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Stoichiometry - The mole concept

Grade 11IGCSEChemistry

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 6.02imes10236.02 imes 10^{23} elementary entities (Avogadro's constant, NAN_A).

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Relative atomic mass (ArA_r) is the average mass of naturally occurring atoms of an element on a scale where the 12C^{12}C atom has a mass of exactly 1212 units.

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Relative molecular mass (MrM_r) or relative formula mass is the sum of the relative atomic masses of the elements in a formula.

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Molar mass (MM) is the mass of one mole of a substance, expressed in g/molg/mol. It is numerically equal to the ArA_r or MrM_r.

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Molar gas volume: One mole of any gas occupies a volume of 24extdm324 ext{ dm}^3 (or 24,000extcm324,000 ext{ cm}^3) at room temperature and pressure (r.t.p.).

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The empirical formula is the simplest whole-number ratio of the different atoms or ions in a compound.

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The molecular formula is the actual number of atoms of each element in one molecule of a compound.

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Stoichiometric coefficients in a balanced chemical equation represent the molar ratio in which reactants react and products are formed.

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Percentage yield compares the actual mass of product obtained to the theoretical maximum mass: Actual YieldTheoretical Yield×100%\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%.

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Percentage purity is the mass of the pure substance divided by the total mass of the impure sample, multiplied by 100%100\%.

📐Formulae

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

n=NNAn = \frac{N}{N_A}

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\%

💡Examples

Problem 1:

Calculate the number of moles in 10extg10 ext{ g} of Calcium Carbonate (CaCO3CaCO_3).

Solution:

Mr(CaCO3)=40+12+(3×16)=100extg/molM_r(CaCO_3) = 40 + 12 + (3 \times 16) = 100 ext{ g/mol}. Using n=mMn = \frac{m}{M}, n=10100=0.1extmoln = \frac{10}{100} = 0.1 ext{ mol}.

Explanation:

First find the molar mass by summing the relative atomic masses, then divide the given mass by the molar mass.

Problem 2:

What volume of Hydrogen gas (H2H_2) at r.t.p. is produced when 0.5extmol0.5 ext{ mol} of Magnesium reacts completely with excess Hydrochloric acid? Reaction: Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g).

Solution:

From the equation, 1extmol1 ext{ mol} of MgMg produces 1extmol1 ext{ mol} of H2H_2. Therefore, 0.5extmol0.5 ext{ mol} of MgMg produces 0.5extmol0.5 ext{ mol} of H2H_2. Volume V=n×24=0.5×24=12extdm3V = n \times 24 = 0.5 \times 24 = 12 ext{ dm}^3.

Explanation:

Use the stoichiometric ratio from the balanced equation (1:1) to find the moles of gas, then multiply by the molar volume at r.t.p.

Problem 3:

A compound contains 40.0%40.0\% Carbon, 6.7%6.7\% Hydrogen, and 53.3%53.3\% Oxygen by mass. Determine its empirical formula.

Solution:

  1. Moles of C=40.012=3.33C = \frac{40.0}{12} = 3.33. 2. Moles of H=6.71=6.7H = \frac{6.7}{1} = 6.7. 3. Moles of O=53.316=3.33O = \frac{53.3}{16} = 3.33. Ratio C:H:O=3.333.33:6.73.33:3.333.33=1:2:1C:H:O = \frac{3.33}{3.33} : \frac{6.7}{3.33} : \frac{3.33}{3.33} = 1:2:1. Empirical formula is CH2OCH_2O.

Explanation:

Divide the percentage of each element by its ArA_r to find the molar ratio, then divide by the smallest value to find the simplest whole-number ratio.

The mole concept - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry