Formulae, equations and amounts of substance

Atoms are far too small to count one at a time, so chemists count them by weighing, by measuring gas volumes and by measuring solution volumes. This topic builds the idea of the mole, the three equations that link it to things you can measure, and the balanced equations that tell you how much of each substance reacts and forms. Almost every calculation in A level chemistry rests on what is here.

1. The mole and the Avogadro constant

A grain of salt holds about 10¹⁸ ions, so counting particles directly is hopeless. Instead chemists use a counting unit, the mole (symbol mol), in the same way that a baker counts eggs in dozens. One mole of anything contains the same number of particles: the Avogadro constant, \( L = 6.02 \times 10^{23}\ \mathrm{mol^{-1}} \).

The number was chosen so that the mass of one mole of a substance in grams is numerically the same as its relative mass. One mole of carbon-12 atoms has a mass of exactly 12 g, one mole of water molecules (relative mass 18.0) has a mass of 18.0 g, and so on. That is why weighing is a way of counting.

• Relative atomic mass, \(A_\mathrm{r}\): the weighted mean mass of an atom of an element, measured relative to one twelfth of the mass of an atom of carbon-12. It has no units.
• Relative molecular mass, \(M_\mathrm{r}\): the mean mass of a molecule relative to one twelfth of the mass of a carbon-12 atom. Add up the \(A_\mathrm{r}\) values of every atom in the formula. For ionic compounds, which have no molecules, the same sum is called the relative formula mass.
• Molar mass, \(M\): the mass of one mole of a substance, in g mol⁻¹. It has the same number as \(M_\mathrm{r}\): for CaCO₃, \(M = 40.1 + 12.0 + 3 \times 16.0 = 100.1\ \mathrm{g\,mol^{-1}}\).
• Number of particles: \( N = n \times L \), where \(n\) is the amount in moles.

Always say which particle you are counting. One mole of water molecules contains two moles of hydrogen atoms, one mole of oxygen atoms and three moles of atoms altogether. One mole of sodium chloride contains one mole of Na⁺ ions and one mole of Cl⁻ ions, so two moles of ions in total.

2. Amount from mass

For any pure solid, liquid or gas, the amount in moles comes from its mass and its molar mass:

\[ n = \frac{m}{M} \qquad \text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1})} \]

The mass must be in grams. Convert first: 1 g = 1000 mg, 1 kg = 1000 g and 1 tonne = 10⁶ g. Rearranging gives \( m = n \times M \) and \( M = m / n \), the last of which is how you identify an unknown substance from a measured mass and amount.

This page uses these relative atomic masses from the periodic table: H 1.0, C 12.0, N 14.0, O 16.0, Na 23.0, Mg 24.3, Al 27.0, S 32.1, Cl 35.5, K 39.1, Ca 40.1, Fe 55.8, Cu 63.5.

Hydrated salts and water of crystallisation

Many salts crystallise from water with water molecules built into their crystal lattice. This is water of crystallisation, and the formula shows it after a dot. Blue hydrated copper(II) sulfate is CuSO₄·5H₂O. Heating drives the water off and leaves white anhydrous CuSO₄. The water is part of the compound, so it counts in the molar mass.

To find the number of water molecules, \(x\), in a hydrated salt, heat a weighed sample in a crucible to drive off the water, then weigh the anhydrous solid that is left. The same method measures the mass lost in other thermal decompositions, or the mass gained when a metal such as magnesium is burned to its oxide.

• Method: weigh the clean, dry crucible and lid; add about 2 g of the hydrated salt and weigh again; heat strongly for a few minutes; let it cool and reweigh; then reheat, cool and reweigh until two readings agree. This is heating to constant mass, and it shows that all the water has gone.
• Sample size: about 2 g is right. A much larger sample may not lose all its water. A tiny sample gives a mass change so small that the balance's uncertainty becomes a large percentage of it.
• Dry crucible: any water already on the crucible is lost on heating and counted as water of crystallisation, so \(x\) comes out too big.
• Cooling: cool the anhydrous solid in a desiccator, because many anhydrous salts reabsorb water from the air.

