### Learning Objectives

By the end of this section, you will be able to:

- Use the ideal gas law to compute gas densities and molar masses
- Perform stoichiometric calculations involving gaseous substances
- State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures

The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.”^{2} Much of the knowledge we do have about Lavoisier's contributions is due to his wife, Marie-Anne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study.

As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles (*n*). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.

### Gas Density and Molar Mass

The ideal gas law described previously in this chapter relates the properties of pressure *P*, volume *V*, temperature *T*, and molar amount *n*. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:

$$PV=nRT$$

The density *d* of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.

$$d=\frac{m}{V}$$

Rearranging the ideal gas equation to isolate *V* and substituting into the density equation yields

$$d=\frac{mP}{nRT}=\left(\frac{m}{n}\right)\frac{P}{RT}$$

The ratio *m*/*n* is the definition of molar mass, *ℳ*:

$$\mathcal{\mathcal{M}}=\frac{m}{n}$$

The density equation can then be written

$$d=\frac{\mathcal{\mathcal{M}}P}{RT}$$

This relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 8.11.

### Example 8.11

#### Measuring Gas Density

What is the density of molecular nitrogen gas at STP?

#### Solution

The molar mass of molecular nitrogen, N_{2}, is 28.01 g/mol. Substituting this value along with standard temperature and pressure into the gas density equation yields

$$d=\frac{\mathcal{\mathcal{M}}P}{RT}=\frac{(28.01\phantom{\rule{0ex}{0ex}}\text{g/mol})(1.00\phantom{\rule{0ex}{0ex}}\text{atm})}{(0.0821\phantom{\rule{0ex}{0ex}}\text{L}\phantom{\rule{0ex}{0ex}}\text{atm}\phantom{\rule{0ex}{0ex}}{\text{mol}}^{-1}{\text{K}}^{-1})(273\phantom{\rule{0ex}{0ex}}\text{K})}=1.25\phantom{\rule{0ex}{0ex}}\text{g}\phantom{\rule{0ex}{0ex}}\text{/}\phantom{\rule{0ex}{0ex}}\text{L}$$

#### Check Your Learning

What is the density of molecular hydrogen gas at 17.0 °C and a pressure of 760 torr?

### Answer:

d = 0.0847 g/L

When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation

$$PV=nRT$$

and the definition of molar mass

$$\mathcal{M}=\frac{m}{n}$$

yields the following equation:

$$\mathcal{M}=\frac{mRT}{PV}$$

Determining the molar mass of a gas via this approach is demonstrated in Example 8.12.

### Example 8.12

#### Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula

Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?

#### Solution

First determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:

$$\text{85.7 g C}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\text{1 mol C}}{\text{12.01 g C}}\phantom{\rule{0ex}{0ex}}=\text{7.136 mol C}\phantom{\rule{0ex}{0ex}}\frac{7.136}{7.136}\phantom{\rule{0ex}{0ex}}=\text{1.00 mol C}$$

$$\text{14.3 g H}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\text{1 mol H}}{\text{1.01 g H}}\phantom{\rule{0ex}{0ex}}=\text{14.158 mol H}\phantom{\rule{0ex}{0ex}}\frac{14.158}{7.136}\phantom{\rule{0ex}{0ex}}=\text{1.98 mol H}$$

Empirical formula is CH_{2} [empirical mass (EM) of 14.03 g/empirical unit].

Next, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:

$$\mathcal{M}=\frac{mRT}{PV}=\frac{(1.56\phantom{\rule{0ex}{0ex}}\text{g})(0.0821\phantom{\rule{0ex}{0ex}}\text{L}\phantom{\rule{0ex}{0ex}}\text{atm}\phantom{\rule{0ex}{0ex}}{\text{mol}}^{-1}{\text{K}}^{-1})(323\phantom{\rule{0ex}{0ex}}\text{K})}{(0.984\phantom{\rule{0ex}{0ex}}\text{atm})(1.00\phantom{\rule{0ex}{0ex}}\text{L})}=42.0\phantom{\rule{0ex}{0ex}}\text{g/mol}$$

Comparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:

$$\frac{\mathcal{M}}{EM}=\frac{42.0\phantom{\rule{0ex}{0ex}}\text{g/mol}}{14.0\phantom{\rule{0ex}{0ex}}\text{g/mol}}=3$$

The molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:

$${({\text{CH}}_{2})}_{3}={\text{C}}_{3}{\text{H}}_{6}$$

#### Check Your Learning

Acetylene, a fuel used welding torches, is composed of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?

