UNIT 2 CHEMISTRY · PEARSON AAQ · D1.1
The Mole and Quantitative Chemistry
FREE CHEMISTRY REVISION · TOPIC 32
The Mole and Quantitative Chemistry
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.1
What you will learn
Use the mole to connect the microscopic number of atoms, molecules, ions or formula units with masses, gas volumes and solution concentrations. Use balanced equations to calculate reacting quantities, identify limiting reagents and calculate percentage yield. This lesson covers the complete Pearson D1.1 scope, with progressive worked examples and practice.
1. The mole counts specified entities
Amount of substance has symbol n and unit mole, mol. One mole contains an Avogadro number of specified entities. For Pearson calculations use the rounded Avogadro constant Nₐ = 6.02 × 10²³ mol⁻¹. A mole is a counting unit, not a mass unit; a mole of different substances usually has a different mass.
Always name what is being counted. One mole of He contains one mole of helium atoms. One mole of O₂ contains one mole of oxygen molecules but two moles of oxygen atoms. One mole of NaCl refers to one mole of formula units, not NaCl molecules in a giant ionic solid. It corresponds to one mole of Na⁺ ions and one mole of Cl⁻ ions when fully dissociated.
Pearson links the mole to the number of atoms in 12.00 g of carbon-12. In modern SI the mole is defined by an exact count, 6.02214076 × 10²³ entities; 6.02 × 10²³ is the rounded value specified for these calculations. Do not describe the rounded value as an exact experimental count.
N = nNₐ and n = N ÷ Nₐ
N is a number of entities, not a number of moles. Multiplying mol by mol⁻¹ leaves a count; dividing a count by mol⁻¹ gives mol. Use a calculator’s exponent key carefully: 6.02 × 10²³ means 6.02E23, not 6.02 × 10 × 23.
2. Relative mass and molar mass
Relative atomic mass, Aᵣ, and relative formula or molecular mass, Mᵣ, are relative quantities with no units. Molar mass, M, is mass per mole, usually in g mol⁻¹ for these calculations. Its numerical value corresponds to the relative atomic/formula mass at the precision used in this course.
Mᵣ(CO₂) = 12.0 + 2(16.0) = 44.0
M(CO₂) = 44.0 g mol⁻¹
Add all atoms in a formula. Brackets multiply everything inside them: M(Ca(OH)₂) = 40.0 + 2(16.0 + 1.0) = 74.0 g mol⁻¹. For a hydrate, include the stated water of crystallisation in the formula mass. Use the relative atomic masses supplied by the question; rounding conventions can slightly change final values.
n = m ÷ M m = nM M = m ÷ n
Use compatible units. If M is in g mol⁻¹, convert mass to g before dividing. For example 250 mg = 0.250 g and 1.20 kg = 1200 g. Keeping a numerical answer without its unit can hide a factor-of-1000 error.
3. A map for choosing the calculation
4. Worked examples: mass and particles
Example 1 — calculate moles and check the unit
Question: How many moles are in 5.85 g NaCl? Use Na = 23.0 and Cl = 35.5.
1. M(NaCl) = 23.0 + 35.5 = 58.5 g mol⁻¹. 2. n = m/M = 5.85/58.5. 3. n = 0.100 mol NaCl. Since 5.85 g is one tenth of 58.5 g, the answer is one tenth of a mole.
Example 2 — reverse the mass calculation
Question: What mass is 0.250 mol CO₂? M = 44.0 g mol⁻¹.
m = nM = 0.250 mol × 44.0 g mol⁻¹ = 11.0 g
Do not divide by M when the unknown is mass. Dimensional checking confirms mol × g mol⁻¹ = g.
Example 3 — molecules versus atoms
Question: How many O₂ molecules and oxygen atoms are in 0.200 mol O₂?
N(O₂ molecules) = 0.200 × 6.02 × 10²³ = 1.204 × 10²³ ≈ 1.20 × 10²³
N(O atoms) = 2 × 1.204 × 10²³ = 2.408 × 10²³ ≈ 2.41 × 10²³
Keep the unrounded value for the second step. Each molecule contains two oxygen atoms; the number of molecules and the number of atoms are not equal.
Example 4 — particles back to moles
Question: A sample contains 3.01 × 10²³ CO₂ molecules. Find its amount and mass.
n = (3.01 × 10²³)/(6.02 × 10²³) = 0.500 mol
m = 0.500 × 44.0 = 22.0 g
The powers of ten cancel. One half of an Avogadro number is one half of a mole.
