Chemistry · Simulation and virtual lab Free
The mole and gas volumes: collect hydrogen, weigh gases
Collect hydrogen from weighed metals, find the volume of one mole of gas, and weigh a litre of cooking gas, oxygen or carbon dioxide.
- You will learn to
- Convert between mass, moles and gas volume.
- Determine the molar gas volume from a metal–acid reaction.
- Show that equal volumes of gases contain equal numbers of moles (Avogadro’s law).
- You will measure
- Mass of metal (g), volume of hydrogen (cm³), and the mass of 1 dm³ of each gas (g).
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: magnesium ribbon, balance (0.001 g), conical flask, gas syringe or measuring cylinder over water, dilute HCl.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To determine the molar volume of a gas at room temperature and pressure by reacting a known mass of metal with excess acid, and to show that equal volumes of gases contain equal numbers of moles.
What to do
- Choose magnesium, 0.040 g. Tap Drop the metal into the acid. When the fizzing stops, tap Record.
- Calculate the moles of Mg (mass ÷ 24.3) yourself. The equation shows 1 mol Mg gives 1 mol H₂.
- Repeat with 0.020, 0.060 and 0.080 g. Look at the graph of volume against moles: find the gradient.
- Try zinc and aluminium. Use the equation for aluminium: 2 mol Al give 3 mol H₂. Do the points stay on the same line?
- Change the room temperature to 35 °C. Does the molar volume change?
- Choose Mass of 1 dm³ of a gas. Weigh several gases and plot mass against molar mass.
Results
Questions
What is the molar gas volume at room temperature and pressure?
About 24 dm³ (24 000 cm³) per mole for any gas.
How many moles are in 0.048 g of magnesium? (Mg = 24)
n = 0.048 ÷ 24 = 0.002 mol.
What volume of hydrogen does 0.002 mol of magnesium produce at RTP?
Mg + 2HCl → MgCl₂ + H₂, so 0.002 mol H₂ × 24 000 cm³ = 48 cm³.
Why must the acid be in excess?
So that all the metal reacts; the volume of hydrogen then depends only on the mass of metal.
Why does 1 dm³ of carbon dioxide weigh more than 1 dm³ of oxygen?
Both contain the same number of moles (Avogadro’s law), but CO₂ has a larger molar mass (44 g/mol against 32 g/mol).
Why does the molar volume increase at a higher temperature?
Gas particles move faster and spread out more at the same pressure, so one mole takes up more space (V ∝ T).
For teachers: expected results and how the model works
Molar volume in the model: Vm = 24.0 × (T + 273) ÷ 293 dm³/mol at 101 kPa, so 24.0 at 20 °C and 24.4 at 25 °C. At 25 °C: 0.040 g Mg (1.65 mmol) gives 40.2 cm³; 0.080 g Zn (1.22 mmol) gives 29.9 cm³; 0.040 g Al (2.22 mmol H₂) gives 54.3 cm³. The syringe reads to 120 cm³ (0.080 g Al at 35 °C gives 110 cm³). Zinc reacts more slowly than magnesium; aluminium in the model starts without its oxide layer. Masses of 1.00 dm³ at 25 °C: H₂ 0.08 g, CH₄ 0.65 g, N₂ 1.15 g, O₂ 1.31 g, CO₂ 1.80 g, butane 2.37 g. The gradient of mass against molar mass is 1 ÷ Vm.
Exam practice Exam set
National-exam style questions on this simulation (14 marks). Try each one on paper first, then open the marking scheme.
0.12 g of magnesium reacts with excess dilute hydrochloric acid. (a) Write the equation. (b) Calculate the volume of hydrogen at RTP. (Mg = 24, molar volume 24 dm³) (5 marks)
Marking scheme: (a) Mg + 2HCl → MgCl₂ + H₂ (2). (b) n(Mg) = 0.12 ÷ 24 = 0.005 mol (1); n(H₂) = 0.005 mol (1); V = 0.005 × 24 000 = 120 cm³ (1).
Calculate the mass of 600 cm³ of carbon dioxide at RTP. (C = 12, O = 16) (3 marks)
Marking scheme: n = 600 ÷ 24 000 = 0.025 mol (1); M = 44 g/mol (1); mass = 0.025 × 44 = 1.1 g (1).
A student collected 66 cm³ of hydrogen from 0.054 g of aluminium. Use 2Al + 6HCl → 2AlCl₃ + 3H₂ to find the molar volume she measured. (Al = 27) (4 marks)
Marking scheme: n(Al) = 0.054 ÷ 27 = 0.002 mol (1); n(H₂) = 0.002 × 3 ÷ 2 = 0.003 mol (1); molar volume = 66 ÷ 0.003 = 22 000 cm³ (1) = 22 dm³/mol (1).
Suggest two reasons why a measured molar volume may be smaller than 24 dm³. (2 marks)
Marking scheme: Some gas escaped before the bung was replaced / leaks (1); the metal had an oxide coating so less metal reacted / the syringe stuck (1).
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
The mole and gas volumes: collect hydrogen, weigh gases
Aim: To determine the molar volume of a gas at room temperature and pressure by reacting a known mass of metal with excess acid, and to show that equal volumes of gases contain equal numbers of moles.
Apparatus (real laboratory): magnesium ribbon, balance (0.001 g), conical flask, gas syringe or measuring cylinder over water, dilute HCl.
