000 | 98.89 |
[latex]{}_{6}^{13}\text{C}[/latex] | 6 | 6 | 7 | 13.0034 | 1.11 | |
[latex]{}_{6}^{14}\text{C}[/latex] | 6 | 6 | 8 | 14.0032 | — (trace) | |
nitrogen | [latex]{}_{7}^{14}\text{N}[/latex] | 7 | 7 | 7 | 14.0031 | 99.63 |
[latex]{}_{7}^{15}\text{N}[/latex] | 7 | 7 | 8 | 15.0001 | 0.37 | |
oxygen | [latex]{}_{8}^{16}\text{O}[/latex] | 8 | 8 | 8 | 15.9949 | 99.757 |
[latex]{}_{8}^{17}\text{O}[/latex] | 8 | 8 | 9 | 16.9991 | 0.038 | |
[latex]{}_{8}^{18}\text{O}[/latex] | 8 | 8 | 10 | 17.9992 | 0.205 | |
fluorine | [latex]{}_{9}^{19}\text{F}[/latex] | 9 | 9 | 10 | 18.9984 | 100 |
neon | [latex]{}_{10}^{20}\text{Ne}[/latex] | 10 | 10 | 10 | 19.9924 | 90.48 |
[latex]{}_{10}^{21}\text{Ne}[/latex] | 10 | 10 | 11 | 20.9938 | 0.27 | |
[latex]{}_{10}^{22}\text{Ne}[/latex] | 10 | 10 | 12 | 21.9914 | 9.25 |
[latex]\text{average mass}=\sum _{i}{\left(\text{fractional abundance}\times \text{isotopic mass}\right)}_{i}[/latex]
For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are ^{10}B with a mass of 10.0129 amu, and the remaining 80.1% are ^{11}B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:[latex]\begin{array}{cc}\hfill \text{boron average mass}& =\left(0.199\times \text{10.0129 amu}\right)+\left(0.801\times \text{11.0093 amu}\right)\hfill \\ & =\text{1.99 amu}+\text{8.82 amu}\hfill \\ & =\text{10.81 amu}\hfill \end{array}[/latex]
It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.[latex]\begin{array}{cc}\hfill \text{average mass}& =\left(0.9184\times \text{19.9924 amu}\right)+\left(0.0047\times \text{20.9940 amu}\right)+\left(0.0769\times \text{21.9914 amu}\right)\hfill \\ & =\left(18.36+0.099+1.69\right)\text{amu}\hfill \\ & =\text{20.15 amu}\hfill \end{array}[/latex]
The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.) [/hidden-answer][latex]\text{average mass}=\left(\text{fraction of}{}_{}^{35}\text{Cl}\times \text{mass of}{}_{}^{35}\text{Cl}\right)+\left(\text{fraction of}{}_{}^{35}\text{Cl}\times \text{mass of}{}_{}^{35}\text{Cl}\right)[/latex]
If we let x represent the fraction that is ^{35}Cl, then the fraction that is ^{37}Cl is represented by 1.00 - x. (The fraction that is ^{35}Cl + the fraction that is ^{37}Cl must add up to 1, so the fraction of ^{37}Cl must equal 1.00 - the fraction of ^{35}Cl.) Substituting this into the average mass equation, we have:[latex]\begin{array}{ccc}\hfill \text{35.453 amu}& =& \left(x\times \text{34.96885 amu}\right)+\left[\left(1.00-x\right)\times \text{36.96590 amu}\right]\hfill \\ \hfill 35.453& =& 34.96885x+36.96590-36.96590x\hfill \\ \hfill 1.99705x& =& 1.513\hfill \\ \hfill x& =& \frac{1.513}{1.99705}=0.7576\hfill \end{array}[/latex]
So solving yields: x = 0.7576, which means that 1.00 - 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% ^{35}Cl and 24.24% ^{37}Cl. [/hidden-answer]Table 1. Common Polyatomic Ions | ||||||
---|---|---|---|---|---|---|
Charge | Name | Formula | Charge | Name | Formula | |
1+ | ammonium | [latex]{\text{NH}}_{4}{}^{+}[/latex] | 1− | permanganate | [latex]{\text{MnO}}_{4}{}^{-}[/latex] | |
1− | acetate | [latex]{\text{C}}_{2}{\text{H}}_{3}{\text{O}}_{2}{}^{-}[/latex] | 1− | hydrogen carbonate, or bicarbonate | [latex]{\text{HCO}}_{3}{}^{-}[/latex] | |
1− | cyanide | [latex]\text{CN}^-[/latex] | 2− | carbonate | [latex]{\text{CO}}_{3}{}^{2-}[/latex] | |
1− | hydroxide | [latex]\text{OH}^-[/latex] | 2− | peroxide | [latex]{\text{O}}_{2}{}^{2-}[/latex] | |
1− | nitrate | [latex]{\text{NO}}_{3}{}^{-}[/latex] | 1− | hydrogen sulfate, or bisulfate | [latex]{\text{HSO}}_{4}{}^{-}[/latex] | |
1− | nitrite | [latex]{\text{NO}}_{2}{}^{-}[/latex] | 2− | sulfate | [latex]{\text{SO}}_{4}{}^{2-}[/latex] | |
1− | perchlorate | [latex]{\text{ClO}}_{4}{}^{-}[/latex] | 2− | sulfite | [latex]{\text{SO}}_{3}{}^{2-}[/latex] | |
1− | chlorate | [latex]{\text{ClO}}_{3}{}^{-}[/latex] | 1− | dihydrogen phosphate | [latex]{\text{H}}_{2}{\text{PO}}_{4}{}^{-}[/latex] | |
1− | chlorite | [latex]{\text{ClO}}_{2}{}^{-}[/latex] | 2− | hydrogen phosphate | [latex]{\text{HPO}}_{4}{}^{2-}[/latex] | |
1− | hypochlorite | [latex]\text{ClO}^-[/latex] | 3− | phosphate | [latex]{\text{PO}}_{4}{}^{3-}[/latex] |
Table 1. Names of Some Ionic Compounds | |
---|---|
NaCl, sodium chloride | Na_{2}O, sodium oxide |
KBr, potassium bromide | CdS, cadmium sulfide |
CaI_{2}, calcium iodide | Mg_{3}N_{2}, magnesium nitride |
CsF, cesium fluoride | Ca_{3}P_{2}, calcium phosphide |
LiCl, lithium chloride | Al_{4}C_{3}, aluminum carbide |
Table 2. Names of Some Polyatomic Ionic Compounds | ||
---|---|---|
KC_{2}H_{3}O_{2}, potassium acetate | (NH_{4})Cl, ammonium chloride | |
NaHCO_{3}, sodium bicarbonate | CaSO_{4}, calcium sulfate | |
Al_{2}(CO_{3})_{3}, aluminum carbonate | Mg_{3}(PO_{4})_{2}, magnesium phosphate |
Table 3. Everyday Ionic Compounds | |
---|---|
Ionic Compound | Use |
NaCl, sodium chloride | ordinary table salt |
KI, potassium iodide | added to “iodized” salt for thyroid health |
NaF, sodium fluoride | ingredient in toothpaste |
NaHCO_{3}, sodium bicarbonate | baking soda; used in cooking (and as antacid) |
Na_{2}CO_{3}, sodium carbonate | washing soda; used in cleaning agents |
NaOCl, sodium hypochlorite | active ingredient in household bleach |
CaCO_{3} calcium carbonate | ingredient in antacids |
Mg(OH)_{2}, magnesium hydroxide | ingredient in antacids |
Al(OH)_{3}, aluminum hydroxide | ingredient in antacids |
NaOH, sodium hydroxide | lye; used as drain cleaner |
K_{3}PO_{4}, potassium phosphate | food additive (many purposes) |
MgSO_{4}, magnesium sulfate | added to purified water |
Na_{2}HPO_{4}, sodium hydrogen phosphate | anti-caking agent; used in powdered products |
Na_{2}SO_{3}, sodium sulfite | preservative |
Table 4. Names of Some Transition Metal Ionic Compounds | |
---|---|
Transition Metal Ionic Compound | Name |
FeCl_{3} | iron(II) chloride |
Hg_{2}O | mercury(I) oxide |
HgO | mercury(II) oxide |
Cu_{3}(PO_{4})_{2} | copper(II) phosphate |
Table 5. Nomenclature Prefixes | ||||
---|---|---|---|---|
Number | Prefix | Number | Prefix | |
1 (sometimes omitted) | mono- | 6 | hexa- | |
2 | di- | 7 | hepta- | |
3 | tri- | 8 | octa- | |
4 | tetra- | 9 | nona- | |
5 | penta- | 10 | deca- |
Table 6. Names of Some Molecular Compounds Composed of Two Elements | ||||
---|---|---|---|---|
Compound | Name | Compound | Name | |
SO_{2} | sulfur dioxide | BCl_{3} | boron trichloride | |
SO_{3} | sulfur trioxide | SF_{6} | sulfur hexafluoride | |
NO_{2} | nitrogen dioxide | PF_{5} | phosphorus pentafluoride | |
N_{2}O_{4} | dinitrogen tetroxide | P_{4}O_{10} | tetraphosphorus decaoxide | |
N_{2}O_{5} | dinitrogen pentoxide | IF_{7} | iodine heptafluoride |
Table 7. Names of Some Simple Acids | |
---|---|
Name of Gas | Name of Acid |
HF(g), hydrogen fluoride | HF(aq), hydrofluoric acid |
HCl(g), hydrogen chloride | HCl(aq), hydrochloric acid |
HBr(g), hydrogen bromide | HBr(aq), hydrobromic acid |
HI(g), hydrogen iodide | HI(aq), hydroiodic acid |
H_{2}S(g), hydrogen sulfide | H_{2}S(aq), hydrosulfuric acid |
Table 8. Names of Common Oxyacids | ||
---|---|---|
Formula | Anion Name | Acid Name |
HC_{2}H_{3}O_{2} | acetate | acetic acid |
HNO_{3} | nitrate | nitric acid |
HNO_{2} | nitrite | nitrous acid |
