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S1.4

Counting particles by mass: the mole

Structure 1 · Models of the particulate nature of matter · SL and HL

🎯What you need to be able to do

  • Convert between amount of substance in moles and the number of specified elementary entities.
  • Determine relative formula masses from relative atomic masses, and explain why neither has units.
  • Solve problems relating the number of particles, the amount in moles and the mass in grams.
  • Interconvert percentage composition by mass and empirical formula, and determine a molecular formula from an empirical formula and a molar mass.
  • Solve problems involving molar concentration, amount of solute and volume of solution, in both g dm−3 and mol dm−3.
  • State Avogadro’s law and use it to solve problems involving gas volumes and mole ratios.

📚The chemistry

The mole, and what it is for

Chemistry has a scale problem: reactions are between individual particles, but laboratories weigh grams. The mole is the bridge. It is the SI unit of amount of substance, and one mole contains exactly the number of elementary entities given by the Avogadro constant, \( N_{\mathrm{A}} = 6.02 \times 10^{23}\ \mathrm{mol^{-1}} \) (data booklet).

\[ N = n \times N_{\mathrm{A}} \]

An elementary entity can be an atom, a molecule, an ion, an electron, any other particle, or a specified group of particles — and the syllabus wants you to say which. “A mole of oxygen” is ambiguous: a mole of O atoms is \( 6.02 \times 10^{23} \) atoms, while a mole of O2 molecules is \( 6.02 \times 10^{23} \) molecules but \( 1.204 \times 10^{24} \) atoms.

Forgetting to multiply up for the particles inside a formula unit. One mole of \( \mathrm{Al_2(SO_4)_3} \) contains 2 mol of \( \mathrm{Al^{3+}} \) ions, 3 mol of \( \mathrm{SO_4^{2-}} \) ions, 3 mol of sulfur atoms and 12 mol of oxygen atoms. Questions that ask for “the number of oxygen atoms” are testing exactly this, and the answer is never just \( n \times N_{\mathrm{A}} \).

Relative atomic mass, relative formula mass, molar mass

Masses of atoms are compared on a scale relative to carbon-12. Because they are ratios of masses, relative atomic mass \( A_{\mathrm{r}} \) and relative formula mass \( M_{\mathrm{r}} \) have no units. Find \( M_{\mathrm{r}} \) by adding the \( A_{\mathrm{r}} \) values of every atom in the formula, using the two-decimal-place values in the data booklet, not rounded whole numbers.

Molar mass \( M \) is numerically the same but does have units, g mol−1. That numerical coincidence is the whole point of the mole: it is defined so that the mass of one mole in grams equals the relative mass.

\[ n = \frac{m}{M} \]

Empirical and molecular formulas

The empirical formula is the simplest whole-number ratio of atoms of each element. The molecular formula is the actual number of atoms in a molecule. Ethane is empirical \( \mathrm{CH_3} \), molecular \( \mathrm{C_2H_6} \); benzene is empirical \( \mathrm{CH} \), molecular \( \mathrm{C_6H_6} \). Note that ionic compounds have only empirical formulas, because there are no discrete molecules — see S2.1.

The empirical-formula procedure never changes, whether you are given percentages or masses:

  1. write down the mass (or percentage — treat as grams in 100 g) of each element;
  2. divide each by that element’s \( A_{\mathrm{r}} \) to get moles;
  3. divide every result by the smallest of them;
  4. if the ratios are not close to whole numbers, multiply all of them by 2, 3 or 4 until they are.

To get from empirical to molecular, divide the molar mass of the compound by the mass of the empirical formula unit, and multiply every subscript by that whole number:

\[ k = \frac{M}{M(\text{empirical unit})} \qquad \text{molecular} = (\text{empirical})_k \]

Combustion analysis is the standard experimental route. Burn a known mass of a hydrocarbon or a compound of C, H and O in excess oxygen and measure the \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \) produced. Every carbon ends up in \( \mathrm{CO_2} \) (so \( n(\mathrm{C}) = n(\mathrm{CO_2}) \)) and every hydrogen in \( \mathrm{H_2O} \) (so \( n(\mathrm{H}) = 2 \times n(\mathrm{H_2O}) \)). If the compound also contains oxygen, find the oxygen by difference: total sample mass minus the mass of C minus the mass of H.

