Skip to content
Cambridge IGCSE0620

Formulae

Chemistry 0620 Chapter Notes

What this chapter covers

FormulaeRelative masses of atoms and moleculesThe mole and the Avogadro constant
ShareWhatsAppPost
Formulae notes

Read online free. Download a watermarked copy with a free account.

Read the notes

The full Formulae notes as text: skim, search, and jump between subtopics.

~24 min read

In this chapter

  1. 1Formulae of elements and compounds
  2. 2Molecular formula: counting the atoms
  3. 3Word equations and symbol equations
  4. 4Balancing and state symbols
  5. 5Empirical formula
  6. 6Formula of an ionic compound
  7. 7Ionic equations
  8. 8Deducing a symbol equation
  9. 9Relative atomic mass
  10. 10Relative atomic mass from isotopes
  11. 11Relative molecular mass and formula mass
  12. 12Formulas with brackets
  13. 13Reacting masses in simple proportion
  14. 14Units of concentration
  15. 15The mole, molar mass and Avogadro constant
  16. 16Reacting masses from an equation
  17. 17Gas volumes at r.t.p.
  18. 18Limiting reactants
  19. 19Concentration and titration
  20. 20Empirical and molecular formulae
  21. 21Percentage yield, composition and purity

1. Formulae of elements and compounds

A formula uses the symbol of each element, with a small subscript number for how many atoms of it there are. Water is H₂O, carbon dioxide is CO₂ and sodium chloride is NaCl.

Some elements are written as molecules. Oxygen is diatomic, so it is O₂. Nitrogen is N₂ and chlorine is Cl₂.

Examiner insight

“Presumably these candidates had not noticed that oxygen was diatomic.” “A few candidates used superscripts to show the numbers of atoms instead of subscripts.”

Worked example 11 mark

What is the formula of potassium oxide?

A P₂O · B PO₂ · C KO · D K₂O

  1. 1

    D: K₂O, two potassium atoms for each oxygen atom. [1]

2. Molecular formula: counting the atoms

The molecular formula of a compound is the number and type of different atoms in one molecule. Ethanol, C₂H₆O, has two carbon atoms, six hydrogen atoms and one oxygen atom: nine atoms in all.

Write the numbers as subscripts, never a ‘+’ between the symbols. To deduce a formula from a model or a diagram, count each sort of atom. A single atom has no number.

Examiner insight

“The molecular formula gives the total number of each type of atom and so it was anticipated that candidates could count up each atom type even though options included some unfamiliar compounds.” “Some candidates put ‘+’ between the elements which is not allowed in a molecular formula.”

Worked example 11 mark

Which compound contains the greatest number of atoms in its formula?

A copper(II) sulfate · B ethane · C ethanol · D sulfuric acid

  1. 1

    C: ethanol, C₂H₆O, has 2 + 6 + 1 = 9 atoms. Copper(II) sulfate CuSO₄ has 6, ethane C₂H₆ has 8 and sulfuric acid H₂SO₄ has 7. [1]

3. Word equations and symbol equations

A word equation names the reactants and the products. A symbol equation gives their formulae and must be balanced: each element has the same number of atoms on both sides.

Aqueous copper(II) chloride and aqueous sodium hydroxide react to form copper(II) hydroxide and sodium chloride. The names give the formulae CuCl₂ and Cu(OH)₂.

CuCl₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaCl(aq)

Examiner insight

“Core candidates are not expected to be able to define oxidation number but should be able to recognise the use of Roman numerals in a chemical name and to write correct formula for the products of precipitation reactions in qualitative tests.” “Most candidates worked out that KCl was formed but many made errors in the rest of the equation such as Br instead of Br2 for bromine.”

Worked example 11 mark

Aqueous copper(II) chloride reacts with aqueous sodium hydroxide. What is the equation for this reaction?

A Cu₂Cl + 2NaOH → Cu(OH)₂ + 2NaCl

B Cu₂Cl + NaOH → Cu₂OH + NaCl

C CuCl₂ + 2NaOH → Cu(OH)₂ + 2NaCl

D CuCl₂ + NaOH → Cu₂OH + NaCl

  1. 1

    C: CuCl₂ + 2NaOH → Cu(OH)₂ + 2NaCl. Each side has 1 Cu, 2 Cl, 2 Na, 2 O and 2 H. [1]

4. Balancing and state symbols

Balance an equation by writing numbers in front of formulae. A state symbol follows each formula:(s) solid,(l) liquid,(g) gas and (aq) aqueous, which means dissolved in water.