HClO_{4} | perchlorate | perchloric acid |
H_{2}CO_{3} | carbonate | carbonic acid |
H_{2}SO_{4} | sulfate | sulfuric acid |
H_{2}SO_{3} | sulfite | sulfurous acid |
H_{3}PO_{4} | phosphate | phosphoric acid |
n = 4 | –0.1361 × 10^{–18} J |
n = 3 | –0.2420 × 10^{–18} J |
n = 2 | –0.5445 × 10^{–18} J |
n = 1 | –2.178 × 10^{–18} J |
Element | Average Atomic Mass (amu) | Molar Mass (g/mol) | Atoms/Mole |
---|---|---|---|
C | 12.01 | 12.01 | [latex]6.022\times {10}^{23}[/latex] |
H | 1.008 | 1.008 | [latex]6.022\times {10}^{23}[/latex] |
O | 16.00 | 16.00 | [latex]6.022\times {10}^{23}[/latex] |
Na | 22.99 | 22.99 | [latex]6.022\times {10}^{23}[/latex] |
Cl | 33.45 | 33.45 | [latex]6.022\times {10}^{23}[/latex] |
[latex]4.7\cancel{\text{g}}\text{K}\left(\frac{\text{mol K}}{39.10\cancel{\text{g}}}\right)=0.12\text{mol K}[/latex]
The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol. [/hidden-answer][latex]9.2\times {10}^{-4}\cancel{\text{mol}}\text{Ar}\left(\frac{39.95\text{g}}{\cancel{\text{mol}}\text{Ar}}\right)=0.037\text{g Ar}[/latex]
The result is in agreement with our expectations as noted above, around 0.04 g Ar. [/hidden-answer][latex]5.00\cancel{\text{g}}\text{Cu}\left(\frac{\cancel{\text{mol}}\text{Cu}}{63.55\cancel{\text{g}}}\right)\left(\frac{6.022\times {10}^{23}\text{atoms}}{\cancel{\text{mol}}}\right)=4.74\times {10}^{22}\text{atoms of copper}[/latex]
The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10^{22} as expected. [/hidden-answer][latex]28.35\cancel{\text{g}}\text{glycine}\left(\frac{\text{mol glycine}}{75.07\cancel{\text{g}}}\right)=0.378\text{ mol glycine}[/latex]
This result is consistent with our rough estimate. [/hidden-answer][latex]1.42\times {10}^{-4}\cancel{\text{mol}}\text{vitamin C}\left(\frac{176.124\text{g}}{\cancel{\text{mol}}\text{vitamin C}}\right)=0.0250\text{g vitamin C}[/latex]
This is consistent with the anticipated result. [/hidden-answer][latex]0.0400\cancel{\text{g}}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}\left(\frac{\cancel{\text{mol}}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}{183.18\cancel{\text{g}}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}\right)\left(\frac{6.022\times {10}^{23}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S molecules}}{1\cancel{\text{mol}}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}\right)=1.31\times {10}^{20}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S molecules}[/latex]
The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:[latex]1.31\times {10}^{20}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S molecules}\left(\frac{7\text{C atoms}}{1{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S molecule}}\right)=9.20\times {10}^{21}\text{C atoms}[/latex]
[/hidden-answer][latex]\frac{3104\cancel{\text{carats}}\times \frac{200\cancel{\text{mg}}}{1\cancel{\text{carat}}}\times \frac{1\cancel{\text{g}}}{1000\cancel{\text{mg}}}}{12.011\cancel{\text{g}}\cancel{{\text{mol}}^{-1}}\left(6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\right)}=3.113\times {10}^{25}\text{C atoms}[/latex]
29. Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g/mol; Then [latex]0.0278\text{mol}\times 342.300\text{g/mol}=9.52\text{g sugar.}[/latex] This 9.52 g of sugar represents [latex]\frac{11.0}{60.0}[/latex] of one serving or[latex]\frac{60.0\text{g serving}}{11.0\cancel{\text{g sugar}}}\times 9.52\cancel{\text{g sugar}}=51.9\text{g cereal.}[/latex]
This amount is [latex]\frac{51.9\text{g cereal}}{60.0\text{g serving}}=0.865[/latex] servings, or about 1 serving. 31. Calculate the number of moles of each species, then remember that 1 mole of anything [latex]=6.022\times {10}^{23}[/latex] species.[latex]\%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%[/latex]
[latex]\%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%[/latex]
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:[latex]\%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%[/latex]
[latex]\%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%[/latex]
[latex]\begin{array}{c}\\ \%\text{C}=\frac{7.34\text{g C}}{12.04\text{g compound}}\times 100\%=61.0\%\\ \%\text{H}=\frac{1.85\text{g H}}{12.04\text{g compound}}\times 100\%=15.4\%\\ \%\text{N}=\frac{2.85\text{g N}}{12.04\text{g compound}}\times 100\%=23.7\%\end{array}[/latex]
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass. [/hidden-answer][latex]\begin{array}{c}\\ \%\text{N}=\frac{14.01\text{amu N}}{17.03\text{amu}{\text{NH}}_{3}}\times 100\%=82.27\%\\ \%\text{H}=\frac{3.024\text{amu N}}{17.03\text{amu}{\text{NH}}_{3}}\times 100\%=17.76\%\end{array}[/latex]
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated in the example problem below. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.[latex]\%\text{C}=\frac{9\text{ mol C}\times \text{ molar mass C}}{\text{molar mass }{\text{C}}_{9}{\text{H}}_{18}{\text{O}}_{4}}\times 100=\frac{9\times 12.01\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{108.09\text{g/mol}}{180.159\text{g/mol}}\times 100[/latex]
%C =60.00% C
[latex]\%\text{H}=\frac{8\text{ mol H }\times \text{ molar mass H}}{\text{molar mass }{\text{C}}_{9}{\text{H}}_{18}{\text{O}}_{4}}\times 100=\frac{8\times 1.008\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{8.064\text{g/mol}}{180.159\text{g/mol}}\times 100[/latex] %H = 4.476% H [latex]\%\text{O}=\frac{4\text{mol O}\times \text{molar mass O}}{\text{molar mass }{\text{C}}_{9}{\text{H}}_{18}{\text{O}}_{4}}\times 100=\frac{4\times 16.00\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{64.00\text{g/mol}}{180.159\text{g/mol}}\times 100[/latex] %O = 35.52% O Note that these percentages sum to equal 100.00% when appropriately rounded. [/hidden-answer][latex]\begin{array}{l}1.17\text{g C}\times \frac{1\text{mol C}}{12.01\text{g C}}=0.142\text{mol C}\\ 0.287\text{g H}\times \frac{1\text{mol H}}{1.008\text{g H}}=0.284\text{mol H}\end{array}[/latex]
Thus, we can accurately represent this compound with the formula C_{0.142}H_{0.248}. Of course, per accepted convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:[latex]\displaystyle{\text{C}}_{\frac{0.142}{0.142}}{\text{H}}_{\frac{0.248}{0.142}}\text{ or }{\text{CH}}_{2}[/latex]
(Recall that subscripts of “1” are not written but rather assumed if no other number is present.) The empirical formula for this compound is thus CH_{2}. This may or not be the compound’s molecular formula as well; however, we would need additional information to make that determination (as discussed later in this section). Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:[latex]{\text{C1}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\frac{0.150}{0.150}}{\text{O}}_{\frac{0.525}{0.150}}={\text{ClO}}_{3.5}[/latex]
In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl_{2}O_{7} as the final empirical formula. In summary, empirical formulas are derived from experimentally measured element masses by:[latex]\begin{array}{lll}\\ 34.97\text{g Fe}\left(\frac{\text{mol Fe}}{55.85\text{g}}\right)& =& 0.6261\text{mol Fe}\hfill \\ 15.03\text{g O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& =& 0.9394\text{mol O}\hfill \end{array}[/latex]
Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:[latex]\begin{array}{l}\frac{0.6261}{0.6261}=1.000\text{mol Fe}\\ \frac{0.0394}{0.6261}=1.500\text{mol O}\end{array}[/latex]
The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe_{1}O_{1.5}). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:2(Fe_{1}O_{1.5}) = Fe_{2}O_{3}
The empirical formula is Fe_{2}O_{3}. [/hidden-answer][latex]\begin{array}{lll}27.29\%\text{C}& =\hfill & \frac{27.29\text{g C}}{100\text{g compound}}\hfill \\ 72.71\%\text{O}& =\hfill & \frac{72.71\text{g O}}{100\text{g compound}}\hfill \end{array}[/latex]
The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:[latex]\begin{array}{lll}27.29\text{g C}\left(\frac{\text{mol C}}{12.01\text{g}}\right)& =\hfill & 2.272\text{mol C}\hfill \\ 72.71\text{g O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& =\hfill & 4.544\text{mol O}\hfill \end{array}[/latex]
Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:[latex]\begin{array}{l}\frac{2.272\text{ mol C}}{2.272}=1\\ \frac{4.544\text{ mol O}}{2.272}=2\end{array}[/latex]
Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO_{2}. [/hidden-answer][latex]\displaystyle\frac{\text{molecular or molar mass}\left(\text{amu or}\frac{\text{g}}{\text{mol}}\right)}{\text{empirical formula mass}\left(\text{amu or}\frac{\text{g}}{\text{mol}}\right)}=n\text{ formula units/molecule}[/latex]
The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown below for the generic empirical formula A_{x}B_{y}:[latex]{{\text{(A}}_{\text{x}}{\text{B}}_{\text{y}}\text{)}}_{\text{n}}={\text{A}}_{\text{nx}}{\text{B}}_{\text{nx}}[/latex]
For example, consider a covalent compound whose empirical formula is determined to be CH_{2}O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:[latex]\displaystyle\frac{180\text{amu/molecule}}{30\frac{\text{amu}}{\text{formula unit}}}=6\text{ formula units/molecule}[/latex]
Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula: [latex]{\text{(CH}}_{\text{2}}{\text{O)}}_{\text{6}}={\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}[/latex] Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.[latex]\begin{array}{lll}\\ \left(74.02\text{g C}\right)\left(\frac{1\text{mol C}}{12.01\text{g C}}\right)& =\hfill & 6.163\text{mol C}\hfill \\ \left(8.710\text{g H}\right)\left(\frac{1\text{mol H}}{1.01\text{g H}}\right)& =\hfill & 8.624\text{mol H}\hfill \\ \left(17.27\text{g N}\right)\left(\frac{1\text{mol N}}{14.01\text{g N}}\right)& =\hfill & 1.233\text{mol N}\hfill \end{array}[/latex]
Next, we calculate the molar ratios of these elements. The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C_{5}H_{7}N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit. We calculate the molar mass for nicotine from the given mass and molar amount of compound:[latex]\frac{40.57\text{g nicotine}}{0.2500\text{mol nicotine}}=\frac{162.3\text{g}}{\text{mol}}[/latex]
Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:[latex]\displaystyle\frac{162.3\text{g/mol}}{81.13\frac{\text{g}}{\text{ formula unit}}}=2\text{formula units/molecule}[/latex]
Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two: [latex]{{\text{(C}}_{\text{5}}{\text{H}}_{\text{7}}\text{N)}}_{\text{6}}={\text{C}}_{\text{10}}{\text{H}}_{\text{14}}{\text{N}}_{\text{2}}[/latex] [/hidden-answer][latex]\begin{array}{l}\text{C:}\frac{92.3\text{g}}{12.011\text{g}{\text{mol}}^{-1}}=7.68\text{mol}\\ \text{H:}\frac{7.7\text{g}}{1.00794\text{g}{\text{mol}}^{-1}}=7.6\text{mol}\end{array}[/latex]
This operation established the relative ration of carbon to hydrogen in the formula. Step 2: To establish a whole-number ratio of carbon to hydrogen, divide each factor by the smallest factor. In this case, both factors are essentially equal; thus the ration of atoms is 1 to 1:[latex]\begin{array}{l}\text{C:}\frac{7.68}{7.6}=1\\ \text{H:}\frac{7.6}{7.6}=1\end{array}[/latex]
The empirical formula is CH. Since the molecular mass of the compound is 78.1 amu, some integer times the sum of the mass of 1C and 1H in atomic mass units (12.011 amu + 1.00794 amu = 13.019 amu) must be equal to 78.1 amu. To find this number, divide 78.1 amu by 13.019 amu:[latex]\frac{78.1\text{amu}}{13.019\text{amu}}=5.9989\rightarrow 6[/latex]
The molecular formula is (CH)_{6} = C_{6}H_{6}. 7. The formulas can be found as follows:[latex]\begin{array}{ccc}\left(28.03\text{g Mg}\right)\left(\frac{1\text{mol Mg}}{24.30\text{g}}\right)=1.153\text{mol Mg}& & \frac{1.153}{0.769}=1.512\text{mol Mg}\\\left(21.60\text{g Si}\right)\left(\frac{1\text{mol Si}}{28.09\text{g Si}}\right)=0.769\text{mol Si}& &\frac{0.769}{0.769}=1.00\text{mol Si}\\\left(1.16\text{g H}\right)\left(\frac{1\text{mol H}}{1.01\text{g H}}\right)=1.149\text{mol H}& & \frac{1.149}{0.769}=1.49\text{mol H}\\\left(49.21\text{g O}\right)\left(\frac{1\text{mol O}}{16.00\text{g O}}\right)=3.076\text{mol O}& & \frac{3.076}{0.769}=4.00\text{mol O}\end{array}[/latex]
(2)(Mg_{1.5}Si_{1}H_{1.5}O_{4}) = Mg_{3}Si_{2}H_{3}O_{8} (empirical formula), empirical mass of 260.1 g/unit [latex]\frac{\text{MM}}{\text{EM}}=\frac{520.8}{260.1}=2.00,[/latex] so (2)(Mg_{3}Si_{2}H_{3}O_{8}) = Mg_{6}Si_{4}H_{6}O_{16} 8. Assume 100.0 g; the percentages of the elements are then the same as their mass in grams. Divide each mass by the molar mass to find the number of moles. Step 1: [latex]\begin{array}{l}\\ \frac{75.95\cancel{\text{g}}}{12.011\cancel{\text{g}}{\text{mol}}^{-1}}=6.323\text{mol C}\\ \frac{17.72\cancel{\text{g}}}{14.0067\cancel{\text{g}}{\text{mol}}^{-1}}=1.265\text{mol N}\\ \frac{6.33\cancel{\text{g}}}{1.00794\cancel{\text{g}}{\text{mol}}^{-1}}=6.28\text{mol H}\end{array}[/latex] Step 2: Divide each by the smallest number. The answers are 5C, 1N, and 5H. The empirical formula is C_{5}H_{5}N, which has a molar mass of 79.10 g/mol. To find the actual molecular formula, divide 240, the molar mass of the compound, by 79.10 to obtain 3. So the formula is three times the empirical formula, or C_{15}H_{15}N_{3}. [/hidden-answer][latex]\displaystyle{M}=\frac{\text{mol solute}}{\text{L solution}}[/latex]
[latex]M=\frac{\text{mol solute}}{\text{L solution}}=\frac{0.133\text{mol}}{355\text{mL}\times \frac{1\text{L}}{1000\text{mL}}}=0.375M[/latex]
[/hidden-answer][latex]\begin{array}{rcl}\\ M&=&\frac{\text{mol solute}}{\text{L solution}}\\ \text{mol solute}&=&M\times \text{L solution}\\ \text{mol solute}&=&0.375\frac{\text{mol sugar}}{\text{L}}\times \left(10\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\right)=0.004\text{mol sugar}\end{array}[/latex]
[/hidden-answer][latex]M=\frac{\text{mol solute}}{\text{L solution}}=\frac{25.2 g{\text{CH}}_{2}{\text{CO}}_{2}\text{H}\times \frac{1{\text{mol CH}}_{2}{\text{CO}}_{2}\text{H}}{{\text{60.052 g CH}}_{2}{\text{CO}}_{2}\text{H}}}{\text{0.500 L solution}}=0.839M[/latex]
[latex]\begin{array}{l}\\ M=\frac{\text{mol solute}}{\text{L solution}}=0.839M\\ M=\frac{0.839\text{mol solute}}{1.00\text{L solution}}\end{array}[/latex]
[/hidden-answer][latex]\begin{array}{rcl}\\ M&=&\frac{\text{mol solute}}{\text{L solution}}\\ \text{mol solute}&=&M\times \text{L solution}\\ \text{mol solute}&=&5.30\frac{\text{mol NaCl}}{\text{L}}\times 0.250\text{L}=1.325\text{mol NaCl}\end{array}[/latex]
