9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions - Chemistry 2e | OpenStax (2023)

Measuring Gas Density

What is the density of molecular nitrogen gas at STP?

Solution

The molar mass of molecular nitrogen, N₂, is 28.01 g/mol. Substituting this value along with standard temperature and pressure into the gas density equation yields

$$d=\frac{\mathcal{ℳ}P}{RT}=\frac{(28.01\text{g/mol})(1.00\text{atm})}{(0.0821{\text{L·atm·mol}}^{\text{−1}}{\text{K}}^{\text{−}1})(273\text{K})}=1.25\text{g/L}$$

Check Your Learning

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

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:

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

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

Empirical formula is CH₂ [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:

$$ℳ=\frac{mRT}{PV}=\frac{(1.56\text{g})(0.0821{\text{L·atm·mol}}^{\text{−}1}{\text{K}}^{-1})(323\text{K})}{(0.984\text{atm})(1.00\text{L})}=42.0\text{g/mol}$$

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

$$\frac{ℳ}{EM}=\frac{42.0\text{g/mol}}{14.0\text{g/mol}}=3$$

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?

Determining the Molar Mass of a Volatile Liquid

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

1. 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
2. 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
3. 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 9.19)

Figure 9.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At $t_{l⟶g},$ 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 cm³ at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

Solution

Since $\text{ℳ}=\frac{m}{n}$ and $n=\frac{PV}{RT},$ substituting and rearranging gives $\text{ℳ}=\frac{mRT}{PV},$

then

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

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?

The Pressure of a Mixture of Gases

A 10.0-L vessel contains 2.50 $\times$ 10⁻³ mol of H₂, 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 $P=\frac{nRT}{V}$:

$$P_{{\text{H}}_2}=\frac{\left(2.50\times {10}^{\text{−3}}\sout{\text{mol}}\right)\left(0.08206\sout{\text{L}}\text{atm}\sout{{\text{mol}}^{\text{−1}}{\text{K}}^{\text{−1}}}\right)\left(308\sout{\text{K}}\right)}{10.0\sout{\text{L}}}=6.32\times {10}^{\text{−3}}\text{atm}$$

$$P_{\text{He}}=\frac{\left(1.00\times {10}^{\text{−3}}\sout{\text{mol}}\right)\left(0.08206\sout{\text{L}}\text{atm}\sout{{\text{mol}}^{\text{−1}}{\text{K}}^{\text{−1}}}\right)\left(308\sout{\text{K}}\right)}{10.0\sout{\text{L}}}=2.53\times {10}^{\text{−3}}\text{atm}$$

$$P_{\text{Ne}}=\frac{\left(3.00\times {10}^{\text{−4}}\sout{\text{mol}}\right)\left(0.08206\sout{\text{L}}\text{atm}\sout{{\text{mol}}^{\text{−1}}{\text{K}}^{\text{−1}}}\right)\left(308\sout{\text{K}}\right)}{10.0\sout{\text{L}}}=7.58\times {10}^{\text{−4}}\text{atm}$$

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

$$P_{\text{T}}=P_{{\text{H}}_2}+P_{\text{He}}+P_{\text{Ne}}=\left(0.00632+0.00253+0.00076\right)\text{atm}=9.61\times {10}^{\text{−3}}\text{atm}$$

Check Your Learning

A 5.73-L flask at 25 °C contains 0.0388 mol of N₂, 0.147 mol of CO, and 0.0803 mol of H₂. What is the total pressure in the flask in atmospheres?

The Pressure of a Mixture of Gases

A gas mixture used for anesthesia contains 2.83 mol oxygen, O₂, and 8.41 mol nitrous oxide, N₂O. The total pressure of the mixture is 192 kPa.

(a) What are the mole fractions of O₂ and N₂O?

(b) What are the partial pressures of O₂ and N₂O?

Solution

The mole fraction is given by $X_A=\frac{n_A}{n_{Total}}$ and the partial pressure is P_A = X_A $\times$ P_Total.