5. Empirical formulae: the simplest mole ratio
An empirical formula gives the simplest whole-number ratio of atoms of each element. Mass ratios are not atom ratios because different atoms have different masses. Convert each mass to an amount in moles, divide by the smallest result and convert the ratio into small whole numbers.
- If percentages are given, assume a 100 g sample so that each percentage becomes a mass in grams.
- If one element is obtained by mass increase, subtract carefully: mass of oxygen gained = final oxide mass − initial metal mass, provided the experiment supports that interpretation.
- If the ratio is near 1:1.5, multiply all entries by 2; near 1:1.333, multiply by 3; near 1:1.25, multiply by 4.
- Allow for reasonable experimental uncertainty, but do not round a clear 1.50 to 2.00 and change the chemistry.
- The empirical formula may differ from a molecular formula. For C₆H₁₂O₆ the empirical formula is CH₂O. A molecular formula needs additional molar-mass information.
6. Worked examples: empirical formulae
Example 5 — percentage composition
Question: A compound is 40.0% C, 6.7% H and 53.3% O by mass. Use C = 12.0, H = 1.0, O = 16.0.
| Element | Mass in 100 g | Moles | Divide by 3.33125 |
|---|---|---|---|
| C | 40.0 g | 40.0/12.0 = 3.333… | 1.001 ≈ 1 |
| H | 6.7 g | 6.7/1.0 = 6.7 | 2.011 ≈ 2 |
| O | 53.3 g | 53.3/16.0 = 3.33125 | 1 |
The simplest ratio is approximately 1:2:1, so the empirical formula is CH₂O. Small departures from exact integers reflect the rounded percentages. This result alone does not prove the molecular formula is CH₂O.
Example 6 — oxygen gained by a metal
Question: 2.70 g Al forms 5.10 g of an oxide. Find the empirical formula, assuming the mass gain is oxygen and the product is pure. Use Al = 27.0 and O = 16.0.
m(O) = 5.10 − 2.70 = 2.40 g
n(Al) = 2.70/27.0 = 0.100 mol
n(O atoms) = 2.40/16.0 = 0.150 mol
Al:O = 1:1.5 = 2:3 → Al₂O₃
Use 16.0 g mol⁻¹ for moles of oxygen atoms in the formula ratio, not 32.0 for O₂ molecules. The final product contains oxygen atoms in an oxide, not intact O₂ molecules.
7. Balanced equations give mole ratios
A balanced equation conserves atoms and gives relative amounts of reacting entities and moles. It does not directly give equal masses. For 2Mg + O₂ → 2MgO, the ratio Mg:O₂:MgO is 2:1:2 in moles. A 1:1 mole ratio between Mg and MgO does not mean a 1:1 mass ratio.
moles of required substance = moles of known substance × (required coefficient / known coefficient)
For a mass-to-mass calculation: write and balance the equation → convert known mass to moles → apply the coefficient ratio → convert product moles to product mass. Check purity assumptions, limiting reagent information and whether yield is theoretical or actual.
8. Worked example: reacting masses
Example 7 — magnesium to magnesium oxide
Question: What mass of MgO can form from 4.80 g Mg in excess oxygen? Use Mg = 24.0 and O = 16.0.
2Mg + O₂ → 2MgO
n(Mg) = 4.80/24.0 = 0.200 mol
n(MgO) = 0.200 × (2/2) = 0.200 mol
M(MgO) = 24.0 + 16.0 = 40.0 g mol⁻¹
m(MgO) = 0.200 × 40.0 = 8.00 g
The product mass is larger than the initial magnesium mass because oxygen joins the magnesium. Mass is conserved when the oxygen reactant is included; 3.20 g of oxygen is incorporated.
9. Limiting reagent and excess reactant
The limiting reagent is used up first according to the balanced equation and restricts the amount of product. An excess reactant is present in more than the required stoichiometric amount. The smallest mass is not automatically limiting: compare amounts in moles with the equation coefficients.
Compare n / coefficient for every reactant.
The smallest value sets the reaction extent and identifies the limiting reagent.
Alternatively calculate how many moles of the other reagent are required and compare with those available. Calculate theoretical product from the limiting reagent, not from whichever reactant gives the bigger result. To find leftover excess, subtract the amount consumed from the amount initially available.
10. Worked example: limiting reagent and amount left over
Example 8 — magnesium with hydrochloric acid
Question: 4.80 g Mg reacts with 0.300 mol HCl. Find the limiting reagent, moles of H₂ and mass of Mg left. Use Mg = 24.0.
Mg + 2HCl → MgCl₂ + H₂
n(Mg) = 4.80/24.0 = 0.200 mol
n(Mg)/1 = 0.200; n(HCl)/2 = 0.150
HCl is limiting. The available acid reacts with 0.300/2 = 0.150 mol Mg and forms 0.150 mol H₂.
Mg remaining = 0.200 − 0.150 = 0.0500 mol
mass Mg remaining = 0.0500 × 24.0 = 1.20 g
The 0.200 mol magnesium would need 0.400 mol HCl, but only 0.300 mol is supplied. This is an independent check on the limiting-reagent decision.
11. Gas volumes at room temperature and pressure
For these Pearson calculations, take the molar gas volume at room temperature and pressure, r.t.p., as 24.0 dm³ mol⁻¹. It is a useful approximation for gas behaviour under the stated conditions. Equal amounts in moles of different gases occupy equal volumes at the same temperature and pressure in this model.
V = 24.0n n = V/24.0 [V in dm³]
1 dm³ = 1000 cm³
24.0 dm³ mol⁻¹ = 24 000 cm³ mol⁻¹
Do not use 24.0 with a volume still in cm³, and do not apply the gas molar-volume rule to liquid water or a dissolved substance. It is not a universal value at every temperature and pressure. When comparing reacting gas volumes at common temperature and pressure, balanced-equation mole ratios also give volume ratios for the gaseous species.
12. Worked examples: gas volumes
Example 9 — gas volume to particles
Question: How many moles and molecules are in 600 cm³ CO₂ at r.t.p.?
V = 600/1000 = 0.600 dm³
n = 0.600/24.0 = 0.0250 mol
N = 0.0250 × 6.02 × 10²³ = 1.505 × 10²² ≈ 1.51 × 10²² molecules
Using 600/24.0 would make the mole result 1000 times too large.
Example 10 — gas from a reacting mass
Question: Calculate the volume of CO₂ at r.t.p. from complete decomposition of 5.00 g CaCO₃. Use Ca = 40.0, C = 12.0 and O = 16.0.
CaCO₃(s) → CaO(s) + CO₂(g)
M(CaCO₃) = 100.0 g mol⁻¹
n(CaCO₃) = 5.00/100.0 = 0.0500 mol
n(CO₂) = 0.0500 mol
V(CO₂) = 0.0500 × 24.0 = 1.20 dm³ = 1200 cm³
The carbonate-to-carbon-dioxide ratio is 1:1. Assume complete decomposition and a gas volume measured at r.t.p.; do not assign the hot furnace gas the same volume without accounting for its conditions.
13. Concentration in mol dm⁻³
Amount concentration c is the amount of solute per volume of final solution. Its unit mol dm⁻³ means moles in each dm³ of solution. The volume is the final solution volume, not simply the volume of water initially added.
c = n/V n = cV V = n/c
Use V in dm³ when c is in mol dm⁻³.
Convert cm³ to dm³ by dividing by 1000: 25.0 cm³ = 0.0250 dm³ and 250 cm³ = 0.250 dm³. A concentration of 0.100 mol dm⁻³ is not a total amount of 0.100 mol unless the solution volume is 1.00 dm³.
For dilution without reaction or loss of solute, solute moles are conserved, so c₁V₁ = c₂V₂. The same volume units may be used on both sides of that dilution equation, but n = cV still requires a volume compatible with the concentration unit.
14. Worked examples: solution calculations
Example 11 — mass needed for a solution
Question: What mass of anhydrous Na₂CO₃ is required to make 250 cm³ of 0.200 mol dm⁻³ solution? Use Na = 23.0, C = 12.0, O = 16.0.
V = 250/1000 = 0.250 dm³
n = cV = 0.200 × 0.250 = 0.0500 mol
M(Na₂CO₃) = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol⁻¹
m = nM = 0.0500 × 106.0 = 5.30 g
The question specifies anhydrous sodium carbonate. A hydrated formula would have a different molar mass. The final solution must have the stated total volume.
Example 12 — concentration from a neutralisation
Question: 25.0 cm³ H₂SO₄ is exactly neutralised by 30.0 cm³ of 0.100 mol dm⁻³ NaOH. Find the acid concentration for full neutralisation.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
n(NaOH) = 0.100 × 0.0300 = 0.00300 mol
n(H₂SO₄) = 0.00300 × (1/2) = 0.00150 mol
c(H₂SO₄) = 0.00150/0.0250 = 0.0600 mol dm⁻³
Do not assume equal acid and alkali mole amounts. The 1:2 ratio comes from the balanced equation.
Example 13 — dilution
Question: 20.0 cm³ of 0.500 mol dm⁻³ solution is diluted to a final volume of 100.0 cm³. Find the final concentration.
c₂ = c₁V₁/V₂ = 0.500 × 20.0/100.0 = 0.100 mol dm⁻³
The final volume is five times the original, so the concentration becomes one fifth. Solute moles stay 0.500 × 0.0200 = 0.0100 mol.
15. Concentration in g dm⁻³
Mass concentration is mass of solute per volume of final solution. Use a different label, such as ρₘ, to avoid confusing it with amount concentration c. A solution can have both descriptions, related by the solute’s molar mass.
mass concentration (g dm⁻³) = m(g)/V(dm³)
m = mass concentration × V
mass concentration = cM
c = mass concentration/M
The conversion cM is dimensionally consistent: (mol dm⁻³)(g mol⁻¹) = g dm⁻³. It applies to the stated solute formula, not indiscriminately to each ion after dissociation.
16. Worked example: convert between concentration units
Example 14 — sodium chloride solution
Question: A solution contains 5.85 g NaCl in 500 cm³ of solution. Find its concentration in g dm⁻³ and mol dm⁻³, and the mass in a 25.0 cm³ sample. M(NaCl) = 58.5 g mol⁻¹.
V = 0.500 dm³
mass concentration = 5.85/0.500 = 11.7 g dm⁻³
c = 11.7/58.5 = 0.200 mol dm⁻³
mass in 25.0 cm³ = 11.7 × 0.0250 = 0.2925 g ≈ 0.293 g
The original solution concentration is not 5.85 g dm⁻³ because the 5.85 g occupies only half a dm³.
17. Theoretical yield, actual yield and percentage yield
Theoretical yield is the maximum amount of the specified product predicted by the balanced equation from the limiting reagent, assuming complete reaction to that product and no loss. Actual yield is the amount of product obtained in the experiment. Both must refer to the same product and use comparable units.
percentage yield = (actual yield / theoretical yield) × 100%
actual yield = (percentage yield / 100) × theoretical yield
theoretical yield = actual yield / (percentage yield / 100)
Actual yield can be lower because reaction is incomplete, side reactions occur, or product is lost during transfer, separation or purification. A reported yield above 100% does not mean atoms were created: wet or impure product, an incorrect identity or an error in measurements or calculation should be considered. Percentage yield is not percentage purity or atom economy.
18. Worked examples: yield and an integrated calculation
Example 15 — percentage yield
Question: 5.00 g CaCO₃ is heated and 2.24 g CaO is collected. Find the percentage yield. Use CaCO₃ = 100.0 g mol⁻¹ and CaO = 56.0 g mol⁻¹.
CaCO₃ → CaO + CO₂
n(CaCO₃) = 5.00/100.0 = 0.0500 mol
theoretical m(CaO) = 0.0500 × 56.0 = 2.80 g
percentage yield = (2.24/2.80) × 100 = 80.0%
The denominator is the calculated maximum mass of CaO, not the initial 5.00 g carbonate mass. For a theoretical yield of 2.80 g at 75.0%, the expected actual yield would be 2.10 g.
Example 16 — solution, limiting reagent, gas volume and yield
Question: 2.40 g Mg is added to 100 cm³ of 1.50 mol dm⁻³ HCl. The collected H₂ volume at r.t.p. is 1.44 dm³. Find the limiting reagent, theoretical H₂ volume and gas percentage yield. Use Mg = 24.0.
Mg + 2HCl → MgCl₂ + H₂
n(Mg) = 2.40/24.0 = 0.100 mol
n(HCl) = 1.50 × 0.100 = 0.150 mol
compare n/coefficient: Mg 0.100; HCl 0.0750
HCl limits → theoretical n(H₂) = 0.0750 mol
theoretical V(H₂) = 0.0750 × 24.0 = 1.80 dm³
percentage yield = (1.44/1.80) × 100 = 80.0%
Comparing gas volumes gives yield only because both volumes refer to the same gas at the same temperature and pressure. Magnesium consumed is 0.0750 mol, leaving 0.0250 mol = 0.600 g Mg.
19. Progressive independent practice
Use Nₐ = 6.02 × 10²³ mol⁻¹, molar gas volume 24.0 dm³ mol⁻¹ at r.t.p., and the relative atomic masses H 1.0, C 12.0, O 16.0, Na 23.0, Mg 24.0, Al 27.0, Cl 35.5 and Ca 40.0.
| Task | Question |
|---|---|
| A — mass and moles | Find moles in 3.60 g H₂O and the mass of 0.150 mol MgO. |
| B — particle identity | Find molecules and total oxygen atoms in 0.0500 mol CO₂. |
| C — empirical formula | An oxide contains 4.80 g Mg and 3.20 g O. Find its empirical formula. |
| D — reacting masses | Find MgO made from 1.20 g Mg in excess oxygen. |
| E — limiting reagent | 0.100 mol Mg reacts with 0.150 mol HCl. Find the limiting reagent and H₂ amount. |
| F — gas volumes | Find moles in 480 cm³ gas at r.t.p. and volume of 0.125 mol gas at r.t.p. |
| G — solution concentration | Find moles in 40.0 cm³ of 0.250 mol dm⁻³ solution. |
| H — mass concentration | 2.00 g NaOH is in 250 cm³ solution. Find g dm⁻³ and mol dm⁻³. |
| I — yield | Theoretical product is 6.40 g and actual product 5.12 g. Find percentage yield. |
| J — integrated | What r.t.p. CO₂ volume is obtained at 75.0% yield from 10.0 g CaCO₃? Assume a 1:1 carbonate-to-CO₂ ratio. |
Reveal practice answers and key steps
- A: n(H₂O) = 3.60/18.0 = 0.200 mol; m(MgO) = 0.150 × 40.0 = 6.00 g.
- B: CO₂ molecules = 0.0500 × 6.02 × 10²³ = 3.01 × 10²²; oxygen atoms = twice this = 6.02 × 10²².
- C: Mg 4.80/24.0 = 0.200 mol; O 3.20/16.0 = 0.200 mol; ratio 1:1, MgO.
- D: n(Mg) = 1.20/24.0 = 0.0500 mol; n(MgO) = 0.0500 mol; mass = 2.00 g.
- E: Mg + 2HCl → MgCl₂ + H₂; HCl is limiting; n(H₂) = 0.150/2 = 0.0750 mol.
- F: 480 cm³ = 0.480 dm³; n = 0.480/24.0 = 0.0200 mol. V = 0.125 × 24.0 = 3.00 dm³.
- G: V = 0.0400 dm³; n = 0.250 × 0.0400 = 0.0100 mol.
- H: 2.00/0.250 = 8.00 g dm⁻³; M(NaOH) = 40.0; c = 8.00/40.0 = 0.200 mol dm⁻³.
- I: (5.12/6.40) × 100 = 80.0%.
- J: n(CaCO₃) = 10.0/100.0 = 0.100 mol; theoretical V(CO₂) = 2.40 dm³; actual V = 0.750 × 2.40 = 1.80 dm³.
Keep working values until the end and round final answers appropriately to the precision of the data. For counting atoms, also include the number of those atoms in each molecule or formula unit.
Common misconceptions
- “One mole always has a mass of 1 g.” Its mass is the substance’s molar mass.
- “Mᵣ has units g mol⁻¹.” Mᵣ is relative and unitless; M has units g mol⁻¹.
- “Moles of O₂ equal moles of O atoms.” Each molecule contains two atoms.
- “Equal equation coefficients mean equal masses.” They mean equal mole amounts; molar masses may differ.
- “The lighter reactant is limiting.” Compare moles with the equation coefficients.
- “24.0 is the volume of a mole of every substance at all conditions.” It is the specified r.t.p. molar gas volume approximation.
- “Use cm³ directly with mol dm⁻³ in n = cV.” Convert volume to dm³ first.
- “Yield is actual product mass divided by reactant mass.” Divide by the theoretical yield of that same product.
Exam tips
- Write the balanced equation before using a reacting mole ratio.
- List known values with units and identify the unknown.
- Convert mg to g and cm³ to dm³ before substitution where needed.
- Write the specified entity: atoms, molecules, ions or formula units.
- Check the limiting reagent whenever quantities of more than one reactant are supplied.
- Show formula, substitution, coefficient ratio and final unit.
- Retain extra digits during intermediate steps; round the final answer sensibly.
- For percentage yield ensure actual and theoretical quantities refer to the same product and conditions.
Check your understanding
Select one option and Check Answer. Read the explanation, then use Try Again to reset.