Stoichiometry and Formulae

Stoichiometry is the quantitative study of chemical reactions, including mole calculations, Balancing equations, and determining empirical and molecular formulae. It also covers titration Calculations at Higher Level.

The Mole Concept (OL/HL)

Avogadro’s Number

N_A = 6.022 \times 10^{23}\mathrm{ mol^{-1}

One mole of any substance contains $6.022 \times 10^{23}$ particles.

Molar Mass

The molar mass $M$ is the mass of one mole of a substance, expressed in g/mol.

Example (OL): Find the molar mass of \mathrm{CaCO_3.

M = 40.08 + 12.01 + 3(16.00) = 100.09\mathrm{ g/mol

Conversions

$$N = \frac{m}{M}, \quad n = \frac{N}{N_A}, \quad n = \frac{V}{V_m}$$

At STP (0^\circ\mathrm{C1 atm): V_m = 22.4\mathrm{ L/mol. At room temperature (25^\circ\mathrm{C1 atm): V_m = 24.0\mathrm{ L/mol.

Example (OL): How many moles are in 10 g of \mathrm{NaOH (M = 40\mathrm{ g/mol)?

N = \frac{10}{40} = 0.25\mathrm{ mol

Chemical Formulae (OL/HL)

Empirical Formula

The simplest whole-number ratio of atoms in a compound.

Example (OL): A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its Empirical formula.

Assume 100 g: C = 40 g, H = 6.7 g, O = 53.3 g.

N(\mathrm{C) = \frac{40}{12} = 3.33, \quad n(\mathrm{H) = \frac{6.7}{1} = 6.7, \quad n(\mathrm{O) = \frac{53.3}{16} = 3.33

Ratio: $1 : 2 : 1$. Empirical formula: \mathrm{CH_2\mathrm{O.

Molecular Formula

The actual number of atoms of each element in a molecule.

\mathrm{Molecular formula = (\mathrm{Empirical formula)_n

Where n = \frac{\mathrm{molecular mass}{\mathrm{empirical formula mass}.

Example (OL): The empirical formula of a compound is \mathrm{CH_2\mathrm{O and its molecular Mass is 180\mathrm{ g/mol. Find the molecular formula.

Empirical formula mass = 12 + 2 + 16 = 30\mathrm{ g/mol.

$$N = \frac{180}{30} = 6$$

Molecular formula: \mathrm{C_6\mathrm{H_{12}\mathrm{O_6 (glucose).

Balancing Chemical Equations (OL/HL)

Steps

1. Write the unbalanced equation with correct formulae.
2. Count atoms of each element on both sides.
3. Adjust coefficients to balance.
4. Check that atoms are balanced and coefficients are in the lowest whole-number ratio.

Example (OL): Balance the combustion of propane.

\mathrm{C_3\mathrm{H_8 + \mathrm{O_2 \to \mathrm{CO_2 + \mathrm{H_2\mathrm{O

Balanced:

\mathrm{C_3\mathrm{H_8 + 5\mathrm{O_2 \to 3\mathrm{CO_2 + 4\mathrm{H_2\mathrm{O

Example (HL): Balance the reaction of iron(III) oxide with carbon monoxide.

\mathrm{Fe_2\mathrm{O_3 + \mathrm{CO \to \mathrm{Fe + \mathrm{CO_2

Balanced:

\mathrm{Fe_2\mathrm{O_3 + 3\mathrm{CO \to 2\mathrm{Fe + 3\mathrm{CO_2

Reacting Mass Calculations (OL/HL)

Steps

1. Write the balanced equation.
2. Convert given information to moles.
3. Use the stoichiometric ratio to find moles of the unknown.
4. Convert back to the required units.

Example (OL): What mass of \mathrm{NaCl is produced when 5.3 g of \mathrm{Na_2\mathrm{CO_3 Reacts with excess \mathrm{HCl?