Dilute hydrochloric acid is HCl(aq), not HCl(l). Water is H₂O(l), not H₂O(aq).

Write state symbols as lower-case letters in brackets.

Examiner insight

“All candidates knew the correct state symbols for the solid MgCO3 and gaseous CO2, but many erroneously assumed it was HCl(l) or H2O(aq).” “A number of candidates used capital letters for the state symbols; these are not accepted.”

Worked example 13 marks

Solid magnesium carbonate reacts with dilute hydrochloric acid to form aqueous magnesium chloride, water and carbon dioxide gas. Write the balanced symbol equation, with state symbols.

  1. 1

    formulae: MgCO₃ + HCl → MgCl₂ + H₂O + CO₂ [1]

  2. 2

    balance: MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂ [1]

  3. 3

    state symbols: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g) [1]

5. Empirical formula

The empirical formula of a compound is the simplest whole number ratio of the different atoms or ions in a compound. The molecular formula C₆H₁₂O₂ has the empirical formula C₃H₆O: divide each number by 2.

From percentages by mass, divide each percentage by the relative atomic mass. Divide the results by the smallest. Then multiply every value up until all are whole numbers, so a ratio of 1 : 2.5 becomes 2 : 5.

Never round a ratio such as 2.5 to the nearest whole number.

Examiner insight

“In option A, candidates identified the structural formula of the ester but did not reduce the formula to the simplest whole number ratio to give the empirical formula.” “Instead of doubling the partial whole number ratio of 1 : 2.5 to make the ratio 2 : 5, many candidates incorrectly changed it 1 : 3 leading to an empirical formula of IO3.”

Worked example 11 mark

Organic compound X has the following percentage composition by mass: carbon 60%; hydrogen 8%; oxygen 32%. What is the empirical formula of compound X?

A C₅H₄O · B C₅H₈O₂ · C C₁₀H₈O₄ · D C₁₅H₂O₈

  1. 1

    A_r: C = 12, H = 1, O = 16. Divide each percentage by its A_r: C = 60 ÷ 12 = 5, H = 8 ÷ 1 = 8, O = 32 ÷ 16 = 2.

  2. 2

    The ratio 5 : 8 : 2 has no common factor, so the empirical formula is C₅H₈O₂: B. [1]

6. Formula of an ionic compound

In an ionic compound the charges on the ions must cancel. Aluminium ions are Al³⁺ and nitrate ions are NO₃⁻, so one aluminium ion needs three nitrate ions: Al(NO₃)₃.

Put brackets round an ion that has more than one atom when you need more than one of it, as in Zn(NO₃)₂. Iron(II) bromide is FeBr₂.

total positive charge = total negative charge

Examiner insight

“A minority of candidates could determine the formula of magnesium nitrate correctly with MgNO3 being a very common error.”

Worked example 11 mark

The formula of an aluminium ion is Al³⁺. The formula of a nitrate ion is NO₃⁻. What is the formula of aluminium nitrate?

A AlNO₃ · B Al(NO₃)₂ · C Al₃NO₃ · D Al(NO₃)₃

  1. 1

    D: charge 3+ is balanced by 3 × (1-), so Al(NO₃)₃. [1]

7. Ionic equations

An ionic equation shows only the ions that react. An ion that is the same on both sides, a spectator ion, is left out.

In a precipitation reaction the two ions in aqueous solution that react are on the left-hand side. The solid precipitate is on the right-hand side.

Use state symbols: the pattern is (aq) + (aq) → (s).

Examiner insight

“In an ionic equation for a precipitation reaction, the two ions in aqueous solution that react to form the precipitate are aqueous and on the left-hand side; the solid precipitate is on the right-hand side.” “Of those who attempted ionic equations, some left spectator ions in their final answer.”

Worked example 13 marks

Aqueous silver nitrate reacts with aqueous sodium chloride to form a precipitate of silver chloride and aqueous sodium nitrate. Write the ionic equation, with state symbols.

