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9. The molar gas volume in calculations, application of Avogadro's Law

This page describes and explains, with fully worked out examples, how to calculate the volume of gas formed from given masses of reactants. You need to know the formula connecting moles, mass and formula mass AND know how to use the molar volume in these gas volume calculation methods. (updated Feb 23rd)

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9. The molar gas volume in calculations, moles, gas volumes and Avogadro's Law

Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules. So the volumes have equal moles of separate particles in them. One mole of any gas (or the formula mass in g), at the same temperature and pressure occupies the same volume . This is 24dm³ (24 litres) or 24000 cm³, at room temperature of 25^oC/298K and normal pressure of 101.3kPa/1atmosphere (both = RTP). The molar volume for s.t.p is 22.4 dm³ (22.4 litres) at 0^oC and 1atmosphere pressure. Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25^oC/298K is usually considered the standard temperature (RTP).

• Some handy relationships for substance Z below:

• moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)

  • mass of Z in g = moles of Z x atomic or formula mass of Z

  • atomic or formula mass of Z = mass of Z / moles of Z

  • 1 mole = formula mass of Z  in g.

• gas volume of Z = moles of Z x molar volume

  • moles of Z = gas volume of Z / molar volume

• Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25^oC and 1 atmosphere pressure.

• Note (ii):

  • Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ...

  • It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles!

• Example 9.1: What is the volume of 3.5g of hydrogen? [A_r(H) = 1]

  • common thinking: hydrogen exists as H₂ molecules, so M_r(H₂) = 2, so 1 mole or formula mass in g = 2g

  • method (a)

    • so moles of hydrogen  = 3.5/2 = 1.75 mol H₂ 

    • so volume H₂ = mol H₂ x molar volume = 1.75 x 24 = 42 dm³ (or 42000 cm³)

  • method (b):

    • 2g occupies 24 dm³, so scaling up for the volume of hydrogen ...

    • 3.5 g will have a volume of 3.5/2  x 24 = 42 dm³ (or 42000 cm³)

• Example 9.2: Given the equation

  • MgCO_3(s) + H₂SO_4(aq) ==> MgSO_4(aq) + H₂O_(l) +CO_2(g)

  • What mass of magnesium carbonate is needed to make 6 dm³ of carbon dioxide? [A_r's: Mg = 24, C = 12, O = 16, H =1 and S = 32]

  • method (a):

    • since 1 mole = 24 dm³, 6 dm³ is equal to 6/24 = 0.25 mol of gas

    • From the equation, 1 mole of MgCO₃ produces 1 mole of CO₂, which occupies a volume of 24 dm³.

    • so 0.25 moles of MgCO₃ is need to make 0.25 mol of CO₂ 

    • formula mass of MgCO₃ = 24 + 12 + 3x16 = 84,

    • so required mass of MgCO₃ = mol x formula mass = 0.25 x 84 = 21g

  • method (b):

    • converting the equation into the required reacting masses ..

    • formula masses: MgCO₃ = 84 (from above), CO₂ = 12 + 2x16 = 44

    • MgCO₃ : CO₂ equation ratio is 1 : 1

    • so 84g of MgCO₃ will form 44g of CO₂ 

    • 44g of CO₂ will occupy 24dm³

    • so scaling down, 6 dm³ of CO₂ will have a mass of 44 x 8/24 = 11g

    • if 84g MgCO₃ ==> 44g of CO₂, then ...

    • 21g MgCO₃ ==> 11g of CO₂ by solving the ratio, scaling down by factor of 4

• Example 9.3: 6g of a hydrocarbon gas had a volume of 4.8 dm³. Calculate its molecular mass.

  • method (a):

    • 1 mole = 24 dm³, so moles of gas = 4.8/24 = 0.2 mol

    • molecular mass = mass in g / moles of gas

    • M_r = 6 / 0.2 = 30

    • i.e. if 6g = 0.2 mol, 1 mol must be equal to 30g by scaling up

  • method (b):

    • 6g occupies a volume of 4.8 dm³

    • the formula mass in g occupies 24 dm³

    • so scaling up the 6g in 4.8 dm³

    • there will be 6 x 24/4.8 = 30g in 24 dm³ 

    • so the molecular or formula mass = 30

• Example 9.4: Given the equation ... (and A_r's Ca = 40, H = 1, Cl = 35.5)

  • Ca_(s) + 2HCl_(aq) ==> CaCl_2(aq) + H_2(g)

  • What volume of hydrogen is formed when ...

    • (i) 3g of calcium is dissolved in excess hydrochloric acid?

    • (ii) 0.25 moles of hydrochloric acid reacts with calcium?

