Gas calculations
Gas calculations
Molar gas volume at RTP
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- Molar gas volume, Vₘ
- The gas volume per mole, V/n, at a stated temperature and pressure.
At room temperature and pressure, use a molar gas volume of 24.0 dm³ mol⁻¹, unless the question supplies a different value.
To find moles, divide the gas volume by the molar gas volume. To find volume, multiply moles by it. Make the units match first: 24.0 dm³ is 24,000 cm³.
At the same temperature and pressure, equal volumes of ideal gases contain equal mole amounts. You can therefore use balanced coefficients as volume ratios when all the substances being compared are gases under those conditions.
V gas volume; match its unit to Vm (dm³ or cm³); Vm molar gas volume: 24.0 dm³ mol⁻¹ or 24,000 cm³ mol⁻¹ at the stated room temperature and pressure; n amount of the specified substance or particles, in mol.
Hydrogen from magnesium
0.486 g Mg reacts with excess HCl at RTP. M(Mg) = 24.3 g mol−1.
Calculate the volume of hydrogen formed, in dm³ and cm³. Use a molar gas volume of 24.0 dm³ mol⁻¹ at RTP.
Worked answer
| Quantity | Mg | H2 |
|---|---|---|
| Mass / g | 0.486 | Not needed |
| M / g mol−1 | 24.3 | Not needed |
| Amount / mol | n = m/M 0.486/24.3 = 0.0200 | 0.0200 (1 : 1) |
V(H2) = 0.0200 × 24.0
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Use a molar gas volume of 24.0 dm3 mol−1 at RTP.
(a) Calculate the volume of 0.0375 mol gas at RTP, in cm3. (2 marks)
(b) Calculate the amount in moles of gas in 360 cm3 at RTP. (2 marks)
(c) For 2CO(g) + O2(g) → 2CO2(g), determine the oxygen volume needed to react with 40.0 cm3 CO. All volumes are measured at the same temperature and pressure. (1 mark)
Total: 5 marks
Show answer & marking points
(a) V = nVₘ
V = 0.0375 × 24.0 = 0.900 dm3 [1] = 9.00 × 102 cm3. [1]
(b) n = V(cm³)/(1000Vₘ)
n = 360/24000 [1] = 0.0150 mol. [1]
(c) 2CO + O₂ → 2CO₂. At the same temperature and pressure, volume follows the coefficient ratio.
| Quantity | CO | O₂ required |
|---|---|---|
| Coefficient | 2 | 1 |
| Volume ÷ coefficient / cm³ | 40.0/2 = 20.0 | Same ratio value: 20.0 |
| Volume / cm³ | 40.0 | 20.0 × 1 = 20.0 [1] |
A 1.250 g limestone sample contains CaCO3 and inert material. It reacts completely with excess hydrochloric acid, releasing 254 cm3 CO2 at RTP. Use M(CaCO3) = 100.1 g mol−1 and molar gas volume 24.0 dm3 mol−1. Determine the percentage by mass of CaCO3. (5 marks)
Total: 5 marks
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Percentage purity = (mass of pure substance ÷ mass of sample) × 100
n(CO₂) = V(cm³)/(1000Vₘ)
n(CO2) = 254/24000 = 0.0105833… mol. [1] CaCO3 + 2HCl → CaCl2 + CO2 + H2O gives a 1:1 ratio, so n(CaCO3) = 0.0105833… mol. [1] Pure CaCO3 mass = 0.0105833… × 100.1 = 1.05939… g. [1] Purity = (1.05939…/1.250) × 100 [1] = 84.8% to 3 significant figures. [1]
The ideal gas equation
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The ideal gas equation connects pressure, volume, temperature and amount. Write pV = nRT, then rearrange for the quantity you need.
- Use pressure in Pa, volume in m³ and temperature in K when using R = 8.314 J mol⁻¹ K⁻¹.