Finally, this molar amount is used to derive the mass of NaCl:[latex]\text{1.325 mol NaCl}\times \frac{58.44\text{g NaCl}}{\text{mol NaCl}}=77.4\text{g NaCl}[/latex]
[/hidden-answer][latex]\text{mol solute}\times \frac{\text{L solution}}{\text{mol solute}}=\text{L solution}[/latex]
Combining these two steps into one yields:[latex]\text{g solute}\times \frac{\text{mol solute}}{\text{g solute}}\times \frac{\text{L solution}}{\text{mol solute}}=\text{L solution}[/latex]
[latex]75.6\text{g}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(\frac{\text{mol}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}}{60.05\text{g}}\right)\left(\frac{\text{L solution}}{0.839\text{mol}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}}\right)=1.50\text{L solution}[/latex]
[/hidden-answer][latex]{n}_{1}={M}_{1}{L}_{1}[/latex] [latex]{n}_{2}={M}_{2}{L}_{2}[/latex]
where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, n_{1} = n_{2}. Thus, these two equations may be set equal to one another:[latex]{M}_{1}{L}_{1}={M}_{2}{L}_{2}[/latex]
This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:[latex]{C}_{1}{V}_{1}={C}_{2}{V}_{2}[/latex]
where C and V are concentration and volume, respectively.[latex]\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\{C}_{2}=\frac{{C}_{1}{V}_{1}}{{V}_{2}}\end{array}[/latex]
Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:[latex]{C}_{2}=\frac{0.850\text{L}\times 5.00\frac{\text{mol}}{\text{L}}}{1.80 L}=2.36M[/latex]
This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M). [/hidden-answer][latex]\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\{V}_{2}=\frac{{C}_{1}{V}_{1}}{{C}_{2}}\end{array}[/latex]
Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original concentration, or around 44 mL. Substituting the given values and solving for the unknown volume yields:[latex]\begin{array}{c}\\ {V}_{2}=\frac{\left(0.45M\right)\left(0.011\text{L}\right)}{\left(0.12M\right)}\\ {V}_{2}=0.041\text{L}\end{array}[/latex]
The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate. [/hidden-answer][latex]\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\ {V}_{1}=\frac{{C}_{2}{V}_{2}}{{C}_{1}}\end{array}[/latex]
Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:[latex]\begin{array}{c}{V}_{1}=\frac{\left(0.100M\right)\left(5.00\text{L}\right)}{1.59M}\\ {V}_{1}=0.314\text{L}\end{array}[/latex]
Thus, we would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our rough estimate. [/hidden-answer][latex]M=\frac{\text{mol}}{\text{liter}}\,\,\,\,\text{or}\,\,\,\,\text{mol}=M\times \text{liter}[/latex]
[latex]\text{mol HCl}=\frac{434.4\cancel{g}}{36.4606\cancel{g}{\text{mol}}^{-1}}=11.91\text{mol}[/latex]
This HCl is present in 1.00 L, so the molarity is 11.9 M. 24. [latex]57\text{g}{\text{K}}_{2}{\text{SO}}_{4}\times \frac{1\text{mol}}{174.26\text{g}}\times \frac{1\text{L}}{0.20\text{mol}}=1.6\text{L}[/latex] [/hidden-answer][latex]\text{mass percentage}=\frac{\text{mass of component}}{\text{mass of solution}}\times 100\%[/latex]
We are generally most interested in the mass percentages of solutes, but it is also possible to compute the mass percentage of solvent. Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section). Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 1) cites the concentration of its active ingredient, sodium hypochlorite (NaOCl), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl. [caption id="" align="aligncenter" width="649"] Figure 1. Liquid bleach is an aqueous solution of sodium hypochlorite (NaOCl). This brand has a concentration of 7.4% NaOCl by mass.[/caption][latex]\%\text{glucose}=\frac{3.75\text{mg glucose}\times \frac{1\text{g}}{1000\text{mg}}}{5.0\text{g spinal fluid}}=0.075\%[/latex]
The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%). Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, we converted the units of solute in the numerator from mg to g to match the units in the denominator. We could just as easily have converted the denominator from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct. [/hidden-answer][latex]\text{500 mL solution}\left(\frac{1.19\text{g solution}}{\text{mL solution}}\right)\left(\frac{37.2\text{g HCl}}{100\text{g solution}}\right)=221\text{g HCl}[/latex]
This mass of HCl is consistent with our rough estimate of approximately 200 g. [/hidden-answer][latex]\text{volume percentage}=\frac{\text{volume solute}}{\text{volume solution}}\times 100\%[/latex]
[latex]\left(355\text{ mL solution}\right)\left(\frac{70\text{ mL isopropyl alcohol}}{100\text{ mL solution}}\right)\left(\frac{0.785\text{ g isopropyl alcohol}}{1\text{ mL isopropyl alcohol}}\right)=195\text{ g isopropyl alchol}[/latex]
[/hidden-answer][latex]\text{ppm}=\frac{\text{mass solute}}{\text{mass solution}}\times {10}^{6}\text{ppm}[/latex]
[latex]\text{ppb}=\frac{\text{mass solute}}{\text{mass solution}}\times {10}^{9}\text{ppb}[/latex]
Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure 3). [caption id="" align="alignnone" width="880"] Figure 3. (a) In some areas, trace-level concentrations of contaminants can render unfiltered tap water unsafe for drinking and cooking. (b) Inline water filters reduce the concentration of solutes in tap water. (credit a: modification of work by Jenn Durfey; credit b: modification of work by “vastateparkstaff”/Wikimedia commons)[/caption][latex]15\text{ppb}\times \frac{1\text{ppm}}{{10}^{3}\text{ppb}}=0.015\text{ppm}[/latex]
The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:[latex]\begin{array}{lll}\\ \text{ppb}=\frac{\text{mass solute}}{\text{mass solution}}\times {10}^{9}\text{ppb}\\ \\ \text{mass solute}=\frac{\text{ppb}\times \text{mass solution}}{{10}^{9}\text{ppb}}\end{array}[/latex]
[latex]\text{mass solute}=\frac{15\text{ ppb}\times 300\text{ mL}\times \frac{1.00\text{ g}}{\text{mL}}}{{10}^{9}\text{ ppb}}=4.5\times {10}^{-6}\text{ g}[/latex]
Finally, convert this mass to the requested unit of micrograms:[latex]4.5\times {10}^{-6}\text{g}\times \frac{1\mu\text{g}}{{10}^{-6}\text{g}}=4.5\mu\text{g}[/latex]
[/hidden-answer][latex]M{\text{Ca}}^{2+}=\frac{\text{mol}{\text{CaCO}}_{3}}{\text{L}}=\frac{175\cancel{\text{mg}}\times \left(\frac{1\text{mol}}{100.0792\cancel{\text{g}}}\right)\times \left(\frac{1\cancel{\text{g}}}{1000\text{mg}}\right)}{1\text{L}}=1.75\times {10}^{-3}M[/latex]
7. 1 mg/dL = 0.01 g/L and 1 L = 10 dL[latex]5.3\cancel{\text{mmol}}\text{/L}\times 180.158\text{mg/}\cancel{\text{mmol}}=9.5\times {10}^{2}\text{mg/L}[/latex] [latex]9.5\times {10}^{2}\text{mg/L}\times \frac{1\text{L}}{10\text{dL}}=95\text{mg/dL}[/latex]
9. 0.0100% of 454 g is [latex]\left(0.000100\times 454\text{g}\right)=0.0454\text{g;}[/latex] Molar mass of CuI = 63.546 + 126.90447 = 190.450 g/mol [latex]\text{mol CuI}=\frac{0.0454\text{g}}{190.450\text{g}{\text{mol}}^{-1}}=0.000238\text{mol}=2.38\times {10}^{-4}\text{mol}[/latex] 11. The molar mass of C_{6}H_{12}O_{6} is [latex]6\times 12.011+12\times 1.00794+6\times 15.9994=180.2\text{g/mol.}[/latex] In 1.000 L, there are:[latex]\begin{array}{l}\left(1000\cancel{\text{mL}}\times 1.029\text{g}{\cancel{\text{mL}}}^{-1}\right)=1029\text{g}\\ \text{mol dextrose}=1029\cancel{\text{g}}\times 0.050\times \frac{1\text{mol}}{180.2\cancel{\text{g}}}=0.29\text{mol}\end{array}[/latex] C_{6}H_{12}O_{6}.
