IGCSE Chemistry · Syllabus 0620 / 0971 · Section 3 of 12

Stoichiometry

Stoichiometry is the arithmetic of chemistry. This topic covers formulae and balanced equations, relative atomic and molecular mass, the mole and the Avogadro constant, and the calculations that link mass, gas volume, solution volume and concentration, including empirical formulae, limiting reactants, percentage yield and purity.

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Syllabus coverage

  • Define the molecular formula of a compound as the number and type of different atoms in one molecule, and deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagram
  • Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound
  • Deduce the formula of an ionic compound from the relative numbers of the ions present in a model or a diagram, or from the charges on the ions
  • Construct word equations and symbol equations, including state symbols, to show how reactants form products
  • Construct symbol equations with state symbols, including ionic equations, and deduce the symbol equation with state symbols for a chemical reaction given relevant information
  • Describe relative atomic mass, Ar, as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12
  • Define relative molecular mass, Mr, as the sum of the relative atomic masses; relative formula mass, Mr, is used for ionic compounds
  • Calculate reacting masses in simple proportions, without using the mole concept
  • State that concentration can be measured in g/dm3 or mol/dm3, and that the mole, mol, is the unit of amount of substance containing 6.02 x 10^23 particles, the Avogadro constant
  • Use the relationship amount of substance (mol) = mass (g) / molar mass (g/mol) to calculate amount of substance, mass, molar mass, relative atomic or molecular mass, and number of particles using the Avogadro constant
  • Use the molar gas volume, taken as 24 dm3 at room temperature and pressure, in calculations involving gases, and calculate stoichiometric reacting masses, limiting reactants, volumes of solutions and concentrations expressed in g/dm3 and mol/dm3, including conversion between cm3 and dm3
  • Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution, and calculate empirical formulae, molecular formulae, percentage yield, percentage composition by mass and percentage purity given appropriate data

Part of the IGCSE Chemistry syllabus 0620 / 0971.

The mole map: the three calculations that meet at the moleThe mole is the unit of amount of substance. One mole of any substance contains 6.02 x 10^23 particles (the Avogadro constant).Every arrow below is a formula: go from the unit you are given to moles, then out to the unit you are asked for.Massn = m / Mrn in mol, m in g, Mr in g/molWorked exampleMr of CO2 = 12 + 16 + 16 = 44Gas volume at r.t.p.n = V / 24 dm3n in mol, V in dm3 (room temperature andpressure)Worked exampleV of 6.0 dm3 of O2 at r.t.p.Concentrationn = c x V / 1000n in mol, c in mol/dm3, V in cm3Worked example25.0 cm3 of 0.100 mol/dm3 NaOHAMOUNT IN MOLES n (mol)The mole is the crossroads: convert into moles first, then convert out to whatever the question asks for.n = m / Mrback to massn = V / 24 dm3back to gas volume at r.t.p.n = c x V / 1000back to concentrationWorked example: moles from massWhat is the mass of 0.25 mol of NaOH?Mr = 23 + 16 + 1 = 40m = n x Mr = 0.25 x 40m = 10 g of NaOHWorked example: moles from volumeWhat volume does 1.5 mol of H2 occupyat r.t.p.?V = n x 24 = 1.5 x 24V = 36 dm3 of H2Worked example: moles from titrationWhat mass of NaOH is in 25.0 cm3 of0.100 mol/dm3 solution?n = 0.0025 mol, m = n x Mr = 0.0025 x 40m = 0.10 g of NaOHCheck the units every time: g and g/mol for mass, dm3 for gas volume, mol/dm3 and cm3 for concentration. Divide cm3 by 1000 to reach dm3.
The mole map: the relationships that link the amount in moles to mass, to the volume of a gas at r.t.p., to the volume and concentration of a solution, and to the number of particles through the Avogadro constant.

Formulae: from a model to a chemical formula

A chemical formula tells you which elements are in a compound and how many atoms of each are present. The molecular formula of a compound is the number and type of different atoms in one molecule, so C2H6 means that one molecule contains two carbon atoms and six hydrogen atoms and nothing else.

You can deduce the formula of a simple compound from a model or a diagram. Count the atoms of each element shown in the model, then write them as a formula using element symbols and subscript numbers. If a model shows two carbon atoms and four hydrogen atoms joined together, the formula is C2H4.

