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Cambridge AS & A Level9701

Relative masses of atoms and molecules

Chemistry 9701 Chapter Notes

What this chapter covers

Relative masses of atoms and moleculesThe mole and the Avogadro constantFormulasReacting masses and volumes (of solutions and gases)
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In this chapter

  1. 1Relative Atomic Mass and Isotopes
  2. 2Calculating Ar from Mass Spectra
  3. 3Relative Molecular and Formula Mass (Mr)
  4. 4The Mole and Avogadro's Constant
  5. 5Empirical and Molecular Formulae
  6. 6Calculating Empirical Formulae
  7. 7Calculations from Chemical Equations

1. Relative Atomic Mass and Isotopes

Atoms are incredibly tiny, so we can't weigh them individually. Instead, we compare their masses to a standard: the Carbon-12 isotope, which is assigned a mass of exactly 12. The relative atomic mass (Ar) of an element is the weighted average mass of its naturally occurring isotopes compared to this standard. Most elements exist as a mixture of isotopes – atoms with the same number of protons but different numbers of neutrons. For example, chlorine exists as Chlorine-35 and Chlorine-37. Because Ar is a weighted average that accounts for the abundance of each isotope, its value is often not a whole number (e.g., Ar of Chlorine is 35.5).

Ar = (weighted average mass of an atom of the element) / (1/12 × mass of one atom of Carbon-12)

Key term

Relative Atomic Mass (Ar): The weighted average mass of naturally occurring atoms of an element on a scale where an atom of carbon-12 has a mass of exactly 12 units.

Examiner insight

Examiners reward clear definitions of relative atomic mass that correctly mention both 'weighted average mass' and the 'carbon-12 scale'.

Common pitfall

Assuming the relative atomic mass is just a simple average of the isotope masses, without considering their different abundances.

Fun fact

Until 1961, both physicists and chemists used different standards for atomic mass (Oxygen-16 and natural oxygen, respectively). The adoption of Carbon-12 unified the scale.

Worked example 12 marks

The relative atomic mass of chlorine is 35.5. Explain why this value is not a whole number. [2]

  1. 1

    Step 1: State that chlorine exists as a mixture of isotopes. The main isotopes are chlorine-35 and chlorine-37.

  2. 2

    Step 2: Explain that the relative atomic mass is a weighted average of the masses of these isotopes, which takes into account their relative abundances. Natural chlorine is approximately 75% chlorine-35 and 25% chlorine-37, so the average mass is closer to 35 than 37, resulting in the value of 35.5.

Recap

  • Relative masses are all measured against the carbon-12 isotope, which has a mass of exactly 12.
  • Isotopes are atoms of the same element with different numbers of neutrons.
  • Relative atomic mass (Ar) is a weighted average, which accounts for the different masses and abundances of an element's isotopes.
  • The presence of isotopes is why the Ar of many elements is not a whole number.

Quick check

  1. Define the term 'isotope'.1 mark
  2. Which isotope is the standard for the relative atomic mass scale?1 mark

2. Calculating Ar from Mass Spectra

A mass spectrometer is an instrument that determines the relative isotopic mass and the relative abundance of each isotope in a sample of an element. It produces a mass spectrum, which is a chart plotting relative abundance (y-axis) against the mass-to-charge ratio (m/z) (x-axis). For singly charged ions, the m/z value is simply the isotopic mass. To calculate the relative atomic mass from a mass spectrum, you multiply each isotopic mass by its relative abundance, sum these values, and then divide by the total abundance.

Ar = Σ (isotopic mass × relative abundance) / Σ (relative abundances)

Key term

Mass Spectrum: A graph produced by a mass spectrometer that shows the relative abundance of isotopes plotted against their mass-to-charge ratio.

Examiner insight

Marks are awarded for clearly showing the full calculation step-by-step. A correct final answer with no working may not receive full credit.

Common pitfall

If the abundances are given as percentages, students sometimes forget to divide by 100 at the end.

Worked example 12 marks

A sample of boron is analysed in a mass spectrometer. The mass spectrum shows two peaks: one at m/z = 10.0 with a relative abundance of 19.9, and another at m/z = 11.0 with a relative abundance of 80.1. Calculate the relative atomic mass of boron to one decimal place. [2]

  1. 1

    Step 1: Multiply each isotopic mass by its relative abundance. (10.0 × 19.9) = 199. (11.0 × 80.1) = 881.1.

