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.

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

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

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

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

$$M = 40.08 + 12.01 + 3(16.00) = 100.09\mathrm{ g/mol}$$$$N = \frac{m}{M}, \quad n = \frac{N}{N_A}, \quad n = \frac{V}{V_m}$$

At STP ($0^\circ\mathrm{C}$1 atm): $V_m = 22.4\mathrm{ L/mol}$. At room temperature ($25^\circ\mathrm{C}$1 atm): $V_m = 24.0\mathrm{ L/mol}$.

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

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

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: $\mathrm{CH_2\mathrm{O}}$.

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).

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:

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

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 $\mathrm{HCl}$?

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

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.

$$N(\mathrm{MgO)} = n(\mathrm{Mg)} = 0.123\mathrm{ mol}$$ $$M(\mathrm{MgO)} = 0.123 \times 40.3 = 4.96\mathrm{ g}$$\%\mathrm{ yield = \frac{\mathrm{actual yield}{\mathrm{theoretical yield} \times 100\%\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.

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

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?

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

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

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 $\mathrm{HCl}$.

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

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%.

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: $\mathrm{CuSO_4 \cdot 5\mathrm{H_2\mathrm{O}}}$.

Using the ideal gas equation:

$$PV = nRT$$

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

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

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.

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?

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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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 $R = 8.314 \mathrm{ J mol^{-1}\mathrm{K^{-1}}}$.

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.

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

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.

$$n_{\mathrm{total}} = 0.40 + 0.10 = 0.50 \mathrm{ mol}$$ $$x_{\mathrm{N_2}} = 0.40/0.50 = 0.80, \quad x_{\mathrm{O_2}} = 0.10/0.50 = 0.20$$ $$p_{\mathrm{N_2}} = 0.80 \times 120 = 96 \mathrm{ kPa, \quad p_{\mathrm{O_2}}} = 0.20 \times 120 = 24 \mathrm{ kPa}$$

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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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).

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).

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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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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$20.0 \mathrm{ mL}$ 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 $\mathrm{H_2\mathrm{SO_4}}$.

$$\mathrm{H_2\mathrm{SO_4}} + 2\mathrm{NaOH} \to \mathrm{Na_2\mathrm{SO_4}} + 2\mathrm{H_2\mathrm{O}}$$ $$n(\mathrm{NaOH)} = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol}$$ $$n(\mathrm{H_2\mathrm{SO_4)}} = \frac{0.00500}{2} = 0.00250 \mathrm{ mol}$$ $$c(\mathrm{H_2\mathrm{SO_4)}} = \frac{0.00250}{0.0200} = 0.125 \mathrm{ M}$$

$12.0 \mathrm{ g}$ 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 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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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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1. A compound contains 69.9% iron and 30.1% oxygen by mass. Find its empirical formula.
2. Balance: $\mathrm{Al} + \mathrm{O_2} \to \mathrm{Al_2\mathrm{O_3}}$.
3. Find the molar mass of $\mathrm{H_2\mathrm{SO_4}}$.
4. $2.5 \mathrm{ g}$ of $\mathrm{CaCO_3}$ is heated until completely decomposed. Calculate the volume of $\mathrm{CO_2}$ produced at RTP.
5. 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.
6. 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.
7. $6.30 \mathrm{ g}$ 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 $x$.
8. 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.
9. What mass of $\mathrm{AgCl}$ precipitate would be formed when excess $\mathrm{AgNO_3}$ is added to $25.0 \mathrm{ cm^3}$ of $0.150 \mathrm{ M}$ $\mathrm{MgCl_2}$?
10. A gas mixture at $100 \mathrm{ kPa}$ and $298 \mathrm{ K}$ contains $\mathrm{N_2}$, $\mathrm{O_2}$ And $\mathrm{CO_2}$ 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 $10.0 \mathrm{ L}$.
11. 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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$$c_1 V_1 = c_2 V_2$$

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.

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

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

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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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 $n(\text{CaCl}_2) = 0.100\,\text{mol}$.

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

Chemistry explains how atoms combine to form the substances that make up everything around you. Bonding is about how electrons are shared or transferred between atoms, determining properties like melting point and conductivity. Chemical reactions are rearrangements of atoms, where old bonds break and new ones form. The mole concept bridges the atomic world and the laboratory, letting you predict exactly how much product a reaction will produce. Understanding these principles lets you predict the behavior of matter before you even enter the lab.

• 1 Atomic Structure
• 2 Bonding
• 1 Cell
• 2 Ecology