  1. 1

    full equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) [1]

  2. 2

    Na⁺ and NO₃⁻ are on both sides: spectator ions, so leave them out [1]

  3. 3

    Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1]

8. Deducing a symbol equation

To deduce an equation, take every formula the question gives and use it as given. Write the formula of any other reactant or product, then balance.

A reactant may be named without its formula. Sodium burning in oxygen needs O₂ as a reactant. Read each state from the information in the question.

Examiner insight

“Others gave an equation with correct formulae but balanced it incorrectly.” “Some candidates opted to give their own formula of sodium oxide, despite Na2O being given.”

Worked example 13 marks

Sodium burns in oxygen to form sodium oxide, Na₂O. Sodium and sodium oxide are solids. Write the balanced symbol equation, with state symbols.

  1. 1

    formulae: Na + O₂ → Na₂O [1]

  2. 2

    balance: 4Na + O₂ → 2Na₂O [1]

  3. 3

    state symbols: 4Na(s) + O₂(g) → 2Na₂O(s) [1]

Recap

  • A formula has symbols and subscripts. Oxygen is O₂, not O.
  • Molecular formula = number and type of different atoms in one molecule.
  • A symbol equation must be balanced. State symbols: (s), (l), (g), (aq).
  • Dilute acids and salt solutions are (aq). Water is (l).
  • Extended: empirical formula = simplest whole number ratio of the different atoms or ions.
  • Extended: the charges on the ions cancel: Al³⁺ with NO₃⁻ gives Al(NO₃)₃.
  • Extended: an ionic equation leaves out the spectator ions: (aq) + (aq) gives (s).

9. Relative atomic mass

Relative atomic mass, Ar, is the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of ^{12}C.

Two words are easy to lose. It is the average mass, not the total mass. The reference is the carbon-12 atom, not just carbon.

Examiner insight

“Most candidates recalled that definitions of relative mass were compared to the carbon-12 isotope but linked it to the total mass, not the average mass of the isotopes.” “The least well-known answer was that the scale is based upon a ‘carbon-12’ atom with just ‘carbon’ being the most frequent incorrect answer.”

Worked example 11 mark

What is the definition of the relative atomic mass, Ar, of an element?

A the average mass of the isotopes of the element compared to 1/12th of the mass of an atom of ^{12}C

B the total mass of the isotopes of the element compared to 1/12th of the mass of an atom of ^{12}C

C the average mass of the isotopes of the element compared to the mass of an atom of ^{1}H

D the total mass of the isotopes of the element compared to the mass of an atom of ^{1}H

  1. 1

    A: the average mass of the isotopes compared to 1/12th of the mass of an atom of ^{12}C [1]

  2. 2

    B uses the total mass. C and D compare with hydrogen, not carbon-12.

10. Relative atomic mass from isotopes

An element with more than one isotope has an Ar that is an average, but not a simple one. Each relative mass counts as often as its percentage abundance.

Multiply each relative mass by its abundance. Add the results. Divide by the total abundance, which is 100 for percentages. Show every line of working.

Ar = ((m₁ × a₁) + (m₂ × a₂) + …)/(total abundance)

Examiner insight

“Many candidates found the average of both atoms without considering the weighted average giving an answer of 64.0.” “Candidates should also be mindful of the number of decimal places required for a response.”

Worked example 12 marks

A sample of magnesium contains three isotopes. The table shows the percentage abundances and the relative masses of the three isotopes of magnesium in this sample. Calculate the relative atomic mass of magnesium to one decimal place.

percentage abundancerelative mass
8024
1025
1026
  1. 1

    (80 × 24) + (10 × 25) + (10 × 26) = 2430 [1]

  2. 2

    2430 ÷ 100 = 24.3 [1]

11. Relative molecular mass and formula mass

Relative molecular mass, Mr, is the sum of the relative atomic masses of the atoms in the formula.

Relative formula mass, also written Mr, is used for ionic compounds.

Multiply each Ar by the number of atoms of that element, then add. Give the value of Mr, not just the sum.

