2.1.3(e), 2.1.3(g) - Reacting masses, solutions and stoichiometry

The mole bridge for calculations

In a reaction calculation, the central quantity is usually amount of substance, n, in mol. Mass, gas volume and solution concentration are different ways of measuring a substance, but each can be converted into mol.

The three main conversion routes are:

The volume unit is a common source of lost marks:

So:

For concentrations, OCR questions may use either mol dm⁻³ or g dm⁻³.

You can connect the two concentration units using molar mass:

For example, a 0.200 mol dm⁻³ solution of NaOH has mass concentration:

because the molar mass of NaOH is 40.0 g mol⁻¹.

Balanced equations give mole ratios

A balanced equation tells you the ratio of amounts in mol. It does not directly give a ratio of masses.

For example:

This means:

• 2 mol of Mg react with 1 mol of O₂.
• 2 mol of Mg produce 2 mol of MgO.
• The mole ratio Mg : MgO is 2 : 2, which simplifies to 1 : 1.

The reliable route is:

1. Convert the given quantity to amount in mol.
2. Use the balanced-equation mole ratio.
3. Convert the required amount into the requested quantity.

Worked example: mass of product from mass of reactant

Calculate the mass of magnesium oxide formed when 1.20 g of magnesium reacts completely with oxygen.

Equation:

Use M(Mg) = 24.3 g mol⁻¹ and M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹.

First find the amount of Mg:

The ratio Mg : MgO is 2 : 2, so the amount of MgO is the same:

Now convert amount of MgO to mass:

To three significant figures, the mass of MgO is:

Gas volume calculations at RTP

For gases at RTP, the data sheet value is:

This route only applies when the question is using molar gas volume at RTP. The ideal gas equation pV = nRT belongs to a separate calculation method and is not needed in this lesson.

Worked example: mass to gas volume

Calcium carbonate decomposes on heating:

Calculate the volume of carbon dioxide formed at RTP from 0.500 g of calcium carbonate.

Use M(CaCO₃) = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹.

First convert the mass of CaCO₃ to amount:

The equation ratio CaCO₃ : CO₂ is 1 : 1, so:

Convert amount of CO₂ to gas volume at RTP:

To three significant figures:

This is also:

Why the unit matters

If a gas volume is given in cm³, convert it to dm³ before using n = V/24.0. For example, 96.0 cm³ = 0.0960 dm³.

Worked example: a direct gas-volume ratio

For $\ce{2CO(g) + O2(g) -> 2CO2(g)}$, gases measured at the same temperature and pressure have volumes in the 2:1:2 mole ratio. Thus 72.0 cm³ of CO needs $72.0/2=36.0$ cm³ of O₂ and forms 72.0 cm³ of CO₂. There is no need to convert each volume into mol when the common molar gas volume cancels. This shortcut does not compare a gas volume with a liquid volume or gases under different conditions.

Solution volume and concentration

For solutions, concentration in mol dm⁻³ tells you the amount of solute in one cubic decimetre of solution.

So:

where volume must be in dm³.

Worked example: reacting volumes of solutions

Silver nitrate solution reacts with sodium chloride solution:

Calculate the volume of 0.150 mol dm⁻³ sodium chloride solution needed to react exactly with 20.0 cm³ of 0.100 mol dm⁻³ silver nitrate solution.

First calculate the amount of AgNO₃:

The equation ratio AgNO₃ : NaCl is 1 : 1, so:

Rearrange n = cV to find the volume of NaCl solution:

Convert to cm³:

Dilution conserves the amount of solute

Adding water changes solution volume but not the amount of dissolved solute, provided none is lost and no reaction occurs. Therefore $c_1V_1=c_2V_2$.

Worked example: dilution to a final volume

20.0 cm³ of 0.500 mol dm⁻³ solution is diluted to 250.0 cm³. The amount transferred is $0.500\times0.0200=0.0100$ mol. Divide by the final solution volume: $c_2=0.0100/0.2500=0.0400\,\mathrm{mol\,dm^{-3}}$.

Equivalently, $c_2=c_1V_1/V_2=0.500\times20.0/250.0$; equal volume units cancel in this ratio. The final volume is 250.0 cm³, not 270.0 cm³. A 12.5-fold increase in volume gives a 12.5-fold decrease in concentration, a useful check.

Concentration in g dm⁻³

If the concentration is given in g dm⁻³, convert to either mass or mol depending on what the question asks.

For example, a sodium chloride solution contains 11.7 g dm⁻³ of NaCl. Use M(NaCl) = 58.5 g mol⁻¹.

In mol dm⁻³:

So 25.0 cm³ of this solution contains:

Multi-step stoichiometry

A longer question is still built from the same small moves. Read the question for:

• the balanced equation
• the quantity you are given
• the substance you must find
• the unit required in the final answer

Then follow the route:

When both reactant amounts are given

The limiting reactant runs out first and sets the maximum product amount. Compare each available amount divided by its coefficient in the balanced equation; the smaller value limits the reaction. Comparing masses or raw amounts alone can give the wrong result.

For $\ce{N2 + 3H2 -> 2NH3}$, suppose 0.100 mol N₂ and 0.240 mol H₂ are available. Compare $0.100/1=0.100$ with $0.240/3=0.0800$: H₂ limits the reaction. At complete conversion it produces $2(0.0800)=0.160$ mol NH₃, leaving $0.100-0.0800=0.0200$ mol N₂. This is the stoichiometric maximum, before any yield loss or equilibrium limitation.

Worked example: solution concentration to gas volume

Hydrogen peroxide decomposes as shown:

Calculate the volume of oxygen formed at RTP when 100 cm³ of 0.500 mol dm⁻³ hydrogen peroxide decomposes completely.

Convert the solution volume:

Calculate the amount of H₂O₂:

Use the equation ratio. 2 mol of H₂O₂ form 1 mol of O₂:

Convert amount of O₂ to volume at RTP:

So the volume of oxygen is:

or:

Quick Recap

Convert the measured quantity into mol, use the balanced-equation ratio, then convert to the requested quantity. Match concentration and volume units. Gas volumes at the same temperature and pressure can use the equation ratio directly; dilution conserves solute amount. When both reactants are supplied, compare amount divided by coefficient to find the limiting reactant.