1.43Relative formula mass
The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula.
In a balanced chemical equation, the sum of the relative formula masses of the reactants in the quantities shown equals the sum of the relative formula masses of the products in the quantities shown. That is conservation of mass again, and it gives you a free check on any equation you have balanced.
Worked example
Calculate the relative formula mass of calcium carbonate, CaCO₃. (Ar: Ca = 40, C = 12, O = 16.)
- List the atoms: 1 calcium, 1 carbon, 3 oxygen.
- Calcium: 1 × 40 = 40.
- Carbon: 1 × 12 = 12.
- Oxygen: 3 × 16 = 48.
- Add them: 40 + 12 + 48.
Answer: Mr = 100
You should be able to calculate the percentage by mass of an element in a compound, given the relative formula mass and the relative atomic masses.
percentage by mass = (Ar of the element × number of atoms of it) ÷ Mr of the compound × 100
Worked example
Calculate the percentage by mass of oxygen in calcium carbonate, CaCO₃.
- Mass of oxygen in the formula: 3 × 16 = 48.
- Relative formula mass of the compound: 100.
- Divide: 48 ÷ 100 = 0.48.
- Multiply by 100 to get a percentage.
Answer: 48%
Common mistake
Use the total mass of that element, not the mass of one atom. In CaCO₃ the oxygen contributes 48, not 16 — there are three oxygen atoms in the formula.
- Relative formula mass
- Mr: the sum of the relative atomic masses of the atoms in the numbers shown in the formula of a compound
1.44–1.46Empirical and molecular formulae
Two kinds of formula are used, and the difference matters.
The molecular formula gives the actual number of atoms of each element in one molecule. Ethane is C₂H₆; glucose is C₆H₁₂O₆.
The empirical formula gives the simplest whole-number ratio of those atoms. For ethane that is CH₃; for glucose it is CH₂O. Divide every number in the molecular formula by their highest common factor and you have it. For an ionic compound there is no molecule at all, so the formula written is always the empirical one.
The empirical formula can be worked out from experimental data — either masses that reacted or a percentage composition — and the method is the same either way. Set the numbers out in a column for each element and work down:
• write the mass in grams, or the percentage, of each element • divide each by that element's relative atomic mass • divide every answer by the smallest of them • round to whole numbers, and if you are left with a .5, double everything.
Worked example
A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Calculate its empirical formula. (Aᵣ: C = 12, H = 1)
- Divide by the relative atomic masses: carbon 2.4 ÷ 12 = 0.2, hydrogen 0.8 ÷ 1 = 0.8.
- Divide both by the smaller, 0.2: carbon 1, hydrogen 4.
- Both are already whole numbers, so no doubling is needed.
Answer: CH₄
The empirical formula of magnesium oxide can be found directly. A known mass of magnesium ribbon is cleaned with emery paper to remove the oxide layer, coiled, and heated strongly in a crucible with the lid lifted briefly and often — lifted, so that oxygen can get in; briefly, so that white smoke does not escape. Heating is repeated to a constant mass, which is how you know all the magnesium has reacted. The mass gained is the mass of oxygen that combined with it, and the two masses go into the method above.
Common mistake
Losing the white smoke, or stopping before the mass is constant, both make the mass of oxygen come out too low — and the answer then suggests more magnesium per oxygen than there really is. If a question asks why the result was not MgO, that is the answer it wants.
1.47–1.48Conservation of mass and balanced chemical equations
The law of conservation of mass states that no atoms are lost or made during a chemical reaction, so the mass of the products equals the mass of the reactants.
This means that chemical reactions can be represented by symbol equations which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation — a balanced equation is the law of conservation of mass written down for one particular reaction.
You should understand the use of the multipliers in equations in normal script before a formula, and in subscript within a formula. They do different jobs.
A number in normal script in front of a formula multiplies everything in that formula: 2H₂O means two water molecules, so four hydrogen atoms and two oxygen atoms.
A number in subscript inside a formula applies only to the symbol it follows: H₂O means one molecule containing two hydrogen atoms and one oxygen atom. Changing it changes the substance.
Worked example
How many atoms of each element are there in 3Ca(OH)₂?
- Work inside the brackets first: OH is one oxygen and one hydrogen.
- The subscript 2 outside the bracket applies to everything inside it: 2 oxygen and 2 hydrogen, plus 1 calcium.
- So one formula unit contains 1 Ca, 2 O and 2 H.
- The 3 in normal script multiplies all of it by three.
Answer: 3 calcium, 6 oxygen and 6 hydrogen atoms.
- Law of conservation of mass
- no atoms are lost or made during a chemical reaction, so the mass of the products equals the mass of the reactants
1.47Mass changes when a reactant or product is a gas
Some reactions may appear to involve a change in mass, but this can usually be explained because a reactant or product is a gas and its mass has not been taken into account.
