Topic 1 · page 4 of 10

Masses from equations

Masses are not in the ratio of the coefficients; moles are. Go into moles, use the equation, come back out, and the same idea turns analysis data into a formula.

15 min read1.71.6

Read first: Relative mass and the mole, Writing equations

Reacting masses 1.7

MethodThree steps for any reacting-mass question

  1. Mass to moles for the substance you know: n = m ÷ M.
  2. Use the equation ratio to find moles of the substance you want.
  3. Moles to mass: m = n × M.

Worked exampleMass of lime from limestone

What mass of calcium oxide forms when 25.0 g of calcium carbonate is heated?

CaCOX3(s)→CaO(s)+COX2(g) \ce{CaCO3(s) -> CaO(s) + CO2(g)}
  1. n(CaCO₃) = 25.0 ÷ 100.1 = 0.2498 mol
  2. Ratio CaCO₃ : CaO = 1 : 1, so n(CaO) = 0.2498 mol
  3. m(CaO) = 0.2498 × 56.1 = 14.0 g

"Vice versa" means running the same three steps from the product back to the reactant.

Worked exampleMass of reactant needed

What mass of iron(III) oxide is needed to make 10.0 g of iron?

FeX2OX3(s)+3 CO(g)→2 Fe(s)+3 COX2(g) \ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)}
  1. n(Fe) = 10.0 ÷ 55.8 = 0.1792 mol
  2. Ratio Fe₂O₃ : Fe = 1 : 2, so n(Fe₂O₃) = 0.1792 ÷ 2 = 0.08961 mol
  3. M(Fe₂O₃) = 2(55.8) + 3(16.0) = 159.6 g mol⁻¹, so m = 0.08961 × 159.6 = 14.3 g

Limiting reactant 1.7

When amounts of two reactants are given, one will usually run out first. This limiting reactant decides how much product forms; the other is in excess.

Worked exampleMagnesium burning in a limited amount of oxygen

4.86 g of magnesium is heated with 4.00 g of oxygen. What mass of magnesium oxide can form?

2 Mg(s)+OX2(g)→2 MgO(s) \ce{2Mg(s) + O2(g) -> 2MgO(s)}
  1. n(Mg) = 4.86 ÷ 24.3 = 0.200 mol. n(O₂) = 4.00 ÷ 32.0 = 0.125 mol.
  2. 0.200 mol Mg needs 0.200 ÷ 2 = 0.100 mol O₂. There is 0.125 mol, so oxygen is in excess and magnesium is limiting.
  3. n(MgO) = n(Mg) = 0.200 mol, so m = 0.200 × 40.3 = 8.06 g

Formulae from experimental data 1.6

Analysis tells you the mass, or the percentage by mass, of each element in a compound. Converting to moles gives the ratio of atoms, which is the empirical formula.

MethodEmpirical formula from masses or percentages

  1. Write the mass of each element. For percentages, assume 100 g, so 40.0% becomes 40.0 g.
  2. Divide each mass by that element's ArA_\text{r} to get moles of atoms.
  3. Divide every answer by the smallest one.
  4. If any result is not close to a whole number, multiply all of them by the same small whole number.

Worked exampleGlucose from percentage composition

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its MrM_\text{r} is 180. Find its empirical and molecular formulae.

C H O
mass in 100 g / g 40.0 6.7 53.3
÷ ArA_\text{r} (moles) 3.33 6.7 3.33
÷ smallest (3.33) 1.00 2.0 1.00

Empirical formula CH₂O, one unit of mass 30.0. Then n = 180 ÷ 30.0 = 6, so the molecular formula is C₆H₁₂O₆.

Worked exampleA metal oxide with a fractional ratio

An oxide of iron is 72.4% iron by mass.

  1. Oxygen = 100 − 72.4 = 27.6 g.
  2. Fe: 72.4 ÷ 55.8 = 1.297 mol. O: 27.6 ÷ 16.0 = 1.725 mol.
  3. ÷ 1.297: Fe = 1.00, O = 1.33.
  4. × 3: Fe = 3, O = 3.99, so the formula is Fe₃O₄.

The data can also arrive as masses from a reaction. For example, 2.70 g of aluminium forms 13.35 g of a chloride: subtract to find the chlorine, 13.35 − 2.70 = 10.65 g, then use the same table.

Worked exampleEmpirical formula from combustion products

0.600 g of a compound containing only carbon, hydrogen and oxygen burns completely to give 1.32 g of CO₂ and 0.720 g of H₂O.

  1. All the carbon ends up in CO₂: n(C) = n(CO₂) = 1.32 ÷ 44.0 = 0.0300 mol, mass of C = 0.0300 × 12.0 = 0.360 g.
  2. Each H₂O holds 2 H: n(H) = 2 × (0.720 ÷ 18.0) = 0.0800 mol, mass of H = 0.0800 g.
  3. Oxygen is what is left: 0.600 − 0.360 − 0.0800 = 0.160 g, so n(O) = 0.160 ÷ 16.0 = 0.0100 mol.
  4. Ratio C : H : O = 0.0300 : 0.0800 : 0.0100 = 3 : 8 : 1, so the empirical formula is C₃H₈O.

Check yourself

  1. What mass of magnesium oxide forms when 4.86 g of magnesium burns completely in oxygen?
  2. What mass of iron can be made from 80.0 g of Fe₂O₃? (Fe₂O₃ + 3CO → 2Fe + 3CO₂)
  3. What mass of calcium carbonate must be heated to make 28.0 g of calcium oxide?
  4. 2.70 g of aluminium reacts with chlorine to form 13.35 g of a chloride with MrM_\text{r} = 267. Find its empirical and molecular formulae.
  5. A hydrocarbon contains 85.7% carbon by mass and has MrM_\text{r} = 56. Find its molecular formula.
  6. 6.54 g of zinc is added to 0.150 mol of hydrochloric acid. Which reactant is limiting, and what mass of zinc chloride can form? (Zn + 2HCl → ZnCl₂ + H₂)

Answers cover these until you have tried the questions

  1. n(Mg) = 4.86 ÷ 24.3 = 0.200 mol = n(MgO). m = 0.200 × 40.3 = 8.06 g
  2. n(Fe₂O₃) = 80.0 ÷ 159.6 = 0.5013 mol. n(Fe) = 2 × 0.5013 = 1.003 mol. m = 1.003 × 55.8 = 55.9 g
  3. n(CaO) = 28.0 ÷ 56.1 = 0.4991 mol = n(CaCO₃). m = 0.4991 × 100.1 = 50.0 g
  4. Chlorine = 13.35 − 2.70 = 10.65 g. Al: 2.70 ÷ 27.0 = 0.100 mol. Cl: 10.65 ÷ 35.5 = 0.300 mol. Ratio 1 : 3, so empirical formula AlCl₃ (mass 133.5). 267 ÷ 133.5 = 2, so molecular formula Al₂Cl₆.
  5. Hydrogen = 100 − 85.7 = 14.3%. C: 85.7 ÷ 12.0 = 7.14. H: 14.3 ÷ 1.0 = 14.3. Ratio 1 : 2, empirical formula CH₂ (mass 14.0). 56 ÷ 14.0 = 4, so molecular formula C₄H₈.
  6. n(Zn) = 6.54 ÷ 65.4 = 0.100 mol, which needs 0.200 mol HCl. Only 0.150 mol is present, so hydrochloric acid is limiting. n(ZnCl₂) = 0.150 ÷ 2 = 0.0750 mol. M(ZnCl₂) = 65.4 + 71.0 = 136.4 g mol⁻¹, so m = 0.0750 × 136.4 = 10.2 g