  2. 2

    Step 2: Sum these values: 199 + 881.1 = 1080.1.

  3. 3

    Step 3: Sum the total relative abundances: 19.9 + 80.1 = 100.

  4. 4

    Step 4: Divide the sum of (mass × abundance) by the total abundance: Ar = 1080.1 / 100 = 10.801. The answer to one decimal place is 10.8.

Recap

  • A mass spectrometer separates isotopes based on their mass.
  • The x-axis of a mass spectrum shows the mass/charge ratio, which is the isotopic mass for +1 ions.
  • The y-axis shows the relative abundance of each isotope.
  • To calculate Ar, multiply each isotopic mass by its abundance, add these up, and divide by the total abundance.

Quick check

  1. On a mass spectrum, what quantity is plotted on the y-axis?1 mark

3. Relative Molecular and Formula Mass (Mr)

While Ar refers to single atoms, we need a way to describe the mass of a whole compound. For covalently bonded substances (molecules), we use 'relative molecular mass'. For ionic compounds, we use 'relative formula mass'. Both are given the symbol Mr and are calculated in exactly the same way: by adding up the relative atomic masses (Ar) of all the atoms shown in the chemical formula. Pay close attention to the small numbers (subscripts) in the formula, and remember to multiply everything inside a bracket by the number outside it.

Mr = Sum of (Number of atoms of an element × Ar of the element) for all elements in the formula.

Key term

Relative Formula Mass (Mr): The sum of the relative atomic masses of all the atoms or ions present in the formula unit of a substance, based on the carbon-12 scale.

Examiner insight

Examiners look for systematic working. Listing the calculation for each element separately before summing them is a good way to avoid errors and gain method marks.

Common pitfall

Forgetting to multiply the number of atoms inside a bracket by the subscript outside it, for example in Ca(OH)2 counting only one O and one H.

Worked example 12 marks

Using the Periodic Table, calculate the relative formula mass (Mr) of ammonium sulfate, (NH4)2SO4. (Ar values: N=14.0, H=1.0, S=32.1, O=16.0) [2]

  1. 1

    Step 1: Identify the number of atoms of each element. The bracket (NH4) is multiplied by 2, so we have 2 Nitrogen atoms and 2 × 4 = 8 Hydrogen atoms. We also have 1 Sulfur atom and 4 Oxygen atoms.

  2. 2

    Step 2: Multiply the number of each atom by its Ar and sum the results.

  3. 3

    N: 2 × 14.0 = 28.0

  4. 4

    H: 8 × 1.0 = 8.0

  5. 5

    S: 1 × 32.1 = 32.1

  6. 6

    O: 4 × 16.0 = 64.0

  7. 7

    Step 3: Add them all together: Mr = 28.0 + 8.0 + 32.1 + 64.0 = 132.1.

Worked example 23 marks

Calculate the Mr of hydrated magnesium nitrate, Mg(NO3)2·6H2O. (Ar values: Mg=24.3, N=14.0, O=16.0, H=1.0) [3]

  1. 1

    Step 1: Calculate the mass of the anhydrous salt part, Mg(NO3)2. Mg: 1 × 24.3 = 24.3. N: 2 × 14.0 = 28.0. O: (3 × 2) × 16.0 = 96.0. Total for Mg(NO3)2 = 24.3 + 28.0 + 96.0 = 148.3.

  2. 2

    Step 2: Calculate the mass of the water of crystallisation part, 6H2O. Mr of H2O = (2 × 1.0) + 16.0 = 18.0. Mass of 6 waters = 6 × 18.0 = 108.0.

  3. 3

    Step 3: Add the two parts together: Total Mr = 148.3 + 108.0 = 256.3.

Recap

  • Mr is found by summing the Ar values of all atoms in a formula.
  • 'Relative molecular mass' is for molecules; 'relative formula mass' is for ionic compounds, but both are calculated the same way.
  • Always be careful with subscripts and brackets in chemical formulae.
  • For hydrated salts (e.g., ·xH2O), calculate the mass of the water molecules and add it to the mass of the salt.

