1. Overview
This topic introduces how chemists measure the mass of atoms and molecules. Because individual atoms are far too small to weigh on a scale, we use a relative scale to compare their masses against a standard (Carbon-12). This allows us to calculate exactly how much of a substance is needed or produced in a chemical reaction.
Key Definitions
- Relative Atomic Mass ($A_r$): The average mass of the isotopes of an element compared to 1/12th of the mass of an atom of $^{12}C$.
- Relative Molecular Mass ($M_r$): The sum of the relative atomic masses of all atoms present in a covalent molecule.
- Relative Formula Mass ($M_r$): The sum of the relative atomic masses of all atoms in the formula of an ionic compound.
- Isotope: Atoms of the same element with the same number of protons but different numbers of neutrons.
Core Content
Understanding Relative Atomic Mass ($A_r$)
Atoms are compared to the Carbon-12 isotope, which is assigned a mass of exactly 12 units.
- An atom of Hydrogen is 1/12th the mass of Carbon-12, so its $A_r = 1$.
- An atom of Magnesium is twice as heavy as Carbon-12, so its $A_r = 24$.
- $A_r$ values are found on the Periodic Table (the larger number in the element box).
Calculating Relative Molecular/Formula Mass ($M_r$)
To find the $M_r$, you add up the $A_r$ values of every atom in the chemical formula.
- For Covalent Molecules ($M_r$):
- Example: Water ($H_2O$)
- Contains 2 x H ($A_r = 1$) and 1 x O ($A_r = 16$)
- $M_r = (2 \times 1) + 16 = 18$
- For Ionic Compounds ($M_r$):
- Example: Magnesium Chloride ($MgCl_2$)
- Contains 1 x Mg ($A_r = 24$) and 2 x Cl ($A_r = 35.5$)
- $M_r = 24 + (2 \times 35.5) = 95$
Calculating Reacting Masses in Simple Proportions
In a balanced equation, the ratio of the masses of reactants and products is constant. You can use the total $M_r$ of each part of the equation to predict masses.
Worked Example: Calculate the mass of Magnesium oxide ($MgO$) produced when $48g$ of Magnesium ($Mg$) is burned in excess oxygen.
Step 1: Write the balanced equation. Magnesium + Oxygen → Magnesium oxide $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$
Step 2: Calculate the relative masses for the substances involved.
- $2 \times Mg = 2 \times 24 = 48$
- $2 \times MgO = 2 \times (24 + 16) = 2 \times 40 = 80$
Step 3: Use simple proportions. The equation shows that $48g$ of Magnesium will produce $80g$ of Magnesium oxide.
- Answer: $80g$ of $MgO(s)$
Worked Example: Multi-step reacting masses How much iron can be produced from $480\text{ g}$ of iron(III) oxide? $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$
- Mr of Fe\u2082O\u2083 = $(2 \times 56) + (3 \times 16) = 160$
- Mr of Fe = $56$
- From the equation: $160\text{ g}$ of Fe\u2082O\u2083 produces $2 \times 56 = 112\text{ g}$ of Fe
- So $480\text{ g}$ of Fe\u2082O\u2083 produces: $\frac{480}{160} \times 112 = 3 \times 112 = 336\text{ g}$ of Fe
The key is to scale everything by the same factor. Here, $480/160 = 3$, so we get 3 times as much product.
Extended Content (Extended Only)
There is no supplement curriculum for this specific sub-topic (3.2).
Key Equations
| Equation | Meaning | Units |
|---|---|---|
| $M_r = \sum A_r$ | $M_r$ is the sum of all Relative Atomic Masses in the formula. | No units (Relative) |
| $\text{Mass Ratio} = \frac{\text{Mass A}}{\text{Mass B}}$ | The ratio of masses in a reaction is constant. | Grams ($g$) |
Common Mistakes to Avoid
- ❌ Wrong: Giving $A_r$ or $M_r$ units like grams ($g$).
- ✓ Right: These are "relative" values; they have no units.
- ❌ Wrong: Using the Atomic Number (the smaller number on the Periodic Table) instead of the Mass Number ($A_r$) for calculations.
- ✓ Right: Always use the larger number (the mass number) from the Periodic Table.
- ❌ Wrong: Forgetting to multiply by the subscript in a formula (e.g., saying $M_r$ of $O_2$ is 16).
- ✓ Right: Multiply the $A_r$ by the number of atoms (e.g., $M_r$ of $O_2$ is $16 \times 2 = 32$).
