Cambridge IGCSE0620

Formulae

Chemistry 0620 Chapter Notes

What this chapter covers

FormulaeRelative masses of atoms and moleculesThe mole and the Avogadro constant
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In this chapter

  1. 1Chemical Formulae and Relative Mass
  2. 2The Mole: A Chemist's Dozen
  3. 3Reacting Mass Calculations
  4. 4Empirical and Molecular Formulae
  5. 5Calculations Involving Gas Volumes
  6. 6Concentration of Solutions
  7. 7Percentage Yield and Purity

1. Chemical Formulae and Relative Mass

A chemical formula tells you which elements are in a compound and the ratio of their atoms. For example, H₂O means two hydrogen atoms are bonded to one oxygen atom. To write a formula for an ionic compound, you balance the charges of the ions. For example, a magnesium ion is Mg²⁺ and a chloride ion is Cl⁻, so you need two Cl⁻ ions to balance one Mg²⁺ ion, giving the formula MgCl₂. The Relative Atomic Mass (Ar) of an element is the average mass of its atoms, found in the periodic table. The Relative Formula Mass (Mr) of a compound is the sum of the Ar values of all the atoms in its formula.

Mr = Σ (Ar of each atom)

Key term

Relative Formula Mass (Mr): The sum of the relative atomic masses of all the atoms shown in the chemical formula of a substance.

Examiner insight

Examiners award marks for showing the breakdown of the calculation, so write out the sum (e.g., (2 × 27) + (3 × 32) + (12 × 16)) before giving the final answer.

Common pitfall

Forgetting to multiply the atoms inside a bracket by the number outside the bracket, for example in Ca(NO₃)₂, calculating with only one nitrogen and three oxygen atoms instead of two nitrogen and six oxygen atoms.

Worked example 12 marks

Calculate the relative formula mass (Mr) of aluminium sulfate, Al₂(SO₄)₃. (Ar: Al = 27, S = 32, O = 16)

  1. 1

    Step 1: Identify the number of each type of atom in the formula. There are 2 Aluminium atoms, 3 Sulfur atoms, and 4x3 = 12 Oxygen atoms.

  2. 2

    Step 2: Find the Ar for each element from the periodic table or question data. Al = 27, S = 32, O = 16.

  3. 3

    Step 3: Multiply the Ar of each element by the number of atoms of that element. Al: 2 × 27 = 54. S: 3 × 32 = 96. O: 12 × 16 = 192.

  4. 4

    Step 4: Add these values together to find the Mr. Mr = 54 + 96 + 192 = 342.

Recap

  • A chemical formula shows the type and number of atoms in a substance.
  • For ionic compounds, the positive and negative charges must balance to zero.
  • Relative Atomic Mass (Ar) is the mass of an atom from the periodic table.
  • Relative Formula Mass (Mr) is calculated by adding up the Ar of all atoms in a formula.
  • Remember to multiply the Ar by the subscript for each element, including those outside brackets.

Quick check

  1. What is the relative formula mass (Mr) of magnesium hydroxide, Mg(OH)₂? (Ar: Mg=24, O=16, H=1)2 marks

2. The Mole: A Chemist's Dozen

In chemistry, we deal with huge numbers of atoms, so we use a unit called the mole (mol) to count them. One mole of any substance contains the same number of particles: 6.02 × 10²³. This giant number is called the Avogadro Constant. The beauty of the mole is its link to mass: one mole of a substance has a mass in grams that is exactly equal to its relative formula mass (Mr). This is called the molar mass, and its unit is grams per mole (g/mol).

Number of particles = moles × Avogadro's constant (6.02 × 10²³)

Key term

Mole: The amount of substance that contains 6.02 × 10²³ particles (e.g., atoms, molecules, or ions), which has a mass in grams equal to the substance's relative formula mass.

Common pitfall

Confusing relative formula mass (Mr), which has no units, with molar mass, which has units of g/mol. They have the same numerical value but represent different concepts.

Fun fact

If you had a mole of pennies and gave away one million dollars per second to every person on Earth, it would take you over 300 years to give it all away.

Worked example 12 marks

What is the mass of 0.25 moles of carbon dioxide, CO₂? (Ar: C=12, O=16)

  1. 1

    Step 1: Calculate the molar mass of CO₂. Mr = 12 + (2 × 16) = 44. So, the molar mass is 44 g/mol.

  2. 2

    Step 2: Use the formula: mass = moles × molar mass.

  3. 3

    Step 3: Substitute the values: mass = 0.25 mol × 44 g/mol = 11 g.

Worked example 22 marks

How many moles are there in 27 g of water, H₂O? (Ar: H=1, O=16)

  1. 1

    Step 1: Calculate the molar mass of H₂O. Mr = (2 × 1) + 16 = 18. So, the molar mass is 18 g/mol.

  2. 2

    Step 2: Use the formula: moles = mass / molar mass.

  3. 3

    Step 3: Substitute the values: moles = 27 g / 18 g/mol = 1.5 moles.

Recap

  • A mole is a specific quantity: 6.02 × 10²³ particles.
  • The number 6.02 × 10²³ is the Avogadro Constant.
  • The mass of one mole of a substance in grams is its molar mass, which is numerically equal to its Mr.
  • You can convert between mass and moles using the molar mass.

Quick check

  1. What is the mass of 1 mole of sodium chloride, NaCl? (Ar: Na=23, Cl=35.5)1 mark
  2. How many atoms are in 1 mole of helium (He)?1 mark

3. Reacting Mass Calculations

A balanced chemical equation doesn't just tell you what reacts; it tells you how much reacts. The large numbers in front of each formula (the stoichiometry) represent the ratio of moles. 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 can use this mole ratio to calculate the mass of a reactant or product if you know the mass of another substance in the reaction.

moles = mass / Mr

mass = moles × Mr

Key term

Stoichiometry: The ratio of the amounts in moles of each substance in a balanced chemical equation.

Examiner insight

Examiners look for the four key steps: moles of known, mole ratio, moles of unknown, mass of unknown. Laying your work out clearly following these steps will help you secure all the marks.

Fun fact

The principles of stoichiometry are used to formulate the precise fuel-to-air mixture in a car engine to ensure complete combustion and maximum efficiency.

Worked example 14 marks

What mass of iron can be produced from 16 tonnes of iron(III) oxide, Fe₂O₃? The equation is: Fe₂O₃ + 3CO → 2Fe + 3CO₂. (Ar: Fe=56, O=16). (1 tonne = 1,000,000 g)

  1. 1

    Step 1: Calculate moles of the known substance (Fe₂O₃). First find its Mr: (2 × 56) + (3 × 16) = 112 + 48 = 160. Molar mass = 160 g/mol.

  2. 2

    Step 2: Convert mass to grams: 16 tonnes = 16,000,000 g. Moles of Fe₂O₃ = 16,000,000 g / 160 g/mol = 100,000 moles.

  3. 3

    Step 3: Use the mole ratio from the balanced equation. The ratio of Fe₂O₃ : Fe is 1 : 2. So, moles of Fe = 100,000 moles × 2 = 200,000 moles.

  4. 4

    Step 4: Calculate the mass of the target substance (Fe). Ar of Fe = 56. Mass of Fe = moles × Ar = 200,000 mol × 56 g/mol = 11,200,000 g.

  5. 5

    Step 5: Convert back to tonnes if needed. Mass = 11,200,000 g / 1,000,000 g/tonne = 11.2 tonnes.

Recap

  • Always start with a balanced chemical equation.
  • The core steps are: Mass A → Moles A → Moles B → Mass B.
  • Convert the mass of the known substance into moles.
  • Use the stoichiometric ratio from the equation to find the moles of the target substance.
  • Convert the moles of the target substance back into mass.

Quick check

  1. In the reaction C + O₂ → CO₂, how many moles of CO₂ are made from 6g of Carbon? (Ar: C=12)2 marks

4. Empirical and Molecular Formulae

Compounds have two types of formulae. The empirical formula is the simplest whole-number ratio of atoms in the compound. The molecular formula gives the actual number of atoms of each element in one molecule. Sometimes they are the same (e.g., H₂O), but often they are different. For example, ethene has a molecular formula of C₂H₄, but its simplest ratio is 1 carbon to 2 hydrogens, so its empirical formula is CH₂. You can find the empirical formula from experimental data (masses or percentages of elements) and then use the compound's Mr to find the true molecular formula.

Molecular Formula = (Empirical Formula)n

n = Mr (Molecular) / Mr (Empirical)

Key term

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

Examiner insight

Setting out your empirical formula calculation in a clear table with headings (Element, Mass/%, Moles, Ratio) is a good way to avoid errors and show your method clearly.

