Chemistry Numericals for Intermediate: Moles, Gas Laws and Equilibrium

Numericals are where many intermediate chemistry students lose confidence, yet they follow a small number of patterns. Almost every calculation in FSc and A Level chemistry is built on the mole, and once you can move comfortably between grams, moles, particles and volumes, gas laws and equilibrium problems become straightforward. This guide shows the methods with fully worked examples.

The Mole: Chemistry's Counting Unit

One mole contains Avogadro's number of particles, \( N_A = 6.022 \times 10^{23} \). The mass of one mole of a substance in grams equals its relative molecular (or formula) mass. Three relationships connect everything:

$$ n = \frac{m}{M}, \qquad N = n \times N_A, \qquad V = n \times 22.4 \text{ dm}^3 \text{ (gas at STP)} $$

Here \( n \) is the number of moles, \( m \) the mass in grams, \( M \) the molar mass in g mol\(^{-1}\), and \( N \) the number of particles. The molar volume of 22.4 dm\(^3\) applies to an ideal gas at 0 °C and 1 atm; some courses use 22.7 dm\(^3\) for 1 bar, so follow the value your syllabus states.

Worked Example 1: Grams to Molecules

How many molecules are present in 4.4 g of carbon dioxide?

Step 1. Molar mass of CO\(_2\): \( 12 + 2(16) = 44 \text{ g mol}^{-1} \).

Step 2. Moles: \( n = \frac{4.4}{44} = 0.10 \text{ mol} \).

Step 3. Molecules: \( N = 0.10 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \).

Bonus: each molecule contains 3 atoms, so there are \( 1.807 \times 10^{23} \) atoms in total.

Stoichiometry: Using Balanced Equations

A balanced equation gives the mole ratio in which substances react. The standard route is always the same:

1. Convert the given mass to moles.
2. Use the mole ratio from the equation.
3. Convert the moles of the required substance back to mass or volume.

Worked Example 2: Burning Methane

Methane burns according to

$$ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} $$

What mass of CO\(_2\) is produced when 8.0 g of methane burns completely, and what volume of oxygen at STP is needed?

Step 1. Molar mass of CH\(_4\) = \( 12 + 4 = 16 \text{ g mol}^{-1} \). Moles of CH\(_4\) = \( \frac{8.0}{16} = 0.50 \text{ mol} \).

Step 2. The ratio CH\(_4\) : CO\(_2\) is 1 : 1, so 0.50 mol of CO\(_2\) forms. Mass = \( 0.50 \times 44 = 22 \text{ g} \).

Step 3. The ratio CH\(_4\) : O\(_2\) is 1 : 2, so 1.0 mol of O\(_2\) is needed. Volume at STP = \( 1.0 \times 22.4 = 22.4 \text{ dm}^3 \).

Check with conservation of mass: O\(_2\) mass = 32 g and water formed = \( 1.0 \times 18 = 18 \) g. Reactants: \( 8 + 32 = 40 \) g. Products: \( 22 + 18 = 40 \) g. Balanced.

Limiting Reactant and Percentage Yield

When amounts of two reactants are given, find which one runs out first by comparing the moles available with the moles required. The product is calculated from the limiting reactant only. If the experiment produces less than the theoretical amount,

$$ \text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 $$

For instance, if only 19.8 g of CO\(_2\) were collected in Example 2, the yield would be \( \frac{19.8}{22} \times 100 = 90\% \).

Gas Laws

Gases are described by pressure \( P \), volume \( V \), temperature \( T \) and amount \( n \). Temperature must always be in kelvin: \( T(\text{K}) = T(^\circ\text{C}) + 273 \).

• Boyle's law: \( P_1V_1 = P_2V_2 \) at constant \( T \).
• Charles's law: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) at constant \( P \).
• General gas equation: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).
• Ideal gas equation: \( PV = nRT \), with \( R = 0.0821 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1} \) or \( 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \).

Worked Example 3: Changing Conditions

A gas occupies 2.0 dm\(^3\) at 27 °C and 1.0 atm. Find its volume at 127 °C and 1.5 atm.

Step 1. Convert temperatures: \( T_1 = 300 \text{ K} \), \( T_2 = 400 \text{ K} \).

Step 2. Rearrange the general gas equation:

$$ V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{1.0 \times 2.0 \times 400}{300 \times 1.5} = \frac{800}{450} = 1.78 \text{ dm}^3 $$

Sanity check: heating alone would increase the volume by a factor of \( \frac{4}{3} \), while the higher pressure alone would reduce it by a factor of \( \frac{2}{3} \). The overall factor is \( \frac{8}{9} \), slightly less than 1, so a volume a little under 2.0 dm\(^3\) makes sense.

Worked Example 4: The Ideal Gas Equation

What pressure is exerted by 0.50 mol of nitrogen in a 10 dm\(^3\) container at 300 K?

$$ P = \frac{nRT}{V} = \frac{0.50 \times 0.0821 \times 300}{10} = \frac{12.315}{10} = 1.23 \text{ atm} $$

Chemical Equilibrium

In a reversible reaction at equilibrium, the forward and backward rates are equal and concentrations stay constant. For a general reaction \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \), the equilibrium constant is

$$ K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} $$

For gases, \( K_p \) uses partial pressures, and the two are linked by \( K_p = K_c(RT)^{\Delta n} \), where \( \Delta n \) is moles of gaseous products minus moles of gaseous reactants.

Worked Example 5: The ICE Table Method

For \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \), \( K_c = 49 \) at a certain temperature. If 1.0 mol of H\(_2\) and 1.0 mol of I\(_2\) are placed in a 1.0 dm\(^3\) flask, find the equilibrium concentrations.

Step 1. Set up the ICE table (Initial, Change, Equilibrium), in mol dm\(^{-3}\):

Step 2. Substitute into \( K_c \):

$$ 49 = \frac{(2x)^2}{(1.0 - x)^2} $$

Step 3. Take the square root of both sides: \( 7 = \frac{2x}{1.0 - x} \).

Step 4. Solve: \( 7 - 7x = 2x \), so \( x = \frac{7}{9} = 0.778 \).

Step 5. Equilibrium concentrations: [H\(_2\)] = [I\(_2\)] = 0.222 mol dm\(^{-3}\) and [HI] = 1.556 mol dm\(^{-3}\).

Check: \( \frac{1.556^2}{0.222^2} = \frac{2.420}{0.0494} \approx 49 \). Correct.

Le Chatelier's principle then predicts how equilibrium shifts: adding reactant, removing product, or changing pressure for reactions where \( \Delta n \neq 0 \). Only a change in temperature changes the value of \( K_c \) itself.

Did You Know?

• Avogadro proposed his hypothesis about equal volumes of gases in 1811, but it was largely ignored until Stanislao Cannizzaro championed it around 1860.
• A mole of grains of sand would cover a large country to a depth of many metres, showing how huge \( 6.022 \times 10^{23} \) really is.
• The Haber process for ammonia is a real-world balancing act of equilibrium and rate, feeding a large share of the world's population through fertilisers.

Key Takeaways

• Always convert to moles first; the mole is the bridge between mass, particles and gas volume.
• Use the mole ratio from a balanced equation, and identify the limiting reactant when two amounts are given.
• Temperatures in gas law calculations must be in kelvin.
• ICE tables organise equilibrium problems and make \( K_c \) calculations systematic.
• Only temperature changes the value of an equilibrium constant.

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