Mr = sum of Ar for every atom in the formula

Examiner insight

“A few candidates did not use the correct number of atoms given in the formula and would benefit from more examples of this type of calculation.” “Many candidates left the relative formula mass calculation of ammonium nitrate (NH4NO3) as 14 + 4 × 1 + 14 + 16 × 3 without providing a value for the relative formula mass.”

Worked example 12 marks

Complete the table to calculate the relative formula mass of Ca₃P₂O₈.

type of atomnumber of atomsrelative atomic mass
calcium3403 × 40 = 120
phosphorus31
oxygen16
  1. 1

    phosphorus: 2 × 31 = 62. oxygen: 8 × 16 = 128 [1]

  2. 2

    relative formula mass = 120 + 62 + 128 = 310 [1]

12. Formulas with brackets

In a formula such as Co(NO₃)₂, the small number after the bracket multiplies every atom inside it. There are two NO₃ groups, so two nitrogen atoms and six oxygen atoms.

Use the relative atomic mass, never the atomic number. The question gives the values you need.

Examiner insight

“There were some common errors in the calculation of the relative formula mass – having just one hydroxide ion being the most common.” “A common error was to use the atomic number for oxygen, even though the relative atomic mass had been given in the table.”

Worked example 12 marks

Cobalt forms a compound with the nitrate ion. The formula of the compound is Co(NO₃)₂. Complete the table to calculate the relative formula mass of Co(NO₃)₂.

type of atomnumber of atomsrelative atomic mass
cobalt1591 × 59 = 59
nitrogen14
oxygen16
  1. 1

    nitrogen: 2 × 14 = 28. oxygen: 6 × 16 = 96 [1]

  2. 2

    relative formula mass = 59 + 28 + 96 = 183 [1]

13. Reacting masses in simple proportion

The masses of the substances in a reaction are in simple proportion. If the mass of one substance halves, the mass of every other substance halves.

Work out the scaling factor from the two masses of the same substance. Then multiply the mass you want by that factor. The calculation does not involve the mole.

new mass = given mass × (new mass of the other substance)/(given mass of the other substance)

Examiner insight

“The commonest incorrect answers were 9.55 and 6.25 due to incorrect multiplication or incorrect use of the simple proportion method.” “Candidates should be advised that there are no mole calculations in this sort of question in the core paper, only calculations involving simple proportion.”

Worked example 11 mark

13.2 g of potassium reacts with excess iodine to produce 56.2 g of potassium iodide. Calculate the mass of potassium iodide produced when 6.6 g of potassium reacts with excess iodine.

  1. 1

    6.6 g is half of 13.2 g, so the mass of potassium iodide is half of 56.2 g

  2. 2

    56.2 × 6.6/13.2 = 28.1 g [1]

Recap

  • Ar = average mass of the isotopes compared to 1/12th of the mass of an atom of ^{12}C.
  • Ar from isotopes: add (mass × abundance) for each isotope, then divide by the total abundance.
  • Mr = sum of the relative atomic masses. Use relative formula mass for ionic compounds.
  • Multiply each Ar by its number of atoms. A number after a bracket multiplies every atom inside.
  • Use relative atomic masses, never atomic numbers. Write the value of Mr.
  • Reacting masses are in proportion: find the scaling factor, then multiply. No moles.

14. Units of concentration

Concentration can be measured in g / dm³ or mol / dm³.

Both units are a quantity of solute per dm³ of solution. A unit with the volume on top, such as dm³/mol, is not a concentration.

Examiner insight

“Some candidates did not read the questions carefully and assumed the unit was dm3.”

Worked example 11 mark

Four units are listed.

1 dm³/mol

2 g/dm³

3 mol/dm³

4 g/mol

Which units are units of concentration?

A 1 and 2B 1 and 4C 2 and 3D 3 and 4

  1. 1

    g/dm³ and mol/dm³ are amounts of solute per dm³. The answer is C. [1]

15. The mole, molar mass and Avogadro constant

The mole, mol, is the unit of amount of substance. One mole contains 6.02 × 10²³ particles, for example atoms, ions or molecules. This number is the Avogadro constant.

Molar mass is the mass of one mole, in g/mol. Rearrange the relationship to find an amount, a mass or a molar mass. A molar mass gives the relative atomic mass or the relative molecular / formula mass.

