CP 1 Molar Volume of a Gas
Core Practical 1 (Edexcel International A Level Chemistry): measuring the molar volume of carbon dioxide from calcium carbonate and ethanoic acid, with the apparatus and its reasons.
It covers weighing by difference, the volume against mass graph, the full mole calculation, the excess check, and the errors, uncertainties and improvements examiners ask about.
Molar Volume of CO2 Bench
Collect the carbon dioxide made when calcium carbonate reacts with ethanoic acid, repeat with different masses, and use a graph to find the volume of one mole of gas at room temperature and pressure.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Moles, Gas Collection and 24 dm³
Three quick questions on the GCSE ideas this practical builds on: moles from mass, which gases can be collected over water, and the molar gas volume.
Measuring Gas Volumes
Gas syringe and collection over water side by side, with the errors of each method and the reminder to record room temperature and pressure.
There are two ways to measure a gas volume in the laboratory. A gas syringe collects the gas directly: the plunger is pushed out as gas enters and the volume is read from the barrel.
Collecting over water uses a measuring cylinder that is filled with water, inverted in a trough and clamped vertically.
Gas bubbling in through a delivery tube collects at the closed end and pushes the water level down, and the volume is read from the graduations at the water level.
This practical uses a 100 cm³ measuring cylinder over water, and its capacity sets the largest mass of solid that can be used.
Collecting over water only works for gases that are almost insoluble and do not react with water. Hydrogen, oxygen and nitrogen are fine; ammonia, hydrogen chloride and sulfur dioxide are not.
Carbon dioxide is slightly soluble, so a little of it dissolves in the trough and the measured volume is slightly too low.
Working in the other direction, any gas collected over water is saturated with water vapour, which adds to the volume and makes the reading slightly too high.
Exam focus: The two effects partly cancel, but an examiner expects you to name both and give their directions.
Three rules for the reading itself. Wait until the gas has returned to room temperature, because the reaction is exothermic and warm gas occupies more space. Keep the cylinder vertical and read at eye level to avoid parallax.
There is no need to level the water inside and outside the cylinder, because it is the volume of the gas space, not its pressure, that is being read.
The small pressure difference is one reason the answer is quoted only to two significant figures.
Exam wording: “Carbon dioxide is slightly soluble in water, so some is lost and the volume collected is too low; a gas syringe avoids this.” Name the gas, the direction and the fix.
Check: Which Gases Can Be Collected Over Water
Decide which gases suit which method and which way each error pushes the reading.
Safety and Apparatus
The assembled apparatus: clamped boiling tube of ethanoic acid, airtight bung and delivery tube, water trough and inverted 100 cm³ measuring cylinder read at the water level.
Every item has a job and, where it is used for a measurement, a precision that you should be able to quote.
The balance and the measuring cylinder are the two instruments whose uncertainties carry into the result; the acid volume does not matter because the acid is in excess.
Once the bung is fitted the whole train of apparatus is a closed system with one outlet: the end of the delivery tube under the mouth of the cylinder.
Gas produced in the boiling tube pushes the air already in the tube ahead of it, so the first bubbles are air.
Keep them: at the end of the run an equal volume of carbon dioxide is left behind in the tube and the space above the acid, so the two exactly cancel and the volume in the cylinder is the volume of carbon dioxide made.
Safety: 1 mol dm⁻³ ethanoic acid is an irritant: wear eye protection and rinse splashes at once. Calcium carbonate is low hazard.
Never let the delivery tube become blocked or pinched while gas is being produced: pressure builds and the bung can be ejected. Clamp the boiling tube so it cannot tip when the powder is added.
Method: Step by Step
Weighing by difference: the test tube is weighed with the powder (14.87 g), the powder is tipped into the acid, and the tube is reweighed (14.82 g), so 0.05 g reacted.
The method is built around two ideas. The first is knowing exactly how much solid reacted, which is why the test tube is weighed before and after tipping.
The second is losing no gas, which is why the bung goes on the instant the powder is in.
Every step below carries its reason, because the reason is what the mark scheme rewards.
Why stop at 0.40 g: 0.40 g of CaCO₃ is 0.40 ÷ 100.1 = 0.0040 mol, which would give about 0.0040 × 24 000 = 96 cm³ of CO₂, almost filling the 100 cm³ cylinder. A larger mass would push gas out of the mouth of the cylinder and the run would be lost.
Check: The Order of the Method
Put the steps of a hydrogen version of this practical in a workable order.
Results and the Graph
The results table, the graph with its line of best fit through the origin read at 0.25 g to give 60 cm³, and the calculation to 24 dm³ mol⁻¹.
Seven runs give the results below. Experimental data are scattered by random error: small differences in how quickly the bung went on, how long the gas was left to cool and how the water level was read.