For O₂,

$$X_{O_2}=\frac{n_{O_2}}{n_{Total}}=\frac{\text{2.83 mol}}{\left(2.83+8.41\right)\text{mol}}=0.252$$

and $P_{O_2}=X_{O_2}\times P_{Total}=0.252\times \text{192 kPa}=\text{48.4 kPa}$

For N₂O,

$$X_{N_{2}O}=\frac{n_{N_{2}O}}{n_{\mathrm{Total}}}=\frac{\text{8.41 mol}}{\left(2.83+8.41\right)\text{mol}}=0.748$$

and

$P_{N_{2}O}=X_{N_{2}O}\times P_{\mathrm{Total}}=0.748\times \text{192 kPa}=\text{144 kPa}$

Check Your Learning

What is the pressure of a mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

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 9.21, 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:

$$P_{\text{Ar}}=750\text{torr}-25.2\text{torr}=725\text{torr}$$

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 from this sample have under the same conditions of temperature and pressure?

Reaction of Gases

Propane, C₃H₈(g), is used in gas grills to provide the heat for cooking. What volume of O₂(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 C₃H₈ and O₂ will be equal to the ratio of their coefficients in the balanced equation for the reaction:

$$\begin{array}{l}{\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g)⟶3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}\left(l\right)\\ \text{1 volume}+\text{5 volumes}\text{3 volumes}+\text{4 volumes}\end{array}$$

From the equation, we see that one volume of C₃H₈ will react with five volumes of O₂:

$$2.7\sout{\text{L}{\text{C}}_3{\text{H}}_8}\times \frac{\text{5 L}{\text{O}}_2}{1\sout{\text{L}{\text{C}}_3{\text{H}}_8}}=\text{13.5 L}{\text{O}}_2$$

A volume of 13.5 L of O₂ will be required to react with 2.7 L of C₃H₈.

Check Your Learning

An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C₂H₂, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 $\times$ 10³ L of O₂ at 0 °C and 1 atm, will be required to burn the acetylene?

$$2{\text{C}}_2{\text{H}}_2+5{\text{O}}_2⟶4{\text{CO}}_2+2{\text{H}}_2\text{O}$$

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 H₂(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N₂?

$${\text{N}}_2(g)+3{\text{H}}_2(g)⟶2{\text{NH}}_3(g)$$

Solution

Because equal volumes of H₂ and NH₃ contain equal numbers of molecules and each three molecules of H₂ that react produce two molecules of NH₃, the ratio of the volumes of H₂ and NH₃ will be equal to 3:2. Two volumes of NH₃, in this case in units of billion ft³, will be formed from three volumes of H₂:

$$683\sout{\text{billion}{\text{ft}}^3{\text{NH}}_3}\times \frac{\text{3 billion}{\text{ft}}^3{\text{H}}_2}{2\sout{\text{billion}{\text{ft}}^3{\text{NH}}_3}}=1.02\times {10}^3\text{billion}{\text{ft}}^3{\text{H}}_2$$

The manufacture of 683 billion ft³ of NH₃ required 1020 billion ft³ of H₂. (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 O₂(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C₂H₄(g), measured under the same conditions of temperature and pressure? The products are CO₂ and water vapor.

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?

$$2\text{Ga}\left(s\right)+6\text{HCl}\left(aq\right)⟶2{\text{GaCl}}_3\left(aq\right)+3{\text{H}}_2(g)$$

Solution

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

$$8.88\sout{\text{g Ga}}\times \frac{1\sout{\text{mol Ga}}}{69.723\sout{\text{g Ga}}}\times \frac{\text{3 mol}{\text{H}}_2}{2\sout{\text{mol Ga}}}=0.191{\text{mol H}}_2$$

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:

$$V=\left(\frac{nRT}{P}\right)=\frac{0.191\sout{\text{mol}}\times \text{0.08206 L}\sout{\text{atm}{\text{mol}}^{\text{−1}}{\text{K}}^{\text{−1}}}\times \text{300 K}}{0.951\sout{\text{atm}}}=\text{4.94 L}$$

Check Your Learning

Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO₂ at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in excess oxygen?

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 if 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 9.24).

Figure 9.24 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 CO₂ are caused by human activity, with fossil fuel burning accounting for about $\frac{3}{4}$ of the recent increase in CO₂. Reliable data from ice cores reveals that CO₂ 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 CO₂ concentration has increased preindustrial levels of ~280 ppm to more than 400 ppm today (Figure 9.25).

Figure 9.25 CO₂ levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.

Susan Solomon

Atmospheric and climate scientist Susan Solomon (Figure 9.26) 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 9.26 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)