### Answer:

Empirical formula, CH; Molecular formula, C_{2}H_{2}

### Example 8.13

#### Determining the Molar Mass of a Volatile Liquid

The approximate molar mass of a volatile liquid can be determined by:

- Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that may escape through the hole
- Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
- Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see Figure 8.19)

Figure 8.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At ${t}_{l\phantom{\rule{0ex}{0ex}}\u27f6\phantom{\rule{0ex}{0ex}}g},$ the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)

Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm^{3} at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

#### Solution

Since $\text{\mathcal{M}}=\phantom{\rule{0ex}{0ex}}\frac{m}{n}$ and $n=\phantom{\rule{0ex}{0ex}}\frac{PV}{RT},$ substituting and rearranging gives $\text{\mathcal{M}}=\phantom{\rule{0ex}{0ex}}\frac{mRT}{PV},$

then

$$\text{\mathcal{M}}=\phantom{\rule{0ex}{0ex}}\frac{mRT}{PV}\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}\frac{\left(\text{0.494 g}\right)\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{0.08206 L\xb7atm/mol K}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{372.8 K}}{\text{0.976 atm}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{0.129 L}}\phantom{\rule{0ex}{0ex}}=120\phantom{\rule{0ex}{0ex}}\text{g/mol}.$$

#### Check Your Learning

A sample of phosphorus that weighs 3.243 $\times $ 10^{−2} g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?

### Answer:

124 g/mol P_{4}

### The Pressure of a Mixture of Gases: Dalton’s Law

Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 8.20). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: *The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases*:

$${P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+\mathrm{...}={\text{\Sigma}}_{\text{i}}{P}_{\text{i}}$$

In the equation *P _{Total}* is the total pressure of a mixture of gases,

*P*is the partial pressure of gas A;

_{A}*P*is the partial pressure of gas B;

_{B}*P*is the partial pressure of gas C; and so on.

_{C}Figure 8.20 If equal-volume cylinders containing gasses at pressures of 300 kPa, 450 kPa, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa.

The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (*X*), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

$${P}_{A}={X}_{A}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{P}_{Total}\phantom{\rule{0ex}{0ex}}\text{where}\phantom{\rule{0ex}{0ex}}{X}_{A}=\phantom{\rule{0ex}{0ex}}\frac{{n}_{A}}{{n}_{Total}}$$

where *P _{A}*,

*X*, and

_{A}*n*are the partial pressure, mole fraction, and number of moles of gas A, respectively, and

_{A}*n*is the number of moles of all components in the mixture.

_{Total}### Example 8.14

#### The Pressure of a Mixture of Gases

A 10.0-L vessel contains 2.50 $\times $ 10^{−3} mol of H_{2}, 1.00 $\times $ 10^{−3} mol of He, and 3.00 $\times $ 10^{−4} mol of Ne at 35 °C.

(a) What are the partial pressures of each of the gases?

(b) What is the total pressure in atmospheres?

#### Solution

The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\phantom{\rule{0ex}{0ex}}\frac{nRT}{V}$:

$${P}_{{\text{H}}_{2}}=\phantom{\rule{0ex}{0ex}}\frac{\left(2.50\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0ex}{0ex}}\overline{)\text{mol}}\right)\left(0.08206\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}\phantom{\rule{0ex}{0ex}}\text{atm}\phantom{\rule{0ex}{0ex}}\overline{){\text{mol}}^{\text{\u22121}}\phantom{\rule{0ex}{0ex}}{\text{K}}^{\text{\u22121}}}\right)\left(308\phantom{\rule{0ex}{0ex}}\overline{)\text{K}}\right)}{10.0\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}}\phantom{\rule{0ex}{0ex}}=6.32\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0ex}{0ex}}\text{atm}$$

