Mixtures and average molar mass

Why a mixture needs one molar mass

Air is not one gas. It is mostly nitrogen and oxygen, with a little argon and carbon dioxide. Yet a chemist often treats air as one substance, with one molar mass. Exam problems do the same with fuel gases, biogas and the gas in a cylinder.

The average molar mass of a mixture is its total mass divided by its total number of moles:

Here $x_1$ and $x_2$ are the mole fractions. Each gas counts in proportion to the number of its molecules, not in proportion to its own molar mass.

Air as one gas

Dry air has these mole fractions, rounded: nitrogen $0.78$, oxygen $0.21$ and argon $0.01$.

1. Multiply each fraction by its molar mass. $0.78 \times 28 = 21.84$, $0.21 \times 32 = 6.72$ and $0.01 \times 40 = 0.40$.
2. Add. $M_{\text{avg}} = 21.84 + 6.72 + 0.40 = 28.96\ \text{g mol}^{-1}$, about $29\ \text{g mol}^{-1}$.
3. Use it. A gas whose molar mass is below $29\ \text{g mol}^{-1}$ is lighter than air. Methane ($16\ \text{g mol}^{-1}$) rises; carbon dioxide ($44\ \text{g mol}^{-1}$) sinks.

Vapour density

Many exam papers give the vapour density of a gas instead of its molar mass. It compares the gas with hydrogen at the same temperature and pressure:

So $M = 2 \times \text{vapour density}$. Air has a vapour density of about $14.5$.

From the average back to the composition

For two gases, write $x$ for the mole fraction of the first. The second then has $1 - x$:

Worked example. A cylinder of cooking gas holds propane ($\text{C}_3\text{H}_8$, $44\ \text{g mol}^{-1}$) and butane ($\text{C}_4\text{H}_{10}$, $58\ \text{g mol}^{-1}$). Its vapour density is $26.5$. Find the mole fraction and the mass percent of propane.

1. Molar mass. $M_{\text{avg}} = 2 \times 26.5 = 53\ \text{g mol}^{-1}$.
2. Mole fraction of propane. $x = \dfrac{58 - 53}{58 - 44} = \dfrac{5}{14} = 0.357$.
3. Mass percent of propane. One mole of the gas holds $0.357 \times 44 = 15.7\ \text{g}$ of propane in $53\ \text{g}$, so $\dfrac{15.7}{53} \times 100 = 29.6\%$.

A mole percent and a mass percent are different numbers. Always say which one you mean.

When the percentages are by mass, change them to moles first. Take $100\ \text{g}$ of the mixture:

For example, a gas that is $40\%$ helium and $60\%$ argon by mass has $\dfrac{40}{4} = 10\ \text{mol}$ of helium and $\dfrac{60}{40} = 1.5\ \text{mol}$ of argon in $100\ \text{g}$. So $M_{\text{avg}} = \dfrac{100}{11.5} = 8.70\ \text{g mol}^{-1}$, far below the plain average of $22\ \text{g mol}^{-1}$.

Mixtures of two solids that react

When a mixture of two solids reacts, each solid follows its own equation. Two unknowns need two equations. One comes from the mass of the mixture. The other comes from a product that you can measure.

Worked example. $30.08\ \text{g}$ of a mixture of magnesium carbonate and calcium carbonate is heated strongly. It gives $14.08\ \text{g}$ of carbon dioxide. How much of each carbonate was there?

1. Name the unknowns. Let the mixture hold $a$ mol of $\text{MgCO}_3$ ($84\ \text{g mol}^{-1}$) and $b$ mol of $\text{CaCO}_3$ ($100\ \text{g mol}^{-1}$).
2. Moles of gas. Each carbonate gives one $\text{CO}_2$, so $a + b = \dfrac{14.08}{44} = 0.32$.
3. Mass. $84a + 100b = 30.08$.
4. Solve. Put $b = 0.32 - a$: $84a + 32 - 100a = 30.08$, so $16a = 1.92$ and $a = 0.12$. Then $b = 0.20$.
5. Masses. $\text{MgCO}_3$: $0.12 \times 84 = 10.08\ \text{g}$. $\text{CaCO}_3$: $0.20 \times 100 = 20.0\ \text{g}$.

Check. $10.08 + 20.0 = 30.08\ \text{g}$, the mass of the mixture.

Try it in Python

The program finds the average molar mass of dry air from four gases, then its vapour density and its density at STP. At STP ($273\ \text{K}$ and $1\ \text{bar}$) one mole of gas takes $22.7\ \text{L}$.

Practise with new numbers

Common mistakes

• Adding the molar masses and halving. That works only for equal moles of two gases. Weight each molar mass by its mole fraction.
• Mixing up mole percent and mass percent. Change a mass percent into moles first.
• Forgetting that oxygen and nitrogen are diatomic. $\text{O}_2$ is $32\ \text{g mol}^{-1}$, not $16\ \text{g mol}^{-1}$.
• Using the vapour density as the molar mass. Double it first.
• One equation for two unknowns. A mixture of two solids needs two facts: its mass, and the amount of one product.

Remember these

• $M_{\text{avg}} = x_1M_1 + x_2M_2 + \dots$, with mole fractions.
• From mass percents: $M_{\text{avg}} = \dfrac{100}{\sum w_i / M_i}$.
• $M = 2 \times \text{vapour density}$.
• For two gases, $x_1 = \dfrac{M_2 - M_{\text{avg}}}{M_2 - M_1}$.
• Two unknowns need two equations: one for the mass and one for the moles.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A mixture holds equal numbers of moles of methane and oxygen. What is its average molar mass?

2. A gas has a vapour density of $22$. Name a gas that it could be.

3. A balloon is filled with a mixture of helium and air. When does it rise?

4. In the carbonate example, how would you know at once that $a = 0.40$ is wrong?