To count particles, multiply the amount in moles by the Avogadro constant. Count the type of particle the question asks for: one molecule of CH₄ holds five atoms.

amount (mol) = (mass (g))/(molar mass (g/mol))

number of particles = n × 6.02 × 10²³

Examiner insight

“Most candidates were able to correctly recall the number of particles in one mole.” “Option A was the most common incorrect answer. This is the number of molecules rather than the number of atoms.”

Worked example 11 mark

What is the total number of atoms in 0.0100 mol of methane, CH₄?

(The value of the Avogadro constant is 6.02 × 10²³.)

A 6.02 × 10²¹B 6.02 × 10²²C 3.01 × 10²²D 3.01 × 10²⁴

  1. 1

    molecules = 0.0100 × 6.02 × 10²³ = 6.02 × 10²¹

  2. 2

    each molecule of CH₄ has 1 + 4 = 5 atoms

  3. 3

    atoms = 5 × 6.02 × 10²¹ = 3.01 × 10²², so the answer is C [1]

16. Reacting masses from an equation

The coefficients in a balanced equation give the ratio of moles. Work in three steps.

First turn the known mass into the number of moles. Then use the ratio in the equation to find the moles of the substance you want. Last, turn those moles into a mass with the molar mass.

moles of B = moles of A × (coefficient of B)/(coefficient of A)

Examiner insight

“Many candidates were unable to use mass ratio to determine the mass of the magnesium oxide product.” “An occasional error was to use a mole ratio of 2:1 instead of 1:2.”

Worked example 13 marks

Oxygen is also produced by thermal decomposition of potassium manganate(VII), KMnO₄.

The equation for the reaction is:

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

Calculate the mass of KMnO₄ used to produce 0.64 g of O₂.

[M_r: KMnO₄, 158; O₂, 32]

  1. 1

    moles of O₂ = 0.64 ÷ 32 = 0.020 [1]

  2. 2

    moles of KMnO₄ = 0.020 × 2 = 0.040 [1]

  3. 3

    mass of KMnO₄ = 0.040 × 158 = 6.32 ≈ 6.3 g [1]

17. Gas volumes at r.t.p.

The molar gas volume is 24 dm³ at room temperature and pressure, r.t.p. One mole of any gas fills this volume.

To find a volume, multiply the moles of gas by 24. To find the moles, divide the volume by 24. The molar mass of the gas is not used for a volume.

Many volumes are given in cm³. Divide by 1000 to convert to dm³.

V (dm³) = n × 24

n = (V (dm³))/24

Examiner insight

“This question required candidates to multiply their previous answer by 24 dm3.” “24 dm3 was rarely used to calculate the number of moles.”

Worked example 13 marks

Calcium carbonate decomposes on heating.

CaCO₃(s) → CaO(s) + CO₂(g)

Calculate the volume of carbon dioxide produced at r.t.p. from 25.0 g of calcium carbonate.

[M_r: CaCO₃, 100]

  1. 1

    moles of CaCO₃ = 25.0 ÷ 100 = 0.250 [1]

  2. 2

    the ratio CaCO₃ : CO₂ is 1 : 1, so moles of CO₂ = 0.250 [1]

  3. 3

    volume = 0.250 × 24 = 6.0 dm³ [1]

18. Limiting reactants

The limiting reactant is the reactant that is used up first. It decides how much product forms. The other reactant is in excess.

Find the moles of each reactant. Check how much of one the equation needs for the other. The one that runs out is limiting.

Always use the moles of the limiting reactant for the product.

moles of product = moles of limiting reactant × (coefficient of product)/(coefficient of limiting reactant)

Examiner insight

“In this reaction, an excess of zinc was used.” “Candidates choosing option B did not determine the moles of acid present and used the moles of zinc in their calculation.”

Worked example 11 mark

4.55 g of zinc is reacted with 50 cm³ of 2.25 mol/dm³ dilute hydrochloric acid.

The equation for the reaction is:

Zn + 2HCl → ZnCl₂ + H₂

Which volume of hydrogen gas, at room temperature and pressure, is produced in the reaction?

