CNC Illustration by Gwendolyn Say (@saydrawings)

Stoichiometry

Stoichiometry empowers us with the mathematical tools to know precisely the number of particles present in a sample before us. It also allows us to gaze into the future as we predict the amount of products without a crystal ball.

SummaryResources

SUMMARY

Mole

1 mole is 6×1023 particles. Stoichiometry means the mole ratio of substances in a reaction.

no of moles = no of particles ÷ 6×1023

Relative Molecular Mass

The mass of a molecule compared to 1/12th of a carbon-12 atom.

Mr of AnBm = (n × Ar of A) + (m × Ar of B)

Molecule

The mass of 1 mol of a substance, numerically equal to the Ar or Mr.

molar mass = mass in g ÷ no of moles

Percentage Mass

The percentage by mass of a compound that is made up of an element.

% mass of A in AnBm= nAr ÷ Mr × 100%

Empirical Formula

The simplest ratio of atoms of each element in a compound.

1. Write the masses of each element
2. Divide by Ar
3. Divide by smallest number

Molecular Formula

Shows the number of atoms of each element in one simple molecule.

1. Find n = Mr ÷ mass of empirical formula
2. Multiply numbers in empirical formula by n

Molar Volume

A mole of gas occupies 24 dm3 of volume at room temperature and pressure (rtp).

no of moles = gas volume in dm3 ÷ 24

Solution Concentration

The number of moles or mass of solutes per dm3 of solvent.

concentration = no of moles ÷ volume in dm3
concentration = mass ÷ volume in dm3

Titration

Carefully adding exact amount of one solution to completely react with another.

1. Find no of moles of titrant = concentration × volume
2. Use mole ratio to find no of moles of unknown
3. Find conc of unknown = no of mol ÷ volume

Limiting Reactant

The reactant that is completely used up, limiting the amount of product.

1. Find the no of moles of reactants A and B provided
2. Find the no of moles of B needed if A were used up
3. A is limiting if there is less B needed than provided

Percentage Yield

The extent and efficiency of converting reactants to products.

% yield = expected yield ÷ theoretical yield × 100%

Percentage Purity

The percentage by mass of an impure mixture that comprises the pure substance.

% purity = mass of pure substance ÷ mass of impure mixture × 100%

Resources

Stoichiometry & the mole concept prelim questions

Learn stoichiometry and the mole concept by solving problems taken from O Level prelim papers. Answers and explanations are included.

Solution concentration & titration

Sorry, we don't have 25% sugar here. How about 100 g/dm3 or 0.292 mol/dm3?

Stoichiometry: Mole ratio & limiting reactant

Look into the future and predict the amount of product formed, by identifying the mole ratio and limiting reactant

Relative atomic mass & molecular mass

Unit-less but meaningful: making sense of the mass of an atom with ratio

Stoichiometry: Percentage yield & percentage purity

Be 100% sure about how to calculate percentage yield and percentage purity

Percentage mass & empirical formula

Calculate how each element in a compound is pulling its weight, and use this to find the chemical formula empirically

Salt preparation TYS questions

It's big brain time, as you think backwards from the salt in question to infer the reagents and method

Stoichiometry TYS questions – Combined Chemistry

There is no stoichiometry without try: try out questions from the O Level Combined Chemistry papers

The mole & molar mass

Learn how how to count, but in moles

Molar volume of gases

Gases of different masses taking up the same spaces

Practise how to write chemical formula

Can you infer the formulae of ionic compounds, based on the charges of their constituent ions?