9. The molar gas volume and Avogadro's Law calculations (re-edit)

Doc Brown's Chemistry - GCSE, IGCSE, GCE (basic A level), O Level, US grade 10-12  online Chemical Calculations

9. The molar gas volume in calculations, application of Avogadro's Law to chemical reactions involving gases

Note on two acronyms and associated units:

STP or s.t.p. is a largely defunct abbreviation for 'standard temperature and pressure', 0^oC/273K temperature and a pressure of 1 atm/101.3 kPa (101325 Pa). (In bygone times 760 mm of Hg of a 'mercury' barometer/manometer)

NTP or RTP is a frequently used abbreviation for 'normal' laboratory room temperature and pressure, taken as 20^oC/293K temperature and pressure of 1 atm/101.3 kPa pressure (101325 Pa), however many standard pieces of chemical data are based on 25^oC/298K, so take care on units! (temperature in K = ^oC + 273)

See also chemistry calculations Part 10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants to products)

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

It is better to know how to rearrange an equation than use a formula triangle

A summary of how to do basic molar gas volume calculations and rearrangement of the formula needed to solve problems

• Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules (irrespective of the nature of the gaseous particles)

  • This means equal amounts of moles of gases occupy the same volume under the same conditions of temperature and pressure.

  • So the volumes have equal moles of separate particles (molecules or individual atoms) in them.

  • Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the same volume .

  • This is 24 dm³ (24 litres) or 24000 cm³, at room temperature of 20^oC/293K and normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to as RTP/NTP).

  • 20^oC (293K) is often treated as room temperature, and the molar gas volume is 24.0 dm³ at normal pressure.

  • 25^oC (298K) is treated as a standard temperature, at which the molar volume of the gas is 24.4 dm³ at normal pressure.

  • The molar volume for s.t.p is 22.4 dm³ (22.4 litres) at 0^oC and 101 kPa1 atmosphere pressure.

  • Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25^oC/298K is usually considered the standard temperature for data e.g. thermochemical and thermodynamic data at a higher academic level.

  • Unfortunately, on the internet, both 20^oC and 25^oC are referred to as 'room temperature'.

• Reminder about the Avogadro Number:

  • The Avogadro number is 6.02 x 10²³ particles/mole of defined chemical 'species'.

  • Therefore, e.g. at 20^oC, 24.0 dm³ of a gas contains 6.03 x 10²³ molecules.

  • 6.03 x 10²³ /  24.0 = 2.5 x 10²² = number of molecules in 1 dm³ of gas.

  • 2.5 x 10²² / 1000 = 2.5 x 10¹⁹ = number of molecules in 1 cm³ of gas.

• Some handy (essential) relationships for substance Z below:

• It is much better to be able to rearrange an equation than use a triangle.

• 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

  • moles = V(dm³) / 24   (at RTP)

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

  • moles = mass / molecular mass = gas volume / volume of 1 mole

  • mass / molecular mass = gas volume / volume of 1 mole

  • molecular mass = mass x volume of 1 mole/volume

  • therefore at RTP: M_r = mass(g) x 24 / V(dm³)

  • 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. 25^oC and 1 atmosphere pressure so the molar volume is 24dm³ or 24000cm³.

• 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!

• Methods of measuring how much gas is formed in a reaction

  • (volume can be compared with theoretical prediction!)

  • (a)

    • You can collect the gases in a calibrated gas syringe.

    • You must make sure too much gas isn't produced and too fast!

    • A gas syringe is more accurate than collecting the gas in an inverted measuring cylinder under water shown below, but its still only accurate to the nearest cm³.

    • You can collect any gas by this method.

  • (b)

    • The gas is collected in a measuring cylinder filled with water and inverted over a trough of water.

    • You can get a more accurate result by using an inverted burette instead of a measuring cylinder.

    • However, this method is no good if the gas is soluble in water!

    • Burettes are calibrated in 0.10 cm³ intervals. measuring cylinders to the nearest cm³ or worse!

    • In both methods the reaction is carried out in conical flask fitted with a sealing rubber bung, but a tube enabling the gas evolved to be collected in some suitable container.

  • (c)

    • A third method is to measure the mass loss due to gas loss by carrying out the reaction in a flask set up on an accurate one-pan electronic balance.

    • You need to put a cotton wool plug in the neck of the conical flask in case you lose any of the solution in a spray as the gas bubbles up - effervescence can produce an aerosol.

