Topic 1 · page 7 of 10

Core Practical 1: molar volume of a gas

React known masses of calcium carbonate with excess acid, collect the carbon dioxide, and the gradient of volume against moles is the molar volume.

13 min read1.111.10

Read first: Gas volumes

Aim and reaction 1.11

To measure the volume occupied by one mole of carbon dioxide at room temperature and pressure, and compare it with the accepted value of 24.0 dm³ mol⁻¹.

CaCOX3(s)+2 CHX3COOH(aq)→(CHX3COO)X2Ca(aq)+HX2O(l)+COX2(g) \ce{CaCO3(s) + 2CH3COOH(aq) -> (CH3COO)2Ca(aq) + H2O(l) + CO2(g)}

Because the acid is in excess, every mole of calcium carbonate gives one mole of carbon dioxide. Measure the mass of CaCO₃ and the volume of CO₂, and you have both n and V.

Method 1.11

Apparatus: a conical flask containing ethanoic acid, with a small test tube of calcium carbonate standing upright inside it, above the acid. A bung in the flask leads through a delivery tube to a horizontal gas syringe marked from 0 to 100 cubic centimetres.
The solid is kept out of the acid until the bung is in, so no gas escapes at the start.

PracticalMethod

  1. Measure about 50 cm³ of dilute ethanoic acid (an excess) into a conical flask.
  2. Weigh about 0.05 g of calcium carbonate into a small test tube on a balance reading to at least 0.01 g. Record the mass.
  3. Stand the test tube upright inside the flask, without spilling any solid into the acid. Fit the bung and connect the gas syringe. Check the syringe reads zero.
  4. Tip the flask so the calcium carbonate falls into the acid. Swirl gently.
  5. When the fizzing has stopped and the plunger no longer moves, record the volume of gas.
  6. Repeat with masses up to about 0.30 g, so the volume stays within the 100 cm³ syringe.
Step Why
Acid in excess All the CaCO₃ reacts, so n(CO₂) = n(CaCO₃)
Solid kept apart until the bung is in No CO₂ escapes before the apparatus is sealed
Ethanoic acid, a weak acid The reaction is gentle, so the gas is given off steadily
Small masses only 0.30 g makes about 70 cm³ of gas; more would overfill the syringe
Wait for the plunger to stop The reaction is complete and the gas is back at room temperature

Results and processing 1.11

The results below are illustrative, chosen to show what a typical class set looks like. They are not measured data.

Mass of CaCO₃ / g Amount of CaCO₃ / mmol Volume of CO₂ / cm³
0.05 0.500 10.9
0.10 0.999 22.7
0.15 1.499 34.1
0.20 1.998 46.1
0.25 2.498 57.6
0.30 2.997 69.4

Amount = mass ÷ 100.1, shown in millimoles (1 mmol = 10⁻³ mol) to keep the numbers readable.

A graph of volume of carbon dioxide in cubic centimetres against amount of calcium carbonate in millimoles. Six points lie close to a straight best-fit line with gradient 23.4, which crosses the volume axis just below zero. A dashed line shows the expected gradient of 24.0 for comparison.
Illustrative results. The gradient of the best-fit line is the molar volume.

Plot volume against amount and draw the line of best fit. Because V = n × VmV_\text{m}, the gradient is the molar volume:

gradient=ΔVΔn=23.4 cm3 mmol−1=23.4 dm3 mol−1 \text{gradient} = \frac{\Delta V}{\Delta n} = 23.4\ \text{cm}^3\ \text{mmol}^{-1} = 23.4\ \text{dm}^3\ \text{mol}^{-1}

Evaluating the result 1.111.10

percentage difference=24.0−23.424.0×100=2.5% low \text{percentage difference} = \frac{24.0 - 23.4}{24.0} \times 100 = 2.5\%\ \text{low}
Source of error Effect on the volume Improvement
CO₂ dissolves slightly in the water in the acid too low use acid already saturated with CO₂
Gas escapes at the bung or syringe joints too low check the seals; grease the joints
Some gas lost while fitting the bung too low keep the solid apart until sealed (already in the method)
Balance uncertainty is large for 0.05 g random scatter, worst for the smallest masses use a 3 d.p. balance, or larger masses with a larger syringe
Reading taken while the gas is still warm too high wait until the volume stops changing

The balance uncertainty is a good example of why the smallest masses are the least reliable. A 2 d.p. balance reads to ±0.005 g, and a mass by difference uses two readings, so the uncertainty is about ±0.01 g:

percentage uncertainty in 0.05 g=0.010.05×100=20% \text{percentage uncertainty in } 0.05\ \text{g} = \frac{0.01}{0.05} \times 100 = 20\%

For 0.30 g the same uncertainty is only about 3%. This matches the illustrative data: the first point is the furthest from the line.

Check yourself

  1. Why must the ethanoic acid be in excess?
  2. Explain why the test tube of calcium carbonate is stood inside the flask before the bung is fitted, rather than tipped in first.
  3. 0.240 g of calcium carbonate gives 55.4 cm³ of CO₂. Calculate the molar volume from this single result.
  4. A student's best-fit line has a gradient of 22.8 dm³ mol⁻¹. Calculate the percentage difference from 24.0 dm³ mol⁻¹, and suggest one reason the value is low.
  5. Why is the gradient of the graph a better value than the result of any single run?

Answers cover these until you have tried the questions

  1. So that all the calcium carbonate reacts. Then the amount of CO₂ equals the amount of CaCO₃ weighed out.
  2. If the solid is added first, the reaction starts before the bung is in and some CO₂ escapes. The measured volume would be too low.
  3. n = 0.240 ÷ 100.1 = 2.398 × 10⁻³ mol. VmV_\text{m} = 55.4 ÷ 2.398 × 10⁻³ = 23 100 cm³ mol⁻¹ = 23.1 dm³ mol⁻¹.
  4. (24.0 − 22.8) ÷ 24.0 × 100 = 5.0% low. Possible reasons: CO₂ dissolving in the aqueous acid, or gas leaking from the bung or syringe.
  5. It uses all the runs, so random errors partly cancel, and a fixed loss in every run shifts the intercept rather than the gradient.