Common pitfall

Stopping after finding the empirical formula and not proceeding to find the molecular formula when the question asks for it.

Worked example 14 marks

A hydrocarbon is found to contain 92.3% carbon and 7.7% hydrogen by mass. Its relative molecular mass (Mr) is 78. Find its empirical and molecular formulae. (Ar: C=12, H=1)

  1. 1

    Step 1: Assume 100g of the compound, so we have 92.3g of Carbon and 7.7g of Hydrogen.

  2. 2

    Step 2: Convert these masses into moles. Moles C = 92.3g / 12 g/mol = 7.69 mol. Moles H = 7.7g / 1 g/mol = 7.7 mol.

  3. 3

    Step 3: Find the simplest ratio by dividing both by the smallest number of moles (7.69). Ratio C = 7.69 / 7.69 = 1. Ratio H = 7.7 / 7.69 ≈ 1.

  4. 4

    Step 4: The simplest whole number ratio is 1:1. So, the empirical formula is CH.

  5. 5

    Step 5: Find the mass of the empirical formula. Mr of CH = 12 + 1 = 13.

  6. 6

    Step 6: Find how many empirical units fit into the molecular formula. n = (Mr of molecule) / (Mr of empirical formula) = 78 / 13 = 6.

  7. 7

    Step 7: Multiply the empirical formula by n to get the molecular formula. Molecular formula = (CH)₆ = C₆H₆.

Recap

  • The empirical formula is the simplest ratio; the molecular formula is the actual number of atoms.
  • To find the empirical formula: convert masses/percentages to moles, then divide by the smallest mole value.
  • To find the molecular formula, you need the empirical formula and the relative molecular mass (Mr).
  • Calculate 'n' by dividing the molecular Mr by the empirical Mr.
  • The molecular formula is the empirical formula multiplied by 'n'.

Quick check

  1. A compound's empirical formula is P₂O₅ and its Mr is 284. What is its molecular formula? (Ar: P=31, O=16)2 marks

5. Calculations Involving Gas Volumes

A surprising fact in chemistry is that one mole of *any* gas occupies the same volume as one mole of *any other* gas, as long as the temperature and pressure are the same. At Room Temperature and Pressure (r.t.p.), defined as 20°C and 1 atmosphere, this volume is 24 decimetres cubed (dm³), or 24,000 cm³. This is called the molar volume. This gives us a simple way to relate the moles of a gas to its volume, allowing us to perform calculations similar to reacting masses but with volumes of gas instead.

Volume of gas (dm³) = moles × 24

moles = Volume of gas (dm³) / 24

1 dm³ = 1000 cm³

Key term

Molar Volume: The volume occupied by one mole of any gas, which is 24 dm³ at room temperature and pressure (r.t.p.).

Examiner insight

Examiners will test your ability to convert between units. Always convert cm³ to dm³ (by dividing by 1000) before using the `moles = volume / 24` formula, unless the question asks for the answer in cm³.

Common pitfall

Using the molar mass (Mr) in gas volume calculations instead of the molar volume (24 dm³). Remember, volume calculations for gases at r.t.p. use the number 24.

Worked example 13 marks

What volume of carbon dioxide gas, measured at r.t.p., is produced when 25 g of calcium carbonate is heated? CaCO₃(s) → CaO(s) + CO₂(g). (Ar: Ca=40, C=12, O=16)

  1. 1

    Step 1: Calculate the moles of the known substance (CaCO₃). Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100. Moles = mass / Mr = 25 g / 100 g/mol = 0.25 moles.

  2. 2

    Step 2: Use the mole ratio from the balanced equation. The ratio of CaCO₃ : CO₂ is 1 : 1. So, moles of CO₂ produced = 0.25 moles.

  3. 3

    Step 3: Calculate the volume of the gas. Volume = moles × molar volume = 0.25 mol × 24 dm³/mol = 6 dm³.

  4. 4

    Step 4: The volume of CO₂ produced is 6 dm³ (or 6000 cm³).

Recap

  • One mole of any gas occupies 24 dm³ (24,000 cm³) at r.t.p.
  • This value (24 dm³/mol) is the molar volume at r.t.p.
  • You can convert between moles of a gas and its volume using the molar volume.
  • Always check your units: dm³ or cm³.
  • Gas volume calculations follow the same mole ratio logic as reacting mass calculations.

