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Gas volume experiments (Core Practical 1)

Edexcel IAL Chemistry · WCH13 Practical Skills in Chemistry I

Core Practical 1 measures the molar volume of a gas. Unit 3 questions use the same skills for any reaction that gives off a gas: calculating amounts, spotting where gas is lost and using the results to identify a substance. The method is taught in the Unit 1 lesson on gas volumes.

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The method in brief

A weighed mass of a solid (such as calcium carbonate or a metal) reacts with excess acid in a flask connected to a gas syringe, or to an inverted measuring cylinder of water. The final gas volume is recorded.

n(solid) = mass / M, and the equation gives n(gas).
Molar volume = volume of gas / n(gas), in dm³ mol^-1 (divide cm³ by 1000).
The acid is in excess so that all the solid reacts; its volume need not be measured accurately.

Where gas is lost, and how to stop it

Gas escapes while the bung is replaced after adding the solid. Better: hold the solid in a small tube inside the flask and tip it over once the bung is in, so the flask is never opened.
Carbon dioxide is slightly soluble in water, so collection over water gives a low volume. A gas syringe avoids this.
With small masses the balance uncertainty is large (0.01 / 0.150 g is almost 7%); use a larger mass or a 3 d.p. balance, within the range of the syringe.
Temperature and pressure affect the volume, so record them; the accepted value at room conditions is about 24.0 dm³ mol^-1.

Using the results

Identify a metal: from volume of H2 find n(H2), use the equation to find n(metal), then Ar = mass / n. Make sure the answer fits any clue given (a Group 2 metal, not aluminium).
Ideal gas equation: pV = nRT with p in Pa, V in m³ (1 cm³ = 1 x 10^-6 m³), T in K and R = 8.31 J mol^-1 K^-1.
Check your answer: a molar volume of 2.4 or 240 dm³ mol^-1 means a unit slip.

Worked example

0.200 g of calcium carbonate reacts with excess hydrochloric acid, giving 46.0 cm³ of CO2. Calculate the molar volume.

M(CaCO3) = 100.1 g mol^-1; n = 0.200 / 100.1 = 0.00200 mol
CaCO3 : CO2 = 1 : 1
Molar volume = 0.0460 dm³ / 0.00200 mol

Answer: 23.0 dm³ mol^-1

Key idea: Molar volume = V(gas) / n(gas). Acid in excess. Stop gas escaping (do not open the flask; use a syringe for CO2). pV = nRT in Pa, m³, K. Check the answer is sensible.

Check yourself

0.150 g of calcium carbonate (M = 100.1 g mol^-1) reacts with excess acid, producing 34.5 cm³ of carbon dioxide. Calculate the molar volume of carbon dioxide in dm³ mol^-1.
Show answer
23.0 dm³ mol^-1. n = 0.150 / 100.1 = 0.001499 mol; 0.0345 / 0.001499 = 23.0 dm³ mol^-1.
0.120 g of a Group 2 metal reacts with excess acid to give 118 cm³ of hydrogen. Take the molar volume as 24.0 dm³ mol^-1. Calculate the relative atomic mass of the metal.
Show answer
24.4. n(H2) = 0.118 / 24.0 = 0.004917 mol = n(metal) (M + 2H+ -> M2+ + H2). Ar = 0.120 / 0.004917 = 24.4, so the metal is magnesium.
Calculate the amount of gas, in mol, in 60.0 cm³ at 101 000 Pa and 298 K. (R = 8.31 J mol^-1 K^-1)
Show answer
0.00245 mol. n = pV / RT = 101000 x 60.0 x 10^-6 / (8.31 x 298) = 2.45 x 10^-3 mol.
Why is the acid used in excess in Core Practical 1?
- So that the gas dissolves
- To keep the temperature constant
- So that the solid reacts completely
- To make the reaction slower and easier to follow
Show answer
So that the solid reacts completely. The amount of gas then depends only on the mass of solid.
The molar volume obtained is lower than expected. Which is the most likely cause?
- Some gas escaped before the bung was replaced
- The acid was too concentrated
- The balance read too low
- The gas syringe was too large
Show answer
Some gas escaped before the bung was replaced. Less gas collected gives a smaller volume per mole.
A student calculates a molar volume of 2.4 dm³ mol^-1. What is the most likely mistake?
- Using too much acid
- Collecting the gas in a syringe
- A unit conversion error between cm³ and dm³
- Using calcium carbonate
Show answer
A unit conversion error between cm³ and dm³. The answer is a factor of 10 from 24, which points to a slip with powers of ten.
A mass of 0.150 g is found by difference on a balance reading to 0.01 g (total uncertainty ±0.01 g). Calculate its percentage uncertainty.
Show answer
6.7 %. 0.01 / 0.150 x 100 = 6.7%.

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Last updated 8 November 2026. Spotted a mistake? Tell us.