Two standard examples:
• When a metal reacts with oxygen, the mass of the oxide produced is greater than the mass of the metal — because oxygen from the air has been added to the solid and is now included when it is weighed. • In thermal decompositions of metal carbonates, carbon dioxide is produced and escapes into the atmosphere, leaving the metal oxide as the only solid product — so the solid left behind weighs less than the carbonate you started with.
You should be able to explain any observed changes in mass in non-enclosed systems during a chemical reaction, given the balanced symbol equation, and explain these changes in terms of the particle model.
A full answer has three parts: which substance is a gas, whether it enters or leaves the container, and — in particle terms — that its particles have crossed the boundary of the open system rather than been created or destroyed.
Worked example
A student heats 10.0 g of copper carbonate in an open crucible. CuCO₃ → CuO + CO₂. The solid left has a mass of 6.4 g. Explain the change.
- Identify the gas from the equation: carbon dioxide, CO₂.
- The crucible is open, so the carbon dioxide escapes into the atmosphere.
- Only the copper oxide is left on the balance, so the measured mass falls.
- In particle terms, the carbon and oxygen particles are still there — they have left the crucible as gas particles, not been destroyed.
- The 3.6 g difference is the mass of carbon dioxide released.
Answer: Mass appears to fall by 3.6 g because carbon dioxide gas escapes from the open crucible; no atoms are lost, so if the gas were collected the total mass would be unchanged.
Common mistake
The law of conservation of mass is never broken in these questions. The mass on the balance changes; the mass of the system does not. Always say the gas escaped or was gained, rather than that mass was lost.
Chemical measurements
Whenever a measurement is made there is always some uncertainty about the result obtained.
You should be able to represent the distribution of results and make estimations of uncertainty, and to use the range of a set of measurements about the mean as a measure of uncertainty.
Worked example
Using the five titres above, calculate the mean and estimate the uncertainty.
- Add the values: 25.6 + 24.9 + 25.1 + 25.0 + 25.2 = 125.8.
- Divide by the number of measurements: 125.8 ÷ 5 = 25.16, so the mean is 25.2 cm³ to 3 significant figures.
- Find the range: largest − smallest = 25.6 − 24.9 = 0.7 cm³.
- The uncertainty is taken as half the range about the mean: 0.7 ÷ 2 = 0.35, about 0.4 cm³.
Answer: Mean 25.2 cm³, with an uncertainty of about ± 0.4 cm³ — so the result would be quoted as 25.2 ± 0.4 cm³.
To represent the distribution of results, plot the repeats on a simple number line rather than only quoting the mean. A tight cluster with one distant value tells you something a mean alone hides: that value is probably an anomaly, and excluding it before averaging gives a more trustworthy result.
Common mistake
Uncertainty is not the same as error. A careful measurement still has uncertainty, because no instrument reads perfectly. Do not write that a spread in results means the student did something wrong.
1.49Concentration of solutions
Many chemical reactions take place in solutions. The concentration of a solution can be measured in mass per given volume of solution, for example grams per dm³ (g/dm³).
concentration in g/dm³ = mass of solute in g ÷ volume of solution in dm³
One dm³ is 1000 cm³, so a volume given in cm³ must be divided by 1000 before it is used.
You should be able to calculate the mass of solute in a given volume of solution of known concentration, in terms of mass per given volume of solution.
Worked example
Calculate the mass of sodium chloride in 250 cm³ of a solution of concentration 20 g/dm³.
- Convert the volume to dm³: 250 ÷ 1000 = 0.25 dm³.
- Rearrange the relationship: mass = concentration × volume.
- Substitute: 20 × 0.25.
- Check the units: g/dm³ × dm³ gives g, which is what was asked for.
Answer: 5 g of sodium chloride
Common mistake
Convert cm³ to dm³ before doing anything else. Forgetting the factor of 1000 is the single most common error in these questions, and it makes the answer wrong by three orders of magnitude.
- Concentration
- the amount of solute in a given volume of solution, measured in g/dm³ as mass of solute divided by volume of solution in dm³
Before you practise
- Relative formula mass is the sum of the relative atomic masses in the formula — multiply out the brackets first.
- percentage by mass = (mass of that element in the formula ÷ relative formula mass) × 100.
- The empirical formula is the simplest whole-number ratio; divide by the relative atomic masses, then by the smallest answer.
- Mass is conserved in every reaction, because no atoms are created or destroyed.
- A mass that appears to fall means a gas escaped; a mass that appears to rise means a gas from the air was taken in.
- concentration in g dm⁻³ = mass in g ÷ volume in dm³, and 1000 cm³ = 1 dm³.
- Quote an answer to a sensible number of significant figures and give the unit.