Quick check

  1. What is the volume, in cm³, of 0.05 moles of hydrogen gas at r.t.p.?1 mark

6. Concentration of Solutions

Concentration tells us how much 'stuff' (solute) is dissolved in a certain volume of a liquid (solvent). A solution with a high concentration is 'strong', and one with a low concentration is 'weak' or dilute. We measure concentration in two main ways: mass concentration in grams per decimetre cubed (g/dm³), and molar concentration in moles per decimetre cubed (mol/dm³). Molar concentration is more common in chemistry calculations as it links directly to mole ratios in reactions.

Concentration (mol/dm³) = moles / volume (dm³)

Concentration (g/dm³) = mass (g) / volume (dm³)

Concentration (g/dm³) = Concentration (mol/dm³) × Mr

Key term

Concentration: The amount of a solute dissolved in a specified volume of a solution, typically measured in mol/dm³ or g/dm³.

Examiner insight

In titration calculations, examiners award separate marks for calculating the moles of the known solution, for using the mole ratio correctly, and for the final concentration calculation. Show each step clearly.

Common pitfall

Forgetting to convert volume from cm³ to dm³. This is the most frequent error in concentration calculations. Remember: divide by 1000.

Worked example 13 marks

Calculate the concentration in mol/dm³ of a solution containing 4.0 g of sodium hydroxide (NaOH) in 200 cm³ of solution. (Ar: Na=23, O=16, H=1)

  1. 1

    Step 1: Calculate the moles of the solute (NaOH). Mr of NaOH = 23 + 16 + 1 = 40. Moles = mass / Mr = 4.0 g / 40 g/mol = 0.1 moles.

  2. 2

    Step 2: Convert the volume to dm³. Volume = 200 cm³ / 1000 = 0.2 dm³.

  3. 3

    Step 3: Calculate the concentration. Concentration = moles / volume = 0.1 mol / 0.2 dm³ = 0.5 mol/dm³.

Worked example 24 marks

25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution was neutralised by 20.0 cm³ of sulfuric acid. Calculate the concentration of the sulfuric acid. The equation is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

  1. 1

    Step 1: Calculate moles of the known substance (NaOH). Volume = 25.0 cm³ = 0.025 dm³. Moles NaOH = conc × vol = 0.100 mol/dm³ × 0.025 dm³ = 0.0025 moles.

  2. 2

    Step 2: Use the mole ratio from the equation. The ratio of NaOH : H₂SO₄ is 2 : 1. So, moles of H₂SO₄ = 0.0025 moles / 2 = 0.00125 moles.

  3. 3

    Step 3: Calculate the concentration of the target substance (H₂SO₄). Volume = 20.0 cm³ = 0.020 dm³. Concentration = moles / volume = 0.00125 mol / 0.020 dm³ = 0.0625 mol/dm³.

Recap

  • Concentration is the amount of solute per unit volume of solution.
  • The standard units are mol/dm³ and g/dm³.
  • Always convert volumes in cm³ to dm³ by dividing by 1000 for calculations.
  • The formula triangle links concentration, moles, and volume.
  • Titration calculations use concentration and volume to find moles, then use the mole ratio.

Quick check

  1. What is the concentration in g/dm³ if 10g of salt is dissolved in 2 dm³ of water?1 mark

7. Percentage Yield and Purity

In the real world, chemical reactions rarely produce the maximum possible amount of product. Some might be lost during transfer, or the reaction may not go to completion. Percentage yield compares the actual mass of product you get (actual yield) with the mass you should have theoretically produced (theoretical yield), calculated from stoichiometry. Separately, percentage purity tells you how much of a sample is the actual chemical you want. For example, a rock might be 80% copper ore and 20% worthless sand. Its purity is 80%.

Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Percentage Purity = (Mass of Pure Substance / Total Mass of Sample) × 100%

Key term

Percentage Yield: The actual amount of product obtained from a reaction, expressed as a percentage of the maximum theoretical amount possible.

Examiner insight

Be clear which value is the actual yield and which is the theoretical yield. A theoretical yield that is smaller than the actual yield indicates a calculation error, as you cannot create matter.

Common pitfall

Mixing up the 'actual yield' and 'theoretical yield' in the percentage yield formula. The theoretical yield is the one you calculate, and it should be the denominator.

Worked example 14 marks

A student heats 12.4 g of copper(II) carbonate, which decomposes according to the equation: CuCO₃ → CuO + CO₂. They collect 6.4 g of copper(II) oxide. Calculate the percentage yield. (Ar: Cu=64, C=12, O=16)

  1. 1

    Step 1: Calculate the theoretical yield of CuO. First, find moles of CuCO₃. Mr of CuCO₃ = 64 + 12 + (3×16) = 124. Moles = 12.4 g / 124 g/mol = 0.1 moles.

