Amount of substance

In a nutshell

Chemists cannot count atoms one by one, so they count them by weighing. The link between a mass you can measure on a balance and the number of particles reacting is the mole.

This subtopic is the toolkit for every quantitative calculation in chemistry: moles from mass, from gas volume and from solution concentration, then using a balanced equation to get from one substance to another. Get this fluent and the calculations in every later topic fall into place.

Assumed knowledge: Atomic structure (relative atomic mass and the mass spectrometer).

Core content

Relative atomic mass and relative molecular mass

Masses of atoms are compared against one standard: the carbon-12 atom.

• Relative atomic mass (A_r) is the average mass of an atom of an element divided by one twelfth of the mass of one atom of ¹²C.
• Relative molecular mass (M_r) is the average mass of a molecule divided by one twelfth of the mass of one atom of ¹²C.

Both are ratios, so they have no units.

You find M_r by adding up the A_r values of every atom in the formula. For example:

$M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$

For an ionic compound there are no molecules, so the same number is called the relative formula mass (still the sum of the A_r values in the formula unit, e.g. for NaCl or CaF₂).

The mole and the Avogadro constant

One mole is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12.

That number is the Avogadro constant (L), equal to 6.022 × 10²³ mol^-1. (You are not required to recall this value; it is given.)

A mole counts whatever particle you name: electrons, atoms, molecules, ions or formula units.

• 1 mol of sodium contains 6.022 × 10²³ atoms.
• 1 mol of water contains 6.022 × 10²³ molecules.
• 1 mol of chloride ions contains 6.022 × 10²³ ions.

Still don't get it? · what a mole really is

Imagine you run a hardware shop and a customer wants 10 000 identical nails. Counting them out would take all day. Instead you weigh ten nails, find they weigh 25 g, and weigh out 25 kg. You have delivered 10 000 nails without counting a single one past the first ten.

A chemist has the same problem, but far worse: atoms are far too small and too numerous to count. So chemists do exactly the hardware-shop trick. They pick a fixed, enormous "counting number", 6.022 × 10²³, and call that many particles one mole. The clever part is that the mass of one mole, in grams, is just the relative atomic or molecular mass. So one mole of carbon (Ar = 12.0) is 12.0 g, and one mole of water (Mr = 18.0) is 18.0 g.

Back to the exam: "amount in moles" is just "how many particles, counted in units of 6.022 × 10²³". Weighing out Mr grams gives you one mole, so the number of moles is $n = \frac{m}{M_r}$, and the number of actual particles is that number of moles times the Avogadro constant.

The three routes to an amount in moles. You measure a substance one of three ways, and each has its own equation:

Once you have an amount in moles, the number of particles is:

$\text{number of particles} = n \times L$

Concentration is the amount of solute, in moles, per dm³ of solution (unit mol dm^-3).

When an ionic solid dissolves it dissociates, so an ion's concentration can differ from the compound's. For example, in 0.10 mol dm^-3 MgCl₂:

$\text{MgCl}_2(\text{s}) \longrightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq})$

the chloride ion concentration is 0.20 mol dm^-3, because each formula unit releases two chloride ions.

Significant figures. Give a final answer to the same number of significant figures as the least precise data in the question, and carry at least that many through the working (never round early). Significant figures and decimal places are not the same thing: 0.0250 has 3 significant figures.

The ideal gas equation

A gas sample is described by the ideal gas equation:

$pV = nRT$

The single thing that costs the most marks here is units. Everything must be in SI units before you substitute:

• pressure p in pascals (Pa): 1 kPa = 1000 Pa
• volume V in cubic metres (m³): divide cm³ by 1 000 000, or dm³ by 1000
• temperature T in kelvin (K): add 273 to a Celsius temperature
• R is the gas constant, 8.31 J K^-1 mol^-1 (given; you need not recall it).

Rearranged to find an amount, $n = \dfrac{pV}{RT}$. Combining this with $n = \dfrac{m}{M_r}$ lets you find the M_r of a volatile liquid from its vapour (see Worked examples).

Empirical and molecular formula

• Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in a molecule.

To find an empirical formula from masses or percentage by mass:

1. Divide each element's mass (or % by mass) by its A_r to get moles.
2. Divide every result by the smallest of them.
3. If needed, scale up to the nearest whole numbers. Those whole numbers are the ratio.

To get the molecular formula, find how many empirical units fit into the M_r:

$n = \frac{M_r}{\text{mass of one empirical unit}}$

then multiply the empirical formula by n. The answer is a formula, not just a ratio of numbers.