A single run could therefore be badly wrong, and a mean of seven runs at different masses is meaningless. Instead the results are plotted and a line of best fit is drawn.
The line averages out the random errors, and a point that lies well off it can be identified as anomalous and ignored.
That is why the practical uses several masses rather than one, and why the final volume is read from the line rather than from any single run.
| Mass of CaCO₃ / g | Volume of CO₂ collected / cm³ | Volume predicted by the line / cm³ |
|---|---|---|
| 0.05 | 11 | 12 |
| 0.11 | 27 | 26 |
| 0.17 | 32 | 41 |
| 0.21 | 50 | 50 |
| 0.24 | 59 | 58 |
| 0.32 | 74 | 77 |
| 0.33 | 80 | 79 |
Mass of calcium carbonate is the independent variable and goes on the x-axis; volume of carbon dioxide is the dependent variable and goes on the y-axis.
The line of best fit is a straight line through the origin, because no solid reacting means no gas produced, and it is positioned so that the points are scattered evenly either side of it.
The third column shows what the line predicts at each mass: six runs lie within 3 cm³ of the line, while the 0.17 g run gave 32 cm³ against a predicted 41 cm³ and is anomalous.
This is most likely because gas escaped before the bung was fitted. It is ignored when the line is drawn.
The gradient of the line is about 240 cm³ g⁻¹, and reading it at 0.25 g gives about 60 cm³.
Graph rule: Independent variable on the x-axis, dependent on the y-axis, units in the axis labels as “mass of CaCO₃ / g” and “volume of CO₂ / cm³”, a straight line through the origin with equal scatter either side, and any anomaly circled and labelled.
Errors, Uncertainty and Improvements
Direction of each error: gas escaping and dissolved CO₂ make the molar volume too low, water vapour and warm gas make it too high, and the first bubbles of displaced air cause no error at all.
Evaluation questions ask for the source of an error, its direction (does it make the molar volume too high or too low?) and an improvement that removes it.
Random errors scatter the points and are dealt with by the line of best fit; systematic errors shift every point the same way and can only be dealt with by changing the method.
Percentage uncertainty is (uncertainty ÷ reading) × 100, using the actual uncertainty of the instrument. The balance reads to ±0.005 g, and a mass by difference uses two readings, so its uncertainty is ±0.01 g. The 100 cm³ cylinder is read to ±0.5 cm³. For the smallest run, 0.05 g and 11 cm³:
mass: 0.01 ÷ 0.05 × 100 = 20%; volume: 0.5 ÷ 11 × 100 = 4.5%; total = 24.5%
For the largest run, 0.33 g and 80 cm³, the same working gives 0.01 ÷ 0.33 × 100 = 3.0% and 0.5 ÷ 80 × 100 = 0.6%, a total of 3.6%.
The mass, not the volume, dominates the uncertainty at every mass, so the useful improvements are a 3 decimal place balance (±0.0005 g, cutting the mass uncertainty tenfold) and larger masses, not a finer measuring cylinder.
The graph is itself an improvement: reading from the line at 0.25 g uses all seven runs, so the random part of these uncertainties is largely averaged out.
Random or systematic: A slow bung, dissolved CO₂ and water vapour shift every run the same way (systematic): the line of best fit cannot remove them. Parallax and timing scatter the points (random): the line does remove them.
Check: Which Way Does Each Error Push the Result
Predict the direction of each error in experiments on hydrogen and oxygen.
What This Practical Is Testing
This practical, Core Practical 1 on the Edexcel International course, measures the molar volume of a gas: the volume occupied by one mole of carbon dioxide at room temperature and pressure.
The value depends on both temperature and pressure, so both are recorded and quoted with the answer.
Carbon dioxide is made by reacting weighed masses of calcium carbonate with an excess of ethanoic acid. The gas is collected over water in a 100 cm³ measuring cylinder, and the volume is plotted against the mass of solid used.
CaCO₃(s) + 2CH₃COOH(aq) → (CH₃COO)₂Ca(aq) + CO₂(g) + H₂O(l)
The equation gives a 1 : 1 ratio between calcium carbonate and carbon dioxide, so the amount of gas in moles equals the amount of solid in moles. Molar volume is simply the volume of gas divided by that amount.
Because the acid is in excess, the volume of gas depends only on the mass of calcium carbonate, which is why the graph is a straight line through the origin.
On the Edexcel International course this practical is examined through the Unit 3 practical skills paper (WCH13), which asks for methods, apparatus choices, data processing, uncertainty and evaluation.
Questions ask you to justify the apparatus, explain why the solid is weighed before and after tipping, read a value from a line of best fit, and complete the mole calculation with units and significant figures.
They also ask you to state the direction of each error, and suggest a gas syringe or a larger mass as an improvement.
The same mole ideas are tested in Core Practical 4 (preparation of a standard solution) and Core Practical 3 (concentration of hcl by titration).