A subscript applies only to the symbol or bracket immediately before it. In Ca(OH)2 the 2 multiplies the whole hydroxide group, so the formula contains one calcium atom, two oxygen atoms and two hydrogen atoms. In Al2(SO4)3 the 3 multiplies the sulfate group, giving two aluminium atoms, three sulfur atoms and twelve oxygen atoms.

Ionic compounds are not made of molecules, so they have no molecular formula. Their formula is the simplest whole number ratio of the ions present, worked out from the charges on the ions: the total positive charge must equal the total negative charge.

  • Molecular formula = the number and type of different atoms in one molecule.
  • Empirical formula = the simplest whole number ratio of the different atoms or ions in a compound.
  • A subscript multiplies only the symbol or bracket immediately before it.
  • For an ionic compound, balance the charges: Mg2+ with Cl- gives MgCl2, and Al3+ with O2- gives Al2O3.
For an ionic compound we work out the relative formula mass, not a relative molecular mass, because there are no molecules - only a giant lattice of ions.

Writing balanced symbol equations with state symbols

A word equation names the reactants and the products. A symbol equation replaces those names with formulae. A symbol equation must be balanced: the number of atoms of each element on the left must equal the number on the right, because atoms are not created or destroyed in a chemical reaction.

You balance an equation by adding large numbers in front of formulae, never by changing a subscript. Changing H2O to H2O2 gives hydrogen peroxide, a different substance; writing 2H2O doubles the amount of water. Start with the element that appears in the fewest formulae, then work through the rest, and re-count every element when you have finished.

State symbols show the physical state of each substance: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water. They are part of the equation and are worth marks, so include them whenever the question asks for them.

In aqueous solutions you can also write an ionic equation, which shows only the ions that actually change. Ions that stay dissolved and unchanged are called spectator ions and are left out. When silver nitrate solution is added to sodium chloride solution, the ionic equation is Ag+(aq) + Cl-(aq) -> AgCl(s); the nitrate and sodium ions are spectators.

State symbolMeaningExample
(s)SolidMg(s) - a piece of magnesium ribbon
(l)LiquidH2O(l) - water
(g)GasCO2(g) - carbon dioxide
(aq)Aqueous: dissolved in waterNaCl(aq) - sodium chloride solution
Never balance an equation by changing a subscript. Change the large number in front of a formula, then check that every element still balances.

Relative atomic mass and relative molecular mass

The mass of a single atom is far too small to weigh in grams, so chemists compare masses instead. The relative atomic mass, Ar, of an element is the average mass of the isotopes of that element compared to 1/12th of the mass of an atom of carbon-12. Carbon-12 is the standard, and its relative atomic mass is exactly 12.

Ar is an average because most elements are mixtures of isotopes. Chlorine is about 75% chlorine-35 and 25% chlorine-37, so its relative atomic mass is 35.5, not a whole number. The average is weighted by how much of each isotope is present in the element.

The relative molecular mass, Mr, is the sum of the relative atomic masses of the atoms in one molecule. For an ionic compound there are no molecules, so the same sum is called the relative formula mass. Both use the symbol Mr and neither has units, because Mr is a comparison of masses.

Worked example: the relative formula mass of calcium carbonate, CaCO3. Using Ca = 40, C = 12 and O = 16, Mr = 40 + 12 + (3 x 16) = 40 + 12 + 48 = 100.

Worked example: the relative formula mass of aluminium sulfate, Al2(SO4)3. Using Al = 27, S = 32 and O = 16, Mr = (2 x 27) + 3 x (32 + (4 x 16)) = 54 + 3 x 96 = 54 + 288 = 342.

  • Ar = the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12.
  • Mr = the sum of the relative atomic masses of the atoms in the formula.
  • Mr has no units, because it is a comparison of masses.
  • Ar values are printed in the question - use those exact values, and do not round them.
Watch the bracket: in Al2(SO4)3 the 3 multiplies everything inside the bracket, so the formula contains three sulfur atoms and twelve oxygen atoms.

The mole and the Avogadro constant

A mole is the unit of amount of substance, and its symbol is mol. One mole of any substance contains 6.02 x 10^23 particles, which may be atoms, ions or molecules. This number is the Avogadro constant, and it is the same for every substance.

