3.2Stoichiometry
Relative Masses of Atoms and Molecules — Cambridge O Level Chemistry 5070 Revision Notes & Topical Questions
5 min read 12 parts Lesson 2 of 3
Lesson summary
Define relative atomic, molecular and formula mass, calculate them and use them in reacting-mass calculations.
Section 1 of 12Relative atomic mass
Symbol
- Ar, shown beneath the symbol in the Periodic Table and larger than the atomic number.
- (except hydrogen where they are equal)
Why needed
- Why needed atoms are too small to weigh, so masses are compared.
- Standard
- carbon-12 has a fixed mass of 12 units.
- Definition
- Ar is the average mass of the isotopes of an element compared with 1/12 of the mass of a carbon-12 atom.
- Examples
- carbon 12, magnesium 24 (twice as heavy), hydrogen 1 (one twelfth).
Section 2 of 12Relative molecular and formula mass
Mr
- Mr the total of the relative atomic masses of all atoms in the formula.
Molecules
- Molecules called relative molecular mass.
- Ionic compounds
- called relative formula mass.
Method
- Method multiply each Ar by the number of atoms, then add.
Water
- H2O is 2 × 1 + 16 = 18.
- Calcium hydroxide
- Ca(OH)2 is 40 + 2 × (16 + 1) = 74.
Section 3 of 12Conservation of mass
Law
- Law mass cannot be created or destroyed, so reactant mass equals product mass.
Use
- Use with Ar and Mr values the balanced equation gives reacting masses.
- Ratio
- the mass ratio of reactants and products is always the same whatever the units.
Section 4 of 12Reacting masses method
Reacting masses method
Write the balanced equation.
Calculate the relative mass of each substance multiplied by its coefficient.
Scale the ratio to the mass given.
Example 2Ca + O2 → 2CaO
- 2 × 40 = 80 units of calcium react with 32 units of oxygen to give 112 units of calcium oxide.
Grams
- 80 g Ca + 32 g O2 gives 112 g CaO.
- Tonnes
- 40 tonnes of calcium gives 56 tonnes of calcium oxide.
Section 5 of 12Worked pattern: methane
Equation
- CH4 + 2O2 → CO2 + 2H2O.
Relative masses
- 16 + 64 = 44 + 36.
- Ratio
- 16 g methane gives 44 g carbon dioxide.
- Scaling
- 32 g methane gives 44 × 2 = 88 g carbon dioxide.
Section 6 of 12Worked method for Mr
1
Write out the formula.
2
Multiply the Ar of each element by the number of atoms of it.
3
Where brackets appear, multiply everything inside by the number outside.
4
Add the totals.
Carbon dioxide
- CO2 is 12 + 2 × 16 = 44.
Sodium carbonate
- Na2CO3 is 2 × 23 + 12 + 3 × 16 = 106.
- Magnesium nitrate
- Mg(NO3)2 is 24 + 2 × (14 + 48) = 148.
Section 7 of 12Why the Periodic Table value is not always whole
Cause
- Ar is an average of isotopes weighted by abundance.
Chlorine
- 35.5 because natural chlorine is a mixture of Cl-35 and Cl-37.
- Exam use
- round to the values given in the data sheet and use them as stated.
Section 8 of 12Using reacting masses in practice
Scaling
- Scaling the mass ratio holds for any quantity, so grams can be scaled up to kilograms or tonnes.
Example
- Example if 80 g of calcium needs 32 g of oxygen, 8 g of calcium needs 3.2 g.
- Industry
- chemical firms use these ratios to decide how much raw material to buy.
Checking
- Checking the total mass of reactants should equal the total mass of products.
Gases
- Gases a product that escapes explains an apparent loss of mass.
- Example
- heating calcium carbonate loses mass as carbon dioxide leaves.
Section 9 of 12Typical exam questions
Define
- Define state Ar and Mr using the carbon-12 reference.
Calculate
- Calculate add up atoms carefully, including brackets.
Apply
- Apply use the ratio from the balanced equation to find an unknown mass, as in finding how much oxide forms from a given mass of metal.
Layout
- Layout show the equation, the relative masses, the ratio and the final scaled answer.
Section 10 of 12Worked examples
Calculate the Mr of ammonium sulfate, (NH4)2SO4. Ar: N = 14, H = 1, S = 32, O = 16. [2]2 marks
- (NH4)2: 2 × (14 + 4) = 36.
- SO4: 32 + 64 = 96.
- Mr = 132.
Calculate the mass of carbon dioxide produced when 32 g of methane burns completely in excess oxygen. [3]3 marks
Worked examplesCH4 + 2O2 → CO2 + 2H2O.16 g CH4 gives 44 g CO2.32 g gives 88 g.What mass of calcium oxide forms from 20 g of calcium in excess oxygen? [3]3 marks
Worked examples2Ca + O2 → 2CaO: 80 g Ca gives 112 g CaO.20 g is one quarter of 80 g.Mass = 112 / 4 = 28 g.Magnesium has an Ar of 24. How many times heavier than a carbon atom is it? [1]1 mark
- 24 / 12 = 2 times.
Section 11 of 12Key terms
| Term | Meaning |
|---|---|
| Relative atomic mass (Ar) | Average mass of isotopes of an element compared with 1/12 of carbon-12. |
| Relative molecular mass (Mr) | Sum of the Ar values of all atoms in a molecule. |
| Relative formula mass | Sum of the Ar values in the formula of an ionic compound. |
| Conservation of mass | Total mass of reactants equals total mass of products. |
| Reacting masses | The masses of substances that react or form in the ratio of the balanced equation. |
Section 12 of 12Exam technique
1
Saying Ar has units — it is a ratio with no units.
2
Forgetting to multiply by coefficients when using the equation.
3
Not multiplying the atoms inside brackets, such as the 2 in Ca(OH)2.
4
Using mass numbers instead of Ar for chlorine, which is 35.5.
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