Exam Tips
- Command Words: If asked to "Calculate," always show your addition steps (e.g., $12 + (16 \times 2)$). This earns marks even if your final answer is wrong.
- Expected Questions: You will often be asked to find the $M_r$ of a compound containing brackets, such as Calcium hydroxide, $Ca(OH)_2$.
- Tip: Everything inside the bracket is multiplied by the small number outside.
- $M_r$ of $Ca(OH)_2 = 40 + [2 \times (16 + 1)] = 74$.
- Real-World Contexts: Questions often use the thermal decomposition of limestone ($CaCO_3$) to ask about reacting masses.
- $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
- Remember that if a gas ($CO_2$) is produced, the mass of the solid will appear to decrease as the gas escapes.
- Periodic Table: You don't need to memorize $A_r$ values; they are always provided in the Periodic Table at the back of the exam paper. Use the specific values provided there (e.g., $Cl = 35.5$).
Exam-Style Questions
Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.
Exam-Style Question 1 — Short Answer [5 marks]
Question:
Lithium reacts with nitrogen to form lithium nitride.
(a) Define the term relative atomic mass. [2]
(b) The relative atomic mass of lithium is 7 and nitrogen is 14. Calculate the relative formula mass of lithium nitride, Li$_3$N. [1]
(c) State what is meant by the term relative formula mass. [2]
Worked Solution:
(a)
- The relative atomic mass is the average mass of the isotopes of an element.
- This mass is compared to 1/12th of the mass of an atom of carbon-12.
How to earn full marks:
- Must mention average mass of isotopes.
- Must mention comparison to 1/12th mass of carbon-12.
(b)
- The relative formula mass of Li$_3$N is the sum of the relative atomic masses of three lithium atoms and one nitrogen atom.
- $M_r(Li_3N) = (3 \times 7) + 14 = 21 + 14 = 35$
How to earn full marks:
- Correctly add the relative atomic masses.
- Final answer: $\boxed{35}$
(c)
- The relative formula mass is the sum of the relative atomic masses of all the atoms
- In the formula unit.
How to earn full marks:
- Must mention sum of relative atomic masses.
- Must refer to the formula unit of the compound.
Common Pitfall: Remember that relative atomic mass is an average that takes into account the different isotopes of an element. Also, make sure you understand the difference between relative atomic mass and relative formula mass.
Exam-Style Question 2 — Short Answer [6 marks]
Question:
Hydrogen reacts with chlorine to form hydrogen chloride, HCl.
(a) The relative atomic mass of hydrogen is 1 and chlorine is 35.5. Calculate the relative molecular mass of hydrogen chloride. [1]
(b) State what is meant by the term relative molecular mass. [2]
(c) In an experiment, 2 g of hydrogen reacts completely with 71 g of chlorine. Calculate the mass of hydrogen chloride produced. [3]
Worked Solution:
(a)
- The relative molecular mass of hydrogen chloride is the sum of the relative atomic masses of one hydrogen atom and one chlorine atom.
- $M_r(HCl) = 1 + 35.5 = 36.5$
How to earn full marks:
- Correctly add the relative atomic masses.
- Final answer: $\boxed{36.5}$
(b)
- The relative molecular mass is the sum of the relative atomic masses of all the atoms
- In the molecule.
How to earn full marks:
- Must mention sum of relative atomic masses.
- Must refer to the molecule.
(c)
- The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction.
- Therefore, the total mass of reactants equals the total mass of products.
- $mass_{hydrogen \ chloride} = 2 + 71 = 73$ g
How to earn full marks:
- Correctly add the mass of hydrogen and chlorine
- Final answer: $\boxed{73 \text{ g}}$
Common Pitfall: The Law of Conservation of Mass is crucial here. Make sure you understand that the total mass of the reactants will always equal the total mass of the products in a chemical reaction.
Exam-Style Question 3 — Extended Response [9 marks]
Question:
Sucrose, C${12}$H${22}$O$_{11}$, can be broken down into glucose, C$6$H${12}$O$_6$, and fructose, C$6$H${12}$O$_6$. The equation for the reaction is:
C${12}$H${22}$O$_{11}$ + H$_2$O $\rightarrow$ C$6$H${12}$O$_6$ + C$6$H${12}$O$_6$
The relative atomic mass of carbon is 12, hydrogen is 1, and oxygen is 16.