Since we selected the volume to be 1.00 L, the molarity of dextrose is [latex]\text{molarity}=\frac{\text{mol}}{\text{L}}=\frac{0.29\text{mol}}{1.00\text{L}}=0.29\text{mol}[/latex]. [/hidden-answer][latex]\left(1{\text{CO}}_{2}\text{ molecule }\times \frac{\text{2 O atoms}}{{\text{CO}}_{2}\text{ molecule }}\right)+\left(2{\text{H}}_{2}\text{O molecule }\times \frac{\text{1 O atom}}{{\text{H}}_{2}\text{O molecule }}\right)=\text{4 O atoms}[/latex]
The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:[latex]{\text{CH}}_{4}+2{\text{O}}_{2}\rightarrow{\text{CO}}_{2}+2{\text{H}}_{2}\text{O}[/latex]
Element | Reactants | Products | Balanced? |
---|---|---|---|
C | 1 × 1 = 1 | 1 × 1 = 1 | 1 = 1, yes |
H | 4 × 1 = 4 | 2 × 2 = 4 | 4 = 4, yes |
O | 2 × 2 = 4 | (1 × 2) + (2 × 1) = 4 | 4 = 4, yes |
[latex]{\text{H}}_{2}\text{O}\rightarrow{\text{H}}_{2}+{\text{O}}_{2}\text{(unbalanced)}[/latex]
Comparing the number of H and O atoms on either side of this equation confirms its imbalance:Element | Reactants | Products | Balanced? |
---|---|---|---|
H | 1 × 2 = 2 | 1 × 2 = 2 | 2 = 2, yes |
O | 1 × 1 = 1 | 1 × 2 = 2 | 1 ≠ 2, no |
[latex]\mathbf{2}\text{H}_{2}\text{O}\rightarrow{\text{H}}_{2}+{\text{O}}_{2}\text{(unbalanced)}[/latex]
Element | Reactants | Products | Balanced? |
---|---|---|---|
H | 2 × 2 = 4 | 1 × 2 = 2 | 4 ≠ 2, no |
O | 2 × 1 = 2 | 1 × 2 = 2 | 2 = 2, yes |
[latex]2{\text{H}}_{2}\text{O}\rightarrow\mathbf{2}{\text{H}}_{2}+{\text{O}}_{2}\text{(balanced)}[/latex]
Element | Reactants | Products | Balanced? |
---|---|---|---|
H | 2 × 2 = 4 | 2 × 2 = 2 | 4 = 4, yes |
O | 2 × 1 = 2 | 1 × 2 = 2 | 2 = 2, yes |
[latex]2{\text{H}}_{2}\text{O}\rightarrow 2{\text{H}}_{2}+{\text{O}}_{2}[/latex]
[latex]{\text{N}}_{2}+{\text{O}}_{2}\rightarrow{\text{N}}_{2}{\text{O}}_{5}\text{(unbalanced)}[/latex]
Next, count the number of each type of atom present in the unbalanced equation.Element | Reactants | Products | Balanced? |
---|---|---|---|
N | 1 × 2 = 2 | 1 × 2 = 2 | 2 = 2, yes |
O | 1 × 2 = 2 | 1 × 5 = 5 | 2 ≠ 5, no |
[latex]{\text{N}}_{2}+\mathbf{5}{\text{O}}_{2}\rightarrow\mathbf{2}{\text{N}}_{2}{\text{O}}_{5}\text{(unbalanced)}[/latex]
Element | Reactants | Products | Balanced? |
---|---|---|---|
N | 1 × 2 = 2 | 2 × 2 = 4 | 2 ≠ 4, no |
O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |
[latex]2{\text{N}}_{2}+5{\text{O}}_{2}\rightarrow 2{\text{N}}_{2}{\text{O}}_{5}[/latex]
Element | Reactants | Products | Balanced? |
---|---|---|---|
N | 2 × 2 = 4 | 2 × 2 = 4 | 4 = 4, yes |
O | 5 × 2 = 10 | 2 × 5 = 10 | 10 = 10, yes |
[latex]{\text{C}}_{2}{\text{H}}_{6}+{\text{O}}_{2}\rightarrow{\text{H}}_{2}\text{O}+{\text{CO}}_{2}\text{(unbalanced)}[/latex]
Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:[latex]{\text{C}}_{2}{\text{H}}_{6}+{\text{O}}_{2}\rightarrow 3{\text{H}}_{2}\text{O}+2{\text{CO}}_{2}\text{(unbalanced)}[/latex]
This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O_{2} reactant to yield an odd number, so a fractional coefficient, [latex]\displaystyle\frac{7}{2}[/latex] , is used instead to yield a provisional balanced equation:[latex]{\text{C}}_{2}{\text{H}}_{6}+\frac{7}{2}{\text{O}}_{2}\rightarrow 3{\text{H}}_{2}\text{O}+2{\text{CO}}_{2}[/latex]
A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:[latex]2{\text{C}}_{2}{\text{H}}_{6}+7{\text{O}}_{2}\rightarrow 6{\text{H}}_{2}\text{O}+4{\text{CO}}_{2}[/latex]
Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,[latex]3{\text{N}}_{2}+9{\text{H}}_{2}\rightarrow 6{\text{NH}}_{3}[/latex]
the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:[latex]{\text{N}}_{2}+3{\text{H}}_{2}\rightarrow 2{\text{NH}}_{3}[/latex]
[latex]2\text{Na(}s\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2\text{NaOH(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}[/latex]
This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water). Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow.[latex]{\text{CaCO}}_{3}\text{(}s\text{)}\stackrel{\Delta}{\rightarrow}\text{CaO(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}[/latex]
Other examples of these special conditions will be encountered in more depth in later chapters.[latex]{\text{CaCl}}_{2}\text{(}aq\text{)}+2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Ca}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+2\text{AgCl(}s\text{)}[/latex]
This balanced equation, derived in the usual fashion, is called a molecular equation, because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:[latex]\begin{array}{l}{\text{CaCl}}_{2}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}\\ 2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow 2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\\ \text{Ca}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\end{array}[/latex]
Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s. Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:[latex]{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}+2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}+2\text{AgCl(}s\text{)}[/latex]
Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca^{2+}(aq) and [latex]{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}[/latex]. These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:[latex]\begin{array}{c}\cancel{{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}}+2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}+\cancel{2{\text{NO}}_{3}{}^{\text{-}}\text{(}aq\text{)}}\rightarrow\cancel{{\text{Ca}}^{\text{2+}}\text{(}aq\text{)}}+\cancel{2{\text{NO}}_{3}{}^{-}\text{(}aq\text{)}}+2\text{AgCl(}s\text{)}\\ 2{\text{Cl}}^{-}\text{(}aq\text{)}+2{\text{Ag}}^{\text{+}}\text{(}aq\text{)}\rightarrow 2\text{AgCl(}s\text{)}\end{array}[/latex]
Following the convention of using the smallest possible integers as coefficients, this equation is then written:[latex]{\text{Cl}}^{\text{-}}\text{(}aq\text{)}+{\text{Ag}}^{+}\text{(}aq\text{)}\rightarrow\text{AgCl(}s\text{)}[/latex]
This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl^{–} and Ag^{+}.[latex]{\text{CO}}_{2}\text{(}aq\text{)}+\text{NaOH(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\text{(unbalanced)}[/latex]
Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:[latex]{\text{CO}}_{2}\text{(}aq\text{)}+2\text{NaOH(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}[/latex]
The two dissolved ionic compounds, NaOH and Na_{2}CO_{3}, can be represented as dissociated ions to yield the complete ionic equation:[latex]{\text{CO}}_{2}\text{(}aq\text{)}+2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}[/latex]
Finally, identify the spectator ion(s), in this case Na^{+}(aq), and remove it from each side of the equation to generate the net ionic equation:[latex]\begin{array}{l}{\text{CO}}_{2}\text{(}aq\text{)}+\cancel{2{\text{Na}}^{\text{+}}\text{(}aq\text{)}}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow 2\cancel{{\text{Na}}^{\text{+}}\text{(}aq\text{)}}+{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\\ {\text{CO}}_{2}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}\rightarrow{\text{CO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\end{array}[/latex]
[/hidden-answer][latex]\text{NaCl(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\,\,\,{\xrightarrow{\text{electricity}}}\,\,\,\text{NaOH(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}[/latex]
Write balanced molecular, complete ionic, and net ionic equations for this process. [reveal-answer q="519484"]Show Answer[/reveal-answer] [hidden-answer a="519484"] [latex]\begin{array}{l}2\text{NaCl(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2\text{NaOH}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(}\text{molecular}\text{)}\\ 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{Cl}}^{\text{-}}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}aq\text{)}+2{\text{OH}}^{\text{-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(}\text{complete ionic}\text{)}\\ 2{\text{Cl}}^{\text{-}}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow 2{\text{OH}}^{\text{-}}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\text{(net ionic)}\end{array}[/latex] [/hidden-answer][latex]2\text{KI(}aq\text{)}+\text{Pb}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}\rightarrow{\text{PbI}}_{2}\text{(}s\text{)}+2{\text{KNO}}_{3}\text{(}aq\text{)}[/latex]
This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts. The net ionic equation representing this reaction is:[latex]{\text{Pb}}^{\text{2+}}\text{(}aq\text{)}+2{\text{I}}^{-}\text{(}aq\text{)}\rightarrow{\text{PbI}}_{2}\text{(}s\text{)}[/latex]
