Stoichiometry Equations

This page collects the mathematical relationships and worked setups for 0620 stoichiometry calculations. Use alongside mole concept explained and the stoichiometry topic page.

Core formulas

Moles from mass

moles = mass (g) / Mr

Rearranged: mass = moles x Mr, or Mr = mass / moles

Moles from concentration and volume

moles = concentration (mol/dm3) x volume (dm3)

To convert cm3 to dm3: divide by 1000.

Rearranged: concentration = moles / volume, or volume = moles / concentration

Moles from gas volume (at r.t.p.)

moles = volume of gas (dm3) / 24

At room temperature and pressure: 1 mole of any gas = 24 dm3 = 24000 cm3.

Rearranged: volume = moles x 24

Relative formula mass (Mr)

Mr is the sum of the relative atomic masses (Ar) of all atoms in the formula.

Compound	Calculation	Mr
H2O	(2 x 1) + 16	18
CO2	12 + (2 x 16)	44
CaCO3	40 + 12 + (3 x 16)	100
H2SO4	(2 x 1) + 32 + (4 x 16)	98
Ca(OH)2	40 + 2(16 + 1)	74
Mg(NO3)2	24 + 2(14 + 48)	148

Empirical formula

The simplest whole-number ratio of atoms.

Method:

1. Write masses or percentages of each element
2. Divide each by its Ar
3. Divide all results by the smallest value
4. Round to nearest whole numbers

Stoichiometry calculation method

1. Write the balanced equation
2. Calculate moles of the known substance
3. Use the mole ratio from the equation
4. Convert moles of the unknown to the required unit (mass, volume, concentration)

Worked setup: mass-to-mass

Question: What mass of CO2 is produced when 10 g of CaCO3 reacts with excess HCl?

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

1. Moles of CaCO3 = 10 / 100 = 0.1 mol
2. Mole ratio: CaCO3 : CO2 = 1 : 1
3. Moles of CO2 = 0.1 mol
4. Mass of CO2 = 0.1 x 44 = 4.4 g

Worked setup: concentration to volume of gas

Question: What volume of H2 (at r.t.p.) is produced when 25 cm3 of 2 mol/dm3 HCl reacts with excess Mg?

Mg + 2HCl -> MgCl2 + H2

1. Moles of HCl = 2 x (25/1000) = 0.05 mol
2. Mole ratio: HCl : H2 = 2 : 1
3. Moles of H2 = 0.025 mol
4. Volume of H2 = 0.025 x 24 = 0.6 dm3 (600 cm3)

Percentage composition

% of element = (total Ar of element in formula / Mr) x 100

Example: % of oxygen in CaCO3 = (3 x 16) / 100 x 100 = 48%

Percentage yield

percentage yield = (actual yield / theoretical yield) x 100

Theoretical yield is calculated from stoichiometry. Actual yield is measured experimentally. Yield is always less than 100% due to: incomplete reactions, side reactions, losses during transfer/purification.

See percentage yield.

Percentage purity

percentage purity = (mass of pure substance / mass of impure sample) x 100

See percentage purity.

Limiting reagent

The limiting reagent is the reactant that is completely used up first. It determines the maximum amount of product formed.

Method:

1. Calculate moles of each reactant
2. Divide each by its coefficient in the balanced equation
3. The smaller value identifies the limiting reagent

Titration calculations

From a titration:

1. Calculate moles of the known solution: moles = concentration x volume (in dm3)
2. Use the mole ratio to find moles of the unknown
3. Calculate concentration or mass of unknown

Water of crystallisation

See water of crystallisation.

Formula: CuSO4.xH2O — find x by calculating moles of anhydrous salt and water separately.

Calculate the mass of magnesium oxide produced when 4.8 g of magnesium is burned in excess oxygen. (Ar: Mg = 24, O = 16) [3 marks]

• 2Mg + O2 -> 2MgO
• Moles of Mg = 4.8 / 24 = 0.2 mol [1]
• Mole ratio Mg : MgO = 1 : 1, so moles of MgO = 0.2 mol [1]
• Mass of MgO = 0.2 x (24 + 16) = 0.2 x 40 = 8.0 g [1]

25.0 cm3 of sodium hydroxide solution of unknown concentration was neutralised by 20.0 cm3 of 0.10 mol/dm3 hydrochloric acid. Calculate the concentration of the sodium hydroxide. [3 marks]

• Moles of HCl = 0.10 x (20.0/1000) = 0.002 mol [1]
• NaOH + HCl -> NaCl + H2O; ratio 1:1, so moles of NaOH = 0.002 mol [1]
• Concentration of NaOH = 0.002 / (25.0/1000) = 0.002 / 0.025 = 0.08 mol/dm3 [1]