$${P}_{\text{He}}=\phantom{\rule{0ex}{0ex}}\frac{\left(1.00\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0ex}{0ex}}\overline{)\text{mol}}\right)\left(0.08206\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}\phantom{\rule{0ex}{0ex}}\text{atm}\phantom{\rule{0ex}{0ex}}\overline{){\text{mol}}^{\text{\u22121}}\phantom{\rule{0ex}{0ex}}{\text{K}}^{\text{\u22121}}}\right)\left(308\phantom{\rule{0ex}{0ex}}\overline{)\text{K}}\right)}{10.0\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}}\phantom{\rule{0ex}{0ex}}=2.53\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0ex}{0ex}}\text{atm}$$

$${P}_{\text{Ne}}=\phantom{\rule{0ex}{0ex}}\frac{\left(3.00\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22124}}\phantom{\rule{0ex}{0ex}}\overline{)\text{mol}}\right)\left(0.08206\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}\phantom{\rule{0ex}{0ex}}\text{atm}\phantom{\rule{0ex}{0ex}}\overline{){\text{mol}}^{\text{\u22121}}\phantom{\rule{0ex}{0ex}}{\text{K}}^{\text{\u22121}}}\right)\left(308\phantom{\rule{0ex}{0ex}}\overline{)\text{K}}\right)}{10.0\phantom{\rule{0ex}{0ex}}\overline{)\text{L}}}\phantom{\rule{0ex}{0ex}}=7.58\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22124}}\phantom{\rule{0ex}{0ex}}\text{atm}$$

The total pressure is given by the sum of the partial pressures:

$${P}_{\text{T}}={P}_{{\text{H}}_{2}}+{P}_{\text{He}}+{P}_{\text{Ne}}=\left(0.00632+0.00253+0.00076\right)\phantom{\rule{0ex}{0ex}}\text{atm}=9.61\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\text{\u22123}}\phantom{\rule{0ex}{0ex}}\text{atm}$$

#### Check Your Learning

A 5.73-L flask at 25 °C contains 0.0388 mol of N_{2}, 0.147 mol of CO, and 0.0803 mol of H_{2}. What is the total pressure in the flask in atmospheres?

### Answer:

1.137 atm

Here is another example of this concept, but dealing with mole fraction calculations.

### Example 8.15

#### The Pressure of a Mixture of Gases

A gas mixture used for anesthesia contains 2.83 mol oxygen, O_{2}, and 8.41 mol nitrous oxide, N_{2}O. The total pressure of the mixture is 192 kPa.

(a) What are the mole fractions of O_{2} and N_{2}O?

(b) What are the partial pressures of O_{2} and N_{2}O?

#### Solution

The mole fraction is given by ${X}_{A}=\phantom{\rule{0ex}{0ex}}\frac{{n}_{A}}{{n}_{Total}}$ and the partial pressure is *P _{A}* =

*X*$\times $

_{A}*P*.

_{Total}For O_{2},

$${X}_{{O}_{2}}=\phantom{\rule{0ex}{0ex}}\frac{{n}_{{O}_{2}}}{{n}_{Total}}\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}\frac{\text{2.83 mol}}{\left(2.83+8.41\right)\phantom{\rule{0ex}{0ex}}\text{mol}}\phantom{\rule{0ex}{0ex}}=0.252$$

and ${P}_{{O}_{2}}={X}_{{O}_{2}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{P}_{Total}=0.252\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{192 kPa}=\text{48.4 kPa}$

For N_{2}O,

$${X}_{{N}_{2}O}=\phantom{\rule{0ex}{0ex}}\frac{{n}_{{N}_{2}O}}{{n}_{\mathrm{Total}}}\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}\frac{\text{8.41 mol}}{\left(2.83+8.41\right)\phantom{\rule{0ex}{0ex}}\text{mol}}\phantom{\rule{0ex}{0ex}}=0.748$$

and

${P}_{{N}_{2}O}={X}_{{N}_{2}O}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{P}_{\mathrm{Total}}=0.748\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{192 kPa}=\text{144 kPa}$

#### Check Your Learning

What is the pressure of a mixture of 0.200 g of H_{2}, 1.00 g of N_{2}, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

### Answer:

1.87 atm

### Collection of Gases over Water

A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 8.21), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

Figure 8.21 When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).

However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 8.22); more detailed information on the temperature dependence of water vapor can be found in Table 8.2, and vapor pressure will be discussed in more detail in the chapter on liquids.

