What is meant by the theoretical mass of a product?

It is the maximum mass of product that could be made if all the reactant were converted and none of the product were lost.

Calcium carbonate, relative formula mass 100, decomposes to calcium oxide, relative formula mass 56, in a 1:1 ratio. Calculate the theoretical mass of calcium oxide from 50.0 g of calcium carbonate.

50.0 ÷ 100 = 0.500 mol, giving 0.500 mol of calcium oxide, so 0.500 × 56 = 28.0 g, given to 3 significant figures.

Magnesium, relative atomic mass 24, burns to magnesium oxide, relative formula mass 40, one mole of metal giving one mole of oxide. Calculate the theoretical mass of oxide from 4.8 g of magnesium.

4.8 ÷ 24 = 0.20 mol, giving 0.20 mol of magnesium oxide, so 0.20 × 40 = 8.0 g, given to 2 significant figures.

One mole of nitrogen, relative formula mass 28, gives two moles of ammonia, relative formula mass 17. Calculate the theoretical mass of ammonia from 28 g of nitrogen.

28 ÷ 28 = 1.0 mol of nitrogen gives 2.0 mol of ammonia, so 2.0 × 17 = 34 g, given to 2 significant figures.

Two moles of hydrogen, relative formula mass 2, give two moles of water, relative formula mass 18. Calculate the theoretical mass of water from 4.0 g of hydrogen.

4.0 ÷ 2 = 2.0 mol of hydrogen gives 2.0 mol of water, so 2.0 × 18 = 36 g, given to 2 significant figures.

OCR GCSE Chemistry A J248

Chemical Reactions

Topic overview Writing formulae of elements and compounds Balanced equations, and deducing formulae from ions State symbols in equations Conservation of mass, and changes of mass in open systems Exothermic and endothermic reactions, and reaction profiles Activation energy and why reactions need it Oxidation and reduction in terms of oxygen Acids, alkalis and neutralisation The ionic equation for neutralisation, and reactions of acids The pH scale, acidity and alkalinity Measuring pH with indicators and pH meters Products at the electrodes, and molten ionic compounds Electrolysis of aqueous solutions Inert and non-inert electrodes

Elements, Compounds and Mixtures

Topic overview Purity, and melting point data Relative formula mass and empirical formulae Formulations, and separating mixtures Paper and thin layer chromatography Interpreting chromatograms and Rf values Choosing a purification method Metals and non-metals in the Periodic Table Electron arrangement and position in the Periodic Table Ionic, covalent and metallic bonding compared Dot and cross diagrams, and the limits of models Reactions, electron arrangement and Mendeleev's table Carbon's four bonds, and the variety of organic compounds Diamond, graphite, fullerenes and graphene Bond strength, intermolecular forces and predicting state Bulk properties of ionic, molecular and giant structures Nanoparticles: size and surface area to volume Uses and risks of nanoparticulate materials

Global Challenges

Monitoring and Controlling Chemical Reactions

Topic overview Carrying out an acid–alkali titration Theoretical mass and percentage yield Atom economy and its calculation Measuring the rate of a reaction, and rate graphs Factors affecting rate, and collision theory Surface area and rate Catalysts and how they work Reversible reactions and dynamic equilibrium Activation energy, and enzymes as catalysts

Particles

Topic overview The particle model and changes of state How the atomic model changed, and the structure of the atom The size of atoms, and the subatomic particles Calculating protons, neutrons and electrons

Predicting and Identifying Reactions

Topic overview Groups 1, 7 and 0, and why they behave as they do Transition metals Predicting reactivity from the Periodic Table Reactivity of metals with water and acids Tests for oxygen, hydrogen, carbon dioxide and chlorine Tests for aqueous cations and anions Flame tests, and identifying metal ions Identifying a species from test results Instrumental methods of analysis

Monitoring and Controlling Chemical Reactions

Theoretical mass and percentage yield

3 things to know
Paper 2

Overview

What a theoretical mass is and the three steps to calculate it, worked examples from calcium carbonate, magnesium, nitrogen and hydrogen, the limiting reactant, percentage yield calculations, and why yields fall short of 100%.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Process

State the three steps used to calculate the theoretical mass of a product from a given mass of reactant.

Key term

If one reactant is present in excess, which reactant must the theoretical mass be calculated from?

Key term

How is the percentage yield of a reaction calculated?

Questions from this subtopic

Each one with the model answer underneath.

1
What is meant by the theoretical mass of a product?
Model answer
It is the maximum mass of product that could be made if all the reactant were converted and none of the product were lost.
2
State the three steps used to calculate the theoretical mass of a product from a given mass of reactant.
Model answer
Convert the mass of reactant to moles, use the balanced equation to find the moles of product, then multiply by the relative formula mass of the product.
3
Calcium carbonate, relative formula mass 100, decomposes to calcium oxide, relative formula mass 56, in a 1:1 ratio. Calculate the theoretical mass of calcium oxide from 50.0 g of calcium carbonate.
Model answer
50.0 ÷ 100 = 0.500 mol, giving 0.500 mol of calcium oxide, so 0.500 × 56 = 28.0 g, given to 3 significant figures.
4
Magnesium, relative atomic mass 24, burns to magnesium oxide, relative formula mass 40, one mole of metal giving one mole of oxide. Calculate the theoretical mass of oxide from 4.8 g of magnesium.
Model answer
4.8 ÷ 24 = 0.20 mol, giving 0.20 mol of magnesium oxide, so 0.20 × 40 = 8.0 g, given to 2 significant figures.
5
One mole of nitrogen, relative formula mass 28, gives two moles of ammonia, relative formula mass 17. Calculate the theoretical mass of ammonia from 28 g of nitrogen.
Model answer
28 ÷ 28 = 1.0 mol of nitrogen gives 2.0 mol of ammonia, so 2.0 × 17 = 34 g, given to 2 significant figures.
6
Two moles of hydrogen, relative formula mass 2, give two moles of water, relative formula mass 18. Calculate the theoretical mass of water from 4.0 g of hydrogen.
Model answer
4.0 ÷ 2 = 2.0 mol of hydrogen gives 2.0 mol of water, so 2.0 × 18 = 36 g, given to 2 significant figures.

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Chemical Reactions

Elements, Compounds and Mixtures

Global Challenges

Monitoring and Controlling Chemical Reactions

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