General Chemistry: free practice, theory and problems

1. Moles and stoichiometry

What is it about?

Chemical reactions occur in fixed numerical ratios between molecules, but in the lab we weigh substances in grams. The mole is the bridge between these two worlds: one mole is exactly $6.022\cdot10^{23}$ particles (Avogadro's number), and the molar mass tells you how many grams one mole of a substance weighs. Stoichiometry is the set of calculations that use the balanced reaction equation to convert between amounts of reactants and products – essential when an engineer has to size a reactor or calculate how much raw material a process needs.

Concepts and formulas

• Molar mass $M$ (g/mol): the sum of the atomic masses in the formula, from the periodic table.
• Number of moles: $n = m/M$, where $m$ is the mass in grams.
• Balanced reaction equation: the coefficients in front of each formula give the mole ratio between the substances, not the mass ratio.
• Limiting reactant: the reactant that is used up first; it determines the maximum possible amount of product (the theoretical yield).
• Procedure for the limiting reactant: convert all reactants to moles, divide by the coefficient in the balanced equation, and see which gives the smallest value – that one is limiting.
• Theoretical yield: the amount of product calculated from the limiting reactant.
• Percent yield: $$\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}}\cdot 100\%$$
• Molarity (concentration): $c = n/V$, with $V$ in liters and $c$ in mol/L (M).
• Dilution: $c_1V_1 = c_2V_2$ (the number of moles is the same before and after dilution).

How to solve the problems

1. Write (or look up) the balanced reaction equation. Check that the number of atoms of each element is equal on both sides.
2. Convert everything given (mass, volume, concentration) to moles.
3. If there are two reactants: divide the number of moles by the coefficient for each, and the one with the smaller ratio is limiting.
4. Use the mole ratio from the equation to find the moles of product, and convert to grams with $m = nM$ if needed.
5. Percent yield: compare the actual (given) yield with the theoretical yield you just calculated.

Example

12 g of hydrogen gas ($M_{H_2}=2.02$ g/mol) reacts with 64 g of oxygen gas ($M_{O_2}=32.00$ g/mol) according to $\mathrm{2H_2 + O_2 \to 2H_2O}$. Which reactant is limiting, and how many grams of water ($M_{H_2O}=18.02$ g/mol) are theoretically formed?

1. Moles: $n_{H_2} = 12/2.02 \approx 5.94$ mol, $n_{O_2} = 64/32.00 = 2.00$ mol.
2. Divide by the coefficients: $5.94/2 \approx 2.97$ versus $2.00/1 = 2.00$. $O_2$ gives the lower ratio, so $O_2$ is limiting.
3. Mole ratio $O_2:H_2O = 1:2$, so $n_{H_2O} = 2\cdot 2.00 = 4.00$ mol.
4. Mass: $m = nM = 4.00\cdot 18.02 \approx 72.1$ g.

Answer: $O_2$ is limiting, and about 72.1 g of water is formed.

Common mistakes

• Using mass ratios directly from the equation instead of mole ratios – the coefficients apply to moles, not grams.
• Forgetting to divide the number of moles by the coefficient when finding the limiting reactant.
• Using the wrong molar mass – look up the correct atomic mass for each element and add the right number of atoms.
• Confusing percent yield with the limiting reactant's share of the total mass.

Concepts in this part

Practise moles and stoichiometry in the app →

2. Acids, bases and equilibrium

What is it about?

Almost every chemical process in water – from corrosion to biological reactions – is affected by how acidic or basic the environment is. The pH scale makes it possible to compare the concentration of H⁺ ions across many orders of magnitude with a single number. Acids and bases also react with each other in equilibria, and Le Chatelier's principle tells you how an equilibrium responds when conditions change – essential for controlling a process or understanding why a buffer keeps pH stable.