Balanced equations and associated calculations

A balanced equation has the same number of atoms of each element on both sides, and the balancing numbers are mole ratios. You should be able to:

• write balanced full equations for reactions you have studied;
• write ionic equations (show only the ions that react, e.g. neutralisation is $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \longrightarrow \text{H}_2\text{O}(\text{l})$);
• balance an unfamiliar equation when the reactants and products are given.

Every reacting-quantity calculation uses the same three-step method:

1. Convert the quantity you are given into moles (using the right route above).
2. Use the mole ratio from the balanced equation to find moles of the substance you want.
3. Convert those moles into the quantity asked for (mass, gas volume, or concentration).

Still don't get it? · using the balanced equation (the mole ratio)

Think of a balanced equation as a recipe. "2 slices of bread + 1 slice of cheese → 1 sandwich" does not mean 2 grams of bread; it means 2 of one thing for every 1 of another. The numbers count items, not weights.

A balanced equation is identical. In $2\text{H}_2 + \text{O}_2 \longrightarrow 2\text{H}_2\text{O}$, the 2, 1 and 2 count moles: 2 mol of hydrogen react with 1 mol of oxygen to make 2 mol of water. You cannot compare masses directly (the molecules weigh different amounts), which is exactly why you convert to moles first.

So the method is always the same three steps: turn what you are given into moles, multiply or divide by the recipe ratio to get moles of what you want, then turn that back into a mass or volume. In the exam, the marks for "use the stoichiometry" are earned on that middle step, so always write the balanced equation down first and read the ratio straight off it.

Percentage yield compares how much product you actually got with the maximum possible:

$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

A yield above 100% is impossible; it means an error (usually an impure or still-wet product).

Percentage atom economy measures how much of the reactant mass ends up in the desired product:

$\% \text{ atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of the } M_r \text{ of all reactants}} \times 100$

Use the balancing numbers when adding up the masses. An addition reaction (one product only) has an atom economy of 100%.

A high atom economy brings real advantages:

• Economic: more product from the same mass of raw material, and less money spent separating and disposing of by-products.
• Environmental: less waste produced, and finite raw materials are used more efficiently.
• Ethical and sustainable: conserving resources and reducing waste is better for society in the long term.

Required practical 1: making a volumetric solution and an acid-base titration

This practical has two parts: preparing a solution of accurately known concentration, then using it in a titration.

Making a standard (volumetric) solution:

1. Weigh the solid by difference (weigh the bottle, tip the solid into a beaker, reweigh the bottle).
2. Dissolve it in less than the final volume of distilled water, stirring with a glass rod.
3. Transfer to a volumetric flask, rinsing the beaker and rod into the flask so no solute is lost.
4. Make up to the graduation mark with distilled water, using a dropping pipette for the last drops so the bottom of the meniscus sits on the line.
5. Stopper and invert several times to mix.

Carrying out the titration:

• Rinse the burette with the solution it will hold (not water), and the pipette with the solution it will measure; rinse the conical flask with distilled water only.
• Pipette a fixed volume of one solution into the conical flask and add a few drops of indicator (methyl orange: red in acid, yellow in alkali; phenolphthalein: colourless in acid, pink in alkali).
• Run in the other solution from the burette, swirling, adding dropwise near the end point, until the indicator just changes colour. Use a white tile to judge the colour.
• Do a rough titration first, then repeat until you have concordant titres (within 0.10 cm³ of each other).

Handling the data (the skills AQA tests here):

• Record every burette reading to 2 decimal places (the nearest 0.05 cm³), e.g. 23.90 cm³, not 23.9.
• A titre is a difference of two readings, so its uncertainty is twice the uncertainty of a single reading.
• Average only the concordant titres; reject an anomalous (outlier) titre before taking the mean.

$\% \text{ uncertainty} = \frac{\text{uncertainty}}{\text{measurement}} \times 100$

Worked examples

Each example below is a different calculation type. Every step is one line; units are carried throughout, and the final answer is given to the significant figures set by the data.

1. Moles and the Avogadro constant. Calculate the number of molecules in 4.00 g of methane, CH₄. (L = 6.022 × 10²³ mol^-1)

$M_r(\text{CH}_4) = 12.0 + (4 \times 1.0) = 16.0$ $n = \frac{m}{M_r} = \frac{4.00}{16.0} = 0.250 \text{ mol}$ $\text{number of molecules} = n \times L = 0.250 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$

(There are four times as many hydrogen atoms: 6.02 × 10²³.)

2. Concentration of a solution. Calculate the concentration, in mol dm^-3, of a solution made by dissolving 2.65 g of sodium carbonate, Na₂CO₃, and making up to 250 cm³.