A 1.35 dm³B 1.68 dm³C 2.70 dm³D 3.36 dm³

  1. 1

    moles of Zn = 4.55 ÷ 65 = 0.0700 (A_r of Zn from the Periodic Table)

  2. 2

    moles of HCl = 0.0500 × 2.25 = 0.1125

  3. 3

    0.0700 mol of Zn needs 2 × 0.0700 = 0.140 mol of HCl. Only 0.1125 mol is present, so HCl is limiting

  4. 4

    moles of H₂ = 0.1125 ÷ 2 = 0.05625

  5. 5

    volume = 0.05625 × 24 = 1.35 dm³, so the answer is A [1]

19. Concentration and titration

Concentration in mol/dm³ is the moles of solute divided by the volume in dm³. Convert cm³ to dm³ by dividing by 1000. To change mol/dm³ into g/dm³, multiply by the molar mass.

In a titration, the volume of one solution that reacts completely with a measured volume of another gives the moles of solute. Find the moles of the solution you know, use the ratio in the equation, then divide by the volume of the other solution.

n = c × V

c = n/V

V (dm³) = (V (cm³))/1000

c (g/dm³) = c (mol/dm³) × M_r

Examiner insight

“Candidates did not realise 2 moles of NaOH neutralise 1 mole of sulfuric acid.” “Many candidates got only part way through this calculation and gave the number of moles of barium hydroxide that were dissolved in 100 cm3. They did not go on to work out the number of moles in 25 cm3.”

Worked example 11 mark

The equation for the reaction of dilute sulfuric acid and aqueous sodium hydroxide is:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

25.0 cm³ of aqueous sodium hydroxide is measured into a conical flask.

12.5 cm³ of 0.100 mol/dm³ sulfuric acid is needed to completely react with the aqueous sodium hydroxide.

What is the concentration of the aqueous sodium hydroxide?

A 0.0250 mol/dm³B 0.0500 mol/dm³C 0.100 mol/dm³D 0.200 mol/dm³

  1. 1

    moles of H₂SO₄ = 0.0125 × 0.100 = 1.25 × 10⁻³

  2. 2

    H₂SO₄ : NaOH is 1 : 2, so moles of NaOH = 2.50 × 10⁻³

  3. 3

    concentration = 2.50 × 10⁻³ ÷ 0.0250 = 0.100 mol/dm³, so the answer is C [1]

20. Empirical and molecular formulae

The empirical formula is the simplest whole number ratio of the different atoms or ions in a compound. The molecular formula is the number and type of different atoms in one molecule.

To find the empirical formula, divide the mass of the element by its relative atomic mass. Percentages work the same way. Divide every answer by the smallest, and keep full calculator values until the end.

For the molecular formula, divide the M_r by the empirical formula mass and multiply the empirical formula by the result.

moles of element = (mass or percentage)/A_r

molecular formula = (empirical formula) × M_r/(empirical formula mass)

Examiner insight

“Some excessive rounding after the first step saw a significant number of candidates end up with the wrong answer.” “Over a third of candidates calculated the ratio of percentages rather than the ratio of moles and chose Distractor D.”

Worked example 14 marks

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its M_r is 180.

Deduce its empirical formula and its molecular formula.

[A_r: C, 12; H, 1; O, 16]

  1. 1

    moles: C 40.0 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33 [1]

  2. 2

    divide by the smallest, 3.33: ratio C : H : O = 1 : 2 : 1 [1]

  3. 3

    empirical formula = CH₂O, with mass 12 + 2 + 16 = 30 [1]

  4. 4

    180 ÷ 30 = 6, so molecular formula = C₆H₁₂O₆ [1]

21. Percentage yield, composition and purity

Percentage yield compares the mass of product actually made with the maximum mass the equation allows. The maximum comes from the moles of the limiting reactant, not from the mass of reactant used.

Percentage composition by mass is the mass of an element in one mole of a compound as a percentage of its M_r. Percentage purity is the mass of pure substance as a percentage of the mass of the sample.

percentage yield = (actual mass of product)/(maximum possible mass) × 100

percentage composition = (mass of element in one mole)/M_r × 100

percentage purity = (mass of pure substance)/(mass of sample) × 100

Examiner insight

“Candidates are reminded that the percentage yield describes the amount of desired product formed and is not calculated using the reacting mass.” “Some candidates mistakenly thought that the question was asking about rate rather than yield of product.”

Worked example 11 mark

A sample of magnesium is burned in air.