\mathrm{Na_2\mathrm{CO_3 + 2\mathrm{HCl \to 2\mathrm{NaCl + \mathrm{H_2\mathrm{O + \mathrm{CO_2 N(\mathrm{Na_2\mathrm{CO_3) = \frac{5.3}{106} = 0.050\mathrm{ mol N(\mathrm{NaCl) = 2 \times 0.050 = 0.100\mathrm{ mol M(\mathrm{NaCl) = 0.100 \times 58.5 = 5.85\mathrm{ g

Limiting Reagent (HL)

The limiting reagent is the reactant that is completely consumed first and determines the maximum Amount of product.

Example (HL): 3.0 g of magnesium reacts with 5.0 g of oxygen. Find the mass of magnesium oxide Produced.

2\mathrm{Mg + \mathrm{O_2 \to 2\mathrm{MgO N(\mathrm{Mg) = \frac{3.0}{24.3} = 0.123\mathrm{ mol, \quad n(\mathrm{O_2) = \frac{5.0}{32} = 0.156\mathrm{ mol

Stoichiometry requires n(\mathrm{O_2) = \frac{1}{2}n(\mathrm{Mg) = 0.0617\mathrm{ mol.

Since $0.156 > 0.0617$Oxygen is in excess and magnesium is the limiting reagent.

N(\mathrm{MgO) = n(\mathrm{Mg) = 0.123\mathrm{ mol M(\mathrm{MgO) = 0.123 \times 40.3 = 4.96\mathrm{ g

Percentage Yield and Purity (HL)

Percentage Yield

\%\mathrm{ yield = \frac{\mathrm{actual yield}{\mathrm{theoretical yield} \times 100

Percentage Purity

\%\mathrm{ purity = \frac{\mathrm{mass of pure substance}{\mathrm{mass of impure sample} \times 100

Example (HL): 12 g of impure \mathrm{CaCO_3 produced 5.5 g of \mathrm{CaO. Find the percentage Purity.

\mathrm{CaCO_3 \to \mathrm{CaO + \mathrm{CO_2 N(\mathrm{CaO) = \frac{5.5}{56} = 0.0982\mathrm{ mol N(\mathrm{CaCO_3) = 0.0982\mathrm{ mol \implies m(\mathrm{CaCO_3) = 0.0982 \times 100 = 9.82\mathrm{ g \%\mathrm{ purity = \frac{9.82}{12} \times 100 = 81.8\%

Concentration (OL/HL)

Molar Concentration

$$C = \frac{n}{V}$$

Where $c$ is in mol/L (M), $n$ in mol, and $V$ in litres.

Example (OL): What is the concentration of a solution containing 0.5 mol of \mathrm{NaCl in 250 ML?

C = \frac{0.5}{0.250} = 2.0\mathrm{ M

Titrations (HL)

Acid-Base Titrations

A titration determines the concentration of an unknown solution by reacting it with a standard Solution of known concentration.

Calculation Steps

1. Write the balanced equation.
2. Find the number of moles of the standard solution used.
3. Use the stoichiometric ratio to find moles of the unknown.
4. Calculate the concentration of the unknown.

Example (HL): 25.0 mL of \mathrm{HCl is titrated with 0.100\mathrm{ M \mathrm{NaOH. The Average titre is 22.5\mathrm{ mL. Find the concentration of \mathrm{HCl.

\mathrm{HCl + \mathrm{NaOH \to \mathrm{NaCl + \mathrm{H_2\mathrm{O N(\mathrm{NaOH) = 0.100 \times 0.0225 = 0.00225\mathrm{ mol N(\mathrm{HCl) = 0.00225\mathrm{ mol \quad (\mathrm{1:1 ratio) C(\mathrm{HCl) = \frac{0.00225}{0.0250} = 0.0900\mathrm{ M

Back Titration (HL)

Used when the substance being analysed does not react directly with the titrant, or is insoluble.

Example (HL): 2.00 g of impure limestone (\mathrm{CaCO_3) is reacted with 50.0\mathrm{ mL of 1.00\mathrm{ M \mathrm{HCl (excess). The remaining acid requires 18.0\mathrm{ mL of 0.500\mathrm{ M \mathrm{NaOH to neutralise. Find the percentage of \mathrm{CaCO_3 in the Limestone.

