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

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

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

$$ℳ=\frac{m}{n}$$

The density equation can then be written $$d=\frac{ℳP}{RT}$$

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

**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

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

**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 molarity $$ℳ=\frac{m}{n}$$

yields the following equation: $$ℳ=\frac{mRT}{PV}$$

Determining the molar mass of a gas via this approach is demonstrated in [link].

**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:

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

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

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:

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

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

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

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

$$(CH_2)_3=C_3H_6$$

**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}

**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 [link])

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

then

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

**Check Your Learning**

A sample of phosphorus that weighs 3.243 ×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}