[caption id="" align="alignright" width="300"] Figure 1. A precipitate of PbI_{2} forms when solutions containing Pb^{2+} and I^{−} are mixed. (credit: Der Kreole/Wikimedia Commons)[/caption] Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure 1). The properties of pure PbI_{2} crystals make them useful for fabrication of X-ray and gamma ray detectors. The solubility guidelines discussed above may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag^{+}, NO^{–}, Na^{+}, and F^{–} ions. Aside from the two ionic compounds originally present in the solutions, AgNO_{3} and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO_{3} and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:[latex]\begin{array}{l}\text{NaF(}aq\text{)}+{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow\text{AgF(}s\text{)}+\text{NaN}{\text{O}}_{3}\text{(}aq\text{)}\text{(molecular)}\\ {\text{Ag}}^{\text{+}}\text{(}aq\text{)}+{\text{F}}^{-}\text{(}aq\text{)}\rightarrow\text{AgF(}s\text{)}\text{(net ionic)}\end{array}[/latex]
[latex]\text{HCl(}aq\text{)}+{\text{H}}_{2}\text{O(}aq\text{)}\rightarrow{\text{Cl}}^{-}\text{(}aq\text{)}+{\text{H}}_{3}{\text{O}}^{\text{+}}\text{(}aq\text{)}[/latex]
The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H_{3}O^{+} ions are produced by a chemical reaction in which H^{+} ions are transferred from HCl molecules to H_{2}O molecules (Figure 2). [caption id="" align="aligncenter" width="700"] Figure 2. When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).[/caption] [caption id="" align="alignright" width="351"] Figure 3. (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. The hydrogen atoms that may be transferred during an acid-base reaction are highlighted in the inset molecular structures. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons)[/caption] The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 1). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:[latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightleftharpoons{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\text{(}aq\text{)}+{\text{H}}_{3}{\text{O}}^{\text{+}}\text{(}aq\text{)}[/latex]
When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, [latex]{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}[/latex] (Figure 3). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)Table 1. Common Strong Acids | |
---|---|
Compound Formula | Name in Aqueous Solution |
HBr | hydrobromic acid |
HCl | hydrochloric acid |
HI | hydroiodic acid |
HNO_{3} | nitric acid |
HClO_{4} | perchloric acid |
H_{2}SO_{4} | sulfuric acid |
[latex]\text{NaOH(}s\text{)}\rightarrow{\text{Na}}^{\text{+}}\text{(}aq\text{)}+{\text{OH}}^{-}\text{(}aq\text{)}[/latex]
This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na^{+} and OH^{–} ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases. Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 4). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:[latex]{\text{NH}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightleftharpoons{\text{NH}}_{4}{}^{\text{+}}\text{(}aq\text{)}+{\text{OH}}^{-}\text{(}aq\text{)}[/latex]
This is, by definition, an acid-base reaction, in this case involving the transfer of H^{+} ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as [latex]{\text{NH}}_{4}{}^{+}[/latex] ions. [caption id="" align="alignnone" width="879"] Figure 4. Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)[/caption] The chemical reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base, the products are often a salt and water, and neither reactant is the water itself:[latex]\text{acid}+\text{base}\rightarrow\text{salt}+\text{water}[/latex]
To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid Mg(OH)_{2}) is ingested to ease symptoms associated with excess stomach acid (HCl):[latex]\text{Mg}{\text{(OH)}}_{2}\text{(}s\text{)}+2\text{HCl(}aq\text{)}\rightarrow{\text{MgCl}}_{2}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}[/latex]
Note that in addition to water, this reaction produces a salt, magnesium chloride.[latex]2\text{Na}(s)+\text{Cl}_2(g)\rightarrow{2}\text{NaCl}(s)[/latex]
It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:[latex]\begin{array}{l}2\text{Na(}s\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}s\text{)}+2{\text{e}}^{-}\\ {\text{Cl}}_{2}\text{(}g\text{)}+2{\text{e}}^{-}\rightarrow 2{\text{Cl}}^{-}\text{(}s\text{)}\end{array}[/latex]
These equations show that Na atoms lose electrons while Cl atoms (in the Cl_{2} molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:[latex]\begin{array}{lll}\hfill \mathbf{\text{oxidation}}& =& \text{loss of electrons}\hfill \\ \hfill \mathbf{\text{reduction}}& =& \text{gain of electrons}\hfill \end{array}[/latex]
In this reaction, then, sodium is oxidized and chlorine is undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.[latex]\begin{array}{lll}\hfill \mathbf{\text{reducing agent}}& =& \text{species that is oxidized}\hfill \\ \hfill \mathbf{\text{oxidizing agent}}& =& \text{species that is reduced}\hfill \end{array}[/latex]
Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:[latex]{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow 2\text{HCl(}g\text{)}[/latex]
The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.[latex]\begin{array}{lll}\hfill \mathbf{\text{oxidation}}& =& \text{increase in oxidation number}\hfill \\ \hfill \mathbf{\text{reduction}}& =& \text{decrease in oxidation number}\hfill \end{array}[/latex]
Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl_{2} to –1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H_{2} to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl_{2} to –1 in HCl). Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:[latex]10\text{Al(}s\text{)}+6{\text{NH}}_{4}{\text{ClO}}_{4}\text{(}s\text{)}\rightarrow 4{\text{Al}}_{2}{\text{O}}_{3}\text{(}s\text{)}+2{\text{AlCl}}_{3}\text{(}s\text{)}+12{\text{H}}_{2}\text{O(}g\text{)}+3{\text{N}}_{2}\text{(}g\text{)}[/latex]
[latex]\text{Zn(}s\text{)}+2\text{HCl}\text{(}aq\text{)}\rightarrow{\text{ZnCl}}_{2}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}[/latex]
Metallic elements may also be oxidized by solutions of other metal salts; for example:[latex]\text{Cu(}s\text{)}+2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+2\text{Ag(}s\text{)}[/latex]
This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu^{2+} ions dissolve in the solution to yield a characteristic blue color (Figure 5). [caption id="" align="alignnone" width="878"] Figure 5. (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)[/caption][latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+{\text{Fe}}^{\text{2+}}\rightarrow{\text{Cr}}^{\text{3+}}+{\text{Fe}}^{\text{3+}}[/latex]
[reveal-answer q="265393"]Show Answer[/reveal-answer] [hidden-answer a="265393"][latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex] [latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow{\text{Cr}}^{\text{3+}}[/latex]
[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex] [latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}[/latex]
[latex]\begin{array}{c}{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}} \\ {\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}\end{array}[/latex]
[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex] [latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]
[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}+{\text{e}}^{-}[/latex] [latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+14{\text{H}}^{+}+6{\text{e}}^{-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]
[latex]{\text{6Fe}}^{\text{2+}}\rightarrow 6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}[/latex] [latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+6{\text{e}}^{-}+14{\text{H}}^{+}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]
[latex]6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+6{\text{e}}^{-}+14{\text{H}}^{+}\rightarrow 6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]
Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:[latex]6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2-}}+14{\text{H}}^{+}\rightarrow 6{\text{Fe}}^{\text{3+}}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]
A final check of atom and charge balance confirms the equation is balanced.Reactants | Products | |
Fe | 6 | 6 |
Cr | 2 | 2 |
O | 7 | 7 |
H | 14 | 14 |
charge | 24+ | 24+ |
[latex]1\text{ cup mix}+\frac{3}{4}\text{cup milk}+1\text{egg}\rightarrow 8\text{pancakes}[/latex]
If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is[latex]24\cancel{\text{pancakes}}\times \frac{1\text{egg}}{8\cancel{\text{pancakes}}}=3\text{eggs}[/latex]
Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:[latex]{\text{N}}_{2}\text{(}g\text{)}+3{\text{H}}_{2}\text{(}g\text{)}\rightarrow 2{\text{NH}}_{3}\text{(}g\text{)}[/latex]
This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:[latex]\displaystyle\frac{2{\text{NH}}_{3}\text{ molecules}}{3{\text{H}}_{2}\text{ molecules}}\text{ or }\frac{\text{2 doz }{\text{NH}}_{3}\text{ molecules}}{\text{3 doz }{\text{H}}_{2}\text{ molecules}}\text{ or }\frac{\text{2 mol}{\text{ NH}}_{3}\text{ molecules}}{\text{3 mol}{\text{ H}}_{2}\text{ molecules}}[/latex]