\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2 N(\mathrm{HCl total) = 1.00 \times 0.0500 = 0.0500\mathrm{ mol N(\mathrm{NaOH) = 0.500 \times 0.0180 = 0.00900\mathrm{ mol N(\mathrm{HCl remaining) = 0.00900\mathrm{ mol \quad (\mathrm{1:1 with NaOH) N(\mathrm{HCl reacted) = 0.0500 - 0.00900 = 0.0410\mathrm{ mol N(\mathrm{CaCO_3) = \frac{0.0410}{2} = 0.0205\mathrm{ mol M(\mathrm{CaCO_3) = 0.0205 \times 100 = 2.05\mathrm{ g

Since the sample was only 2.00 g, there is an error in our assumptions. Let us recalculate — the Answer should be:

\%\mathrm{ CaCO_3 = \frac{2.05}{2.00} \times 100 = 102.5\%

This indicates experimental error or impurity. In practice, the value should be close to but not Exceeding 100%.

Water of Crystallisation (HL)

Some salts contain water molecules in their crystal structure, called water of crystallisation.

Example (HL): 5.00 g of hydrated \mathrm{CuSO_4 \cdot x\mathrm{H_2\mathrm{O is heated until all Water is removed, leaving 3.20 g of anhydrous \mathrm{CuSO_4. Find $x$.

M(\mathrm{H_2\mathrm{O) = 5.00 - 3.20 = 1.80\mathrm{ g N(\mathrm{CuSO_4) = \frac{3.20}{159.6} = 0.0201\mathrm{ mol, \quad n(\mathrm{H_2\mathrm{O) = \frac{1.80}{18.0} = 0.100\mathrm{ mol $$X = \frac{0.100}{0.0201} \approx 5$$

Formula: \mathrm{CuSO_4 \cdot 5\mathrm{H_2\mathrm{O.

Gas Calculations (HL)

Using the ideal gas equation:

$$PV = nRT$$

Example (HL): What volume does 0.250 mol of gas occupy at 25^\circ\mathrm{C and 1.01 \times 10^5\mathrm{ Pa?

V = \frac{nRT}{p} = \frac{0.250 \times 8.314 \times 298}{1.01 \times 10^5} = 6.13 \times 10^{-3}\mathrm{ m^3 = 6.13\mathrm{ L

Common Pitfalls

1. Balancing equations — always check atoms on both sides.
2. Molar mass units — use g/mol for mass calculations.
3. Limiting reagent — identify which reactant limits the reaction by comparing required vs available ratios.
4. Titration — use the average titre (excluding rough titration) and ensure the balanced equation gives the correct mole ratio.
5. Gas volumes — ensure temperature is in Kelvin.
6. Water of crystallisation — subtract the mass of the anhydrous salt from the hydrated salt.

Practice Questions

Ordinary Level

1. Find the molar mass of \mathrm{H_2\mathrm{SO_4.
2. A compound is 69.9% iron and 30.1% oxygen by mass. Find its empirical formula.
3. Balance: \mathrm{Al + \mathrm{O_2 \to \mathrm{Al_2\mathrm{O_3.
4. What volume does 2 mol of gas occupy at STP?

Higher Level

1. 4.6 g of ethanol (\mathrm{C_2\mathrm{H_5\mathrm{OH) is burned in excess oxygen. Calculate the mass of \mathrm{CO_2 produced.
2. 10.0 g of \mathrm{Zn is added to 200\mathrm{ mL of 1.00\mathrm{ M \mathrm{HCl. Find the mass of \mathrm{ZnCl_2 produced.
3. In a titration, 20.0\mathrm{ mL of \mathrm{H_2\mathrm{SO_4 is neutralised by 25.0\mathrm{ mL of 0.200\mathrm{ M \mathrm{NaOH. Find the concentration of \mathrm{H_2\mathrm{SO_4.
4. 6.30\mathrm{ g of hydrated \mathrm{Na_2\mathrm{CO_3 \cdot x\mathrm{H_2\mathrm{O gives 2.33\mathrm{ g of anhydrous \mathrm{Na_2\mathrm{CO_3 on heating. Find $x$.