Figure 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature.

Vapor Pressure of Ice and Water in Various Temperatures at Sea Level

Temperature (°C) | Pressure (torr) | Temperature (°C) | Pressure (torr) | Temperature (°C) | Pressure (torr) | ||
---|---|---|---|---|---|---|---|

–10 | 1.95 | 18 | 15.5 | 30 | 31.8 | ||

–5 | 3.0 | 19 | 16.5 | 35 | 42.2 | ||

–2 | 3.9 | 20 | 17.5 | 40 | 55.3 | ||

0 | 4.6 | 21 | 18.7 | 50 | 92.5 | ||

2 | 5.3 | 22 | 19.8 | 60 | 149.4 | ||

4 | 6.1 | 23 | 21.1 | 70 | 233.7 | ||

6 | 7.0 | 24 | 22.4 | 80 | 355.1 | ||

8 | 8.0 | 25 | 23.8 | 90 | 525.8 | ||

10 | 9.2 | 26 | 25.2 | 95 | 633.9 | ||

12 | 10.5 | 27 | 26.7 | 99 | 733.2 | ||

14 | 12.0 | 28 | 28.3 | 100.0 | 760.0 | ||

16 | 13.6 | 29 | 30.0 | 101.0 | 787.6 |

Table 8.2

### Example 8.16

#### Pressure of a Gas Collected Over Water

If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 8.21, what is the partial pressure of argon?

#### Solution

According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:

$${P}_{\text{T}}={P}_{\text{Ar}}+{P}_{{\text{H}}_{2}\text{O}}$$

Rearranging this equation to solve for the pressure of argon gives:

$${P}_{\text{Ar}}={P}_{\text{T}}-{P}_{{\text{H}}_{2}\text{O}}$$

The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:

$${P}_{\text{Ar}}=750\phantom{\rule{0ex}{0ex}}\text{torr}\phantom{\rule{0ex}{0ex}}-25.2\phantom{\rule{0ex}{0ex}}\text{torr}=725\phantom{\rule{0ex}{0ex}}\text{torr}$$

#### Check Your Learning

A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?

### Answer:

0.583 L

### Chemical Stoichiometry and Gases

Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.

We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.

### Avogadro’s Law Revisited

Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to ${\text{N}}_{2}(g)+3{\text{H}}_{2}(g)\phantom{\rule{0ex}{0ex}}\u27f6\phantom{\rule{0ex}{0ex}}2{\text{NH}}_{3}(g),$ a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in Figure 8.23. According to Avogadro’s law, equal volumes of gaseous N_{2}, H_{2}, and NH_{3}, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N_{2} reacts with three molecules of H_{2} to produce two molecules of NH_{3}, the volume of H_{2} required is three times the volume of N_{2}, and the volume of NH_{3} produced is two times the volume of N_{2}._{}

Figure 8.23 One volume of N_{2} combines with three volumes of H_{2} to form two volumes of NH_{3}.

### Example 8.17

#### Reaction of Gases

Propane, C_{3}H_{8}(*g*), is used in gas grills to provide the heat for cooking. What volume of O_{2}(*g*) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

#### Solution

The ratio of the volumes of C_{3}H_{8} and O_{2} will be equal to the ratio of their coefficients in the balanced equation for the reaction:

$$\begin{array}{l}{\text{C}}_{3}{\text{H}}_{8}(g)+5{\text{O}}_{2}(g)\text{\hspace{1em}}\text{\hspace{1em}}\u27f6\text{\hspace{1em}}\text{\hspace{1em}}3{\text{CO}}_{2}(g)+4{\text{H}}_{2}\text{O}\left(l\right)\\ \text{1 volume}+\text{5 volumes}\phantom{\rule{0ex}{0ex}}\text{3 volumes}+\text{4 volumes}\end{array}$$

From the equation, we see that one volume of C_{3}H_{8} will react with five volumes of O_{2}:

$$2.7\phantom{\rule{0ex}{0ex}}\overline{)\text{L}\phantom{\rule{0ex}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\text{5 L}\phantom{\rule{0ex}{0ex}}{\text{O}}_{2}}{1\phantom{\rule{0ex}{0ex}}\overline{)\text{L}\phantom{\rule{0ex}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}}\phantom{\rule{0ex}{0ex}}=\text{13.5 L}\phantom{\rule{0ex}{0ex}}{\text{O}}_{2}$$

A volume of 13.5 L of O_{2} will be required to react with 2.7 L of C_{3}H_{8}.