Quick check

  1. Calculate the relative molecular mass of glucose, C6H12O6. (Ar: C=12.0, H=1.0, O=16.0)2 marks

4. The Mole and Avogadro's Constant

A 'mole' (mol) is a specific quantity, just like a 'dozen' means 12. In chemistry, one mole is the amount of substance that contains 6.02 × 10^23 particles. This huge number is called the Avogadro Constant (L). The particles can be atoms, molecules, or ions. Conveniently, one mole of a substance has a mass in grams that is numerically equal to its relative formula mass (Mr). For example, as carbon's Ar is 12.0, one mole of carbon atoms has a mass of 12.0 g. This relationship is the bridge that connects the mass you can weigh in a lab to the number of particles involved in a reaction.

Number of moles (n) = mass (m) / Molar mass (M)

mass (m) = Number of moles (n) × Molar mass (M)

Number of particles = Number of moles (n) × Avogadro's Constant (L)

Key term

Mole (mol): The amount of substance that contains the same number of specified particles as there are atoms in exactly 12 g of the carbon-12 isotope.

Examiner insight

When asked for the number of atoms in a molecular substance, a common mistake is to forget the final step of multiplying the number of molecules by the number of atoms in each molecule.

Common pitfall

Confusing moles of atoms with moles of molecules. For example, 1 mole of O2 contains 1 mole of molecules but 2 moles of atoms.

Fun fact

Amedeo Avogadro, for whom the constant is named, didn't actually calculate its value. He proposed the principle that equal volumes of gases contain equal numbers of particles, and the constant was named in his honour later.

Worked example 12 marks

How many moles are present in 90.0 g of water, H2O? (Ar: H=1.0, O=16.0) [2]

  1. 1

    Step 1: Calculate the molar mass (M) of H2O. M = (2 × 1.0) + 16.0 = 18.0 g/mol.

  2. 2

    Step 2: Use the formula n = m / M. n = 90.0 g / 18.0 g/mol = 5.00 mol.

Worked example 23 marks

Calculate the number of molecules in 11 g of carbon dioxide, CO2. (Ar: C=12.0, O=16.0; Avogadro Constant = 6.02 × 10^23 mol⁻¹) [3]

  1. 1

    Step 1: Calculate the molar mass (M) of CO2. M = 12.0 + (2 × 16.0) = 44.0 g/mol.

  2. 2

    Step 2: Calculate the number of moles of CO2. n = m / M = 11 g / 44.0 g/mol = 0.25 mol.

  3. 3

    Step 3: Calculate the number of molecules. Number of molecules = n × L = 0.25 mol × (6.02 × 10^23 molecules/mol) = 1.505 × 10^23 molecules.

Recap

  • A mole is a unit for the amount of substance.
  • One mole of any substance contains 6.02 × 10^23 particles (Avogadro's Constant).
  • The mass of one mole of a substance in grams is its molar mass (M), which is numerically equal to its Mr.
  • The key formula is: moles = mass / molar mass.

Quick check

  1. What is the mass of 2 moles of sodium chloride, NaCl? (Ar: Na=23.0, Cl=35.5)2 marks

5. Empirical and Molecular Formulae

A chemical formula tells us the ratio of elements in a compound, but there are two main types. The Empirical Formula is the simplest possible whole-number ratio of atoms. The Molecular Formula gives the actual number of atoms of each element in one molecule of the substance. For some compounds, like water (H2O), the empirical and molecular formulae are the same. For others, they are different. For example, the sugar glucose has a molecular formula of C6H12O6. To find its empirical formula, we divide the subscripts by the highest common factor (6), giving CH2O.

Molecular Formula = (Empirical Formula)n

n = Relative Molecular Mass (Mr) / Relative Mass of Empirical Formula

Key term

Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.

Examiner insight

Students must give the final answer as a formula (e.g., C4H8), not just the value of 'n' or the ratio.

Common pitfall

Assuming the empirical formula is always the same as the molecular formula. This is only true when the simplest ratio is the actual ratio.

Worked example 13 marks

A compound has the empirical formula CH2 and a relative molecular mass (Mr) of 56.0. Deduce its molecular formula. (Ar: C=12.0, H=1.0) [3]

  1. 1

    Step 1: Calculate the relative mass of the empirical formula (CH2). Mass = 12.0 + (2 × 1.0) = 14.0.

  2. 2

    Step 2: Find the whole-number multiplier 'n' by dividing the Mr by the empirical formula mass. n = Mr / Empirical Mass = 56.0 / 14.0 = 4.