$M_r(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0$ $n = \frac{m}{M_r} = \frac{2.65}{106.0} = 0.0250 \text{ mol}$ $V = 250 \text{ cm}^3 = 0.250 \text{ dm}^3$ $c = \frac{n}{V} = \frac{0.0250}{0.250} = 0.100 \text{ mol dm}^{-3}$

3. Ideal gas equation (M_r of a volatile liquid). 0.238 g of a volatile liquid was vaporised and occupied 97.0 cm³ at 100 kPa and 373 K. Calculate its M_r. (R = 8.31 J K^-1 mol^-1)

$p = 100 \text{ kPa} = 1.00 \times 10^{5} \text{ Pa}$ $V = 97.0 \text{ cm}^3 = 9.70 \times 10^{-5} \text{ m}^3$ $n = \frac{pV}{RT} = \frac{(1.00 \times 10^{5}) \times (9.70 \times 10^{-5})}{8.31 \times 373} = 3.13 \times 10^{-3} \text{ mol}$ $M_r = \frac{m}{n} = \frac{0.238}{3.13 \times 10^{-3}} = 76.1$

The data are quoted to 3 significant figures, so the answer is 76.1 (no units, as M_r is a ratio).

4. Empirical formula from percentage composition. A compound is 40.0% carbon, 6.70% hydrogen and 53.3% oxygen by mass. Find its empirical formula.

$n(\text{C}) = \frac{40.0}{12.0} = 3.33 \qquad n(\text{H}) = \frac{6.70}{1.0} = 6.70 \qquad n(\text{O}) = \frac{53.3}{16.0} = 3.33$ $\text{divide by the smallest } (3.33): \quad \text{C} = 1.00, \quad \text{H} = 2.01, \quad \text{O} = 1.00$ $\text{empirical formula} = \text{CH}_2\text{O}$

5. Molecular formula from the empirical formula and M_r. The compound above (empirical formula CH₂O) has M_r = 180.0. Find its molecular formula.

$\text{mass of one CH}_2\text{O unit} = 12.0 + (2 \times 1.0) + 16.0 = 30.0$ $n = \frac{M_r}{30.0} = \frac{180.0}{30.0} = 6$ $\text{molecular formula} = (\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$

6. Reacting masses. Calculate the mass of magnesium oxide formed when 4.86 g of magnesium burns completely in oxygen. (A_r: Mg 24.3, O 16.0)

$2\text{Mg}(\text{s}) + \text{O}_2(\text{g}) \longrightarrow 2\text{MgO}(\text{s})$ $n(\text{Mg}) = \frac{4.86}{24.3} = 0.200 \text{ mol}$ $2\,\text{Mg} : 2\,\text{MgO} \implies n(\text{MgO}) = 0.200 \text{ mol}$ $m(\text{MgO}) = n \times M_r = 0.200 \times 40.3 = 8.06 \text{ g}$

7. Volume of a gas from an equation. Calculate the volume of carbon dioxide, in dm³, produced at 298 K and 100 kPa when 6.57 g of sodium hydrogencarbonate decomposes completely. (R = 8.31)

$2\text{NaHCO}_3(\text{s}) \longrightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$ $n(\text{NaHCO}_3) = \frac{6.57}{84.0} = 0.0782 \text{ mol}$ $2\,\text{NaHCO}_3 : 1\,\text{CO}_2 \implies n(\text{CO}_2) = \frac{0.0782}{2} = 0.0391 \text{ mol}$ $V = \frac{nRT}{p} = \frac{0.0391 \times 8.31 \times 298}{1.00 \times 10^{5}} = 9.68 \times 10^{-4} \text{ m}^3 = 0.968 \text{ dm}^3$

8. Percentage yield. In the reaction $\text{N}_2 + 3\text{H}_2 \longrightarrow 2\text{NH}_3$, 14.0 g of nitrogen produced 12.8 g of ammonia. Calculate the percentage yield.

$n(\text{N}_2) = \frac{14.0}{28.0} = 0.500 \text{ mol}$ $1\,\text{N}_2 : 2\,\text{NH}_3 \implies \text{theoretical } n(\text{NH}_3) = 1.00 \text{ mol}$ $\text{theoretical mass} = 1.00 \times 17.0 = 17.0 \text{ g}$ $\% \text{ yield} = \frac{12.8}{17.0} \times 100 = 75.3\%$

9. Percentage atom economy. Compare two routes to ethanol (C₂H₅OH, M_r = 46.0).

Hydration of ethene: $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \longrightarrow \text{C}_2\text{H}_5\text{OH}$

$\% \text{ atom economy} = \frac{46.0}{28.0 + 18.0} \times 100 = \frac{46.0}{46.0} \times 100 = 100\%$

Fermentation of glucose: $\text{C}_6\text{H}_{12}\text{O}_6 \longrightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$

$\% \text{ atom economy} = \frac{2 \times 46.0}{180.0} \times 100 = \frac{92.0}{180.0} \times 100 = 51.1\%$

The addition route wastes nothing; fermentation loses almost half its mass as CO₂ (the values plotted above).