Concepts and formulas

• $\mathrm{pH} = -\log[\mathrm{H^+}]$ and $\mathrm{pOH} = -\log[\mathrm{OH^-}]$. At 25 °C: $\mathrm{pH + pOH} = 14$.
• Strong acids/bases ionize completely: $[\mathrm{H^+}] = c_{acid}$ (for monoprotic acids like HCl).
• Weak acids ionize only partially, governed by the acid constant $K_a$: $$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$ The larger $K_a$ (or the smaller $\mathrm{p}K_a = -\log K_a$), the stronger the acid.
• For a weak acid with initial concentration $c$ and a small degree of dissociation, you can often approximate $[\mathrm{H^+}] \approx \sqrt{K_ac}$.
• Buffer: a mixture of a weak acid and its conjugate base (or the reverse) that keeps the pH nearly constant when a little acid or base is added.
• The equilibrium constant $K$ does not change with concentration or pressure – only with temperature.
• Le Chatelier's principle: increasing the concentration of a reactant shifts the equilibrium toward the products; increasing the pressure (in gas reactions) shifts it toward the side with fewer gas molecules; increasing the temperature shifts an endothermic reaction further toward the products.

How to solve the problems

1. Decide whether the acid/base is strong or weak. Strong: $[\mathrm{H^+}]$ (or $[\mathrm{OH^-}]$) equals the given concentration directly.
2. Weak acid: set up the $K_a$ expression, use the approximation $[\mathrm{H^+}] \approx \sqrt{K_ac}$ if the dissociation is small (less than about 5%).
3. Compute $\mathrm{pH} = -\log[\mathrm{H^+}]$, or the reverse: $[\mathrm{H^+}] = 10^{-\mathrm{pH}}$.
4. To find pOH or $[\mathrm{OH^-}]$: use $\mathrm{pH+pOH}=14$ and $\mathrm{pOH}=-\log[\mathrm{OH^-}]$.
5. Equilibrium shift: look at which side of the equation is "boosted" or "relieved" by the change, and remember that only temperature changes $K$ itself.

Example

Acetic acid (CH₃COOH) has $K_a = 1.8\cdot10^{-5}$. What is the pH of a 0.10 M solution?

1. Acetic acid is weak, so we use the approximation: $[\mathrm{H^+}] \approx \sqrt{K_ac} = \sqrt{1.8\cdot10^{-5}\cdot 0.10}$.
2. $[\mathrm{H^+}] \approx \sqrt{1.8\cdot10^{-6}} \approx 1.34\cdot10^{-3}$ M.
3. $\mathrm{pH} = -\log(1.34\cdot10^{-3}) \approx 2.87$.

Answer: pH ≈ 2.87. Note that this is higher (less acidic) than a strong acid of the same concentration would give (pH 1), because only a small fraction of the acetic acid ionizes.

Common mistakes

• Using $[\mathrm{H^+}] = c$ for a weak acid – that only applies to strong acids.
• Mixing up $K_a$ and $\mathrm{p}K_a$: a small $\mathrm{p}K_a$ means a strong acid (large $K_a$).
• Believing that adding a catalyst or changing the pressure of a solution with no gas shifts the equilibrium.
• Forgetting that $\mathrm{pH+pOH}=14$ only holds at 25 °C.

Concepts in this part

Practise acids, bases and equilibrium in the app →

3. Thermochemistry and electrochemistry

What is it about?

Thermochemistry deals with the energy that accompanies chemical reactions – how much heat is released or required, and whether the reaction happens on its own. Electrochemistry links chemical reactions to electric current: in a battery, a spontaneous redox reaction drives the current, while electrolysis uses current to force a reaction that would not otherwise happen. Both matter to an engineer, whether sizing cooling for an exothermic process, choosing the right battery, or understanding why a metal corrodes.

Concepts and formulas

• Reaction enthalpy $\Delta H$: the heat exchanged at constant pressure. $\Delta H<0$ is exothermic (releases heat), $\Delta H>0$ is endothermic.
• Hess's law: $\Delta H$ for a reaction depends only on the initial and final states, not on the path. You can therefore add (and flip the sign of) sub-reactions to find the $\Delta H$ of a reaction you have not measured directly.
• Free energy (Gibbs energy): $\Delta G = \Delta H - T\Delta S$. The reaction is spontaneous (at constant $T$, $p$) when $\Delta G < 0$.
• Redox reaction: oxidation is the loss of electrons, reduction is the gain of electrons (mnemonic: OIL RIG).
• Standard cell potential: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ (reduction potentials from a table). A positive $E^\circ_{cell}$ means a spontaneous (galvanic) reaction.
• Faraday's law: the amount of substance converted by electrolysis is $m = \dfrac{ItM}{zF}$, where $I$ is the current (A), $t$ is time (s), $M$ is the molar mass, $z$ is the number of electrons per ion, and $F = 96\,485$ C/mol.
• Catalyst: lowers the activation energy and increases the reaction rate without being consumed or changing $\Delta G$ or the equilibrium constant.