  3. 3

    Step 3: Multiply the subscripts in the empirical formula by 'n' to find the molecular formula. Molecular Formula = (CH2)₄ = C4H8.

Recap

  • The empirical formula is the simplest ratio of atoms.
  • The molecular formula is the actual number of atoms in a molecule.
  • The molecular formula is always a whole-number multiple of the empirical formula.
  • You can find the molecular formula if you know the empirical formula and the relative molecular mass (Mr).

Quick check

  1. What is the empirical formula of hydrogen peroxide, H2O2?1 mark
  2. Ethene has the molecular formula C2H4. What is its empirical formula?1 mark

6. Calculating Empirical Formulae

You can determine a compound's empirical formula if you know the mass (or percentage by mass) of each element within it. There is a reliable four-step method: 1. List the elements and their masses (if given percentages, assume a 100 g sample). 2. Convert the mass of each element into moles by dividing by its Ar. 3. Divide every mole value by the smallest mole value you calculated. 4. The results give you the ratio. If they are whole numbers, you have the formula. If not, multiply all the numbers by a small integer (e.g., 2, 3) to make them whole numbers.

Key term

Composition by Mass: The breakdown of a compound's mass into the mass of each constituent element, often expressed as a percentage.

Examiner insight

Presenting your working clearly in a table is highly recommended. It helps you stay organised and makes it easy for the examiner to follow your logic and award partial credit.

Common pitfall

Rounding a mole ratio like 1.5 to 2 instead of multiplying the entire ratio by 2 to get 2:3.

Worked example 13 marks

A sample of a hydrocarbon was found to contain 85.7% carbon and 14.3% hydrogen by mass. Calculate its empirical formula. (Ar: C=12.0, H=1.0) [3]

  1. 1

    Step 1: Assume a 100 g sample, so we have 85.7 g of Carbon and 14.3 g of Hydrogen. Set up a table.

  2. 2

    Step 2: Calculate moles of each element. Moles C = 85.7 g / 12.0 g/mol = 7.142 mol. Moles H = 14.3 g / 1.0 g/mol = 14.3 mol.

  3. 3

    Step 3: Divide by the smallest number of moles (7.142). For C: 7.142 / 7.142 = 1. For H: 14.3 / 7.142 = 2.002.

  4. 4

    Step 4: The ratio is C:H = 1:2. The empirical formula is CH2.

Worked example 23 marks

1.35 g of aluminium reacts completely with oxygen to form 2.55 g of aluminium oxide. Find the empirical formula of the oxide. (Ar: Al=27.0, O=16.0) [3]

  1. 1

    Step 1: Find the mass of each element. Mass of Al = 1.35 g. Mass of O = Mass of oxide - Mass of Al = 2.55 g - 1.35 g = 1.20 g.

  2. 2

    Step 2: Calculate moles of each element. Moles Al = 1.35 g / 27.0 g/mol = 0.050 mol. Moles O = 1.20 g / 16.0 g/mol = 0.075 mol.

  3. 3

    Step 3: Divide by the smallest number of moles (0.050). For Al: 0.050 / 0.050 = 1. For O: 0.075 / 0.050 = 1.5.

  4. 4

    Step 4: The ratio is Al:O = 1:1.5. To get whole numbers, multiply both by 2. The ratio becomes 2:3. The empirical formula is Al2O3.

Recap

  • The method for finding an empirical formula is: Mass -> Moles -> Divide by Smallest -> Ratio.
  • If given percentages, assume you have a 100 g sample.
  • If a ratio is not a whole number (e.g., 1.5, 2.33), multiply all numbers in the ratio to get whole numbers.
  • Do not round mole ratios like 1.49 or 1.51 too early; they likely represent 1.5.

Quick check

  1. A compound contains 4.8 g of magnesium and 3.2 g of oxygen. What is its empirical formula? (Ar: Mg=24.0, O=16.0)2 marks

7. Calculations from Chemical Equations

A balanced chemical equation is a recipe. The large numbers in front of each formula, called stoichiometric coefficients, tell you the mole ratio in which substances react and are produced. For example, in 2H₂ + O₂ → 2H₂O, the ratio is 2 moles of hydrogen react with 1 mole of oxygen to make 2 moles of water. You cannot relate masses directly, but you can use this mole ratio to calculate reacting masses. The process is: 1. Convert the mass of your starting substance into moles. 2. Use the mole ratio from the balanced equation to find the moles of your target substance. 3. Convert the moles of your target substance back into mass.