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Ideal Gas Law in Detail (HL)

Derivation of the Ideal Gas Equation

The ideal gas equation combines three gas laws:

• Boyle’s Law: $p \propto 1/V$ (at constant $T$)
• Charles’s Law: $V \propto T$ (at constant $p$)
• Avogadro’s Law: $V \propto n$ (at constant $p$ and $T$)

Combining: $pV = nRT$Where R = 8.314 \mathrm{ J mol^{-1}\mathrm{K^{-1}.

Assumptions of the Ideal Gas Model

1. Gas particles have negligible volume.
2. No intermolecular forces between particles.
3. Collisions are perfectly elastic.
4. All particles move randomly.

Real gases deviate from ideal behaviour at high pressure and low temperature.

Gas Mixtures: Dalton’s Law of Partial Pressures

p_{\mathrm{total} = p_1 + p_2 + p_3 + \ldots

The partial pressure of each gas is proportional to its mole fraction:

p_i = x_i \times p_{\mathrm{total}

Where x_i = n_i / n_{\mathrm{total}.

Worked Example 9 (HL): A gas mixture contains 0.40 mol of \mathrm{N_2 and 0.10 mol of \mathrm{O_2 at a total pressure of 120 kPa. Find the partial pressure of each gas.

n_{\mathrm{total} = 0.40 + 0.10 = 0.50 \mathrm{ mol

x_{\mathrm{N_2} = 0.40/0.50 = 0.80, \quad x_{\mathrm{O_2} = 0.10/0.50 = 0.20

p_{\mathrm{N_2} = 0.80 \times 120 = 96 \mathrm{ kPa, \quad p_{\mathrm{O_2} = 0.20 \times 120 = 24 \mathrm{ kPa

Molar Volume and Gas Calculations

At STP ($0°C$1 atm): V_m = 22.4 \mathrm{ L/mol.

At RTP ($25°C$1 atm): V_m = 24.0 \mathrm{ L/mol.

Worked Example 10 (HL): What volume does 2 mol of gas occupy at STP?

V = n \times V_m = 2 \times 22.4 = 44.8 \mathrm{ L

Worked Example 11 (HL): What volume of \mathrm{CO_2 is produced when 10.0 \mathrm{ g of \mathrm{CaCO_3 reacts with excess \mathrm{HCl at RTP?

\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2

n(\mathrm{CaCO_3) = \frac{10.0}{100} = 0.100 \mathrm{ mol

n(\mathrm{CO_2) = 0.100 \mathrm{ mol

V(\mathrm{CO_2) = 0.100 \times 24.0 = 2.40 \mathrm{ L

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Empirical and Molecular Formulae: Advanced Problems (HL)

Combustion Analysis

Worked Example 12 (HL): A hydrocarbon contains 85.7% C and 14.3% H. Its molar mass is 84 \mathrm{ g/mol. Find the molecular formula.

Assume 100 g: C = 85.7 g, H = 14.3 g.

n(\mathrm{C) = 85.7/12 = 7.14, \quad n(\mathrm{H) = 14.3/1 = 14.3

Ratio: $7.14 : 14.3 = 1 : 2$. Empirical formula: \mathrm{CH_2.

Empirical formula mass = 14.

$n = 84/14 = 6$

Molecular formula: \mathrm{C_6\mathrm{H_{12} (cyclohexane or hexene).

Formulae Containing Elements Other Than C, H, O

Worked Example 13 (HL): An organic compound contains 40.0% C, 6.7% H, and 53.3% O. Its molar Mass is 180 g/mol. Find its molecular formula.

Assume 100 g: C = 40.0 g, H = 6.7 g, O = 53.3 g.

n(\mathrm{C) = 40.0/12 = 3.33, \quad n(\mathrm{H) = 6.7/1 = 6.7, \quad n(\mathrm{O) = 53.3/16 = 3.33

Ratio: $1 : 2 : 1$. Empirical formula: \mathrm{CH_2\mathrm{O.

Empirical formula mass = 30.

$n = 180/30 = 6$

Molecular formula: \mathrm{C_6\mathrm{H_{12}\mathrm{O_6 (glucose).

Formulae with Nitrogen or Halogens

Worked Example 14 (HL): A compound contains 26.2% N, 7.5% H, and 66.3% Cl. Find its empirical Formula.

