Molar volume of a gas: amounts, collection and bias

What you’ll learn

• Use mass and stoichiometry to calculate the amount of gas formed.
• Evaluate gas collection data at a stated temperature and pressure.

Understand the idea

Core practical 1 links a measured gas volume to the amount of a limiting reactant. First write the balanced equation; the coefficient ratio determines how many moles of gas form. Convert a gas volume from $\mathrm{cm^3}$ to $\mathrm{dm^3}$ before reporting a molar volume in $\mathrm{dm^3\,mol^{-1}}$. The result applies to the temperature and pressure of collection, so comparison with an accepted value requires matching conditions.

A gas syringe measures volume directly. Collection over water is unsuitable for a gas that dissolves significantly, and water vapour may require a correction if the question supplies the necessary data. Gas lost through a leak or before the collection system is ready makes the measured volume too small. If the reacting mass is still used in full, the calculated molar volume is too small too. An oxide coating can also make the mass of reactive metal smaller than the mass recorded.

Judge a gas-volume trace by its plateau, rather than by an arbitrary time at which the reaction may be unfinished. In a supervised metal–acid investigation, the stated flammability of hydrogen requires ignition sources to be excluded. Apparatus must permit safe collection without creating a sealed, pressurised reaction vessel. These are reasons for evaluating a teacher-approved setup, not instructions to improvise one.

Key formulae

$$\mathrm{Mg(s)}+2\mathrm{H^{+}(aq)}\rightarrow\mathrm{Mg^{2+}(aq)}+\mathrm{H_2(g)}$$

$$n=\frac{m}{M}$$

$$V_m=\frac{V}{n}$$

Worked example

In a supplied dataset, $0.0486\ \mathrm{g}$ of pure magnesium, $M=24.3\ \mathrm{g\,mol^{-1}}$, produces $48.0\ \mathrm{cm^3}$ of hydrogen at the recorded room conditions. Acid is in excess and the reaction is complete. Calculate the experimental molar volume.

1. The amount of magnesium is $0.0486/24.3=0.00200\ \mathrm{mol}$.
2. The equation has a $1:1$ magnesium-to-hydrogen ratio, so $n(\mathrm{H_2})=0.00200\ \mathrm{mol}$.
3. Convert the volume: $48.0\ \mathrm{cm^3}=0.0480\ \mathrm{dm^3}$, giving $V_m=0.0480/0.00200=24.0\ \mathrm{dm^3\,mol^{-1}}$.

Answer: $24.0\ \mathrm{dm^3\,mol^{-1}}$ at the collection conditions.