Chemistry » Gases » Stoichiometry of Gaseous Substances, Mixtures, and Reactions

# Density of a Gas

## Density of a Gas

Recall that the density of a gas is its mass to volume ratio, $$\rho =\phantom{\rule{0.2em}{0ex}}\cfrac{m}{V}.$$ Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can be used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in the example below.

## Example

### Derivation of a Density Formula from the Ideal Gas Law

Use PV = nRT to derive a formula for the density of gas in g/L

### Solution

1. PV = nRT
2. Rearrange to get (mol/L): $$\cfrac{n}{v}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{P}{RT}$$
3. Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained:$$\left(\text{ℳ}\right)\left(\cfrac{n}{V}\right)=\left(\cfrac{P}{RT}\right)\left(\text{ℳ}\right)$$
4. $$m\text{/V}=\rho =\phantom{\rule{0.2em}{0ex}}\cfrac{P\text{ℳ}}{RT}$$

We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at STP.

## Example

### Empirical/Molecular Formula Problems Using the Ideal Gas Law and Density of a Gas

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

Strategy: First solve the empirical formula problem using methods discussed earlier. 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:

$$\text{85.7 g C}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 mol C}}{\text{12.01 g C}}\phantom{\rule{0.2em}{0ex}}=\text{7.136 mol C}\phantom{\rule{2em}{0ex}}\cfrac{7.136}{7.136}\phantom{\rule{0.2em}{0ex}}=\text{1.00 mol C}$$

$$\text{14.3 g H}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 mol H}}{\text{1.01 g H}}\phantom{\rule{0.2em}{0ex}}=\text{14.158 mol H}\phantom{\rule{2em}{0ex}}\cfrac{14.158}{7.136}\phantom{\rule{0.2em}{0ex}}=\text{1.98 mol H}$$

Empirical formula is CH2 [empirical mass (EM) of 14.03 g/empirical unit].

Next, use the density equation related to the ideal gas law to determine the molar mass:

$$\text{d}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{Pℳ}}{\text{RT}}\phantom{\rule{2em}{0ex}}\cfrac{\text{1.56 g}}{\text{1.00 L}}\phantom{\rule{0.2em}{0ex}}=\text{0.984 atm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{ℳ}}{\text{0.0821 L atm/mol K}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{323 K}$$

ℳ = 42.0 g/mol, $$\cfrac{\text{ℳ}}{\text{Eℳ}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{42.0}{14.03}\phantom{\rule{0.2em}{0ex}}=2.99,$$ so (3)(CH2) = C3H6 (molecular formula)

## Molar Mass of a Gas

Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:

$$\text{ℳ}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{grams of substance}}{\text{moles of substance}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{m}{n}$$

The ideal gas equation can be rearranged to isolate n:

$$n=\phantom{\rule{0.2em}{0ex}}\cfrac{PV}{RT}$$

and then combined with the molar mass equation to yield:

$$\text{ℳ}=\phantom{\rule{0.2em}{0ex}}\cfrac{mRT}{PV}$$

This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.

## Example

### 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 the figure below)

When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At $${t}_{l\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}g},$$ 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 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?

### Solution

Since $$\text{ℳ}=\phantom{\rule{0.2em}{0ex}}\cfrac{m}{n}$$ and $$n=\phantom{\rule{0.2em}{0ex}}\cfrac{PV}{RT},$$ substituting and rearranging gives $$\text{ℳ}=\phantom{\rule{0.2em}{0ex}}\cfrac{mRT}{PV},$$

then

$$\text{ℳ}=\phantom{\rule{0.2em}{0ex}}\cfrac{mRT}{PV}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(\text{0.494 g}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0.08206 L·atm/mol K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{372.8 K}}{\text{0.976 atm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0.129 L}}\phantom{\rule{0.2em}{0ex}}=120\phantom{\rule{0.2em}{0ex}}\text{g/mol}.$$

## 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 (see the figure below). 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:

$${P}_{Total}={P}_{A}+{P}_{B}+{P}_{C}+…={\text{Σ}}_{\text{i}}{P}_{\text{i}}$$

In the equation PTotal is the total pressure of a mixture of gases, PA is the partial pressure of gas A; PB is the partial pressure of gas B; PC is the partial pressure of gas C; and so on.