  • (i) method (a):

    • 3g Ca = 3/40 = 0.075 mol Ca

    • from 1 : 1 ratio in equation, 1 mol Ca produces 1 mol H₂ 

    • so 0.075 mol Ca produces 0.075 mol H₂ 

    • so volume H₂ = 0.075 x 24 = 1.8 dm³ (or 1800 cm³)

  • (i) method (b):

    • from equation 1 Ca ==> 1 H₂ means 40g ==> 2g

    • so scaling down: 3g Ca will produce 2 x 3/40 = 0.15g H₂

    • 2g H₂ has a volume 24 dm³, so scaling down ...

    • 0.15g H₂ has a volume of (0.15/2) x 24 = 1.8 dm³ (or 1800 cm³)

  • (ii) method (a) only:

    • from equation: 2 moles HCl ==> 1 mole H₂ (mole ratio 2:1)

    • so 0.25 mol HCl ==> 0.125 mol H₂, volume 1 mole gas = 24 dm³ 

    • so volume H₂ = 0.125 x 24 = 3 dm³

• Example 9.5: Given the equation ... (and A_r's Mg = 24, H = 1, Cl = 35.5)

  • Mg_(s) + 2HCl_(aq) ==> MgCl_2(aq) + H_2(g)

  • How much magnesium is needed to make 300 cm³ of hydrogen gas?

  • method (a)

    • 300 cm³ = 300/24000 = 0.0125 mol H₂ (since 1 mol of any gas = 24000 cm³)

    • from the equation 1 mole Mg ==> 1 mole H₂ 

    • so 0.0125 mole Mg needed to make 0.0125 mol H₂ 

    • so mass of Mg = mole Mg x A_r(Mg)

    • so mass Mg needed = 0.0125 x 24 = 0.3g

  • method (b)

    • reaction ratio in equation is 1 Mg ==> 1 H₂,

    • so reacting mass ratio is 24g Mg ==> 2g H₂,

    • 2g H₂ has a volume of 24000 cm³ (volume of formula mass in g)

    • so scaling down: mass Mg needed = 24 x (300/24000) = 0.3g

• Example 9.6: A small teaspoon of sodium hydrogencarbonate (baking soda) weighs 4.2g. Calculate the moles, mass and volume of carbon dioxide formed when it is thermally decomposed in the oven. Assume room temperature for the purpose of the calculation.

  • 2NaHCO_3(s) ==> Na₂CO_3(s) + H₂O_(g) + CO_2(g) 

  • Formula mass of NaHCO₃ is 23+1+12+(3x16) = 84 = 84g/mole

  • Formula mass of CO₂ = 12+(2x16) = 44 = 44g/mole (not needed by this method)

    • or a molar gas volume of 24000 cm³ at RTP (definitely needed for this method)

  • In the equation 2 moles of NaHCO₃ give 1 mole of CO₂ (2:1 mole ratio in equation)

  • Moles NaHCO₃ = 4.2/84 = 0.05 moles ==> 0.05/2 = 0.025 mol CO₂ on decomposition.

  • Mass = moles x formula mass, so mass CO₂ = 0.025 x 44 = 1.1g CO₂ 

  • Volume = moles x molar volume = 0.025 x 24000 = 600 cm³ of CO₂ 

• Example 9.7: What volume of carbon dioxide is formed at RTP when 5g of carbon is burned?

  • C(s) + O₂(g) ==> CO₂(g)

  • 1 mole carbon gives 1 mole of carbon dioxide, atomic mass of carbon = 12

  • moles = mass / atomic mass, moles carbon = moles carbon dioxide = 5/12 = 0.417 mol

  • 1 mole of gas at RTP occupies 24 dm³

  • so 0.417 mol occupies a volume of 0.417 x 24 = 10.0 dm³

• Example 9.8: What volume of carbon dioxide gas is formed at RTP if 1Kg of propane gas fuel is burned?

  • C₃H₈(g) + 5O₂(g)  ==> 3CO₂(g) + 4H₂O(l)

  • 1 mole of propane gas gives 3 moles of carbon dioxide gas on complete combustion

  • 1 kg = 1000g, atomic masses: C = 12, H =1

  • M_r(propane) = (3 x 12) + (8 x 1) = 44

  • moles = mass in g / molecular mass, therefore moles propane = 1000/44 = 22.73 mol

  • from equation molar ratio: moles carbon dioxide = 3 x moles of propane

  • mol propane = 3 x 22.73 = 68.18 mol

  • 1 mole of gas at RTP occupies a volume of 24 dm³

  • so 68.18 mol of gas occupies a volume of 68.18 x 24 = 1636 dm³

• See also QA7.1 on the mole introduction page

• Self-assessment Quizzes

  • [mvg] type in answer Honly  or  multiple choice Honly

OTHER CALCULATION PAGES

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law (this page)

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants-products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do volumetric titration calculations e.g. acid-alkali titrations (and diagrams of apparatus)

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

17. Radioactivity & half-life calculations including dating materials

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