A student calculates that 10 g of magnesium oxide should be produced.

After the reaction, the student determines the percentage yield of magnesium oxide.

Which formula is used to determine the percentage yield of magnesium oxide?

A rule[-1.1em]0pt2.9em(relative atomic mass of magnesium)/(relative formula mass of magnesium oxide) × 100

B rule[-1.1em]0pt2.9em(mass of magnesium oxide formed)/(mass of magnesium used) × 100

C rule[-1.1em]0pt2.9em(mass of magnesium oxide formed)/(mass of magnesium oxide expected) × 100

D rule[-1.1em]0pt2.9em(mass of pure magnesium oxide)/(mass of all reaction products) × 100

  1. 1

    The 10 g the student calculates is the mass of magnesium oxide expected. It is the maximum possible mass.

  2. 2

    Percentage yield compares the mass of magnesium oxide formed with this expected mass, not with the mass of magnesium used. The answer is C. [1]

Recap

  • Concentration is measured in g / dm³ or mol / dm³.
  • Extended: the mole is the unit of amount of substance. One mole contains 6.02 × 10²³ particles (the Avogadro constant).
  • Extended: amount (mol) = mass (g) ÷ molar mass (g / mol).
  • Extended: use the ratio in the equation to go from moles of one substance to moles of another.
  • Extended: the molar gas volume is 24 dm³ at r.t.p., so V = n × 24.
  • Extended: the limiting reactant is used up first. Use its moles for the product.
  • Extended: c = n ÷ V with V in dm³. Divide cm³ by 1000.
  • Extended: empirical formula = divide by A_r, divide by the smallest, give the simplest whole number ratio.
  • Extended: percentage yield, composition and purity are each a part divided by a whole, times 100.