If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size cylinder, the total pressure of the 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:

$${P}_{A}={X}_{A}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{P}_{Total}\phantom{\rule{2em}{0ex}}\text{where}\phantom{\rule{2em}{0ex}}{X}_{A}=\phantom{\rule{0.2em}{0ex}}\cfrac{{n}_{A}}{{n}_{Total}}$$

where PA, XA, and nA are the partial pressure, mole fraction, and number of moles of gas A, respectively, and nTotal is the number of moles of all components in the mixture.

## Example

### The Pressure of a Mixture of Gases

A 10.0-L vessel contains 2.50 $$×$$ 10−3 mol of H2, 1.00 $$×$$ 10−3 mol of He, and 3.00 $$×$$ 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 $$P=\phantom{\rule{0.2em}{0ex}}\cfrac{nRT}{V}$$:

$${P}_{{\text{H}}_{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\right)\left(0.08206\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}}\right)\left(308\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)}{10.0\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}}\phantom{\rule{0.2em}{0ex}}=6.32\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{atm}$$

$${P}_{\text{He}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\right)\left(0.08206\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}}\right)\left(308\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)}{10.0\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}}\phantom{\rule{0.2em}{0ex}}=2.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{atm}$$

$${P}_{\text{Ne}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\right)\left(0.08206\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}}\right)\left(308\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{K}}\right)}{10.0\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}}}\phantom{\rule{0.2em}{0ex}}=7.58\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−4}}\phantom{\rule{0.2em}{0ex}}\text{atm}$$

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

$${P}_{\text{T}}={P}_{{\text{H}}_{2}}+{P}_{\text{He}}+{P}_{\text{Ne}}=\left(0.00632+0.00253+0.00076\right)\phantom{\rule{0.2em}{0ex}}\text{atm}=9.61\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}\phantom{\rule{0.2em}{0ex}}\text{atm}$$

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

## Example

### 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 $${X}_{A}=\phantom{\rule{0.2em}{0ex}}\cfrac{{n}_{A}}{{n}_{Total}}$$ and the partial pressure is PA = XA$$×$$PTotal.

For O2,

$${X}_{{O}_{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{n}_{{O}_{2}}}{{n}_{Total}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{2.83 mol}}{\left(2.83+8.41\right)\phantom{\rule{0.2em}{0ex}}\text{mol}}\phantom{\rule{0.2em}{0ex}}=0.252$$

and $${P}_{{O}_{2}}={X}_{{O}_{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{P}_{Total}=0.252\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{192 kPa}=\text{48.4 kPa}$$

For N2O,

$${X}_{{N}_{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{n}_{{N}_{2}}}{{n}_{\mathrm{Total}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{8.41 mol}}{\left(2.83+8.41\right)\phantom{\rule{0.2em}{0ex}}\text{mol}}\phantom{\rule{0.2em}{0ex}}=0.748$$

and

$${P}_{{N}_{2}}={X}_{{N}_{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{P}_{\mathrm{Total}}=0.748\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{192 kPa}=\text{143.6 kPa}$$

## 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 (see the figure below), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

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 (see the figure below); more detailed information on the temperature dependence of water vapor can be found in the table below, and vapor pressure will be discussed in more detail in the next tutorial on liquids.

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)
–101.95 1815.5 3031.8
–53.01916.53542.2
–23.92017.54055.3
04.62118.75092.5
25.32219.860149.4
46.12321.170233.7
67.02422.480355.1
88.02523.890525.8
109.22625.295633.9
1210.52726.799733.2
1412.02828.3100.0760.0
1613.62930.0101.0787.6

## Example

### 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 the figure above, 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 (see this appendix), so:

$${P}_{\text{Ar}}=750\phantom{\rule{0.2em}{0ex}}\text{torr}\phantom{\rule{0.2em}{0ex}}-25.2\phantom{\rule{0.2em}{0ex}}\text{torr}=725\phantom{\rule{0.2em}{0ex}}\text{torr}$$