- Convert before substituting: add 273 to temperature in °C for these school calculations, and use 1 dm³ = 10⁻³ m³ or 1 cm³ = 10⁻⁶ m³.
- Calculate the amount in moles. If mass is supplied, use M = m/n to find molar mass.
The ideal-gas model is a better approximation at low pressure and high temperature. The particles then occupy a smaller fraction of the volume, and attractions have less effect.
R gas constant, 8.314 J mol⁻¹ K⁻¹, unless the question supplies a different value; T absolute temperature of the gas, in K; V gas volume, in m³ in the ideal gas equation; n amount of the specified substance or particles, in mol; p gas pressure, in Pa in the ideal gas equation.
| Quantity | Unit | Conversion |
|---|---|---|
| p | Pa | kPa × 1000 |
| V | m3 | dm3 ÷ 1000; cm3 ÷ 1000000 |
| n | mol | Use amount, not mass |
| T | K | °C + 273 (or +273.15 when precision requires) |
| R | J mol−1 K−1 | 8.314 |
Find an amount in moles of gas
100 cm3 of gas is measured at 101 kPa and 27 °C. Use R = 8.314 J mol−1 K−1.
Calculate the number of moles of gas.
Worked answer
n = (101000 × 0.000100)/(8.314 × 300)
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A gas occupies 1.25 dm3 at 96.0 kPa and 20.0 °C. Use R = 8.314 J mol−1 K−1 and T/K = temperature in °C + 273.
(a) Calculate the amount in moles of gas. (4 marks)
(b) State two conditions under which a real gas is more closely approximated by the ideal-gas model. (2 marks)
Total: 6 marks
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(a) p = 96000 Pa; V = 1.25 × 10−3 m3; T = 293 K. [1] n = pV/(RT) [1] = (96000 × 1.25 × 10−3)/(8.314 × 293) [1] = 0.0493 mol to 3 significant figures. [1]
(b) Low pressure [1] and high temperature, well above condensation. [1]
Gas measurements and molar mass
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A gas calculation is only as good as the volume measurement. Use a gas-tight apparatus with a syringe plunger that moves freely, and begin collection promptly so gas is not lost.
Measure temperature and pressure consistently. Collection over water is suitable only if the gas is sufficiently insoluble and does not react with the water.
To find molar mass, first calculate moles from the gas data, then use M = m/n.
If gas escapes, measured volume is too small and the calculated mole amount is too small. With the gas mass used in the calculation unchanged, dividing by that smaller n makes the calculated M too high.
Molar mass of a gas
0.184 g gas occupies 100 cm3 at 100 kPa and 27 °C. Use R = 8.314 J mol−1 K−1.
Calculate the molar mass of the gas in g mol⁻¹.
Worked answer
n = (100000 × 0.000100)/(8.314 × 300) = 0.00400946… mol
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A 0.823 g sample of gas occupies 420 cm3 at 98.5 kPa and 315 K. Use R = 8.314 J mol−1 K−1.
(a) Calculate the molar mass. (4 marks)
(b) Explain one reason why collecting a gas over water may underestimate the gas volume. (1 mark)
(c) Some gas escapes, while the mass used in the calculation remains unchanged. Explain the effect on the calculated molar mass. (2 marks)
(d) State one apparatus check needed for reliable gas-volume measurement with a gas syringe. (1 mark)
Total: 8 marks
Show answer & marking points
(a) p = 98500 Pa and V = 420 × 10−6 m3. [1] n = pV/(RT) = 0.0157966… mol. [1] M = m/n
M = 0.823/0.0157966… [1] = 52.1 g mol−1 to 3 significant figures. [1]
(b) Some of the gas may dissolve in or react with the water. [1]
(c) Measured volume, and hence calculated n, is too low. [1] M = m/n is therefore too high for the unchanged mass. [1]
(d) Check gas-tight connections, or that the syringe plunger moves freely, or begin collection promptly without losing gas. [1]