    • This method can be used for any reaction that produces a gas, but the gas is released into the laboratory, ok if its harmless.

    • It is potentially the most accurate method, BUT, the mass loss may be quite small especially hydrogen [M_r(H₂) = 2], better for the 'heavier' gas carbon dioxide [M_rCO₂) = 44]

    • There is also a possible error if the gas dissolves in water, oxygen has a very low solubility, but carbon dioxide is much more soluble and will reduce the measured mass of gas evolved and measured.

• In these gas volume calculations assume the molar gas volume is 24 dm³ (24 000 cm³) at 'normal' temperature and pressure.

• Molar gas volume calculation Example 9.1

  • (a) What is the volume of 3.5 g of hydrogen (H₂)? [A_r(H) = 1]

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

  • method (i)

    • 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 (ii):

    • 2 g 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)

    • -

  • (b) What is the volume of 11.6 g of butane gas at RTP?

    • The formula of butane is C₄H₁₀, atomic masses are C = 12 and H = 1

    • Formula mass of butane = (4 x 12) + (10 x 1) = 58

    • moles butane = mass butane / formula mass = 11.6/58 = 0.20 mol

    • volume = moles x molar volume

    • volume butane = 0.20 x 24 = 4.8 dm³

    • -

  • (c) What mass of chlorine (Cl₂) does 6.0 dm³ of the gas contain at rtp?

    • Atomic mass of Cl = 35.5, molecular mass of chlorine = 2 x 35.5 = 71

    • 1 mol gas = 24 dm³, so mol Cl₂ = 6/24 = 0.25 mol

    • mass = moles x molecular mass = 0.25 x 71 = 17.75 g (4sf, 2dp)

    • -

  • (d) What mass of methane (CH₄) is contained in a 10 dm³ cylinder at three times normal atmospheric pressure at room temperature.

    • Atomic masses: C = 12,  H = 1, molar gas volume = 24 dm³.

    • Formula mass of methane = 16 + (4 x 1) = 16

    • At normal pressure mol of CH₄ would be 10/24 = 0.4166

    • However, in 3x normal pressure there will be 3x as many methane particles in the same volume.

    • So the actual number of moles at this higher pressure will be 3 x 0.4166 = 12.5 mol

    • mass CH₄ = mol CH₄ x formula mass of CH₄ = 12.5 x 16 = 200 g (or 0.20 kg)

    • -

  • From now on the calculations involve interpreting an equation in terms of moles e.g.

  • (e) If 10 g of calcium carbonate (limestone) is thermally decomposed what volume of carbon dioxide is formed at room temperature and pressure?

    • The equation is: CaCO₃(s)  ===>  CaO(s)  +  CO₂(g)

    • Atomic masses: Ca = 40,  C = 12,  O =16. Formula mass CaCO₃ = 40 + 12 + (3 x 16) = 100

    • 1 mol of calcium carbonate gives 1 mol of carbon dioxide.

    • Therefore mol CaCO₃ = mol CO₂ = 10/100 = 0.10 mol

    • Therefore volume CO₂ = mol CO₂ x molar gas volume

    • Volume of CO₂ = 0.10 x 24.0 = 2.4 dm³ (or 2.4 x 1000 = 2400 cm³)

    • -

  • AND from now on, you have to the questions - ANSWERS at bottom of page

  • AND remember it is better to be able to rearrange a formula than use a triangle.

  • AND do not assume all atomic masses quoted are needed to solve the problem.

• Molar gas volume calculation Example 9.2 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

  • Given the equation

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

  • This equation is read as 1 mole of magnesium carbonate reacts with 1 mole of sulfuric acid to form 1 mole of magnesium sulfate, 1 mole of water and 1 mole of carbon dioxide gas

  • 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]

  • The important mole ratio is 1 MgCO₃ ==> 1 CO₂

  • Fully worked out ANSWERS at the bottom of the page

    • -

• Molar gas volume calculation Example 9.3 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

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

  • Fully worked out ANSWERS at the bottom of the page

    -

• Molar gas volume calculation Example 9.4 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

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

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

  • The equation is read as 1 mole of calcium atoms reacts with 2 moles of hydrochloric acid to form 1 mole of calcium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

  • What volume of hydrogen is formed when ...

