CCEA A-Level Life and Health Sciences 0008
Experimental Techniques
Human Body Systems
Physical Chemistry in Industrial Processes
Brain ScienceDouble Award
Material ScienceDouble Award
Medicine, Drugs and Clinical TrialsDouble Award
Scientific Method, Investigation, Analysis and Evaluation
Organic Chemistry
Medical PhysicsA2 option
Sound and LightA2 option
Genetics, Stem Cell Research and CloningA2 option
MicrobiologyDouble Award
Oral Health and DentistryDouble Award
Histology and PathologyDouble Award
Analytical Chemistry TechniquesDouble Award
Enabling TechnologyDouble Award
Physical Chemistry in Industrial Processes
Moles, reacting masses and percentage yield
- 3 things to know
- AS 3
Overview
Relative formula mass, the mole, balancing equations, reacting masses in industrial processes, theoretical and percentage yield in the Haber process, and atom economy.
What you need to know
Answer each one before you reveal it.
What is the relative formula mass of a substance?
Calculate the relative formula mass of sulfuric acid, H₂SO₄, using Ar values H = 1.0, S = 32.1 and O = 16.0.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What is the relative formula mass of a substance?
Model answer
The relative formula mass is the sum of the relative atomic masses of all the atoms in its formula. Relative formula mass has no units.
- 2
Calculate the relative formula mass of sulfuric acid, H₂SO₄, using Ar values H = 1.0, S = 32.1 and O = 16.0.
Model answer
The relative formula mass of H₂SO₄ is 98.1. The working is (2 × 1.0) + 32.1 + (4 × 16.0) = 98.1.
- 3
What is one mole of a substance?
Model answer
One mole is the amount of substance containing 6.02 × 10²³ particles, the Avogadro constant. A mole of a substance has a mass in grams equal to its relative formula mass.
- 4
State the relationship between amount in moles, mass and molar mass.
Model answer
The amount in moles equals the mass in grams divided by the molar mass in g mol⁻¹. Mass equals moles multiplied by molar mass.
- 5
A water-treatment chemist weighs out 5.30 g of anhydrous sodium carbonate, Na₂CO₃ (Mr = 106.0). How many moles is this?
Model answer
The sample contains 0.0500 mol of sodium carbonate. The working is 5.30 ÷ 106.0 = 0.0500 mol.
- 6
Why must a symbol equation be balanced before it is used to calculate reacting quantities?
Model answer
A balanced equation gives the mole ratio in which substances react and form. An unbalanced equation gives the wrong ratio and so the wrong masses.
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