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study examples carefully9. The molar gas volume in calculations, application of Avogadro's Lawstudy examples carefully

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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study examples carefully9. The molar gas volume in calculations, moles, gas volumes and Avogadro's Lawstudy examples carefully

  • 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 25oC/298K and normal pressure of 101.3 kPa or 1 atmosphere (both = RTP).

    • The molar volume for s.t.p is 22.4 dm3 (22.4 litres) at 0oC and 1atmosphere pressure.

    • Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25oC/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 volume of 1 mole

    • rearranging this equation gives ...

    • moles of Z = gas volume of Z / volume of 1 mole

    • The latter form of the equation can be used to calculate molecular mass from experimental data because

      • moles = mass / molecular mass

      • molecular mass = mass / moles

      • so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.

      • This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer.

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

  • top sub-indexMolar gas volume calculation Example 9.1

    • What is the volume of 3.5g of hydrogen? [Ar(H) = 1]

    • common thinking: hydrogen exists as H2 molecules, so Mr(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)

  • Molar gas volume calculation Example 9.2

    • Given the equation

    • MgCO3(s) + H2SO4(aq) ==> MgSO4(aq) + H2O(l) +CO2(g)

    • What mass of magnesium carbonate is needed to make 6 dm3 of carbon dioxide? [Ar'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

  • Molar gas volume calculation Example 9.3

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

    • top sub-indexmethod (a):

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

      • molecular mass = mass in g / moles of gas

      • Mr = 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

  • Molar gas volume calculation Example 9.4

    • Given the equation ... (and Ar's Ca = 40, H = 1, Cl = 35.5)

    • Ca(s) + 2HCl(aq) ==> CaCl2(aq) + H2(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

  • top sub-indexMolar gas volume calculation Example 9.5

    • Given the equation ... (and Ar's Mg = 24, H = 1, Cl = 35.5)

    • Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(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 Ar(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

  • Molar gas volume calculation 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.

      • 2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(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 

  • Molar gas volume calculation 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

  • Molar gas volume calculation 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

      • Mr(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

  • top sub-indexSelf-assessment Quizzes


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

  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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