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Stoichiometry - The mole concept

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The mole is the unit for the amount of substance. One mole contains exactly 6.022imes10236.022 imes 10^{23} elementary entities (atoms, molecules, or ions), known as Avogadro's constant (LL or NAN_A).

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Molar mass (MM) is the mass of one mole of a substance, numerically equal to the relative atomic mass (ArA_r) or relative molecular mass (MrM_r) expressed in g/mol\text{g/mol}.

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Molar volume of a gas: At room temperature and pressure (r.t.p.), one mole of any gas occupies approximately 24 dm324\text{ dm}^3 (or 24,000 cm324,000\text{ cm}^3).

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The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in a molecule, calculated as n×(empirical formula)n \times (\text{empirical formula}).

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Stoichiometry involves using the balanced chemical equation to determine the molar ratios of reactants and products.

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The limiting reactant is the substance that is completely consumed in a reaction, determining the maximum amount of product formed. The reactant remaining is in 'excess'.

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Percentage yield compares the actual yield obtained from an experiment to the theoretical yield calculated from the stoichiometry: ActualTheoretical×100%\frac{\text{Actual}}{\text{Theoretical}} \times 100\%.

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Concentration of a solution is often expressed in mol/dm3\text{mol/dm}^3 (molarity) or g/dm3\text{g/dm}^3. To convert cm3\text{cm}^3 to dm3\text{dm}^3, divide by 10001000.

📐Formulae

n=mMn = \frac{m}{M}

n=VVm (where Vm=24 dm3 at r.t.p.)n = \frac{V}{V_m} \text{ (where } V_m = 24\text{ dm}^3 \text{ at r.t.p.)}

c=nVc = \frac{n}{V}

Percentage Yield=(Actual YieldTheoretical Yield)×100%\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%

Percentage Purity=(Mass of pure substanceTotal mass of impure sample)×100%\text{Percentage Purity} = \left( \frac{\text{Mass of pure substance}}{\text{Total mass of impure sample}} \right) \times 100\%

Atom Economy=(Mr of desired productTotal Mr of all reactants)×100%\text{Atom Economy} = \left( \frac{M_r \text{ of desired product}}{\text{Total } M_r \text{ of all reactants}} \right) \times 100\%

💡Examples

Problem 1:

Calculate the mass of CO2CO_2 produced when 10.0 g10.0\text{ g} of calcium carbonate (CaCO3CaCO_3) is completely decomposed by heating. The equation is: CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g).

Solution:

  1. Calculate MrM_r of CaCO3CaCO_3: 40+12+(3×16)=100 g/mol40 + 12 + (3 \times 16) = 100\text{ g/mol}.
  2. Find moles of CaCO3CaCO_3: n=10.0 g100 g/mol=0.1 moln = \frac{10.0\text{ g}}{100\text{ g/mol}} = 0.1\text{ mol}.
  3. Use stoichiometry: Ratio of CaCO3:CO2CaCO_3:CO_2 is 1:11:1, so n(CO2)=0.1 moln(CO_2) = 0.1\text{ mol}.
  4. Convert moles to mass: Mr(CO2)=12+(2×16)=44 g/molM_r(CO_2) = 12 + (2 \times 16) = 44\text{ g/mol}. Mass =0.1 mol×44 g/mol=4.4 g= 0.1\text{ mol} \times 44\text{ g/mol} = 4.4\text{ g}.

Explanation:

First, find the moles of the known substance, use the balanced equation to find the molar ratio, and then convert those moles back into the required units (mass).

Problem 2:

What volume of hydrogen gas (H2H_2) at r.t.p. is produced when 0.46 g0.46\text{ g} of sodium (NaNa) reacts with excess water? 2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g).

Solution:

  1. Moles of NaNa: n=0.46 g23 g/mol=0.02 moln = \frac{0.46\text{ g}}{23\text{ g/mol}} = 0.02\text{ mol}.
  2. Stoichiometric ratio Na:H2Na:H_2 is 2:12:1. Therefore, n(H2)=0.022=0.01 moln(H_2) = \frac{0.02}{2} = 0.01\text{ mol}.
  3. Volume at r.t.p.: V=0.01 mol×24 dm3/mol=0.24 dm3V = 0.01\text{ mol} \times 24\text{ dm}^3/\text{mol} = 0.24\text{ dm}^3 (or 240 cm3240\text{ cm}^3).

Explanation:

Ensure the molar ratio from the balanced equation is applied correctly. Since 22 moles of NaNa produce 11 mole of H2H_2, the moles of H2H_2 are half the moles of NaNa.

Problem 3:

A compound contains 40.0%40.0\% carbon, 6.7%6.7\% hydrogen, and 53.3%53.3\% oxygen by mass. Its relative molecular mass is 180180. Determine its molecular formula.

Solution:

  1. Molar ratio: C=4012=3.33C = \frac{40}{12} = 3.33, H=6.71=6.7H = \frac{6.7}{1} = 6.7, O=53.316=3.33O = \frac{53.3}{16} = 3.33.
  2. Divide by smallest value: C=3.333.33=1C = \frac{3.33}{3.33} = 1, H=6.73.33≈2H = \frac{6.7}{3.33} \approx 2, O=3.333.33=1O = \frac{3.33}{3.33} = 1. Empirical formula =CH2O= CH_2O.
  3. Empirical formula mass: 12+(2×1)+16=3012 + (2 \times 1) + 16 = 30.
  4. Factor n=18030=6n = \frac{180}{30} = 6. Molecular formula =C6H12O6= C_6H_{12}O_6.

Explanation:

The empirical formula gives the ratio of atoms. Dividing the actual MrM_r by the empirical mass gives the multiplier needed to find the molecular formula.

The mole concept - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry