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Chemistry - Stoichiometry

Grade 10IGCSE

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

🔑Concepts

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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) is the sum of the relative atomic masses of the elements in a formula.

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The Mole (molmol) is the unit for the amount of substance. One mole of any substance contains Avogadro's constant (6.02×10236.02 \times 10^{23}) of particles.

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Molar Volume: One mole of any gas at room temperature and pressure (r.t.p.) occupies a volume of 24 dm324\ dm^3 or 24,000 cm324,000\ cm^3.

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Concentration of a solution is often expressed in mol/dm3mol/dm^3 or g/dm3g/dm^3.

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Empirical Formula is the simplest whole-number ratio of atoms of each element in a compound, whereas the Molecular Formula shows the actual number of atoms.

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Stoichiometry involves using the balanced chemical equation to determine the reacting ratios (mole ratios) of reactants and products.

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Limiting Reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product formed.

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

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Percentage Purity is calculated as Mass of pure substanceTotal mass of impure sample×100%\frac{\text{Mass of pure substance}}{\text{Total mass of impure sample}} \times 100\%.

📐Formulae

n=mMrn = \frac{m}{M_r}

n=V24 dm3 (for gases at r.t.p.)n = \frac{V}{24\ dm^3} \text{ (for gases at r.t.p.)}

n=C×V (where V is in dm3)n = C \times V \text{ (where } V \text{ is in } dm^3)

Concentration (g/dm3)=Concentration (mol/dm3)×Mr\text{Concentration } (g/dm^3) = \text{Concentration } (mol/dm^3) \times M_r

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

💡Examples

Problem 1:

Calculate the mass of magnesium oxide (MgOMgO) formed when 2.4 g2.4\ g of magnesium (MgMg) reacts completely with oxygen (O2O_2). The equation is: 2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO. (Ar:Mg=24,O=16A_r: Mg = 24, O = 16)

Solution:

n(Mg)=2.4 g24 g/mol=0.1 moln(Mg) = \frac{2.4\ g}{24\ g/mol} = 0.1\ mol. According to the ratio 2Mg:2MgO2Mg:2MgO (or 1:11:1), n(MgO)=0.1 moln(MgO) = 0.1\ mol. Mass of MgO=0.1 mol×(24+16) g/mol=4.0 gMgO = 0.1\ mol \times (24 + 16)\ g/mol = 4.0\ g.

Explanation:

First, convert the given mass of the known reactant to moles. Use the stoichiometric ratio from the balanced equation to find the moles of the product, then convert those moles back into mass.

Problem 2:

A compound contains 40%40\% carbon, 6.67%6.67\% hydrogen, and 53.33%53.33\% oxygen by mass. Find its empirical formula. (Ar:C=12,H=1,O=16A_r: C = 12, H = 1, O = 16)

Solution:

n(C)=4012=3.33n(C) = \frac{40}{12} = 3.33, n(H)=6.671=6.67n(H) = \frac{6.67}{1} = 6.67, n(O)=53.3316=3.33n(O) = \frac{53.33}{16} = 3.33. Dividing by the smallest value (3.333.33): C=3.333.33=1C = \frac{3.33}{3.33} = 1, H=6.673.33=2H = \frac{6.67}{3.33} = 2, O=3.333.33=1O = \frac{3.33}{3.33} = 1. The empirical formula is CH2OCH_2O.

Explanation:

Divide the percentage of each element by its ArA_r to find the molar ratio. Divide all results by the smallest number of moles to obtain the simplest whole-number ratio.

Problem 3:

What volume of hydrogen gas (H2H_2) at r.t.p. is produced when 0.5 mol0.5\ mol of zinc reacts with excess sulfuric acid (H2SO4H_2SO_4)? Equation: Zn+H2SO4→ZnSO4+H2Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2.

Solution:

From the equation, 1 mol1\ mol of ZnZn produces 1 mol1\ mol of H2H_2. Therefore, 0.5 mol0.5\ mol of ZnZn produces 0.5 mol0.5\ mol of H2H_2. Volume of H2=0.5 mol×24 dm3/mol=12 dm3H_2 = 0.5\ mol \times 24\ dm^3/mol = 12\ dm^3.

Explanation:

Use the mole ratio from the balanced equation to find the moles of gas produced. Multiply the moles of gas by the molar gas volume (24 dm324\ dm^3 at r.t.p.).

Stoichiometry - Revision Notes & Key Formulas | IGCSE Grade 10 Science