3. Empirical and molecular formulae

• Empirical formula: the simplest whole-number ratio of the atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula.

Ethene is C₂H₄ and cyclohexane is C₆H₁₂, but both have the empirical formula CH₂. Analysis of a compound tells you the mass (or percentage by mass) of each element, which only gives the ratio of atoms. You need the relative molecular mass, for example from a mass spectrum, to get the molecular formula.

1. Divide the mass (or percentage) of each element by its \(A_\mathrm{r}\). This gives the amount of each element in moles.
2. Divide every answer by the smallest one.
3. If any result is close to x.5, x.33 or x.25, multiply all of them by 2, 3 or 4 to reach whole numbers.
4. For the molecular formula, divide \(M_\mathrm{r}\) by the empirical formula mass and multiply the empirical formula by the result.

4. Balanced equations and ionic equations

Atoms are not created or destroyed in a chemical reaction, so a balanced equation has the same number of each kind of atom on both sides. You balance it by changing the numbers in front of formulas, never the formulas themselves. The balancing numbers give the mole ratio in which substances react: in N₂(g) + 3H₂(g) → 2NH₃(g), 1 mol of nitrogen reacts with 3 mol of hydrogen to make 2 mol of ammonia. State symbols show the physical state: (s) solid, (l) liquid, (g) gas and (aq) dissolved in water.

Ionic equations

When ionic compounds dissolve, the ions separate and move independently. Many of them take no part in the reaction and are present unchanged at the end. These are spectator ions. An ionic equation leaves them out and shows only the particles that change, which reveals what is really happening. It must balance for charge as well as for atoms.

• Split into ions: soluble ionic compounds in solution (aq), including strong acids such as HCl(aq), H₂SO₄(aq) and HNO₃(aq), and soluble hydroxides such as NaOH(aq).
• Do not split: solids (s), including precipitates, insoluble oxides and carbonates; liquids such as water; gases; and covalent molecules.

Displacement reactions

A more reactive metal displaces a less reactive metal from a solution of its compound, because the more reactive metal gives up electrons more readily. Zinc dropped into blue copper(II) sulfate solution becomes coated in red-brown copper, the blue colour fades and the mixture warms up:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)    ionic: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Halogens behave the same way: a more reactive halogen, which is a stronger oxidising agent, displaces a less reactive one from a solution of its halide ions. The colour change shows the new halogen forming.

You will meet the reasons for these trends in the topic on redox and the halogens.

Reactions of acids

An acid releases hydrogen ions, H⁺(aq), in water. The common strong acids are hydrochloric acid HCl, sulfuric acid H₂SO₄ and nitric acid HNO₃. A base neutralises an acid: metal oxides, metal hydroxides, carbonates and ammonia are all bases. An alkali is a base that dissolves in water to release hydroxide ions, OH⁻(aq), for example NaOH, KOH and aqueous ammonia. A salt forms when the hydrogen of an acid is replaced by a metal ion or an ammonium ion.

Precipitation reactions

Mixing two solutions can bring together a cation and an anion that form an insoluble solid. The solid appears as a cloudy precipitate, and its colour often identifies the ions.

Making salts

• Soluble salt from an insoluble base, metal or carbonate: add the solid to warm acid until no more dissolves, so that it is in excess and all the acid is used up. Filter off the excess solid. Evaporate the filtrate to about half its volume, then leave it to cool and crystallise slowly. Filter off the crystals and dry them between filter papers.
• Soluble salt from an alkali: both reactants are solutions, so excess cannot be filtered off. Do a titration with an indicator to find the exact volumes that react, then repeat with those volumes and no indicator, and crystallise as above.
• Insoluble salt: mix solutions containing its two ions, filter off the precipitate, wash it with distilled water to remove soluble impurities, and dry it.

The yield of crystals is always below 100%, because some salt stays dissolved in the solution left behind and some is lost on the filter paper and in transfers.