10. Titration calculation (Required practical 1). 25.0 cm³ of sodium hydroxide solution was exactly neutralised by 22.5 cm³ of 0.100 mol dm^-3 hydrochloric acid. Calculate the concentration of the sodium hydroxide.

$\text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \longrightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})$ $n(\text{HCl}) = c \times V = 0.100 \times \frac{22.5}{1000} = 2.25 \times 10^{-3} \text{ mol}$ $1\,\text{HCl} : 1\,\text{NaOH} \implies n(\text{NaOH}) = 2.25 \times 10^{-3} \text{ mol}$ $c(\text{NaOH}) = \frac{n}{V} = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900 \text{ mol dm}^{-3}$

Common exam mistakes

• Leaving words out of the A_r definition. Examiners repeatedly report the missing pieces: the top line must say average (or mean) mass, and the bottom line must say one twelfth of the mass of one atom of ¹²C. Mixing "mole" and "atom" in the two lines caps the mark.
• Not converting units in pV = nRT. The commonest single error is leaving pressure in kPa instead of Pa; others leave volume in cm³ or forget to add 273 to a Celsius temperature. Convert everything to Pa, m³ and K before substituting.
• Wrong number of significant figures. Writing every digit off the calculator, confusing decimal places with significant figures, or truncating instead of rounding. If the answer needs 3 significant figures, keep at least 3 all the way through the working.
• Ignoring the mole ratio. Treating moles of reactant as moles of product in a 1:1 way when the equation is not 1:1 (for example forgetting the factor of 2 for NaOH reacting with H₂SO₄). Always read the ratio off a balanced equation.
• Atom economy slips. Forgetting to multiply by 100, not using the balancing numbers in the reactant total, or inverting the fraction. Learn the definition and write it down before you calculate.
• Giving a ratio instead of a formula. The answer to an empirical formula question is a formula (e.g. CH₂O), not "1 : 2 : 1".
• Wrong ionic formulae. Writing MgOH instead of Mg(OH)₂, or similar, wrecks the whole balanced equation and every mark that follows.
• Using A_r = 24 for magnesium (or other rounded values) when the data sheet gives 24.3, then rounding the answer wrongly.
• Reading the burette upside down (recording 25.70 instead of 24.30 cm³), quoting a titre to only 1 decimal place, or doubling the burette uncertainty incorrectly.
• Titration technique errors: rinsing the burette with water instead of the solution, or rinsing the conical flask with the solution instead of water; both change the amount of substance and the result.

Key definitions

• Relative atomic mass (A_r) is the average mass of an atom of an element divided by one twelfth of the mass of one atom of carbon-12.
• Relative molecular mass (M_r) is the average mass of a molecule divided by one twelfth of the mass of one atom of carbon-12.
• Relative formula mass is the same quantity for the formula unit of an ionic compound.
• The Avogadro constant is the number of particles in one mole of a substance, 6.022 × 10²³ mol^-1.
• Mole is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12.
• Concentration is the amount of solute, in moles, dissolved per dm³ of solution (mol dm^-3).
• Empirical formula is the simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula is the actual number of atoms of each element in a molecule.
• Percentage yield = (actual yield / theoretical yield) × 100.
• Percentage atom economy = (M_r of the desired product / sum of the M_r of all reactants) × 100.

Specification

• I can define relative atomic mass and relative molecular mass in terms of ¹²C, and use relative formula mass for ionic compounds.
• I can state that the Avogadro constant is the number of particles in a mole, and apply the mole to electrons, atoms, molecules, ions, formulas and equations.
• I can carry out calculations using the Avogadro constant, using mass, M_r and amount in moles, and using concentration, volume and amount for a solution.
• I can use the ideal gas equation pV = nRT with the variables in SI units.
• I can calculate an empirical formula from composition by mass or percentage by mass.
• I can calculate a molecular formula from the empirical formula and the relative molecular mass.
• I can write and balance full and ionic equations, including unfamiliar ones when reactants and products are given.
• I can use balanced equations to calculate masses, volumes of gases, percentage yields, percentage atom economies, and concentrations and volumes in solution.
• I can explain the economic, ethical and environmental advantages of a high atom economy.
• I can make up a volumetric solution and carry out a simple acid-base titration, selecting concordant titres and finding the uncertainty in a burette reading.

Related notes

• Atomic structure
• Bonding