What to do (with the simulation on www.donatstem.co.rw)
- Choose magnesium, 0.040 g. Tap Drop the metal into the acid. When the fizzing stops, tap Record.
- Calculate the moles of Mg (mass ÷ 24.3) yourself. The equation shows 1 mol Mg gives 1 mol H₂.
- Repeat with 0.020, 0.060 and 0.080 g. Look at the graph of volume against moles: find the gradient.
- Try zinc and aluminium. Use the equation for aluminium: 2 mol Al give 3 mol H₂. Do the points stay on the same line?
- Change the room temperature to 35 °C. Does the molar volume change?
- Choose Mass of 1 dm³ of a gas. Weigh several gases and plot mass against molar mass.
Results
| # | Metal | Mass (g) | Moles of metal (mol) | Moles of H₂ (mol) | Volume of H₂ (cm³) | Molar volume (dm³/mol) |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
Graph
Plot the volume of hydrogen (cm³) against moles of hydrogen (mmol). Draw the best line through the origin and use its gradient to find the molar volume.
Questions
- What is the molar gas volume at room temperature and pressure? [2]
- How many moles are in 0.048 g of magnesium? (Mg = 24) [2]
- What volume of hydrogen does 0.002 mol of magnesium produce at RTP? [2]
- Why must the acid be in excess? [2]
- Why does 1 dm³ of carbon dioxide weigh more than 1 dm³ of oxygen? [2]
- Why does the molar volume increase at a higher temperature? [2]
- 0.12 g of magnesium reacts with excess dilute hydrochloric acid. (a) Write the equation. (b) Calculate the volume of hydrogen at RTP. (Mg = 24, molar volume 24 dm³) [5]
- Calculate the mass of 600 cm³ of carbon dioxide at RTP. (C = 12, O = 16) [3]
- A student collected 66 cm³ of hydrogen from 0.054 g of aluminium. Use 2Al + 6HCl → 2AlCl₃ + 3H₂ to find the molar volume she measured. (Al = 27) [4]
- Suggest two reasons why a measured molar volume may be smaller than 24 dm³. [2]
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: The mole and gas volumes: collect hydrogen, weigh gases
Expected results
Molar volume in the model: Vm = 24.0 × (T + 273) ÷ 293 dm³/mol at 101 kPa, so 24.0 at 20 °C and 24.4 at 25 °C. At 25 °C: 0.040 g Mg (1.65 mmol) gives 40.2 cm³; 0.080 g Zn (1.22 mmol) gives 29.9 cm³; 0.040 g Al (2.22 mmol H₂) gives 54.3 cm³. The syringe reads to 120 cm³ (0.080 g Al at 35 °C gives 110 cm³). Zinc reacts more slowly than magnesium; aluminium in the model starts without its oxide layer. Masses of 1.00 dm³ at 25 °C: H₂ 0.08 g, CH₄ 0.65 g, N₂ 1.15 g, O₂ 1.31 g, CO₂ 1.80 g, butane 2.37 g. The gradient of mass against molar mass is 1 ÷ Vm.
Answers
- What is the molar gas volume at room temperature and pressure? [2]
About 24 dm³ (24 000 cm³) per mole for any gas.
- How many moles are in 0.048 g of magnesium? (Mg = 24) [2]
n = 0.048 ÷ 24 = 0.002 mol.
- What volume of hydrogen does 0.002 mol of magnesium produce at RTP? [2]
Mg + 2HCl → MgCl₂ + H₂, so 0.002 mol H₂ × 24 000 cm³ = 48 cm³.
- Why must the acid be in excess? [2]
So that all the metal reacts; the volume of hydrogen then depends only on the mass of metal.
- Why does 1 dm³ of carbon dioxide weigh more than 1 dm³ of oxygen? [2]
Both contain the same number of moles (Avogadro’s law), but CO₂ has a larger molar mass (44 g/mol against 32 g/mol).
- Why does the molar volume increase at a higher temperature? [2]
Gas particles move faster and spread out more at the same pressure, so one mole takes up more space (V ∝ T).
- 0.12 g of magnesium reacts with excess dilute hydrochloric acid. (a) Write the equation. (b) Calculate the volume of hydrogen at RTP. (Mg = 24, molar volume 24 dm³) [5]
(a) Mg + 2HCl → MgCl₂ + H₂ (2). (b) n(Mg) = 0.12 ÷ 24 = 0.005 mol (1); n(H₂) = 0.005 mol (1); V = 0.005 × 24 000 = 120 cm³ (1).
- Calculate the mass of 600 cm³ of carbon dioxide at RTP. (C = 12, O = 16) [3]
n = 600 ÷ 24 000 = 0.025 mol (1); M = 44 g/mol (1); mass = 0.025 × 44 = 1.1 g (1).
- A student collected 66 cm³ of hydrogen from 0.054 g of aluminium. Use 2Al + 6HCl → 2AlCl₃ + 3H₂ to find the molar volume she measured. (Al = 27) [4]
n(Al) = 0.054 ÷ 27 = 0.002 mol (1); n(H₂) = 0.002 × 3 ÷ 2 = 0.003 mol (1); molar volume = 66 ÷ 0.003 = 22 000 cm³ (1) = 22 dm³/mol (1).
- Suggest two reasons why a measured molar volume may be smaller than 24 dm³. [2]
Some gas escaped before the bung was replaced / leaks (1); the metal had an oxide coating so less metal reacted / the syringe stuck (1).