5. Concentration of solutions

A solution is a solute dissolved in a solvent, usually water. Its concentration tells you how much solute is in each cubic decimetre of solution:

\[ c = \frac{n}{V} \qquad \text{concentration (mol dm}^{-3}) = \frac{\text{amount (mol)}}{\text{volume of solution (dm}^{3})} \]

1 dm³ is a cube 10 cm on each side, so it equals 1000 cm³ (one litre), and 1 m³ = 1000 dm³. Volumes are usually measured in cm³, so divide by 1000 before using the equation. Concentration can also be written in g dm⁻³, the mass of solute in each dm³: multiply the concentration in mol dm⁻³ by the molar mass.

Concentrations of ions

A dissolved ionic compound separates into its ions, so the concentration of an ion can be larger than the concentration of the compound. In 0.200 mol dm⁻³ sodium sulfate, Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq), so [Na⁺] = 0.400 mol dm⁻³ and [SO₄²⁻] = 0.200 mol dm⁻³.

Making a standard solution

A standard solution is one whose concentration is known accurately. The aim is to get every bit of the weighed solid into a known final volume.

• Weigh: weigh a weighing bottle containing the solid, tip the solid into a beaker, and reweigh the bottle. The difference is the mass transferred. (Alternatively, rinse the weighing bottle into the beaker with distilled water.)
• Dissolve: add about 100 cm³ of distilled water and stir with a glass rod until all the solid has dissolved. Warm gently if needed, and let the solution cool before the next step, because volume changes with temperature.
• Transfer: pour the solution through a funnel into a volumetric flask (for example 250 cm³). Rinse the beaker, rod and funnel with distilled water and add the rinsings to the flask, so that no solute is left behind.
• Make up to the mark: add distilled water until the bottom of the meniscus sits on the graduation line, using a dropping pipette for the last few drops, with your eye level with the mark.
• Mix: stopper the flask and invert it several times so that the concentration is the same throughout.

Dilution

Adding water to a solution does not change the amount of solute, it only spreads it through a bigger volume. So \( n = c_1V_1 = c_2V_2 \). The new volume is the original volume plus the volume of water added. To dilute accurately, measure the original solution with a volumetric pipette (it has a much smaller uncertainty than a measuring cylinder) and make it up in a volumetric flask.

6. Parts per million

Very low concentrations, such as pollutants in water or trace gases in the air, are given in parts per million (ppm), which avoids awkward small numbers.

\[ \text{ppm by mass} = \frac{\text{mass of substance}}{\text{total mass of mixture}} \times 10^{6} \qquad \text{ppm by volume} = \frac{\text{volume of gas}}{\text{total volume of gas mixture}} \times 10^{6} \]

1 ppm by mass is 1 mg in every kg. A dilute aqueous solution has a density of about 1 kg dm⁻³, so 1 ppm is also roughly 1 mg in each dm³.

7. Volumes of gases

In a gas the molecules are far apart, so the volume a gas takes up depends almost entirely on how many molecules there are, not on how big they are. This is Avogadro's law: equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. It means one mole of any gas occupies the same volume under the same conditions. At room temperature and pressure (r.t.p.) this molar volume is taken as 24.0 dm³ mol⁻¹.

\[ n = \frac{V}{24.0} \quad (V \text{ in dm}^3) \qquad\text{or}\qquad n = \frac{V}{24\,000} \quad (V \text{ in cm}^3) \]

Reacting volumes of gases

Because volume is proportional to amount, the balancing numbers in an equation give the ratio of gas volumes directly, as long as all the volumes are measured at the same temperature and pressure. There is no need to convert to moles.

The ideal gas equation

The molar volume of 24.0 dm³ only applies at room conditions. For any temperature and pressure use the ideal gas equation, which works for any gas or mixture of gases (then \(n\) is the total amount of all the gases):

\[ pV = nRT \]

For a fixed amount of gas, \(nR\) is constant, so \( \dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2} \). This handles a gas moved to new conditions without finding \(n\) first. At constant temperature it reduces to \(p_1V_1 = p_2V_2\): if 100 cm³ of gas at 100 kPa is let into an empty 500 cm³ flask, its pressure falls to \(100 \times 100 / 500 = 20\) kPa.