#### Check Your Learning

An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C_{2}H_{2}, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 $\times $ 10^{3} L of O_{2} at 0 °C and 1 atm, will be required to burn the acetylene?

$$2{\text{C}}_{2}{\text{H}}_{2}+5{\text{O}}_{2}\phantom{\rule{0ex}{0ex}}\u27f6\phantom{\rule{0ex}{0ex}}4{\text{CO}}_{2}+2{\text{H}}_{2}\text{O}$$

### Answer:

3.34 tanks (2.34 $\times $ 10^{4} L)

### Example 8.18

#### Volumes of Reacting Gases

Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H_{2}(*g*), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N_{2}?

$${\text{N}}_{2}(g)+3{\text{H}}_{2}(g)\phantom{\rule{0ex}{0ex}}\u27f6\phantom{\rule{0ex}{0ex}}2{\text{NH}}_{3}(g)$$

#### Solution

Because equal volumes of H_{2} and NH_{3} contain equal numbers of molecules and each three molecules of H_{2} that react produce two molecules of NH_{3}, the ratio of the volumes of H_{2} and NH_{3} will be equal to 3:2. Two volumes of NH_{3}, in this case in units of billion ft^{3}, will be formed from three volumes of H_{2}:

$$683\phantom{\rule{0ex}{0ex}}\overline{)\text{billion}\phantom{\rule{0ex}{0ex}}{\text{ft}}^{3}\phantom{\rule{0ex}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\text{3 billion}\phantom{\rule{0ex}{0ex}}{\text{ft}}^{3}\phantom{\rule{0ex}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0ex}{0ex}}\overline{)\text{billion}\phantom{\rule{0ex}{0ex}}{\text{ft}}^{3}\phantom{\rule{0ex}{0ex}}{\text{NH}}_{3}}}\phantom{\rule{0ex}{0ex}}=1.02\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{3}\phantom{\rule{0ex}{0ex}}\text{billion}\phantom{\rule{0ex}{0ex}}{\text{ft}}^{3}\phantom{\rule{0ex}{0ex}}{\text{H}}_{2}$$

The manufacture of 683 billion ft^{3} of NH_{3} required 1020 billion ft^{3} of H_{2}. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

#### Check Your Learning

What volume of O_{2}(*g*) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C_{2}H_{4}(*g*), measured under the same conditions of temperature and pressure? The products are CO_{2} and water vapor.

### Answer:

51.0 L

### Example 8.19

#### Volume of Gaseous Product

What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

$$2\text{Ga}\left(s\right)+6\text{HCl}\left(aq\right)\phantom{\rule{0ex}{0ex}}\u27f6\phantom{\rule{0ex}{0ex}}2{\text{GaCl}}_{3}\left(aq\right)+3{\text{H}}_{2}(g)$$

#### Solution

Convert the provided mass of the limiting reactant, Ga, to moles of hydrogen produced:

$$8.88\phantom{\rule{0ex}{0ex}}\overline{)\text{g Ga}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{1\phantom{\rule{0ex}{0ex}}\overline{)\text{mol Ga}}}{69.723\phantom{\rule{0ex}{0ex}}\overline{)\text{g Ga}}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\text{3 mol}\phantom{\rule{0ex}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0ex}{0ex}}\overline{)\text{mol Ga}}}\phantom{\rule{0ex}{0ex}}=0.191\phantom{\rule{0ex}{0ex}}{\text{mol H}}_{2}$$

Convert the provided temperature and pressure values to appropriate units (K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:

$$V=\left(\frac{nRT}{P}\right)=\phantom{\rule{0ex}{0ex}}\frac{0.191\phantom{\rule{0ex}{0ex}}\overline{)\text{mol}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{0.08206 L}\phantom{\rule{0ex}{0ex}}\overline{)\text{atm}\phantom{\rule{0ex}{0ex}}{\text{mol}}^{\text{\u22121}}\phantom{\rule{0ex}{0ex}}{\text{K}}^{\text{\u22121}}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\text{300 K}}{0.951\phantom{\rule{0ex}{0ex}}\overline{)\text{atm}}}\phantom{\rule{0ex}{0ex}}=\text{4.94 L}$$

#### Check Your Learning

Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO_{2} at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in excess oxygen?