A mole is simply a counting unit, like a dozen but enormously larger. Because particles are so small, 6.02 x 10^23 of them make a convenient, weighable amount. The mass of one mole of a substance in grams is numerically equal to its relative formula mass, and this mass is called the molar mass, measured in g/mol.

The key relationship is: amount of substance (mol) = mass (g) / molar mass (g/mol). One mole of water has Mr = 18, so its molar mass is 18 g/mol and one mole of water has a mass of 18 g. You can rearrange the relationship in any direction to find the mass or the molar mass.

Worked example: how many moles are in 11.0 g of carbon dioxide? For CO2, Mr = 12 + (2 x 16) = 44, so the molar mass is 44 g/mol and amount = 11.0 / 44 = 0.250 mol.

Worked example: what is the mass of 0.500 mol of sodium carbonate? For Na2CO3, Mr = (2 x 23) + 12 + (3 x 16) = 106, so the molar mass is 106 g/mol and mass = 0.500 x 106 = 53.0 g.

To find a number of particles, multiply the amount in moles by the Avogadro constant. For example, 0.250 mol of carbon dioxide contains 0.250 x 6.02 x 10^23 = 1.51 x 10^23 molecules.

  • The mole is the unit of amount of substance, and 1 mol contains 6.02 x 10^23 particles.
  • amount (mol) = mass (g) / molar mass (g/mol).
  • The molar mass in g/mol is numerically equal to Mr.
  • To find the number of particles, multiply the number of moles by 6.02 x 10^23.
Check the units before you calculate: the mass must be in g and the molar mass in g/mol, or the answer will be wrong.

Reacting masses and limiting reactants

A balanced equation tells you the ratio in which substances react, measured in moles. The large numbers in front of the formulae give the mole ratio. In 2Mg + O2 -> 2MgO, two moles of magnesium react with one mole of oxygen to give two moles of magnesium oxide.

To calculate a reacting mass, use the same four steps every time. Write the balanced equation. Convert the mass you are given into moles using amount = mass / molar mass. Use the mole ratio from the equation to find the moles of the substance you want. Then convert those moles back into a mass using mass = amount x molar mass.

Worked example: what mass of magnesium oxide forms when 4.8 g of magnesium burns in excess oxygen? Using Mg = 24 and O = 16, the equation is 2Mg + O2 -> 2MgO. Amount of Mg = 4.8 / 24 = 0.20 mol. The ratio Mg : MgO is 2 : 2, which is 1 : 1, so 0.20 mol of MgO forms. Mr of MgO = 24 + 16 = 40, so mass = 0.20 x 40 = 8.0 g.

The limiting reactant is the reactant that runs out first, and it fixes the maximum amount of product that can form. The reactant in excess is left over when the reaction stops. If a question gives you two masses, convert both to moles and compare them with the mole ratio in the equation.

Worked example: 8.0 g of methane is burned in 40.0 g of oxygen, according to CH4 + 2O2 -> CO2 + 2H2O. Amount of CH4 = 8.0 / 16 = 0.50 mol and amount of O2 = 40.0 / 32 = 1.25 mol. The equation needs 2 mol of oxygen for every 1 mol of methane, so 0.50 mol of methane needs 1.00 mol of oxygen. There is 1.25 mol of oxygen, which is more than the 1.00 mol needed, so oxygen is in excess and methane is the limiting reactant.

  • The mole ratio comes from the large numbers in front of the formulae in the balanced equation.
  • Four steps: equation, moles of what you know, mole ratio, moles of what you want, then convert to mass.
  • The limiting reactant runs out first and fixes the amount of product.
  • A reactant in excess is left over when the reaction stops.
If a question gives two masses, do not assume the first one is limiting. Convert both to moles and compare them with the mole ratio.
Weighing out a reactant on a balance before a reaction: the mass used is chosen so that it matches the mole ratio in the balanced equation.

Gases, solutions and titrations

One mole of any gas occupies the same volume at the same temperature and pressure, because the particles are tiny compared with the space between them. At room temperature and pressure, r.t.p., the molar gas volume is taken as 24 dm3 per mole, whatever the gas is.