Solutions and concentration

Concentration is amount per unit volume. Square brackets denote molar concentration, so \( [\mathrm{HCl}] \) means the concentration of hydrochloric acid in mol dm−3.

\[ n = c\,V \qquad (V \text{ in } \mathrm{dm^3}) \]

Two conversions are needed constantly and are the source of most errors on this topic:

  • Volume: \( 1\ \mathrm{dm^3} = 1000\ \mathrm{cm^3} \), so divide a volume in cm3 by 1000 before using \( n = cV \). One dm3 is one litre.
  • Concentration units: to go from mol dm−3 to g dm−3, multiply by the molar mass; to go the other way, divide by it.

A standard solution is one of accurately known concentration, made by weighing the solute precisely, dissolving it in less than the final volume, transferring quantitatively to a volumetric flask and making up to the graduation mark. A serial dilution makes a set of known lower concentrations by repeated fixed-ratio dilution; in every dilution the amount of solute is unchanged, which gives \( c_1V_1 = c_2V_2 \).

Avogadro’s law

Equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules. The practical consequence is powerful: for gases at the same temperature and pressure, volume ratio equals mole ratio, so you can read stoichiometry straight off the balanced equation without converting to moles at all.

In \( \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g) \), 10 cm3 of nitrogen reacts with 30 cm3 of hydrogen to give 20 cm3 of ammonia. The law applies only to gases, and only when they are compared at the same temperature and pressure — and strictly to ideal gases, which is where S1.5 picks the story up.

✏️Worked example

A compound contains carbon, hydrogen and oxygen only. When 1.50 g of it is burned completely in excess oxygen, 2.20 g of carbon dioxide and 0.900 g of water are produced. The molar mass of the compound is 60.0 g mol−1.
(a) Determine the empirical formula.
(b) Determine the molecular formula.
(c) Calculate the number of oxygen atoms in 1.50 g of the compound.
(d) 1.50 g of the compound is dissolved in water and made up to 250.0 cm3 in a volumetric flask. Calculate the concentration in mol dm−3.

(a) Work out the mass of each element in the 1.50 g sample.

Carbon: \( n(\mathrm{CO_2}) = \dfrac{2.20}{44.01} = 0.0500\ \mathrm{mol} \), and every \( \mathrm{CO_2} \) came from one C, so \( n(\mathrm{C}) = 0.0500\ \mathrm{mol} \), mass \( = 0.0500 \times 12.01 = 0.600\ \mathrm{g} \).

Hydrogen: \( n(\mathrm{H_2O}) = \dfrac{0.900}{18.02} = 0.0500\ \mathrm{mol} \), and each water contains two H, so \( n(\mathrm{H}) = 0.100\ \mathrm{mol} \), mass \( = 0.100 \times 1.01 = 0.101\ \mathrm{g} \).

Oxygen by difference: \( 1.50 - 0.600 - 0.101 = 0.799\ \mathrm{g} \), so \( n(\mathrm{O}) = \dfrac{0.799}{16.00} = 0.0499\ \mathrm{mol} \).

Divide through by the smallest, 0.0499:

\[ \mathrm{C} : \mathrm{H} : \mathrm{O} \;=\; \frac{0.0500}{0.0499} : \frac{0.100}{0.0499} : \frac{0.0499}{0.0499} \;=\; 1.00 : 2.00 : 1.00 \]

Empirical formula CH2O.

(b) The empirical unit has mass \( 12.01 + 2(1.01) + 16.00 = 30.03 \). So \( k = \dfrac{60.0}{30.03} = 2.00 \), and the molecular formula is C2H4O2 (ethanoic acid).