### Answer:

1.30 $\times $ 10^{3} L

### How Sciences Interconnect

#### Greenhouse Gases and Climate Change

The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost $\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most of this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the *greenhouse effect*. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than 30 °C (nearly 60 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 8.24).

Figure 8.24 Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.

There is strong evidence from multiple sources that higher atmospheric levels of CO_{2} are caused by human activity, with fossil fuel burning accounting for about $\frac{3}{4}$ of the recent increase in CO_{2}. Reliable data from ice cores reveals that CO_{2} concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO_{2} concentration has increased from preindustrial levels of ~280 ppm to more than 400 ppm today (Figure 8.25).

Figure 8.25 CO_{2} levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.

### Link to Learning

Click here to see a 2-minute video explaining greenhouse gases and global warming.

### Portrait of a Chemist

#### Susan Solomon

Atmospheric and climate scientist Susan Solomon (Figure 8.26) is the author of one of *The New York Times* books of the year (*The Coldest March*, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.

For more information, watch this video about Susan Solomon.

Figure 8.26 Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)

## FAQs

### How to calculate stoichiometry for gases in chemical reactions? ›

To account for these conditions, we use the ideal gas equation **PV=nRT** where P is the pressure measured in atmosphere(atm), V is the volume measured in liters (L), n is the number of moles, R is the gas constant with a value of . 08206 L atm mol^{-}^{1} K^{-}^{1}, and T is the temperature measured in kelvin (K).

**What is the first step to solving a stoichiometry problem? ›**

**We can tackle this stoichiometry problem using the following steps:**

- Step 1: Convert known reactant mass to moles. ...
- Step 2: Use the mole ratio to find moles of other reactant. ...
- Step 3: Convert moles of other reactant to mass.

**What are the 4 steps to solving stoichiometry problems? ›**

**Almost all stoichiometric problems can be solved in just four simple steps:**

- Balance the equation.
- Convert units of a given substance to moles.
- Using the mole ratio, calculate the moles of substance yielded by the reaction.
- Convert moles of wanted substance to desired units.

**What is an example of stoichiometry? ›**

For example, **when oxygen and hydrogen react to produce water, one mole of oxygen reacts with two moles of hydrogen to produce two moles of water**. In addition, stoichiometry can be used to find quantities such as the amount of products that can be produced with a given amount of reactants and percent yield.

**What are the 4 types of stoichiometry problems? ›**

**Stoichiometric difficulties can be classified into four categories:**

- Mass to mass conversion.
- Mass to moles conversion.
- Mole to mass steps conversion.
- Mole to mole steps conversion.

**What is stoichiometry solution example? ›**

Solution Stoichiometry Movie Text

It is defined as the moles of a substance contained in one liter of solution. For instance, **if a solution has a concentration of 1.20 M NaCl, this means that there are 1.20 moles of NaCl per liter of solution**.

**What grade level is stoichiometry? ›**

Stoichiometry concepts taught at secondary level, mostly for **grade 11 and grade 12**, involve problem-solving and under standing of chemical reactions and equations.

**How hard is stoichiometry? ›**

Stoichiometry **might be difficult for students because they often don't see the big picture**. That is because they don't understand how all the concepts fit together and why they are being in the real world.

**Is stoichiometry easy or hard? ›**

Stoichiometry **can be difficult** because it builds upon a number of individual skills. To be successful you must master the skills and learn how to plan your problem solving strategy. Master each of these skills before moving on: Calculating Molar Mass.

**What is stoichiometric formula? ›**

The stoichiometry of a reaction **describes the relative amounts of reactants and products in a balanced chemical equation**. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s).

### What is the rule of stoichiometry? ›

Stoichiometry is founded on the law of conservation of mass where **the total mass of the reactants equals the total mass of the products**, leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers.

**What is the summary of stoichiometry? ›**

The stoichiometry of a reaction **describes the relative amounts of reactants and products in a balanced chemical equation**. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s).