For a gas, volume (dm3) = amount (mol) x 24. Rearranged, amount (mol) = volume (dm3) / 24. Because the molar volume is the same for every gas, you never need to know which gas it is in order to work out its volume at r.t.p.

The concentration of a solution can be measured in g/dm3 or mol/dm3. To change from g/dm3 to mol/dm3, divide by the molar mass: concentration (mol/dm3) = concentration (g/dm3) / molar mass (g/mol). Concentration is also linked to amount and volume by concentration (mol/dm3) = amount (mol) / volume (dm3). Always convert cm3 to dm3 by dividing by 1000, because 1 dm3 = 1000 cm3.

Worked example: what volume of carbon dioxide at r.t.p. forms when 0.50 mol of calcium carbonate decomposes? The equation is CaCO3 -> CaO + CO2. It shows that 1 mol of calcium carbonate gives 1 mol of carbon dioxide, so 0.50 mol of carbon dioxide forms. Volume = 0.50 x 24 = 12 dm3.

Worked example, using titration data: 25.0 cm3 of sodium hydroxide solution is exactly neutralised by 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. For NaOH + HCl -> NaCl + H2O, amount of HCl = 0.100 x (20.0 / 1000) = 0.00200 mol. The equation shows a 1 : 1 ratio, so the amount of NaOH is also 0.00200 mol. Concentration of NaOH = 0.00200 / (25.0 / 1000) = 0.0800 mol/dm3.

  • At r.t.p. one mole of any gas occupies 24 dm3.
  • volume of gas (dm3) = amount (mol) x 24.
  • concentration (mol/dm3) = amount (mol) / volume (dm3), with the volume in dm3.
  • Divide a volume in cm3 by 1000 to convert it to dm3.
The commonest error in titration calculations is forgetting to divide the volume in cm3 by 1000. Convert to dm3 before you divide.

Empirical formulae, yield, composition and purity

The empirical formula is the simplest whole number ratio of the different atoms or ions in a compound. The molecular formula gives the actual number of atoms in one molecule, and it is always a whole number multiple of the empirical formula.

To find an empirical formula from percentage composition by mass, divide the percentage of each element by its relative atomic mass to get the number of moles, then divide every value by the smallest one and round to the nearest whole number.

Worked example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Moles of C = 40.0 / 12 = 3.33, moles of H = 6.7 / 1 = 6.7 and moles of O = 53.3 / 16 = 3.33. Dividing every value by the smallest, 3.33, gives C : H : O = 1 : 2 : 1, so the empirical formula is CH2O.

To find the molecular formula, divide the relative molecular mass by the mass of one empirical formula unit. If a compound has the empirical formula CH2 and Mr = 42, the mass of CH2 is 14, so 42 / 14 = 3 and the molecular formula is C3H6.

Percentage composition by mass is the mass of one element in a formula as a percentage of the whole formula. For water, H2O with Mr = 18, the percentage of hydrogen is (2 x 1) / 18 x 100 = 11.1%. Percentage yield compares the product actually made with the maximum possible: percentage yield = (mass of product actually made / maximum mass of product possible) x 100. Percentage purity compares the mass of the pure substance in a sample with the total mass of the sample, using the same idea.

Yields are always below 100% in practice. Some reactant may not react, the reaction may be reversible and not go to completion, product is lost when it is transferred or filtered, and unwanted by-products may form instead of the one you want.