End-of-chapter exercise

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

  1. State what the molecular formula of a compound shows.1 mark
  2. State the total number of atoms in one molecule of ethanol, C₂H₆O.1 mark
  3. A model of a molecule is made from three carbon atoms, six hydrogen atoms and one oxygen atom. Write its molecular formula using symbols and subscript numbers only.1 mark
  4. Write the word equation for the reaction of aqueous copper(II) chloride with aqueous sodium hydroxide, which forms copper(II) hydroxide and sodium chloride.2 marks
  5. Solid zinc reacts with dilute hydrochloric acid to form aqueous zinc chloride and hydrogen gas. Write the balanced symbol equation, with state symbols.3 marks
  6. Butane has the molecular formula C₄H₁₀. Give its empirical formula.2 marks
  7. The formula of a magnesium ion is Mg²⁺ and the formula of a nitrate ion is NO₃⁻. Deduce the formula of magnesium nitrate.2 marks
  8. A compound contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Calculate its empirical formula. A_r: C = 12, H = 1, O = 16.4 marks
  9. A white solid oxide of phosphorus contains 43.7% phosphorus and 56.3% oxygen by mass. A_r: P = 31, O = 16. (a) Calculate the empirical formula of the oxide. (b) The solid oxide reacts with water. The only product is aqueous phosphoric acid, H₃PO₄. Use your empirical formula as the formula of the oxide. Deduce the balanced symbol equation, with state symbols.6 marks
  10. Aluminium ions are Al³⁺, sulfate ions are SO₄²⁻, barium ions are Ba²⁺ and chloride ions are Cl⁻. Aqueous aluminium sulfate reacts with aqueous barium chloride to form a precipitate of barium sulfate and aqueous aluminium chloride. (a) Deduce the formulae of aluminium sulfate and aluminium chloride. (b) Write the balanced symbol equation, with state symbols. (c) Write the ionic equation, with state symbols.7 marks
  11. State what the 1/12th in the definition of relative atomic mass refers to.1 mark
  12. Define relative molecular mass, Mr.1 mark
  13. Calculate the relative molecular mass of methane, CH₄. (Ar: C = 12, H = 1)2 marks
  14. Calculate the relative formula mass of Na₂CO₃. (Ar: Na = 23, C = 12, O = 16)2 marks
  15. A sample of chlorine contains 75% of ^{35}Cl and 25% of ^{37}Cl. Calculate the relative atomic mass of chlorine to one decimal place.2 marks
  16. Calculate the relative formula mass of calcium nitrate, Ca(NO₃)₂. (Ar: Ca = 40, N = 14, O = 16)2 marks
  17. 6.0 g of magnesium burns in excess oxygen to make 10.0 g of magnesium oxide. Calculate the mass of magnesium oxide made from 18.0 g of magnesium.2 marks
  18. 8.0 g of sulfur reacts with excess iron to make 22.0 g of iron(II) sulfide. Calculate the mass of sulfur needed to make 5.5 g of iron(II) sulfide.2 marks
  19. A sample of 20 atoms of gallium is analysed. 12 of the atoms have a relative mass of 69 and 8 of the atoms have a relative mass of 71. (a) Calculate the relative atomic mass of gallium in this sample to one decimal place. (b) Gallium forms the compound Ga₂(SO₄)₃. Use your answer to (a) to calculate the relative formula mass of Ga₂(SO₄)₃. (Ar: S = 32, O = 16)5 marks
  20. A metal M forms a hydroxide, M(OH)₃, with a relative formula mass of 78. (a) Calculate the relative atomic mass of M. (Ar: O = 16, H = 1) (b) Heating 15.6 g of M(OH)₃ makes 10.2 g of the oxide of M. Calculate the mass of M(OH)₃ that must be heated to make 25.5 g of the oxide. (c) The hydroxide is supplied as a damp powder. 20.0 g of the damp powder contains 15.6 g of M(OH)₃. Calculate the mass of damp powder needed to make 25.5 g of the oxide.6 marks
  21. State the two units in which concentration can be measured.2 marks
  22. State the number of particles in one mole of a substance, and the name of this number.2 marks
  23. Calculate the amount of substance in 11.0 g of carbon dioxide, CO₂. [A_r: C, 12; O, 16]2 marks
  24. Calculate the number of chloride ions in 0.10 mol of magnesium chloride, MgCl₂. (The Avogadro constant is 6.02 × 10²³.)3 marks
  25. Magnesium burns in oxygen. 2Mg + O₂ → 2MgO Calculate the mass of magnesium oxide formed from 6.0 g of magnesium. [A_r: Mg, 24; O, 16]3 marks
  26. Zinc reacts with excess dilute sulfuric acid. Zn + H₂SO₄ → ZnSO₄ + H₂ Calculate the volume of hydrogen produced at r.t.p. from 3.25 g of zinc. [A_r: Zn, 65]3 marks
  27. Sodium reacts with chlorine. 2Na + Cl₂ → 2NaCl 0.30 mol of sodium reacts with 0.10 mol of chlorine. Identify the limiting reactant and calculate the mass of sodium chloride formed. [A_r: Na, 23; Cl, 35.5]4 marks
  28. An oxide of phosphorus contains 43.7% phosphorus and 56.3% oxygen by mass. Its M_r is 284. Deduce its empirical formula and its molecular formula. [A_r: P, 31; O, 16]4 marks
  29. A student dissolves 4.28 g of a metal hydroxide, M(OH)₂, in water and makes the solution up to 250 cm³. 25.0 cm³ of this solution reacts completely with 20.0 cm³ of 0.250 mol/dm³ hydrochloric acid in a titration. M(OH)₂ + 2HCl → MCl₂ + 2H₂O Calculate the concentration of the M(OH)₂ solution in mol/dm³ and in g/dm³. Use these to calculate the relative atomic mass of M. [A_r: H, 1; O, 16]6 marks
  30. A 5.00 g sample of marble chips is 90.0% calcium carbonate by mass. The impurities do not react. The sample is added to 50.0 cm³ of 1.60 mol/dm³ hydrochloric acid. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) 816 cm³ of carbon dioxide is collected at r.t.p. Calculate the percentage yield of carbon dioxide. [M_r: CaCO₃, 100]6 marks

Go deeper

Practise and revise with member-only material for this chapter.

Workbook

Exam-style practice questions with full worked solutions.

Unlock with a plan

Chapter at a Glance

A single-page summary of every key idea for fast revision.

Unlock with a plan

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters

Solids, liquids and gasesElements, compounds and mixturesElectrolysisExothermic and endothermic reactionsPhysical and chemical changesThe characteristic properties of acids and bases