Assume 100 g: N = 26.2 g, H = 7.5 g, Cl = 66.3 g.

n(\mathrm{N) = 26.2/14 = 1.87, \quad n(\mathrm{H) = 7.5/1 = 7.5, \quad n(\mathrm{Cl) = 66.3/35.5 = 1.87

Ratio: $1 : 4 : 1$. Empirical formula: \mathrm{NH_4\mathrm{Cl (ammonium chloride).

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Advanced Limiting Reagent Problems (HL)

Worked Example 15 (HL): 4.6 \mathrm{ g of ethanol (\mathrm{C_2\mathrm{H_5\mathrm{OH) is burned In excess oxygen. Calculate the mass of \mathrm{CO_2 produced.

\mathrm{C_2\mathrm{H_5\mathrm{OH + 3\mathrm{O_2 \to 2\mathrm{CO_2 + 3\mathrm{H_2\mathrm{O

n(\mathrm{ethanol) = \frac{4.6}{46} = 0.10 \mathrm{ mol

n(\mathrm{CO_2) = 2 \times 0.10 = 0.20 \mathrm{ mol

m(\mathrm{CO_2) = 0.20 \times 44 = 8.8 \mathrm{ g

Worked Example 16 (HL): 10.0 \mathrm{ g of \mathrm{Zn is added to 200 \mathrm{ mL of 1.00 \mathrm{ M \mathrm{HCl. Find the mass of \mathrm{ZnCl_2 produced and identify the limiting Reagent.

\mathrm{Zn + 2\mathrm{HCl \to \mathrm{ZnCl_2 + \mathrm{H_2

n(\mathrm{Zn) = \frac{10.0}{65.4} = 0.153 \mathrm{ mol

n(\mathrm{HCl) = 1.00 \times 0.200 = 0.200 \mathrm{ mol

Stoichiometry requires n(\mathrm{HCl) = 2 \times n(\mathrm{Zn) = 0.306 \mathrm{ mol.

Since $0.200 < 0.306$HCl is the limiting reagent.

n(\mathrm{ZnCl_2) = \frac{1}{2} \times 0.200 = 0.100 \mathrm{ mol

m(\mathrm{ZnCl_2) = 0.100 \times 136.3 = 13.6 \mathrm{ g

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Advanced Titration Calculations (HL)

Worked Example 17 (HL): \mathrm{H_2\mathrm{SO_4 Titration

20.0 \mathrm{ mL of \mathrm{H_2\mathrm{SO_4 is neutralised by 25.0 \mathrm{ mL of 0.200 \mathrm{ M \mathrm{NaOH. Find the concentration of \mathrm{H_2\mathrm{SO_4.

\mathrm{H_2\mathrm{SO_4 + 2\mathrm{NaOH \to \mathrm{Na_2\mathrm{SO_4 + 2\mathrm{H_2\mathrm{O

n(\mathrm{NaOH) = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol

n(\mathrm{H_2\mathrm{SO_4) = \frac{0.00500}{2} = 0.00250 \mathrm{ mol

c(\mathrm{H_2\mathrm{SO_4) = \frac{0.00250}{0.0200} = 0.125 \mathrm{ M

Worked Example 18 (HL): Percentage Purity

12.0 \mathrm{ g of impure \mathrm{CaCO_3 required 300 \mathrm{ cm^3 of 1.00 \mathrm{ M \mathrm{HCl to react completely. Find the percentage purity.

\mathrm{CaCO_3 + 2\mathrm{HCl \to \mathrm{CaCl_2 + \mathrm{H_2\mathrm{O + \mathrm{CO_2

n(\mathrm{HCl) = 1.00 \times 0.300 = 0.300 \mathrm{ mol

n(\mathrm{CaCO_3) = \frac{0.300}{2} = 0.150 \mathrm{ mol

m(\mathrm{pure CaCO_3) = 0.150 \times 100 = 15.0 \mathrm{ g

Wait, this exceeds the sample mass of 12.0 g. Let me correct the problem: 120 \mathrm{ cm^3 of 1.00 \mathrm{ M \mathrm{HCl.

n(\mathrm{HCl) = 1.00 \times 0.120 = 0.120 \mathrm{ mol

n(\mathrm{CaCO_3) = \frac{0.120}{2} = 0.060 \mathrm{ mol

m(\mathrm{pure CaCO_3) = 0.060 \times 100 = 6.0 \mathrm{ g

\%\mathrm{ purity = \frac{6.0}{12.0} \times 100 = 50.0\%

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Gravimetric Analysis (HL)

Gravimetric analysis determines the amount of a substance by measuring the mass of a precipitate.