  • Empirical formula = the simplest whole number ratio of the atoms or ions in a compound.
  • Molecular formula = the actual number of atoms in one molecule, a whole number multiple of the empirical formula.
  • percentage yield = (mass of product actually made / maximum mass possible) x 100.
  • percentage purity = (mass of pure substance / mass of the sample) x 100.
If the ratio comes out as 1 : 1.5, multiply every value by 2 to get whole numbers. Empirical formulae always contain whole numbers of atoms.
Finding an empirical formula from percentage compositionWorked example: a compound contains 40.0 per cent carbon, 6.7 per cent hydrogen and 53.3 per cent oxygen by mass. Ar: C = 12, H = 1, O= 16.The empirical formula is the simplest whole number ratio of the atoms in a compound.The methodPercentageMassMolesSimplest ratioFormulaTreat the percentages as masses in grams, divide each by the Ar of the element, divide through by the smallest answer, then write the formula.The workingElementPercentage by massMass in 100 gMoles = mass / ArDivide by smallestCarbon40.0 %40.0 g40.0 / 12 = 3.333.33 / 3.33 = 1Hydrogen6.7 %6.7 g6.7 / 1 = 6.76.7 / 3.33 = 2Oxygen53.3 %53.3 g53.3 / 16 = 3.333.33 / 3.33 = 1Smallest answer = 3.33, so divide every answer by 3.33. The ratio C : H : O is 1 : 2 : 1.Empirical formula =CH2OFrom empirical to molecular formulaThe empirical formula gives the ratio only. The molecularformula is a whole number multiple of it.If the relative molecular mass is known, work out themultiplier:multiplier = Mr / mass of one empirical formula unitMr of CH2O = 12 + (2 x 1) + 16 = 30. If the real Mr is 180,the multiplier is 180 / 30 = 6, so the molecular formula isC6H12O6 (glucose).Where the percentages come fromPercentage composition is found by experiment, usually byburning a known mass of the compound and measuring themass of each product.If you are given masses instead of percentages, use themdirectly in the third column - the method is identical.Check your answer: the percentages must add to 100 per cent(40.0 + 6.7 + 53.3 = 100.0).A ratio that comes out as 1 : 1.5 : 2 must be multiplied by 2 to give 2 : 3 : 4 - an empirical formula must be in whole numbers.
Finding an empirical formula from percentage composition: the percentage of each element is divided by its relative atomic mass, and the resulting mole values are divided by the smallest to give the simplest whole number ratio.

Key terms you must be able to define

Molecular formula
The number and type of different atoms in one molecule of a compound, for example C2H6.
Empirical formula
The simplest whole number ratio of the different atoms or ions in a compound, for example CH2O.
Relative atomic mass, Ar
The average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12.
Relative molecular mass, Mr
The sum of the relative atomic masses of the atoms in one molecule; for an ionic compound the same sum is called the relative formula mass.
Mole
The unit of amount of substance, symbol mol; one mole of any substance contains 6.02 x 10^23 particles.
Avogadro constant
The number of particles in one mole of a substance, 6.02 x 10^23; the particles may be atoms, ions or molecules.
Molar gas volume
The volume occupied by one mole of any gas, taken as 24 dm3 at room temperature and pressure.
Limiting reactant
The reactant that runs out first in a reaction; it determines the maximum amount of product that can form.

Applied to the real world

Cambridge sets chemistry in unfamiliar contexts - an industrial plant, a water-treatment works, a hospital, a laboratory. Use these as models for how to apply the chemistry, then write your own.

In industry - working out how much raw material to buy

A plant making ammonia uses the equation N2 + 3H2 -> 2NH3. The mole ratio tells the plant that three moles of hydrogen are needed for every mole of nitrogen. To make 1000 mol of ammonia it needs 500 mol of nitrogen, so it can order exactly the right amount of raw material instead of guessing and paying for a large excess.

In a hospital - making up a saline drip

A saline drip is sodium chloride solution with a concentration of about 0.15 mol/dm3. To make up 1 dm3 of it, the pharmacist needs 0.15 mol of sodium chloride. Mr of NaCl = 23 + 35.5 = 58.5, so the mass required is 0.15 x 58.5 = 8.8 g. The same calculation is used to check that a bag of drip fluid contains the concentration the patient needs.

In a laboratory - preparing a pure salt

To prepare copper(II) sulfate crystals, copper(II) oxide is added to warm dilute sulfuric acid: CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l). The equation has a 1 : 1 mole ratio, so you can calculate exactly how much copper(II) oxide is needed to react with a measured amount of acid. Adding just the right mass leaves no unreacted oxide to filter off, which makes the crystals purer.

In a power station - the carbon dioxide released by burning coal

Burning the carbon in coal follows C(s) + O2(g) -> CO2(g). One mole of carbon, which is 12 g, produces one mole of carbon dioxide, and one mole of any gas occupies 24 dm3 at r.t.p. A power station burning thousands of tonnes of coal can therefore estimate its carbon dioxide emissions directly from the mass of fuel burned.