These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.[latex]2\text{Al}+3{\text{I}}_{2}\rightarrow 2{\text{AlI}}_{3}[/latex]
[caption id="" align="alignnone" width="879"] Figure 1. Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)[/caption] [reveal-answer q="307193"]Show Answer[/reveal-answer] [hidden-answer a="307193"] Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\displaystyle\frac{3\text{ mol I}_{2}}{2\text{ mol Al}}[/latex]. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:[latex]\begin{array}{ll}\hfill {\text{mol I}}_{2}& =0.429\cancel{\text{mol Al}}\times \frac{\text{3 mol}{\text{I}}_{2}}{2\cancel{\text{mol Al}}}\\ & =\text{0.644 mol}{\text{I}}_{2}\end{array}[/latex]
[/hidden-answer][latex]{\text{C}}_{3}{\text{H}}_{8}+5{\text{O}}_{2}\rightarrow 3{\text{CO}}_{2}+4{\text{H}}_{2}\text{O}[/latex]
[reveal-answer q="159596"]Show Answer[/reveal-answer] [hidden-answer a="159596"] The approach here is the same as for Example 1, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro’s number. The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:[latex]\displaystyle\frac{\text{3 mol}{\text{CO}}_{2}}{\text{1 mol}{\text{C}}_{3}{\text{H}}_{8}}[/latex]
Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,[latex]0.75\cancel{\text{mol}{\text{C}}_{3}{\text{H}}_{8}}\times \frac{3\cancel{\text{mol}{\text{CO}}_{2}}}{1\cancel{\text{mol}{\text{C}}_{3}{\text{H}}_{8}}}\times \frac{6.022\times {10}^{23}{\text{CO}}_{2}\text{molecules}}{\cancel{\text{mol}{\text{CO}}_{2}}}=1.4\times {10}^{24}{\text{CO}}_{2}\text{molecules}[/latex]
[/hidden-answer][latex]{\text{(}{\text{NH}}_{4}\text{)}}_{2}{\text{SO}}_{4}+\text{Ca}{\text{(}\text{OH}\text{)}}_{2}\rightarrow 2{\text{NH}}_{3}+{\text{CaSO}}_{4}+2{\text{H}}_{2}\text{O.}[/latex]
[reveal-answer q="488302"]Show Answer[/reveal-answer] [hidden-answer a="488302"]4.8 × 10^{24} NH_{3} molecules[/hidden-answer][latex]{\text{MgCl}}_{2}\text{(}aq\text{)}+2\text{NaOH}\text{(}aq\text{)}\rightarrow\text{Mg}{\text{(}\text{OH}\text{)}}_{2}\text{(}s\text{)}+2\text{NaCl}\text{(}aq\text{)}[/latex]
[reveal-answer q="973760"]Show Answer[/reveal-answer] [hidden-answer a="973760"] The approach used previously in Example 1 and Example 2 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:[latex]16\cancel{\text{g Mg}{\text{(}\text{OH}\text{)}}_{2}}\times \frac{1\cancel{\text{mol Mg}{\text{(}\text{OH}\text{)}}_{2}}}{58.3\cancel{\text{g Mg}{\text{(}\text{OH}\text{)}}_{2}}}\times \frac{2\cancel{\text{mol NaOH}}}{1\cancel{\text{mol Mg}{\text{(}\text{OH}\text{)}}_{2}}}\times \frac{\text{40.0 g NaOH}}{\cancel{\text{mol NaOH}}}=\text{22 g NaOH}[/latex]
[/hidden-answer][latex]2{\text{C}}_{8}{\text{H}}_{18}+25{\text{O}}_{2}\rightarrow 16{\text{CO}}_{2}+18{\text{H}}_{2}\text{O}[/latex]
[reveal-answer q="976311"]Show Answer[/reveal-answer] [hidden-answer a="976311"] The approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.[latex]702\cancel{\text{g}{\text{C}}_{8}{\text{H}}_{18}}\times \frac{1\cancel{\text{mol}{\text{C}}_{8}{\text{H}}_{18}}}{114.23\cancel{\text{g}{\text{C}}_{8}{\text{H}}_{18}}}\times \frac{25\cancel{\text{mol}{\text{O}}_{2}}}{2\cancel{\text{mol}{\text{C}}_{8}{\text{H}}_{18}}}\times \frac{\text{32.00 g}{\text{O}}_{2}}{\cancel{\text{mol}{\text{O}}_{2}}}=2.46\times {10}^{3}\text{g}{\text{O}}_{2}[/latex]
[/hidden-answer][latex]2{\text{NaN}}_{3}\text{(}s\text{)}\rightarrow 3{\text{N}}_{2}\text{(}g\text{)}+2\text{Na}\text{(}s\text{)}[/latex]
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN_{3} will generate approximately 50 L of N_{2}. For more information about the chemistry and physics behind airbags and for helpful diagrams on how airbags work, go to How Stuff Works' "How Airbags Work" article.[latex]\text{g SiC}\rightarrow\text{mol SiC}\rightarrow\text{mol}{\text{SiO}}_{2}\rightarrow\text{g}{\text{SiO}}_{2}[/latex]
Molar masses: SiO_{2} = 60.0843 g mol^{–1}; SiC = 40.0955 g mol^{–1}[latex]\text{mass SiO}2=3000\cancel{\text{g SiC}}\times \frac{1\cancel{\text{mol SiC}}}{40.955\cancel{\text{g SiC}}}\times \frac{1\cancel{\text{mol}{\text{SiO}}_{2}}}{1\cancel{\text{mol SiC}}}\times \frac{\text{60.843 g}{\text{SiO}}_{2}}{1\cancel{\text{mol}{\text{SiO}}_{2}}}=\text{4496 g}{\text{SiO}}_{2}=\text{4.50 kg}{\text{SiO}}_{2}[/latex]
12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol^{–1}[latex]\text{1 mol C}\rightarrow 1{\text{mol CO}}_{2}\rightarrow 1\text{mol urea}[/latex] [latex]\begin{array}{ll}\hfill \text{mass urea}& =1.00\times {10}^{3}\cancel{\text{kg}}\times \frac{1000\cancel{\text{g}}}{\cancel{\text{kg}}}\times \frac{1\cancel{\text{mol C}}}{12.0\cancel{\text{g C}}}\times \frac{1\cancel{\text{mol urea}}}{1\cancel{\text{mol C}}}\times \frac{\text{60.054 g urea}}{1\cancel{\text{mol urea}}}\\ & =5.00\times {10}^{6}\text{g or}5.00\times {10}^{3}\text{kg}\end{array}[/latex]
14. The balanced chemical equation is [latex]\text{C}\text{(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{CO}}_{2}\text{(}g\text{)}[/latex][latex]\begin{array}{l}500\cancel{\text{miles}}\times \frac{1\cancel{\text{gallon}}}{37.5\cancel{\text{miles}}}\times \frac{3.785\cancel{\text{L}}}{1\cancel{\text{gallon}}}\times \frac{1000\cancel{\text{mL}}}{1\cancel{\text{L}}}\times \frac{0.8205\cancel{\text{g gas}}}{1\cancel{\text{mL}}\text{gas}}\times \frac{84.2\cancel{\text{g C}}}{100\cancel{\text{g gas}}}\times \frac{1\cancel{\text{mol C}}}{12.01\cancel{\text{g C}}}\\ \times \frac{1\cancel{\text{mol}{\text{CO}}_{2}}}{1\cancel{\text{mol C}}}\times \frac{\text{44.01 g}{\text{CO}}_{2}}{1\cancel{\text{mol}{\text{CO}}_{2}}}=1.28\times {10}^{5}\text{g}{\text{CO}}_{2}\end{array}[/latex]
16. Use molarity to convert. This solution involves the following steps:[latex]\text{43.88 mL}\times \frac{\text{1 L}}{\text{1000 mL}}\times \frac{0.3842\cancel{\text{mol Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}}{\text{1 L}}\times \frac{4\cancel{\text{mol KI}}}{2\cancel{\text{mol Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}}\times \frac{\text{1 L KI}}{0.2089\cancel{\text{mol KI}}}=\text{161.4 mL}[/latex]
All of these steps can be shown together, as follows:[latex]\frac{\text{43.88 mL Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}{1}\times \frac{\text{0.3842 mol Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}{\text{1000 mL Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}\times \frac{\text{4 mol KI}}{\text{2 mol Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}\times \frac{\text{1000 mL KI}}{\text{0.2089 mol KI}}=\text{161.40 mL KI solution}[/latex]
18. Find from worked example, check your learning problem[latex]\text{mass of ilmenite}=379\cancel{\text{g ore}}\times \frac{\text{0.883 g}{\text{FeTiO}}_{3}}{1\cancel{\text{g ore}}}=\text{334.6 g}{\text{FeTiO}}_{3}[/latex]
[latex]\text{mass of rutile}=334.6\cancel{\text{g}{\text{FeTiO}}_{3}}\times \frac{1\cancel{\text{mol}{\text{FeTiO}}_{3}}}{151.7\cancel{\text{g}{\text{FeTiO}}_{3}}}\times \frac{2\cancel{\text{mol}{\text{TiO}}_{2}}}{2\cancel{\text{mol}{\text{FeTiO}}_{3}}}\times \frac{\text{79.88 g}{\text{TiO}}_{2}}{1\cancel{\text{mol}{\text{TiO}}_{2}}}=\text{176 g}{\text{TiO}}_{2}[/latex]
[/hidden-answer][latex]\text{1 slice of cheese}+\text{2 slices of bread}\rightarrow\text{1 sandwich}[/latex]
Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess. [caption id="" align="aligncenter" width="879"] Figure 1. Sandwich making can illustrate the concepts of limiting and excess reactants.[/caption] Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:[latex]{\text{H}}_{2}\text{(}s\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow\text{2 HCl}\text{(}g\text{)}.[/latex]
The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H_{2} and 2 moles of Cl_{2}. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted. An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield[latex]\text{mol HCl produced}=3\text{ mol H}_{2}\times\frac{2\text{ mol HCl}}{1\text{ mol H}_{2}}=\text{6 mol HCl}[/latex]
Complete reaction of the provided chlorine would produce[latex]\text{mol HCl produced}=2\text{ mol Cl}_{2}\times\frac{\text{2 mol HCl}}{\text{1 mol Cl}_{2}}=\text{4 mol HCl}[/latex]
The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 2). [caption id="" align="aligncenter" width="881"] Figure 2. When H_{2} and Cl_{2} are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.[/caption][latex]\text{3Si}\text{(}s\text{)}+2{\text{N}}_{2}\text{(}g\text{)}\rightarrow{\text{Si}}_{3}{\text{N}}_{4}\text{(}s\text{)}[/latex]
Which is the limiting reactant when 2.00 g of Si and 1.50 g of N_{2} react? [reveal-answer q="839920"]Show Answer[/reveal-answer] [hidden-answer a="839920"] Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.[latex]\text{mol Si}=2.00\cancel{\text{g Si}}\times \frac{\text{1 mol Si}}{28.09\cancel{\text{g Si}}}=\text{0.0712 mol Si}[/latex]
[latex]\text{mol}{\text{ N}}_{2}=1.50\cancel{\text{g}{\text{ N}}_{2}}\times \frac{\text{1 mol}{\text{ N}}_{2}}{28.09\cancel{\text{g}{\text{ N}}_{2}}}=\text{0.0535 mol}{\text{ N}}_{2}[/latex]