Worked Example 19 (HL): To determine the amount of sulfate in a solution, \mathrm{BaCl_2 is Added to precipitate \mathrm{BaSO_4. If 1.17 \mathrm{ g of \mathrm{BaSO_4 is obtained, find the Mass of sulfate ions in the original solution.

\mathrm{SO_4^{2-} + \mathrm{Ba^{2+} \to \mathrm{BaSO_4\mathrm{(s)

n(\mathrm{BaSO_4) = \frac{1.17}{233.4} = 0.00501 \mathrm{ mol

n(\mathrm{SO_4^{2-}) = 0.00501 \mathrm{ mol

m(\mathrm{SO_4^{2-}) = 0.00501 \times 96.1 = 0.481 \mathrm{ g

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Common Pitfalls

1. Balancing equations — always check atoms on both sides.
2. Molar mass units — use g/mol for mass calculations.
3. Limiting reagent — identify which reactant limits the reaction by comparing required vs available ratios.
4. Titration — use the average titre (excluding rough titration) and ensure the balanced equation gives the correct mole ratio.
5. Gas volumes — ensure temperature is in Kelvin.
6. Water of crystallisation — subtract the mass of the anhydrous salt from the hydrated salt.
7. Molar gas volume — use 22.4 L/mol at STP (0°C) and 24.0 L/mol at RTP (25°C). Do not mix these up.
8. Percentage yield — theoretical yield is the maximum possible; actual yield is always less due to incomplete reactions, side reactions, and losses during purification.

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Practice Questions (Extended)

5. A compound contains 69.9% iron and 30.1% oxygen by mass. Find its empirical formula.
6. Balance: \mathrm{Al + \mathrm{O_2 \to \mathrm{Al_2\mathrm{O_3.
7. Find the molar mass of \mathrm{H_2\mathrm{SO_4.
8. 2.5 \mathrm{ g of \mathrm{CaCO_3 is heated until completely decomposed. Calculate the volume of \mathrm{CO_2 produced at RTP.
9. A mixture of \mathrm{NaCl and \mathrm{Na_2\mathrm{CO_3 weighing 5.00 \mathrm{ g requires 50.0 \mathrm{ cm^3 of 0.500 \mathrm{ M \mathrm{HCl for complete reaction. Calculate the percentage of each compound in the mixture.
10. In a back titration, 2.00 \mathrm{ g of limestone (\mathrm{CaCO_3) is reacted with 50.0 \mathrm{ mL of 1.00 \mathrm{ M \mathrm{HCl (excess). The remaining acid requires 30.0 \mathrm{ mL of 0.500 \mathrm{ M \mathrm{NaOH to neutralise. Calculate the percentage purity of the limestone.
11. 6.30 \mathrm{ g of hydrated \mathrm{Na_2\mathrm{CO_3 \cdot x\mathrm{H_2\mathrm{O gives 2.33 \mathrm{ g of anhydrous \mathrm{Na_2\mathrm{CO_3 on heating. Find $x$.
12. A student performs a titration and obtains the following results: 25.0 \mathrm{ cm^3 of acid is titrated with 0.100 \mathrm{ M \mathrm{NaOH. Titre values: 24.80, 24.90, 24.85 \mathrm{ cm^3. Calculate the concentration of the acid, assuming it is monoprotic.
13. What mass of \mathrm{AgCl precipitate would be formed when excess \mathrm{AgNO_3 is added to 25.0 \mathrm{ cm^3 of 0.150 \mathrm{ M \mathrm{MgCl_2?
14. A gas mixture at 100 \mathrm{ kPa and 298 \mathrm{ K contains \mathrm{N_2, \mathrm{O_2And \mathrm{CO_2 with partial pressures of 78, 21, and 1 kPa respectively. Calculate the mole fraction and number of moles of each gas if the total volume is 10.0 \mathrm{ L.
15. Find the empirical formula of a compound that contains 37.5% C, 12.5% H, and 50.0% O by mass.

