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# CSEC Chemistry: The Mole Concept in Relation to Gases

Updated: Apr 17, 2020

Gases

When dealing with gases, due to their low density, we measure them by volume instead of mass.

CSEC requires that you know Avogadro's Law, which uses the idea of the mole to relate the volume of gases.

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

The volume of a gas which contains one mole of molecules of the gas is referred to as its molar volume. This value is the same for all gases when measured under the same temperature and pressure conditions.

Molar volume is considered under one of two sets of conditions:

1. Standard Temperature and Pressure (s.t.p.), a temperature of 273 K (0 degrees Celsius) and 101 kPa (1 atm). The approximate volume of 1 mole of any gas at s.t.p. is 22.4 dm^3 (22400 cm^3).

2. Room Temperature and Pressure (r.t.p.), a temperature of 298 K (25 degrees Celsius) and pressure of 101 kPa (1 atm). The approximate volume of 1 mole of any gas at r.t.p. is 24 dm^3 (24000 cm^3).

If you are asked to calculate how much volume is occupied by a certain mass of a gas:

(mass given/molar mass of gas molecule) × volume at r.t.p or s.t.p based on the question

Example: What volume is occupied by 8 grams of hydrogen at s.t.p.?

1 mol of hydrgen gas (H2) has a mass of (2×1) = 2 g

2 g of hydrogen occupy 22.4 dm^3 at s.t.p., therefore:

8 g of hydrogen occupy (8 g/2 g) × 22.4 dm^3 = 89.6 dm^3 at s.t.p.

You can also find the number of molecules present in a certain volume of gas:

(volume given/molar volume at s.t.p. or r.t.p.) × Avogadro's number

Example: How many molecules are present in 4.2 dm^3 of carbon dioxide at s.t.p.?

Any gas has a molar volume of 22.4 dm^3 at s.t.p and contains 6.0 × 10^23,

therefore:

(4.2/22.4) × 6.0 × 10^23 = 1.125 × 10^25 CO2 molecules

Finding the molar mass of a gas

Example: 100 cm^3 of nitrogen gas (relative molecular mass= 28) has a mass of 0.14 g.

a) Calculate the molar mass of a gaseous hydrocarbon if 250 cm^3 weighs 0.7 g at the same temperature and pressure as the nitrogen. (In case you're unfamiliar with hydrocarbons, they are just organic compounds made up only of carbon and hydrogen atoms).

number of moles of N2 present in 100 cm^3= 0.14 g/28 gmol^-1= 0.005 mol

According to Avogadro's Law, 250 cm^3 of hydrocarbon at the same temperature and pressure will contain:

(250 cm^3/100 cm^3) × 0.005 mol = 0.0125 mol

Hence, 0.0125 mol of the hydrocarbon weighs 0.7 g and this means that 1 mol of the hydrocarbon weighs:

(1 mol/0.0125 mol) × 0.7 g = 56 g

b) Suggest the likely formula for the hydrocarbon.

Since the molar mass of the hydrocarbon is 56 gmol-1, it corresponds that the compound (containing hydrogen and carbon atoms) must contain 4 carbon atoms (4 × 12 = 48) and 8 hydrogen atoms (8 × 1 = 8). The molecular formula of the hydrocarbon is C4H8 (butene).

Finding the Volumes of Gaseous Reactants and Products in a Balanced Equation

Balanced equations give the number of moles of reactants and products involved in the reaction. Now that we know Avogadro's Law, we can change the number of moles to volume (and vice versa), since the volume occupied by 1 mole of a gas at a specific pressure and temperature will be the same regardless of the type of gas. So, you could think of the number of reactants and products as a ratio of volumes.

Example: What volume of oxygen is required for the complete combustion of 100 cm^3 of propane? What volume of carbon dioxide is produced? (All gas volumes are measured at the same pressure and temperature)

First, we write the balanced chemical equation:

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

From this, we can tell that 1 mole of propane reacts with 5 moles of oxygen gas to form 3 moles of carbon dioxide (and 4 moles of water).

This is the same as saying 1 volume of propane reacts with 5 volumes of oxygen gas to give 3 volumes of carbon dioxide.

We then multiply by the value of '1 volume', 1 volume being 100 cm^3 (propane). So, the volume of oxygen needed is 500 cm^3 and the volume of carbon dioxide produced is 300 cm^3.

Using Gas Volumes to Calculate the Equation for a Reaction

When heated, 40 cm^3 of ammonia gives 20 cm^3 of nitrogen and 60 cm^3 of hydrogen. Work out the decomposition reaction of ammonia by heat.

First, we write the information given in the form of a chemical reaction:

40 cm^3 of NH3 ---> 20 cm^3 of N2 + 60 cm^3 of H2

Then, we divide by the smallest volume (20 cm^3), giving us:

2NH2(g) --->N2(g) + 3H2(g)

So now, you can likely see how chemical equations involving gases could be related to as a ratio of volumes. In the first example, you can see the ratio of the volumes of gases to be 1:5:3. When the volume of propane (1 volume) is shown to be 100 cm^3, we multiply the values in the ratio by 100 cm^3 to find the volumes of the other components of the reaction. We see something similar in the second example, where the ratio given is 40:20:60. By simplifying the ratio, we get the ratio of the compounds in the chemical equation.