The provided Si:N_{2} molar ratio is:[latex]\displaystyle\frac{\text{0.0712 mol Si}}{\text{0.0535 mol}{\text{ N}}_{2}}=\frac{\text{1.33 mol Si}}{\text{1 mol}{\text{ N}}_{2}}[/latex]
The stoichiometric Si:N_{2} ratio is:[latex]\displaystyle\frac{\text{3 mol Si}}{\text{2 mol}{\text{ N}}_{2}}=\frac{\text{1.5 mol Si}}{\text{1 mol}{\text{ N}}_{2}}[/latex]
Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant. Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield[latex]\text{mol Si}_{3}{\text{N}}_{4}\text{ produced}=\text{0.0712 mol Si}\times \frac{1\text{ mol }{\text{Si}}_{3}{\text{N}}_{4}}{\text{3 mol Si}}=0.0237\text{ mol }{\text{Si}}_{3}{\text{N}}_{4}[/latex]
while the 0.0535 moles of nitrogen would produce[latex]\text{mol}{\text{ Si}}_{3}{\text{N}}_{4}\text{ produced}=\text{0.0535 mol}{\text{ N}}_{2}\times \frac{\text{1 mol}{\text{ Si}}_{3}{\text{N}}_{4}}{\text{2 mol}{\text{ N}}_{2}}=\text{0.0268 mol}{\text{ Si}}_{3}{\text{N}}_{4}[/latex]
Since silicon yields the lesser amount of product, it is the limiting reactant. [/hidden-answer][latex]\text{percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\%[/latex]
Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.[latex]{\text{CuSO}}_{4}\text{(}aq\text{)}+\text{Zn}\text{(}s\text{)}\rightarrow\text{Cu}\text{(}s\text{)}+{\text{ZnSO}}_{4}\text{(}aq\text{)}[/latex]
What is the percent yield? [reveal-answer q="115156"]Show Answer[/reveal-answer] [hidden-answer a="115156"] The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:[latex]1.274\cancel{\text{g}{\text{CuSO}}_{4}}\times \frac{1\cancel{\text{mol}{\text{CuSO}}_{4}}}{159.62\cancel{\text{g}{\text{CuSO}}_{4}}}\times \frac{1\cancel{\text{mol Cu}}}{1\cancel{\text{mol}{\text{CuSO}}_{4}}}\times \frac{63.55\text{g Cu}}{1\cancel{\text{mol Cu}}}=\text{0.5072 g Cu}[/latex]
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be[latex]\text{percent yield}=\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right)\times 100[/latex] [latex]\begin{array}{c}\\ \text{percent yield}=\left(\frac{\text{0.392 g Cu}}{\text{0.5072 g Cu}}\right)\times 100\\ =77.3\%\end{array}[/latex]
[/hidden-answer][latex]\text{atom economy}=\frac{\text{mass of product}}{\text{mass of reactants}}\times 100\%[/latex]
Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry. The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 3). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997. [caption id="" align="aligncenter" width="600"] Figure 3. (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)[/caption][latex]2{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\text{(}aq\text{)}+{\text{K}}_{2}{\text{CO}}_{3}\text{(}s\text{)}\rightarrow{\text{KCH}}_{3}{\text{CO}}_{3}\text{(}aq\text{)}+{\text{CO}}_{2}\text{(}g\text{)}+{\text{H}}_{2}\text{O}\text{(}l\text{)}\text{.}[/latex]
The bubbling was due to the production of CO_{2}. The test of vinegar with potassium carbonate is one type of quantitative analysis—the determination of the amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.[latex]\text{HCl}\text{(}aq\text{)}+\text{NaOH}\text{(}aq\text{)}\rightarrow\text{NaCl}\text{(}aq\text{)}+{\text{H}}_{2}\text{O}\text{(}l\text{)}\text{.}[/latex]
What is the molarity of the HCl? [reveal-answer q="521583"]Show Answer[/reveal-answer] [hidden-answer a="521583"] As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants are provided and requested are expressed as solution concentrations. For this exercise, the calculation will follow the following outlined steps: The molar amount of HCl is calculated to be:[latex]35.23\cancel{\text{mL NaOH}}\times \frac{1\cancel{\text{L}}}{1000\cancel{\text{mL}}}\times \frac{0.250\cancel{\text{mol NaOH}}}{1\cancel{\text{L}}}\times \frac{\text{1 mol HCl}}{1\cancel{\text{mol NaOH}}}=8.81\times {10}^{-3}\text{mol HCl}[/latex]
Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is:[latex]\begin{array}{l}\\ M=\frac{\text{mol HCl}}{\text{L solution}}\\ M=\frac{8.81\times {10}^{-3}\text{mol HCl}}{\text{50.00 mL}\times \frac{\text{1 L}}{\text{1000 mL}}}\\ M=0.176M\end{array}[/latex]
Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:[latex]M=\frac{\text{mol solute}}{\text{L solution}}\times \frac{\frac{{10}^{3}\text{mmol}}{\text{mol}}}{\frac{{10}^{3}\text{mL}}{\text{L}}}=\frac{\text{mmol solute}}{\text{mL solution}}\text{.}[/latex]
Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:[latex]\displaystyle\frac{35.23\text{mL NaOH}\times \frac{0.250\text{mmol NaOH}}{\text{mL NaOH}}\times \frac{1\text{mmol HCl}}{1\text{mmol NaOH}}}{50.00\text{mL solution}}=0.176M\text{HCl.}[/latex]
[/hidden-answer][latex]2{\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}+5{\text{H}}_{2}{\text{C}}_{2}{\text{O}}_{4}\text{(}aq\text{)}+6{\text{H}}^{+}\text{(}aq\text{)}\rightarrow 10{\text{CO}}_{2}\text{(}g\text{)}+2{\text{Mn}}^{\text{2+}}\text{(}aq\text{)}+8{\text{H}}_{2}\text{O}\text{(}l\text{)}[/latex]
A volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity? [reveal-answer q="275206"]Show Answer[/reveal-answer] [hidden-answer a="275206"]0.2648 M[/hidden-answer][latex]{\text{CaSO}}_{4}\text{(}aq\text{)}+\text{Ba}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}\rightarrow{\text{BaSO}}_{4}\text{(}s\text{)}+\text{Ca}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}[/latex]
What is the concentration (percent) of CaSO_{4} in the mixture? [reveal-answer q="25361"]Show Answer[/reveal-answer] [hidden-answer a="25361"] The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO_{4} and CaSO_{4} through their stoichiometric factor. Once the mass of CaSO_{4} is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration. The mass of CaSO_{4} that would yield the provided precipitate mass is[latex]0.6168\cancel{{\text{g BaSO}}_{4}}\times \frac{1\cancel{{\text{mol BaSO}}_{4}}}{233.43\cancel{{\text{g BaSO}}_{4}}}\times \frac{1\cancel{{\text{mol CaSO}}_{4}}}{1\cancel{{\text{mol BaSO}}_{4}}}\times \frac{136.14{\text{g CaSO}}_{4}}{1\cancel{{\text{mol CaSO}}_{4}}}=0.3597{\text{g CaSO}}_{4}[/latex]
The concentration of CaSO_{4} in the sample mixture is then calculated to be[latex]\begin{array}{l}\\ {\text{percent CaSO}}_{4}=\frac{{\text{mass CaSO}}_{4}}{\text{mass sample}}\times 100\%\\ \frac{\text{0.3597 g}}{0.4550 g}\times 100\%=79.05\%\end{array}[/latex]
[/hidden-answer][latex]{\text{Ag}}^{+}\text{(}aq\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}\rightarrow\text{AgCl}\text{(}s\text{)}[/latex]
[reveal-answer q="996093"]Show Answer[/reveal-answer] [hidden-answer a="996093"]23.76%[/hidden-answer][latex]{\text{C}}_{\text{x}}{\text{H}}_{\text{y}}\text{(}s\text{)}+{\text{excess O}}_{2}\text{(}g\text{)}\rightarrow x{\text{CO}}_{2}\text{(}g\text{)}+y{\text{H}}_{2}\text{O}\text{(}g\text{)}[/latex]
Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed. First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:[latex]\begin{array}{l}\\ \text{mol C}=0.00394{\text{g CO}}_{2}\times \frac{1{\text{mol CO}}_{2}}{\text{44.01 g/mol}}\times \frac{1\text{mol C}}{1{\text{mol CO}}_{2}}=8.95\times {10}^{-5}\text{mol C}\\ \text{mol H}=0.00161{\text{g H}}_{2}\text{O}\times \frac{1{\text{mol H}}_{2}\text{O}}{18.02\text{g/mol}}\times \frac{2\text{mol H}}{1{\text{mol H}}_{2}\text{O}}=1.79\times {10}^{-4}\text{mol H}\end{array}[/latex]
The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is[latex]\frac{\text{mol H}}{\text{mol C}}=\frac{1.79\times {10}^{-4}\text{mol H}}{8.95\times {10}^{-5}\text{mol C}}=\frac{2\text{mol H}}{1\text{mol C}}[/latex]
and the empirical formula for polyethylene is CH_{2}. [/hidden-answer]All the magic that we know is in the transfer of electrons. Reduction (gaining electrons) and oxidation (the loss of electrons) combine to form Redox chemistry, which contains the majority of chemical reactions. As electrons jump from atom to atom, they carry energy with them, and that transfer of energy is what makes all life on earth possible.
https://youtu.be/lQ6FBA1HM3s ]]>[latex]c=2.998\times {10}^{8}{\text{ms}}^{-1}=\lambda \nu [/latex]
Wavelength and frequency are inversely proportional: As the wavelength increases, the frequency decreases. The inverse proportionality is illustrated in Figure 2. This figure also shows the electromagnetic spectrum, the range of all types of electromagnetic radiation. Each of the various colors of visible light has specific frequencies and wavelengths associated with them, and you can see that visible light makes up only a small portion of the electromagnetic spectrum. Because the technologies developed to work in various parts of the electromagnetic spectrum are different, for reasons of convenience and historical legacies, different units are typically used for different parts of the spectrum. For example, radio waves are usually specified as frequencies (typically in units of MHz), while the visible region is usually specified in wavelengths (typically in units of nm or angstroms). [caption id="" align="aligncenter" width="879"]