(c) \( n(\text{compound}) = \dfrac{1.50}{60.0} = 0.0250\ \mathrm{mol} \). Each molecule contains two oxygen atoms, so \( n(\mathrm{O\ atoms}) = 0.0500\ \mathrm{mol} \), and

\[ N = 0.0500 \times 6.02 \times 10^{23} = 3.01 \times 10^{22}\ \text{oxygen atoms} \]

(d) \( V = 250.0\ \mathrm{cm^3} = 0.2500\ \mathrm{dm^3} \), and \( n = 0.0250\ \mathrm{mol} \), so

\[ c = \frac{n}{V} = \frac{0.0250}{0.2500} = 0.100\ \mathrm{mol\,dm^{-3}} \]
Check it. Two independent checks are available. First, mass balance: the masses of C, H and O must add back to the sample mass, and \( 0.600 + 0.101 + 0.799 = 1.500 \). Second, the value of \( k \) in (b) must come out as a whole number — 2.00 here. If it comes out as, say, 1.6, the empirical formula is wrong, not the molar mass, so go back and check step 3 of the empirical calculation. In (c), the answer must be larger than \( n \times N_{\mathrm{A}} \) because each molecule contributes two atoms; getting \( 1.5 \times 10^{22} \) means you forgot the subscript.
Finding oxygen from the oxygen in the products. The tempting move in (a) is to take the oxygen in \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \) and call it the oxygen in the compound. It is not: most of that oxygen came from the \( \mathrm{O_2} \) burned, and there is no way to tell the two sources apart. Oxygen in a combustion analysis is always found by difference from the sample mass. The related slip is \( n(\mathrm{H}) = n(\mathrm{H_2O}) \) — water has two hydrogens, so the factor of two is compulsory, and leaving it out turns \( \mathrm{CH_2O} \) into \( \mathrm{CHO} \).

📝Practise

Work through these on paper, then reveal the answer.