  2. 2

    Step 2: Use the mole ratio. CuCO₃ : CuO is 1 : 1. So, 0.1 moles of CuO should be produced.

  3. 3

    Step 3: Calculate the theoretical mass of CuO. Mr of CuO = 64 + 16 = 80. Mass = 0.1 mol × 80 g/mol = 8.0 g. This is the theoretical yield.

  4. 4

    Step 4: Calculate the percentage yield. The actual yield given is 6.4 g. % Yield = (Actual / Theoretical) × 100 = (6.4 g / 8.0g) × 100 = 80%.

Worked example 22 marks

An impure sample of magnesium oxide has a mass of 5.0 g. After purification, 4.5 g of pure magnesium oxide is recovered. What was the percentage purity of the original sample?

  1. 1

    Step 1: Identify the mass of the pure substance and the total mass of the impure sample. Mass of pure MgO = 4.5 g. Total mass of sample = 5.0 g.

  2. 2

    Step 2: Use the percentage purity formula. % Purity = (Mass of pure / Total mass) × 100.

  3. 3

    Step 3: Substitute the values: % Purity = (4.5 g / 5.0g) × 100 = 90%.

Recap

  • Percentage yield measures the efficiency of a chemical reaction.
  • Percentage purity measures the cleanness of a chemical sample.
  • To calculate yield, you must first find the theoretical maximum mass using stoichiometry.
  • The actual yield is always given in the question or measured in an experiment.
  • Do not confuse yield (about a process) with purity (about a substance).

Quick check

  1. A reaction has a theoretical yield of 50g. The actual yield was 45g. What is the percentage yield?1 mark

End-of-chapter exercise

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

  1. Calculate the relative formula mass (Mr) of hydrated ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂.6H₂O. (Ar: N=14, H=1, Fe=56, S=32, O=16)3 marks
  2. A sample of methane (CH₄) has a mass of 4.0 g. Calculate the number of methane molecules in the sample. (Ar: C=12, H=1; Avogadro constant = 6.02 x 10²³ mol⁻¹)3 marks
  3. Hydrogen peroxide decomposes to form water and oxygen: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Calculate the mass of oxygen produced when 17 g of hydrogen peroxide decomposes. (Ar: H=1, O=16)3 marks
  4. A compound is found to contain 26.7% phosphorus, 12.1% nitrogen, and 61.2% chlorine by mass. Calculate its empirical formula. (Ar: P=31, N=14, Cl=35.5)4 marks
  5. When 5.4 g of aluminium reacts with excess chlorine gas, aluminium chloride is formed. 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). Calculate the volume of chlorine gas that reacted, measured at r.t.p. (Ar: Al=27)4 marks
  6. A student wants to make a 0.250 mol/dm³ solution of sodium carbonate (Na₂CO₃). What mass of sodium carbonate would they need to dissolve in water to make 250 cm³ of this solution? (Ar: Na=23, C=12, O=16)4 marks
  7. The combustion of propane is given by the equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. If 11 g of propane is burned, a student collects 25 g of carbon dioxide. Calculate the percentage yield of carbon dioxide. (Ar: C=12, H=1, O=16)4 marks
  8. Butane (C₄H₁₀) is a gas at room temperature. Its molecular formula is C₄H₁₀. An unknown gaseous hydrocarbon has an empirical formula of C₂H₅ and a density such that 100 cm³ of the gas has a mass of 0.242 g at r.t.p. Deduce the molecular formula of this unknown hydrocarbon.5 marks
  9. In a titration, 25.0 cm³ of a solution of sodium hydroxide required 22.5 cm³ of 0.100 mol/dm³ sulfuric acid for complete neutralisation. The reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Calculate the concentration of the sodium hydroxide solution in g/dm³. (Ar: Na=23, O=16, H=1)5 marks
  10. A 10.0 g sample of impure zinc carbonate was heated strongly. The carbon dioxide gas produced was collected and found to have a volume of 1.50 dm³ at r.t.p. The equation for the reaction is: ZnCO₃(s) → ZnO(s) + CO₂(g). Calculate the percentage purity of the zinc carbonate sample. (Ar: Zn=65, C=12, O=16)5 marks

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