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Introduction to quantitative chemistry review
Review of Year 11 Chemistry Module 2: a summary of every topic, the key formulas and data sheet values, a worked multi-step problem that links mass, solution and gas calculations, key words, and a mixed HSC-style test.
Module 2 is a toolkit more than a topic. Almost every calculation in Modules 3 to 8 starts by turning something into moles: a mass, a volume of solution or a volume of gas. Use this page to check the tools are sharp before you move on.
Module summary
| Topic | The big ideas |
|---|---|
| 2A Equations and conservation of mass | Coefficients give the mole ratio. Balance with coefficients, never subscripts. Mass is conserved; in an open system a mass change equals the mass of gas lost or gained. |
| 2B The mole | 1 mol = 6.022 × 10²³ entities exactly (2019 SI). n = m ÷ M; N = n × NA. Always state the entity. |
| 2C Formulae from data | Law of definite proportions. % composition. Masses → moles → ÷ smallest → whole numbers. Molecular = empirical × k. Heat to constant mass. Combustion analysis: C from CO₂, H from H₂O, O by difference. |
| 2D Limiting reagents and yield | Limiting reagent has the smallest moles ÷ coefficient. Theoretical yield from the limiting reagent; % yield = actual ÷ theoretical × 100. |
| 2E Concentration units | g L⁻¹, %(w/w), %(w/v), %(v/v), ppm (mg kg⁻¹ ≈ mg L⁻¹), ppb, mol L⁻¹ = g L⁻¹ ÷ M. |
| 2F Molarity and dilution | c = n ÷ V (litres). Standard solutions from primary standards in volumetric flasks. Dilution: c₁V₁ = c₂V₂. |
| 2G Stoichiometry in solution | n = cV, then the mole ratio. Leftover ions are in the combined volume. Continuous variation finds the mole ratio. |
| 2H Gas laws | Kelvin always. Boyle P₁V₁ = P₂V₂; Charles V₁/T₁ = V₂/T₂; Gay-Lussac P₁/T₁ = P₂/T₂; combined P₁V₁/T₁ = P₂V₂/T₂. Explained by kinetic particle theory. |
| 2I Avogadro and PV = nRT | Equal volumes, equal numbers of particles. Vm = 22.71 L (0 °C) or 24.79 L (25 °C) at 100 kPa. PV = nRT with R = 8.314. Real gases deviate at high P and low T. |
Key formulas and data sheet values
| Relationship | Formula | Units |
|---|---|---|
| moles from mass | $n = \dfrac{m}{M}$ | g, g mol⁻¹ |
| number of particles | $N = nN_A$ | $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ |
| concentration | $c = \dfrac{n}{V}$ | mol L⁻¹, L |
| dilution | $c_1V_1 = c_2V_2$ | same volume unit both sides |
| molar volume | $n = \dfrac{V}{V_m}$ | 24.79 L mol⁻¹ (25 °C), 22.71 L mol⁻¹ (0 °C), 100 kPa |
| combined gas law | $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ | T in K |
| ideal gas law | $PV = nRT$ | kPa, L, K; R = 8.314 J K⁻¹ mol⁻¹ |
| percentage yield | $\dfrac{\text{actual}}{\text{theoretical}} \times 100$ | same mass units |
Putting it together
Real problems join several routes on the mole map. Find moles from whatever you are given, use the mole ratio, then convert to whatever is asked.
A 1.00 g antacid tablet is dropped into 50.0 mL of 0.500 mol L⁻¹ HCl. The CO₂ produced occupies 0.218 L at 25 °C and 100 kPa. Assuming only the CaCO₃ in the tablet produces gas, find (a) the percentage by mass of CaCO₃ in the tablet and (b) the concentration of HCl left in the solution.
Gas route: n(CO₂) = 0.218 ÷ 24.79 = 8.794 × 10⁻³ mol.
Mole ratio: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, so n(CaCO₃) = 8.794 × 10⁻³ mol.
Mass route: m(CaCO₃) = 8.794 × 10⁻³ × 100.09 = 0.880 g, which is 88.0% of the tablet.
Solution route: n(HCl) at start = 0.500 × 0.0500 = 0.0250 mol; used = 2 × 8.794 × 10⁻³ = 0.01759 mol; left = 0.00741 mol. c = 0.00741 ÷ 0.0500 = 0.148 mol L⁻¹. The acid was in excess, which confirms that CaCO₃ was limiting and the calculation is valid.
Key words
Stoichiometry: the quantitative relationships between reactants and products. Mole: the amount containing exactly 6.022 140 76 × 10²³ entities. Molar mass: mass of one mole (g mol⁻¹). Empirical formula: simplest whole-number ratio of atoms. Molecular formula: actual numbers of atoms in a molecule. Limiting reagent: the reactant used up first. Theoretical yield: maximum product calculated from the limiting reagent. Solute, solvent, solution. Molarity: mol of solute per L of solution. Standard solution: accurately known concentration. Primary standard: pure, stable solid used to make one. Dilution. Ideal gas: one that obeys the kinetic particle theory assumptions exactly. Absolute zero: 0 K, −273.15 °C. Molar volume: volume of one mole of gas at stated conditions.
Test yourself
How many hydrogen atoms are in 0.200 mol of ammonium sulfate, (NH₄)₂SO₄?
8 H per formula unit: 0.200 × 8 = 1.60 mol × 6.022 × 10²³ = 9.64 × 10²³.
A compound is 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. What is its empirical formula?
Moles per 100 g: C 4.35, H 12.9, O 2.18; ratio 2 : 6 : 1.
4.0 g of hydrogen and 40.0 g of oxygen react to form water. What mass of water forms?
H₂ 1.98 mol (÷2 = 0.99); O₂ 1.25 mol (÷1 = 1.25). H₂ is limiting: 1.98 mol H₂O × 18.02 = 35.7 g.
What is the concentration of a solution made by diluting 15.0 mL of 1.20 mol L⁻¹ NaOH to 250.0 mL?
1.20 × 15.0 ÷ 250.0 = 0.0720 mol L⁻¹.
A river sample contains 3.0 × 10⁻⁵ mol L⁻¹ of copper(II) ions. What is this in ppm?
3.0 × 10⁻⁵ × 63.55 = 1.9 × 10⁻³ g L⁻¹ = 1.9 mg L⁻¹ = 1.9 ppm.
A gas occupies 600 mL at 27 °C and 120 kPa. What volume does it occupy at 0 °C and 100 kPa?
600 × (120 ÷ 100) × (273.15 ÷ 300.15) = 655 mL.
What mass of methane, CH₄, is in a 10.0 L cylinder at 25 °C and 500 kPa?
n = 500 × 10.0 ÷ (8.314 × 298.15) = 2.017 mol; m = 2.017 × 16.04 = 32.4 g.
Which change would double the pressure of a fixed amount of gas?
Boyle's law. Celsius doubling is not kelvin doubling; option 3 leaves P unchanged; option 4 halves P.
Which route to moles would you use first?
Balanced equations are balanced by changing . The reactant that runs out first is the reagent. On dilution, the of solute are unchanged. At 0 °C and 100 kPa one mole of gas occupies L. In every gas law, temperature must be in .
2.50 g of zinc is added to 40.0 mL of 1.50 mol L⁻¹ hydrochloric acid. Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). (a) Identify the limiting reagent. (b) Calculate the volume of hydrogen produced at 22 °C and 101.0 kPa. (c) Calculate the mass of excess reagent left over and the concentration of zinc chloride in the final solution (assume the volume stays 40.0 mL).
(a) n(Zn) = 2.50 ÷ 65.38 = 0.03824 mol; n(HCl) = 1.50 × 0.0400 = 0.0600 mol. Dividing by coefficients: Zn 0.03824; HCl 0.0300. HCl is limiting.
(b) n(H₂) = 0.0600 ÷ 2 = 0.0300 mol. T = 295.15 K. V = nRT ÷ P = 0.0300 × 8.314 × 295.15 ÷ 101.0 = 0.729 L.
(c) Zn used = 0.0300 mol; Zn left = 0.03824 − 0.0300 = 0.00824 mol = 0.00824 × 65.38 = 0.539 g. n(ZnCl₂) = 0.0300 mol; [ZnCl₂] = 0.0300 ÷ 0.0400 = 0.750 mol L⁻¹.
One mole of hydrogen gas and one mole of carbon dioxide gas both occupy about 24.8 L at 25 °C and 100 kPa, but one mole of liquid water occupies 18 mL and one mole of liquid ethanol 58 mL. Explain these observations using the kinetic particle theory, and explain why the molar volume of a gas becomes less reliable at very high pressures.
In a gas the particles are very far apart compared with their size, so almost all of the volume is empty space. The volume a gas occupies therefore depends only on how many particles there are and on the temperature and pressure, not on the size or identity of the particles. That is why a mole of H₂ and a mole of CO₂ occupy the same volume at the same conditions (Avogadro's law).
In a liquid the particles are touching, held close by intermolecular forces. The volume of a mole of liquid depends on the size and shape of the molecules and how strongly they attract each other: ethanol molecules are larger than water molecules and pack differently, so a mole of ethanol occupies more space.
At very high pressure the gas particles are squeezed close together. Their own volume is no longer negligible, and intermolecular attractions start to affect their collisions with the walls. Both assumptions of the ideal gas model fail, so real gases no longer have the same molar volume.