1. Calculate (a) the amount, in mol, in 4.40 g of CO2; (b) the mass of 0.250 mol of Ca(OH)2; (c) the number of ions in 0.100 mol of Na2SO4.
(a) \( M(\mathrm{CO_2}) = 12.01 + 2(16.00) = 44.01 \), so \( n = \dfrac{4.40}{44.01} = \mathbf{0.100\ mol} \). (b) \( M(\mathrm{Ca(OH)_2}) = 40.08 + 2(16.00 + 1.01) = 74.10 \), so \( m = 0.250 \times 74.10 = \mathbf{18.5\ g} \). (c) Each formula unit of \( \mathrm{Na_2SO_4} \) gives three ions: two \( \mathrm{Na^+} \) and one \( \mathrm{SO_4^{2-}} \). So \( n(\text{ions}) = 0.300\ \mathrm{mol} \) and \( N = 0.300 \times 6.02 \times 10^{23} = \mathbf{1.81 \times 10^{23}} \) ions. (The sulfate ion counts as one ion, not five — it is a single polyatomic species.)
2. A compound has percentage composition by mass 40.0% C, 6.7% H and 53.3% O, and a molar mass of 180 g mol−1. Determine its empirical and molecular formulas.
Treat the percentages as grams in 100 g. Moles: C \( = \dfrac{40.0}{12.01} = 3.33 \); H \( = \dfrac{6.7}{1.01} = 6.63 \); O \( = \dfrac{53.3}{16.00} = 3.33 \). Divide by the smallest (3.33): \( 1.00 : 1.99 : 1.00 \), so the empirical formula is CH2O, of mass 30.03. Then \( k = \dfrac{180}{30.03} = 5.99 \approx 6 \), giving the molecular formula C6H12O6 — glucose. Note that this has the same empirical formula as ethanoic acid in the worked example, which is exactly why an empirical formula alone never identifies a compound.
3. Describe how you would prepare 250.0 cm3 of a 0.200 mol dm−3 standard solution of anhydrous sodium carbonate (\( M = 105.99 \) g mol−1), including the mass required.
Amount needed: \( n = cV = 0.200 \times 0.2500 = 0.0500\ \mathrm{mol} \). Mass: \( m = nM = 0.0500 \times 105.99 = \mathbf{5.30\ g} \). Method: weigh 5.30 g accurately on an analytical balance, ideally by difference (weigh the weighing bottle before and after transfer, so nothing left behind is unaccounted for). Dissolve completely in a beaker in a volume of distilled water well under 250 cm3. Transfer quantitatively to a 250.0 cm3 volumetric flask through a funnel, rinsing the beaker, stirring rod and funnel into the flask several times so that no solute is lost. Make up to the graduation mark with distilled water, adding the last few drops with a pipette so the bottom of the meniscus sits on the line at eye level. Stopper and invert repeatedly to mix. The volumetric flask is essential: a beaker or measuring cylinder is nowhere near precise enough for a standard solution.
4. 25.0 cm3 of 0.500 mol dm−3 NaOH is diluted to 100.0 cm3. Calculate the new concentration, and state the concentration in g dm−3.
Dilution does not change the amount of solute: \( n = cV = 0.500 \times 0.0250 = 0.0125\ \mathrm{mol} \). New concentration: \( c = \dfrac{0.0125}{0.1000} = \mathbf{0.125\ mol\,dm^{-3}} \). (Equivalently \( c_1V_1 = c_2V_2 \), so \( c_2 = 0.500 \times \dfrac{25.0}{100.0} \) — the volume ratio can be used in cm3 directly, since the units cancel.) In mass terms, \( M(\mathrm{NaOH}) = 22.99 + 16.00 + 1.01 = 40.00 \), so \( 0.125 \times 40.00 = \mathbf{5.00\ g\,dm^{-3}} \).
5. 50 cm3 of propane, C3H8, is burned in 300 cm3 of oxygen, all volumes measured at the same temperature and pressure, above 100 °C. Determine the composition of the final gas mixture.
Equation: \( \mathrm{C_3H_8}(g) + 5\mathrm{O_2}(g) \rightarrow 3\mathrm{CO_2}(g) + 4\mathrm{H_2O}(g) \). By Avogadro’s law, volume ratios equal mole ratios for gases at the same temperature and pressure, so no conversion to moles is needed. 50 cm3 of propane requires \( 5 \times 50 = 250\ \mathrm{cm^3} \) of oxygen. Only 250 of the 300 cm3 supplied is used, so propane is limiting and 50 cm3 of O2 remains unreacted. Products: \( 3 \times 50 = 150\ \mathrm{cm^3}\ \mathrm{CO_2} \) and \( 4 \times 50 = 200\ \mathrm{cm^3}\ \mathrm{H_2O} \). Above 100 °C the water is a gas and counts, so the mixture is 150 cm3 CO2, 200 cm3 H2O and 50 cm3 O2, total 400 cm3. Had the mixture been cooled to room temperature the water would condense and the gaseous total would be 200 cm3 — questions exploit that difference, so always check the stated temperature.
6. Explain why relative atomic mass has no units but molar mass does, and why one mole of oxygen gas has a mass of 32.00 g while one mole of oxygen atoms has a mass of 16.00 g.
Relative atomic mass is defined as a ratio — the mass of an atom compared with one twelfth the mass of a carbon-12 atom. A ratio of two masses is dimensionless, so the units cancel and \( A_{\mathrm{r}} \) has none. Molar mass is a mass per unit amount, \( M = m/n \), and therefore carries units of g mol−1. They are numerically equal because the mole was defined to make them so, which is the entire convenience of the unit. As for oxygen: the elementary entity must be specified. One mole of oxygen atoms is \( 6.02 \times 10^{23} \) O atoms with a mass of 16.00 g; one mole of oxygen gas means \( 6.02 \times 10^{23} \) O2 molecules, each containing two atoms, so it contains \( 1.20 \times 10^{24} \) atoms and has a mass of 32.00 g. Saying “a mole of oxygen” without specifying which is genuinely ambiguous, and the guide requires the entity to be stated.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • RSC Learn Chemistry — the standard protocol for preparing a standard solution and for weighing by difference, with the reasoning for each step.
  • PhET — Molarity and Concentration, which let you change amount and volume independently and watch the concentration respond.
  • Khan Academy — empirical and molecular formula problems, if you want more combustion analysis practice than one page can hold.