How to solve the problems

1. Thermochemistry/Hess: write down the sub-reactions with known $\Delta H$ values. Flip a reaction (change the sign of $\Delta H$) or multiply it (multiply $\Delta H$ accordingly) so the sub-reactions add up to the overall reaction. Add the $\Delta H$ values.
2. Spontaneity: substitute $\Delta H$, $T$ (in kelvin!) and $\Delta S$ into $\Delta G = \Delta H - T\Delta S$. Remember $\Delta S$ is often given in J/(mol·K) while $\Delta H$ is usually in kJ/mol – convert to the same unit.
3. Electrochemistry: identify the anode (oxidation) and cathode (reduction), find $E^\circ_{cell}$ from table values.
4. Electrolysis: use Faraday's law, making sure $z$ matches the ion's charge.
5. Check the sign and order of magnitude – a large positive $E^\circ_{cell}$ or a large negative $\Delta G$ means a strongly driving reaction.

Example

Given $\Delta H_1$ for $\mathrm{C(s) + O_2(g) \to CO_2(g)}$: $-393.5$ kJ/mol, and $\Delta H_2$ for $\mathrm{CO(g) + \tfrac12O_2(g) \to CO_2(g)}$: $-283.0$ kJ/mol. Find $\Delta H$ for $\mathrm{C(s) + \tfrac12O_2(g) \to CO(g)}$.

1. The target reaction plus reaction 2 gives reaction 1: $[\mathrm{C + \tfrac12O_2 \to CO}] + [\mathrm{CO + \tfrac12O_2 \to CO_2}] = [\mathrm{C + O_2 \to CO_2}]$.
2. Therefore $\Delta H_{target} = \Delta H_1 - \Delta H_2 = -393.5 - (-283.0)$.
3. $\Delta H_{target} = -110.5$ kJ/mol.

Answer: $\Delta H \approx -110.5$ kJ/mol (exothermic).

Common mistakes

• Forgetting to flip the sign of $\Delta H$ when a sub-reaction is reversed.
• Using Celsius instead of kelvin in $\Delta G = \Delta H - T\Delta S$.
• Mixing the units kJ and J for $\Delta H$ and $\Delta S$.
• Believing a catalyst changes $\Delta G$ or the equilibrium constant – it only changes how fast equilibrium is reached.
• Confusing anode and cathode: in a galvanic cell, oxidation happens at the anode.

Concepts in this part

Practise thermochemistry and electrochemistry in the app →

4. Gas laws

What is it about?

Gases in pressure bottles, tyres, engines and ventilation systems follow a few simple laws. If you know three of the quantities pressure, volume, temperature and amount of substance, you can calculate the fourth.

Concepts and formulas

• The ideal gas law (SI units: Pa, m³, K):

• Combined gas law for a fixed amount of gas:

• Special cases: constant temperature gives $p_1V_1 = p_2V_2$ (Boyle). Constant pressure gives $V_1/T_1 = V_2/T_2$. Constant volume gives $p_1/T_1 = p_2/T_2$.
• Molar volume at 0 °C and 1 atm (101.3 kPa): 22.4 L/mol.
• Partial pressure (Dalton): in a mixture, the pressure of each gas equals its fraction times the total pressure. Air is about 21 % oxygen.
• Density of a gas: $\rho = \dfrac{pM}{RT}$, where $M$ is the molar mass in kg/mol.

How to solve the problems

1. Convert to SI: kPa to Pa, L to m³ (divide by 1000), °C to K (add 273.15).
2. Use absolute pressure, not gauge pressure.
3. Insert into the right law and solve for the unknown.

Example

A car tyre has an absolute pressure of 200 kPa at 10 °C. What is the pressure at 40 °C (same volume)?

1. $T_1 = 283.15$ K and $T_2 = 313.15$ K.
2. $p_2 = p_1\cdot T_2/T_1 = 200\cdot 313.15/283.15$.
3. $p_2 \approx 221$ kPa.

Common mistakes

• Using degrees Celsius. That gives a wrong answer, and division by zero at 0 °C.
• Using the gauge pressure from a tyre gauge instead of absolute pressure.
• Using litres and kPa together with $R = 8.314$ without converting.

Concepts in this part

Practise gas laws in the app →