In a factory - why percentage yield matters

The Haber process, N2 + 3H2 -> 2NH3, is reversible, so not all the nitrogen and hydrogen react on one pass through the reactor. Unreacted gases are recycled to improve the yield. The percentage yield tells the factory how much of the raw material actually becomes product, which controls both the cost per tonne of ammonia and the energy used.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is the molecular formula of a compound?
The number and type of different atoms in one molecule.
Q2What is the empirical formula of a compound?
The simplest whole number ratio of the different atoms or ions in a compound.
Q3Define relative atomic mass, Ar.
The average mass of the isotopes of an element compared to 1/12th of the mass of an atom of carbon-12.
Q4Define relative molecular mass, Mr.
The sum of the relative atomic masses of the atoms in one molecule; for an ionic compound it is called the relative formula mass.
Q5What is the relative formula mass of CaCO3? (Ar: Ca = 40, C = 12, O = 16)
40 + 12 + (3 x 16) = 100, with no units.
Q6What is a mole?
The unit of amount of substance; one mole of any substance contains 6.02 x 10^23 particles.
Q7What is the Avogadro constant?
6.02 x 10^23, the number of particles in one mole of a substance.
Q8Give the relationship between amount, mass and molar mass.
amount of substance (mol) = mass (g) / molar mass (g/mol).
Q9What is the molar gas volume at r.t.p.?
24 dm3 per mole, the same for every gas.
Q10How do you convert a volume in cm3 to dm3?
Divide by 1000, because 1 dm3 = 1000 cm3.
Q11Give the equation linking concentration, amount and volume.
concentration (mol/dm3) = amount (mol) / volume (dm3), with the volume in dm3.
Q12What is the limiting reactant?
The reactant that runs out first; it determines the maximum amount of product that can form.
Q13How is percentage yield calculated?
(mass of product actually made / maximum mass of product possible) x 100.
Q14How do you find an empirical formula from percentage composition by mass?
Divide each percentage by the Ar of that element, then divide every value by the smallest to get the simplest whole number ratio.
0 of 14 marked as known

Exam-style questions

9 questions, 28 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

StateAO11 mark
State what is meant by the empirical formula of a compound.

Mark scheme — tick each point you actually made

Examiner tip: The mark is for 'simplest whole number ratio'. Writing 'the number of atoms in a molecule' describes the molecular formula, not the empirical formula.
DefineAO12 marks
Define relative atomic mass, Ar.

Mark scheme — tick each point you actually made

Examiner tip: Both halves are needed: the average of the isotopes, and the comparison with 1/12th of the mass of a carbon-12 atom.
ExplainAO23 marks
Chlorine consists of two isotopes, chlorine-35 and chlorine-37, which are present in a ratio of about 3 : 1. Explain why the relative atomic mass of chlorine is 35.5 and not a whole number.

Mark scheme — tick each point you actually made

Examiner tip: Say that Ar is a weighted average of the isotopes. Simply saying 'chlorine has two isotopes' does not explain why the value is not a whole number.
CalculateAO24 marks
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Calculate its empirical formula, and use Mr = 42 to deduce its molecular formula. (Ar: C = 12, H = 1)

Mark scheme — tick each point you actually made

Examiner tip: Show the moles of each element before you take the ratio. A ratio that does not come out as whole numbers, such as 1 : 1.5, must be multiplied up, never rounded.
CalculateAO25 marks
Magnesium burns in oxygen: 2Mg + O2 -> 2MgO. Calculate the mass of magnesium oxide formed when 4.8 g of magnesium is burned completely in excess oxygen. (Ar: Mg = 24, O = 16)

Mark scheme — tick each point you actually made

Examiner tip: Write down the mole ratio before you convert anything. Most marks are lost by jumping from the mass of magnesium straight to the mass of product without using the equation.
CalculateAO24 marks
Calcium carbonate decomposes on strong heating: CaCO3 -> CaO + CO2. Calculate the volume of carbon dioxide produced at r.t.p. when 0.50 mol of calcium carbonate is completely decomposed. (molar gas volume = 24 dm3 at r.t.p.)

Mark scheme — tick each point you actually made

Examiner tip: The molar gas volume is the same for every gas at r.t.p., so you do not need to know anything about carbon dioxide beyond the number of moles produced.
CalculateAO23 marks
25.0 cm3 of sodium hydroxide solution is exactly neutralised by 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. Calculate the concentration of the sodium hydroxide solution in mol/dm3. NaOH + HCl -> NaCl + H2O

Mark scheme — tick each point you actually made

Examiner tip: Convert both volumes from cm3 to dm3 before you use them. Forgetting the division by 1000 is the single commonest mistake in titration calculations.
WriteAO12 marks
Write the symbol equation, including state symbols, for the reaction between dilute hydrochloric acid and aqueous sodium hydroxide.