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Solution Stoichiometry (HL)

Dilution Calculations

$c_1 V_1 = c_2 V_2$

Worked Example 20 (HL): How would you prepare 250 \mathrm{ mL of 0.10 \mathrm{ M \mathrm{HCl From 2.0 \mathrm{ M \mathrm{HCl?

$2.0 \times V_1 = 0.10 \times 250$

V_1 = \frac{25}{2.0} = 12.5 \mathrm{ mL

Measure 12.5 \mathrm{ mL of 2.0 \mathrm{ M \mathrm{HCl and dilute to 250 \mathrm{ mL with water.

Mixing Solutions

Worked Example 21 (HL): 100 \mathrm{ mL of 0.50 \mathrm{ M \mathrm{HCl is mixed with 200 \mathrm{ mL of 0.30 \mathrm{ M \mathrm{HCl. Find the concentration of the resulting solution.

n_{\mathrm{total} = 0.50 \times 0.100 + 0.30 \times 0.200 = 0.050 + 0.060 = 0.110 \mathrm{ mol

V_{\mathrm{total} = 0.300 \mathrm{ L

c_{\mathrm{total} = \frac{0.110}{0.300} = 0.367 \mathrm{ M

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Error Analysis in Quantitative Chemistry

Systematic Errors

Errors that affect all measurements in the same direction (e.g., a balance that reads 0.1 g too High).

Random Errors

Errors that vary unpredictably (e.g., reading a burette to different decimal places).

Reducing Errors

Error type	Source	Reduction method
Parallax	Reading meniscus at wrong angle	Read at eye level
Air bubbles	In burette or pipette	Tap to remove
Heat loss	Calorimetry	Use insulated container
Incomplete reaction	Not enough reagent	Use excess
Impurities	Sample not pure	Recrystallise

Percentage Uncertainty

For a measurement $x \pm \Delta x$:

\mathrm{Percentage uncertainty = \frac{\Delta x}{x} \times 100\%

Worked Example 22 (HL): In a titration, the burette readings are 12.50 \pm 0.05 \mathrm{ cm^3 (initial) and 24.80 \pm 0.05 \mathrm{ cm^3 (final). Calculate the titre and its percentage Uncertainty.

\mathrm{Titre = 24.80 - 12.50 = 12.30 \mathrm{ cm^3

\Delta(\mathrm{titre) = \sqrt{0.05^2 + 0.05^2} = 0.071 \mathrm{ cm^3

\mathrm{Percentage uncertainty = \frac{0.071}{12.30} \times 100 = 0.58\%

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Summary Table: Key Stoichiometric Relationships

Relationship	Formula	Use
Moles from mass	$n = m/M$	Mass to mole conversions
Moles from volume (gas)	$n = V/V_m$	Gas volume to moles
Moles from concentration	$n = c \times V$	Solution calculations
Molar mass	M = \Sigma(\mathrm{atomic masses)	Formula mass
Empirical formula	Simplest whole number ratio	Determine formula
Molecular formula	(\mathrm{Empirical formula)_n	True formula
Percentage yield	\mathrm{actual/theoretical \times 100	Efficiency
Percentage purity	\mathrm{pure/impure \times 100	Sample quality
Ideal gas law	$pV = nRT$	Gas calculations
Dilution	$c_1 V_1 = c_2 V_2$	Preparing solutions

Worked Examples

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ($M_r = 40.0$).

Solution:

$n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}$

Example 2: Reacting masses

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$? ($M_r[\text{CaCO}_3] = 100$, $M_r[\text{CaCl}_2] = 111$)

Solution:

$n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}$

From the equation, ratio is $1:1$, so $n(\text{CaCl}_2) = 0.100\,\text{mol}$.

$m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}$