Key idea: Molar volume = volume of gas ÷ amount of gas in moles. Quote it with its conditions: about 24 dm³ mol⁻¹ at room temperature and pressure.
Calculating the Molar Volume
The graph is read at 0.25 g, a mass inside the measured range that does not correspond to any single run, so the value comes from the line, not from one measurement.
Every step is written out with its unit; students lose marks by jumping from 60 cm³ to 24 dm³ mol⁻¹ without showing the moles.
amount of CaCO₃ = mass ÷ molar mass = 0.25 g ÷ 100.1 g mol⁻¹ = 0.00250 mol
amount of CO₂ = 0.00250 mol (1 : 1 ratio in the equation)
molar volume = volume ÷ amount = 60 cm³ ÷ 0.00250 mol = 24 000 cm³ mol⁻¹ = 24 dm³ mol⁻¹
The molar mass of CaCO₃ is 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹.
The volume was read from the graph to 2 significant figures, so the answer is quoted to 2 significant figures: 24 dm³ mol⁻¹ at room temperature and pressure (about 20 °C and 101 kPa).
Writing 24.0 dm³ mol⁻¹ claims a precision the reading does not have.
A second route that examiners like uses the gradient directly. The gradient of the line is the volume of gas per gram of solid, so multiplying by the mass of one mole gives the volume of one mole.
Molar volume = gradient × Mr = 240 cm³ g⁻¹ × 100.1 g mol⁻¹ = 24 000 cm³ mol⁻¹.
The two routes must agree because reading at 0.25 g is just the gradient multiplied by 0.25.
Show the units: g ÷ g mol⁻¹ = mol, then cm³ ÷ mol = cm³ mol⁻¹, then ÷ 1000 for dm³ mol⁻¹. A unit chain that works is proof the method is right.
Comparing With the Accepted Value
The accepted molar volume of an ideal gas at room temperature and pressure is 24.0 dm³ mol⁻¹.
A result of 24 dm³ mol⁻¹ agrees with it within the precision of the reading, and the comment “the result agrees with the accepted value to 2 significant figures” earns the evaluation mark.
A stricter reading of the line at 0.25 g gives 59 cm³ rather than 60 cm³, and then:
molar volume = 59 ÷ 0.00250 = 23 600 cm³ mol⁻¹ = 23.6 dm³ mol⁻¹
percentage error = (23.6 − 24.0) ÷ 24.0 × 100 = −1.7%
The negative sign matters: it says the experiment gives a molar volume that is too low.
That is exactly what the two main systematic errors, gas escaping before the bung is fitted and carbon dioxide dissolving in the water, would produce.
A percentage error whose sign matches the expected direction of the systematic errors is strong evidence that the method, not carelessness, is responsible.
The accepted value itself depends on conditions, which is why they are recorded.
The ideal gas equation pV = nRT gives, for one mole at the recorded 21 °C (294 K) and 101 kPa, V = nRT ÷ p = 1 × 8.314 × 294 ÷ 101 000 = 0.0242 m³ = 24.2 dm³.
Comparing 24 dm³ mol⁻¹ with 24.2 dm³ mol⁻¹ is the fairest test of the experiment, and quoting the conditions with the result is what allows anyone else to make the same comparison.
Exam wording: “The result is lower than the accepted value. Gas escaped before the bung was fitted and some CO₂ dissolved in the water, both of which make the volume collected, and so the molar volume, too low.”
Check: The Calculation With New Data
Carry out the full calculation, both routes, on a magnesium carbonate experiment and comment on the error.
Checking That Ethanoic Acid Is in Excess
For the volume of gas to depend only on the mass of solid, calcium carbonate must be the limiting reagent in every run.
So the acid must be in excess even for the largest mass allowed. This check should appear in your write-up and is a common two-mark question.
largest mass of CaCO₃ = 0.40 g, so amount = 0.40 ÷ 100.1 = 0.003996 mol
amount of CH₃COOH available = 1.00 mol dm⁻³ × 30 ÷ 1000 dm³ = 0.030 mol
amount of CH₃COOH needed = 2 × 0.003996 = 0.00799 mol
The acid available, 0.030 mol, is nearly four times the 0.00799 mol needed, so ethanoic acid is in excess and calcium carbonate is the limiting reagent in every run.
That is why the volume of carbon dioxide is proportional to the mass of calcium carbonate and the graph is a straight line through the origin.
The excess is large for one run but not for all of them together. The seven masses add up to 1.43 g, or 0.0143 mol of calcium carbonate, which would need 2 × 0.0143 = 0.0286 mol of acid.
If the same 30 cm³ of acid were reused for every run the acid would still be in excess, but only just, and the last runs would be reacting in a much weaker solution and finishing slowly.
Exam tip: Using fresh acid for each run costs nothing and keeps the excess unquestionable; if you are told the acid was reused, say that the excess is marginal and the largest masses are the least reliable.