**Is stoichiometry a math? ›**

Stoichiometry Definition

**Stoichiometry is math having to do with chemical reactions**. There are different types of calculations you can perform; stoichiometry with moles is the most common, but you can also do math with masses and even percentages.

**What type of math is stoichiometry? ›**

Stoichiometry is **the numerical relationship between the reactants and products of a chemical reaction**. In fact, the word 'stoichiometry' is derived from the Ancient Greek words stoicheion "element" and metron "measure".

**What are the 2 laws of stoichiometry? ›**

The basis of the stoichiometric calculations is the **law of conservation of mass** which states that the mass is neither created nor destroyed in a chemical reaction. Another form of the law states that atoms are neither created nor destroyed in a chemical reaction.

**What is stoichiometry 2 step problems? ›**

A two-step stoichiometry problem involves **converting from the grams of a substance to the moles of that substance (using a molar mass) to the moles of a second substance (using coefficients)**.

**What are everyday examples of stoichiometry? ›**

Stoichiometry is at the heart of the production of many things you use in your daily life. **Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars** are just a few commodities you use that are chemically engineered, or produced through chemical reactions.

**How do you convert mass to mass in stoichiometry? ›**

First, convert moles of substance A into moles of substance B. Then, convert moles of B into mass of B. Moles A → moles B → mass B. The main equation is: **moles A x (mole ratio of B/A) x molar mass of B = mass of B**.

**Why is it called stoichiometry? ›**

Stoichiometry is the study of the quantitative, or measureable, relationships that exist in chemical formulas and chemical reactions. The term is **derived from the Greek words stoicheion, meaning element, and metron, meaing measure**.

**How is stoichiometry used in gas? ›**

The use of the ideal or combined gas law in conjunction with stoichiometry **connects at the calculation of moles in either case**. What is meant by this is you can use a balanced chemical equation to determine the number of moles of gas being produced and then use the ideal gas law to find its pressure or volume etc.

### What is gas reaction stoichiometry? ›

Gas stoichiometry is **the quantitative relationship (ratio) between reactants and products in a chemical reaction with reactions that produce gases**. Gas stoichiometry applies when the gases produced are assumed to be ideal, and the temperature, pressure, and volume of the gases are all known.

**What is gas ratio stoichiometry? ›**

The stoichiometric mixture for a gasoline engine is the ideal ratio of air to fuel that burns all fuel with no excess air. For gasoline fuel, the stoichiometric air–fuel mixture is about **14.7:1** i.e. for every one gram of fuel, 14.7 grams of air are required.

**What is stoichiometry used for answers? ›**

Using the same concept of mole ratios as explained above, stoichiometry is used **to figure out how much reactant is needed to make a desired quantity of product in a laboratory or manufacturing facility**.

**What are 4 ways stoichiometry is used? ›**

Stoichiometry is considered the heart of chemistry with so many practical uses. The composition of all the chemical products we use in our daily lives, such as **shampoos, cleaners, perfumes, soaps, and fertilizers** is formed using stoichiometric calculations. Without stoichiometry, the chemical industry doesn't exist.

**How important is stoichiometry in chemistry? ›**

Given a chemical reaction, stoichiometry tells us what quantity of each reactant we need in order to get enough of our desired product. Because of its real-life applications in chemical engineering as well as research, stoichiometry is **one of the most important and fundamental topics in chemistry**.

**What is stoichiometry in easy words? ›**

Stoichiometry is **a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data**. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.

**How do you calculate gas ratio? ›**

To calculate the oil to gas ratio, **divide the oil volume by the gas volume**.

**What is the difference between stoichiometry and gas stoichiometry? ›**

The main difference is that **both liquids and solids have definite volume, whereas gases do not have a definite volume**.

**What are the 3 laws of gas? ›**

The gas laws consist of three primary laws: **Charles' Law, Boyle's Law and Avogadro's Law** (all of which will later combine into the General Gas Equation and Ideal Gas Law).

**What are the 4 gas laws formulas? ›**

Laws | Formula |
---|---|

Charles Law | V/T=K |

Boyle's Law | PV=K |

Avogadros law | V/n=K |

Ideal Gas Law | PV=nRT |

### What are the 4 gas laws? ›

Gas Laws: **Boyle's Law, Charle's Law, Gay-Lussac's Law, Avogadro's Law**.