Measuring a gas volume

A gas syringe lets you measure the volume of gas given off, which can be turned into an amount with \(n = V/24.0\) or \(pV = nRT\). The same idea is used to measure the molar volume of a gas: react a weighed piece of magnesium ribbon with excess dilute hydrochloric acid and collect the hydrogen. For example, 0.0360 g of Mg (\(0.0360 / 24.3 = 1.48 \times 10^{-3}\) mol) gives the same amount of H₂. If 35.6 cm³ is collected, the molar volume is \(35.6 / 1.48 \times 10^{-3} = 24\,000\ \mathrm{cm^3\,mol^{-1}}\).

• Gas lost before the bung goes in: the volume measured is too small. Keep the reactants apart (for example the solid in a small tube inside the flask) and tip them together after sealing.
• A sticking plunger: the gas is not at room pressure when you read the volume.
• Soluble gases: carbon dioxide and sulfur dioxide partly dissolve in the water in the flask, so less gas is collected than was made.

8. Moving between quantities

Every quantity you can measure links to amount in moles by one of four routes: mass through the molar mass, number of particles through the Avogadro constant, gas volume through the molar volume (or \(pV = nRT\)), and solution volume through the concentration. The amount is the hub that connects them all.

9. Calculations from equations

Most questions give you a quantity of one substance and ask about another substance in the same reaction. They all follow three steps:

1. Turn the quantity you are given into an amount in moles (mass, gas volume or solution).
2. Use the mole ratio from the balanced equation to find the amount of the substance you want.
3. Turn that amount into the quantity asked for.

When one reactant runs out first

If you are given amounts of two reactants, one may be in excess. The other, the limiting reactant, is used up first and decides how much product forms. Work out how much of each is needed for the other, and see which one runs short.

10. Acid–base titrations

A titration finds the concentration of a solution by reacting it with a solution of known concentration. In an acid–alkali titration a measured volume of one solution goes in a conical flask with a few drops of indicator, and the other is added from a burette until the indicator just changes colour. That point is the end point, and it shows that the acid and alkali have exactly neutralised each other.

• Measure the solution in the flask with a volumetric pipette (usually 25.0 cm³), rinsed first with that solution. Rinse the burette with the solution it will hold, so any water left inside does not dilute it.
• Read the burette at eye level, at the bottom of the meniscus, to the nearest 0.05 cm³. The titre is the final reading minus the initial reading.
• Swirl and go slowly: stand the flask on a white tile so the colour change is easy to see, and add the solution drop by drop near the end point.
• Repeat: the first run is a rough titration to find roughly where the end point is. Repeat accurately until you have concordant titres, within 0.10 cm³ of each other, and use the mean of those.

For a strong acid with a strong alkali either indicator works, because the pH jumps by several units with a single drop at the end point. Use phenolphthalein when the acid is weak, such as ethanoic acid, and methyl orange when the alkali is weak, such as aqueous ammonia. Universal indicator is never used: it changes colour gradually, so there is no sharp end point.

11. Percentage yield and atom economy

The amount of product calculated from the equation is the theoretical yield, the most you could possibly get. In practice you get less:

\[ \text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100 \]

Yields fall short because the reaction may not go to completion (it may be reversible, or stopped too soon), side reactions make other products, and some product is lost when it is transferred between containers, filtered or purified.

Atom economy asks a different question: of all the atoms you put in, what fraction ends up in the product you want? It comes straight from the balanced equation, so it is fixed for a given reaction however carefully you work.

\[ \text{percentage atom economy} = \frac{\text{molar mass of the desired product}}{\text{sum of the molar masses of all the reactants}} \times 100 \]

Use the balancing numbers when you add up the molar masses. A process with a high atom economy makes little waste and uses raw materials efficiently, so it is more sustainable. Reactions with only one product, such as additions, have 100% atom economy. When there is a by-product, finding a use for it or selling it improves the economics of the process.

Two ways of making ethanol show the difference. Fermentation, C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, has an atom economy of \( (2 \times 46.0) / 180.0 \times 100 = 51.1\% \). Hydration of ethene, C₂H₄ + H₂O → C₂H₅OH, has an atom economy of 100%, because every atom ends up in the ethanol.

Quick checks

Interactive diagrams