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

      • The important mole ratio is 1 Ca ==> 1 H₂

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

      • The important mole ratio is 1 Ca + 2HCl

• Molar gas volume calculation Example 9.5 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

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

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

  • The equation is read as 1 mole of magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of magnesium chloride salt and 1 mole of hydrogen gas molecules (NOT atoms).

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

  • Fully worked out ANSWERS at the bottom of the page

    • -

• Molar gas volume calculation Example 9.6 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

  • 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) 

    • atomic masses: Na = 23,  H = 1,  C = 12,  O = 16

    • Fully worked out ANSWERS at the bottom of the page

    • -

• Molar gas volume calculation Example 9.7 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

  • What volume of carbon dioxide is formed at RTP when 5g of carbon is completely burned?

    • C(s) + O₂(g) ==> CO₂(g)

    • atomic mass of carbon = 12  O = 16

    • Fully worked out ANSWERS at the bottom of the page

      • -

• Molar gas volume calculation Example 9.8 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

  • 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)

    • atomic masses: C = 12, O = 16  H =1

    • Fully worked out ANSWERS at the bottom of the page

    • -

• Molar gas volume calculation Example 9.9 (At RTP 1 mol gas = 24 dm³ or 24000 cm³)

  • 3.27g of metal dissolved in dilute sulfuric acid to form 1.2 dm³ of hydrogen gas at RTP.

  • It was believed that the equation for the reaction is:

    • M  +  H₂SO₄  ===>  MSO₄  +  H₂

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

  • Calculate the atomic mass of the metal.

  • Fully worked out ANSWERS at the bottom of the page

  • -

• Molar gas volume calculation Example 9.10

  • Sodium hydrogencarbonate reacts with acids to form a salt, water and carbon dioxide e.g.

  • sodium hydrogencarbonate  +  sulfuric acid  ===>  sodium sulfate  +  water  + carbon dioxide

  • The balanced symbol equation:

    • 2NaHCO₃(s)  +  H₂SO₄(aq)  ===>  Na₂SO₄(aq)  +  2H₂O(l)  +  2CO₂(g)

  • Atomic masses: Na =23,   H = 1,   S = 32,   O = 16

  • (a) Calculate the formula masses of sodium hydrogencarbonate, sodium sulfate and carbon dioxide.

    • -

  • (b) What is the maximum volume of carbon dioxide that can be collected if 10.0 g of sodium hydrogencarbonate is dissolved in excess sulfuric acid?

    • Fully worked out ANSWERS at the bottom of the page

    • -

  • (c) What mass of sodium sulfate is made assuming a 100% yield (hardly so in practice!)

    • Fully worked out ANSWERS at the bottom of the page

    • -

• Molar gas volume calculation Example 9.11

  • Atomic masses: Na = 23,  H = 1,  O = 16  and the volume of 1 mole of gas is 24.0 dm³.

  • Sodium reacts with excess water to form sodium hydroxide and hydrogen gas.

  • If 4.6 g of sodium reacts with water, what volume of hydrogen is formed at room temperature and pressure?

  • The equation is: 2Na(s)  +  2H₂O(l)  ===>  2NaOH(aq)  +  H₂(g)

  • Fully worked out ANSWERS at the bottom of the page

Learning objectives for problem solving using the molar gas volume

Know that the volume for 1 mole of gas varies with temperature and pressure.

Know the different units of temperature and pressure and be able to convert from one to the other in the context of calculations of moles and volumes of gases.

Know how to solve problems and do calculations based on the volume of 1 mole of gas (molar volume, usually taken as 24 dm³ at GCSE level).

Know at RTP moles of gas = volume of gas in dm³ / 24 and be able to rearrange this moles of gas' equation.

Know that Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.

Be able to describe and explain how to measure the volume of a gas formed in a chemical reaction e.g. collection in gas syringe or recording loss of mass from a system as a gas is produced and escapes into the atmosphere.

Know how to read equations in terms of (molar) mole ratios e.g. to deduce volumes of gases formed from a given mass of reactant - spot the crucial mole ratio in the equation.

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WHERE NEXT?

Self-assessment Quizzes on reacting gas volume ratios and Avogadro's Law

type in answer QUIZ   

or   multiple choice QUIZ

See also for gas calculations

Notes on the gas laws and P V T gas calculations

Part 10. Reacting gas volume ratios, Avogadro's Law & Gay-Lussac's Law

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