(a) Calculate the relative molecular mass of sucrose, C${12}$H${22}$O$_{11}$. [2]
(b) Calculate the relative molecular mass of glucose, C$6$H${12}$O$_6$. [2]
(c) In an experiment, 34.2 g of sucrose produces 18 g of glucose. Calculate the mass of sucrose required to produce 27 g of glucose assuming the same proportion. [3]
(d) State two reasons why the mass of glucose produced in the experiment might be less than expected. [2]
Worked Solution:
(a)
- The relative molecular mass of sucrose is the sum of the relative atomic masses of twelve carbon atoms, twenty-two hydrogen atoms, and eleven oxygen atoms.
- $M_r(C_{12}H_{22}O_{11}) = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342$
How to earn full marks:
- Correctly add the relative atomic masses.
- Final answer: $\boxed{342}$
(b)
- The relative molecular mass of glucose is the sum of the relative atomic masses of six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.
- $M_r(C_6H_{12}O_6) = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180$
How to earn full marks:
- Correctly add the relative atomic masses.
- Final answer: $\boxed{180}$
(c)
- First determine the amount of sucrose required to produce 1 g of glucose
- $34.2 \text{ g sucrose} \rightarrow 18 \text{ g glucose}$
- $\frac{34.2}{18} \text{ g sucrose} \rightarrow 1 \text{ g glucose}$
- $1.9 \text{ g sucrose} \rightarrow 1 \text{ g glucose}$
- Then multiply by 27 to get the mass of sucrose required to produce 27 g of glucose.
- $1.9 \times 27 = 51.3$
How to earn full marks:
- Determine the amount of sucrose required to produce 1 g of glucose.
- Multiply by 27 to get the mass of sucrose required to produce 27 g of glucose.
- Final answer: $\boxed{51.3 \text{ g}}$
(d)
- Some glucose may have been lost during the experiment due to spillage or incomplete transfer.
- The reaction may not have gone to completion, leaving some sucrose unreacted.
How to earn full marks:
- State two valid reasons for the reduced yield.
Common Pitfall: Always double-check your calculations, especially when dealing with multiple atoms of the same element in a molecule. Also, remember that experimental yields are often less than theoretical yields due to various factors.
Exam-Style Question 4 — Extended Response [8 marks]
Question:
Aluminium reacts with oxygen gas to form aluminium oxide, Al$_2$O$_3$.
The relative atomic mass of aluminium is 27 and oxygen is 16.
(a) Calculate the relative formula mass of aluminium oxide, Al$_2$O$_3$. [2]
(b) In an experiment, 5.4 g of aluminium reacts completely with oxygen. Calculate the mass of aluminium oxide produced. [3]
(c) Aluminium oxide can also be produced by reacting aluminium hydroxide, Al(OH)$_3$, with heat. Write the balanced chemical equation for this reaction. [3]
Worked Solution:
(a)
- The relative formula mass of aluminium oxide is the sum of the relative atomic masses of two aluminium atoms and three oxygen atoms.
- $M_r(Al_2O_3) = (2 \times 27) + (3 \times 16) = 54 + 48 = 102$
How to earn full marks:
- Correctly add the relative atomic masses.
- Final answer: $\boxed{102}$
(b)
- The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction.
- Therefore, the total mass of reactants equals the total mass of products.
- The ratio is 2 Al : 3 O
- Calculate the mass of oxygen needed to react with 5.4g of aluminium
- $2 \times 27 = 54$ (2 Al atoms)
- $3 \times 16 = 48$ (3 O atoms)
- $\frac{5.4}{54} = 0.1$
- Multiply by 48, the mass of oxygen needed
- $0.1 \times 48 = 4.8$
- Add the mass of aluminium and oxygen
- $5.4 + 4.8 = 10.2$
How to earn full marks:
- Correctly calculate the mass of aluminium oxide produced.
- Final answer: $\boxed{10.2 \text{ g}}$
(c)
- The reactant is aluminium hydroxide.
- The products are aluminium oxide and water.
- $2Al(OH)_3(s) \rightarrow Al_2O_3(s) + 3H_2O(g)$
How to earn full marks:
- Correct chemical formulas for all reactants and products.
- Correct balancing of the equation.
- State symbols are not required.
Common Pitfall: In part (b), many students forget to calculate the mass of oxygen that reacted. Remember that to find the mass of the product, you need to consider the masses of all the reactants that combined. Also, make sure your chemical equations are properly balanced.