# 9.3 Stoichiometry of gaseous substances, mixtures, and reactions  (Page 2/13)

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## 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:

$\text{85.7 g C}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol C}}{\text{12.01 g C}}\phantom{\rule{0.2em}{0ex}}=\text{7.136 mol C}\phantom{\rule{2em}{0ex}}\frac{7.136}{7.136}\phantom{\rule{0.2em}{0ex}}=\text{1.00 mol C}$
$\text{14.3 g H}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol H}}{\text{1.01 g H}}\phantom{\rule{0.2em}{0ex}}=\text{14.158 mol H}\phantom{\rule{2em}{0ex}}\frac{14.158}{7.136}\phantom{\rule{0.2em}{0ex}}=\text{1.98 mol H}$

Empirical formula is CH 2 [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:

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

ℳ = 42.0 g/mol, $\frac{\text{ℳ}}{\text{Eℳ}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{42.0}{14.03}\phantom{\rule{0.2em}{0ex}}=2.99,$ so (3)(CH 2 ) = C 3 H 6 (molecular formula)

Acetylene, a fuel used welding torches, is comprised 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?

Empirical formula, CH; Molecular formula, C 2 H 2

## 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 :

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

The ideal gas equation can be rearranged to isolate n :

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

and then combined with the molar mass equation to yield:

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

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

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

then

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

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?

124 g/mol P 4

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