Step 1: Moles of known = Mass of known / Molar mass of known

Step 2: Moles of unknown = Moles of known × (Mole ratio from equation)

Step 3: Mass of unknown = Moles of unknown × Molar mass of unknown

Key term

Stoichiometry: The mole ratio of reactants and products as determined by the balancing numbers in a chemical equation.

Examiner insight

The first mark in a reacting mass calculation is often for correctly calculating the moles of the initial substance. Even if you make a mistake later, you can still get this mark with clear working.

Common pitfall

Trying to use a ratio of masses directly (e.g., 'if 2g of X makes 5g of Y, then 4g of X makes 10g of Y'). This is incorrect and calculations must always go via moles.

Worked example 13 marks

What mass of iron(III) oxide (Fe2O3) is needed to produce 112 g of iron in the blast furnace? The equation is: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g). (Ar: Fe=56.0, O=16.0) [3]

  1. 1

    Step 1: Calculate moles of the known substance (iron, Fe). Moles of Fe = mass / Ar = 112 g / 56.0 g/mol = 2.00 mol.

  2. 2

    Step 2: Use the mole ratio from the equation to find moles of the unknown (Fe2O3). The ratio of Fe2O3 to Fe is 1:2. So, moles of Fe2O3 = (moles of Fe) / 2 = 2.00 mol / 2 = 1.00 mol.

  3. 3

    Step 3: Calculate the mass of the unknown (Fe2O3). First, find its molar mass: M(Fe2O3) = (2 × 56.0) + (3 × 16.0) = 112.0 + 48.0 = 160.0 g/mol. Mass of Fe2O3 = moles × M = 1.00 mol × 160.0 g/mol = 160 g.

Recap

  • A balanced equation provides the mole ratio (stoichiometry) for a reaction.
  • You must always perform reacting mass calculations via moles.
  • Never use a ratio of masses directly from the equation.
  • The three steps are: Mass A -> Moles A -> Moles B -> Mass B.

Quick check

  1. In the reaction C + O2 -> CO2, what mass of carbon is needed to produce 4.4 g of CO2? (Ar: C=12.0, O=16.0)3 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define the term 'relative atomic mass'.2 marks
  2. Calculate the relative formula mass (Mr) of calcium nitrate, Ca(NO3)2. (Ar values: Ca=40.1, N=14.0, O=16.0)2 marks
  3. A sample of magnesium has three stable isotopes: Mg-24 (78.99%), Mg-25 (10.00%), and Mg-26 (11.01%). Calculate the relative atomic mass of magnesium to two decimal places.2 marks
  4. A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. Calculate its empirical formula. (Ar values: C=12.0, H=1.0, O=16.0)3 marks
  5. Calculate the mass of carbon dioxide produced when 6.0 g of carbon is burnt in excess oxygen. The equation for the reaction is C + O2 -> CO2. (Ar values: C=12.0, O=16.0)3 marks
  6. Calculate the total number of atoms in 34.2 g of sucrose, C12H22O11. (Ar values: C=12.0, H=1.0, O=16.0; Avogadro constant = 6.02 x 10^23 mol⁻¹)4 marks
  7. A hydrocarbon has the empirical formula C3H7. Its relative molecular mass is 86.0. What is its molecular formula? (Ar values: C=12.0, H=1.0)3 marks
  8. When 12.7 g of copper reacts with silver nitrate solution, 32.4 g of silver is produced. The equation is: Cu(s) + 2AgNO3(aq) -> Cu(NO3)2(aq) + 2Ag(s). Show by calculation that the copper was the limiting reactant. (Ar values: Cu=63.5, Ag=108.0)4 marks
  9. When 1.20 g of a hydrocarbon is completely combusted, it produces 3.52 g of carbon dioxide and 1.80 g of water. Determine the empirical formula of the hydrocarbon. (Ar values: C=12.0, H=1.0, O=16.0)4 marks
  10. An unknown metal carbonate, XCO3, has a molar mass of 125.0 g/mol. Identify the metal X by calculating its relative atomic mass. (Ar values: C=12.0, O=16.0)3 marks

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