Exam wording: “Moles of acid available (0.030 mol) is greater than moles needed (2 × 0.0040 = 0.0080 mol), so the acid is in excess and CaCO₃ is the limiting reagent.” Show both numbers and the comparison.
Check: Limiting Reagent
Decide whether the acid is in excess for a different carbonate and a different acid.
Common Mistakes
The same errors appear in students’ write-ups and exam answers year after year. Each one costs a mark that the corrected version earns.
- Writing “the volume of one mole of gas is 24 dm³” with no conditions. The value is only meaningful at a stated temperature and pressure.
- Quoting the molar volume as 24.0 dm³ mol⁻¹ from a volume read to 2 significant figures. Match the significant figures to the least precise measurement.
- Using the mass of powder weighed out instead of the mass by difference. Some powder always stays in the test tube.
- Reading a single run instead of the line, or drawing the line through the anomalous point instead of ignoring it.
- Saying the first bubbles should be discarded “because they are air”. They are, and they are matched by gas left in the apparatus at the end.
- Giving “human error” or “the equipment was inaccurate” as an error. Name the source, its direction and a specific improvement.
- Calculating percentage uncertainty with a made-up uncertainty. Use the instrument’s actual value: ±0.005 g per balance reading, ±0.5 cm³ for the cylinder.
- Forgetting the 2 : 1 ratio in the excess check and comparing 0.030 mol of acid with 0.0040 mol of carbonate directly.
Check: Graphs, Gas Syringes and Conditions
Pick the accurate statement in each round; one word usually makes a statement wrong.
Common Exam Points
Say
“Weighing the test tube before and after tipping gives the mass of CaCO₃ that actually reacted.”
“The line passes through the origin because no solid gives no gas, and reading from it at 0.25 g averages out random error.”
“The molar volume is 24 dm³ mol⁻¹ at room temperature and pressure (2 s.f.); CO₂ escaping before the bung was fitted and dissolving in the water make it too low.”
“A gas syringe avoids the loss of CO₂ by dissolving.”
Do not say
“The acid must be exactly the right amount” (it must be in excess so that CaCO₃ is the limiting reagent).
“Discard the first bubbles because they are air” (keep them; an equal volume of CO₂ stays in the apparatus).
“The result was wrong because of human error” (name the source and its direction). “24 dm³” with no conditions.
Watch for
The excess check needs the 2 : 1 ratio. Percentage error uses the accepted value as the denominator. Percentage uncertainty of a mass by difference uses two readings, ±0.01 g.
The independent variable (mass) goes on the x-axis. Water vapour is the one error that makes the volume too high; every loss of gas makes it too low.
FAQs
Quick answers to the questions students ask most about measuring the molar volume of a gas: fitting the bung, the graph, the first bubbles and the conditions.
Why must the bung be fitted so quickly?
The reaction is fastest in the first few seconds, so gas produced before the bung is in escapes to the room.
The volume collected is then too low and so is the calculated molar volume, and because the loss cannot be measured it cannot be corrected for.
A better arrangement suspends the solid inside the sealed tube and tips it in afterwards.
Why is the line of best fit drawn through the origin?
If no calcium carbonate reacts, no carbon dioxide is produced, so the point (0 g, 0 cm³) is a known point on the graph.
The line is fixed at the origin and then angled so that the measured points are scattered evenly either side of it, ignoring any anomaly.
Why read the graph at 0.25 g instead of using one of the runs?
0.25 g lies inside the range of masses measured, so the reading comes from the line of best fit rather than from a single run.
The line averages the random errors of all seven runs and ignores the anomalous one, so the volume read at 0.25 g (about 60 cm³) is more reliable than any single measurement.
Any mass on the line would give the same molar volume, because the line has one gradient.
Why are the first bubbles not discarded?
They are air pushed out of the delivery tube and the space above the acid by the gas being made.
At the end of the run that same space is full of carbon dioxide that never reached the cylinder, and its volume equals the volume of air pushed out at the start.
Keeping the first bubbles lets the two cancel; discarding them makes the volume too low.
Why record room temperature and pressure?
The volume of a fixed amount of gas depends on both. One mole occupies about 24.0 dm³ at 20 °C and 101 kPa but 24.8 dm³ at 25 °C and 100 kPa, so a molar volume without its conditions cannot be compared with anything.
The ideal gas equation pV = nRT gives the expected value at the recorded conditions.
When is a gas syringe better than collecting over water?
Whenever the gas dissolves in water: carbon dioxide slightly, ammonia, hydrogen chloride and sulfur dioxide almost completely.
The syringe collects the gas directly, so nothing is lost by dissolving and no water vapour is added to the volume. Its own weaknesses are a sticking plunger and a scale read only to the nearest 1 cm³.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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