Mark scheme — tick each point you actually made

Examiner tip: The equation is already balanced with one of each formula, so all the marks come from writing the correct formulae and the correct state symbols.
ExplainAO24 marks
8.0 g of methane is burned in 40.0 g of oxygen: CH4 + 2O2 -> CO2 + 2H2O. Use the mole ratio to show which reactant is in excess, and state which reactant is limiting. (Ar: H = 1, C = 12, O = 16)

Mark scheme — tick each point you actually made

Examiner tip: You must compare the two amounts using the mole ratio from the equation. Comparing 8.0 g with 40.0 g directly proves nothing, because the two substances have different molar masses.

Exam technique for this topic

  • For every reacting mass calculation write the four steps: balanced equation, moles of what you are given, mole ratio from the equation, then moles and mass of what you want. Method marks are available even if the arithmetic slips.
  • Always show the molar mass you have calculated and the working that follows it. A bare answer scores less than the same answer with the calculation shown, and clear working can still earn method marks if the final value is wrong.
  • Use the Ar values printed in the question. If you substitute your own rounded values you can reach a different final answer and lose the last mark, even though your method was right.
  • Check the units before you finish: mass in g, gas volume in dm3, solution volume in dm3, and concentration in mol/dm3 or g/dm3. An answer without a unit loses the unit mark.
  • Never balance an equation by changing a subscript. Change only the large number in front of a formula, then re-count the atoms of every element to check the equation still balances.
  • In an empirical formula question, do not round a ratio such as 1.5 to 2. Multiply every value by the same number to reach the simplest whole number ratio.

Why students lose marks here

A student wroteA student balanced an equation by changing H2O to H2O2.
Why it loses marksChanging a subscript changes the substance. Hydrogen peroxide is a different compound from water, so the equation no longer represents the reaction the student was asked about.
The answer that scoresBalance by changing the large number in front of a formula instead. For the combustion of hydrogen, 2H2 + O2 -> 2H2O is balanced: there are four hydrogen atoms and two oxygen atoms on each side, and the substances are unchanged.
Weighing reactants on a balance: the masses you use must match the balanced equation, so a formula must never be altered to make the atoms balance.
A student wroteA student wrote: 'One mole of oxygen gas has a mass of 16 g.'
Why it loses marksThe student used the relative atomic mass of an oxygen atom instead of the relative molecular mass of an oxygen molecule. Oxygen gas exists as O2, so the molar mass is twice the Ar value.
The answer that scoresOxygen gas is made of diatomic molecules, O2, so Mr = 2 x 16 = 32 and one mole of oxygen gas has a mass of 32 g. The same applies to hydrogen, H2, nitrogen, N2 and chlorine, Cl2.
A student wroteA student wrote: 'The volume of 0.50 mol of carbon dioxide at r.t.p. is 0.50 x 24 = 12 cm3.'
Why it loses marksThe arithmetic is correct but the unit is wrong. The molar gas volume is 24 dm3, so the volume must be reported in dm3, not cm3.
The answer that scoresThe volume is 0.50 x 24 = 12 dm3, which is 12000 cm3. Always write the unit that matches the quantity you calculated, and check whether the question asks for dm3 or cm3.
A student wroteA student calculating a reacting mass from 4.8 g of magnesium wrote: 'moles of Mg = 4.8 / 24 = 0.20, so moles of MgO = 0.10 because of the 2 in 2MgO.'
Why it loses marksThe student treated the coefficient as a divisor. The equation 2Mg + O2 -> 2MgO gives a 2 : 2 ratio between magnesium and magnesium oxide, which simplifies to 1 : 1, so the amounts are equal.
The answer that scoresThe mole ratio Mg : MgO is 2 : 2, which is 1 : 1, so 0.20 mol of magnesium gives 0.20 mol of magnesium oxide. Mr of MgO = 24 + 16 = 40, so mass = 0.20 x 40 = 8.0 g.

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