Chemical Calculations and Stoichiometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your NECO exam.

Relative Atomic Mass ($A_r$): Average mass of atoms of an element relative to $\frac{1}{12}$ of Carbon-12.

Relative Molecular Mass ($M_r$): Sum of $A_r$ values of all atoms in a molecule.

Mole: The amount of substance containing $6.02 \times 10^{23}$ particles (Avogadro’s number, $N_A$).

$$\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{\text{mass}}{M_r}$$

Molar Volume: 1 mole of any gas at STP (standard temperature and pressure: $273$ K, $1$ atm) occupies $22.4$ dm³. At room temperature and pressure (rtp): $24$ dm³.

Concentration:

• Molar concentration (mol/dm³): $C = \dfrac{n}{V}$ where $V$ is in dm³
• $C = \dfrac{\text{mass}}{M_r \times V}$ for mass-based calculations

Empirical Formula: Simplest whole-number ratio of atoms in a compound. Molecular Formula: Actual number of atoms in one molecule.

⚡ NECO Tip: Always convert volumes to dm³ (divide cm³ by 1000) before using them in mole calculations. $1 \text{ dm}^3 = 1000 \text{ cm}^3 = 1$ litre. Watch out for “STP” vs “rtp” — they give different molar volumes.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for NECO Chemistry students with a few days to months.

Balancing Chemical Equations

A balanced equation shows conservation of mass. For example: $$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

This means: 2 moles of hydrogen + 1 mole of oxygen → 2 moles of water.

Stoichiometric Calculations (Mole Method):

1. Write the balanced equation
2. Convert given substances to moles
3. Find the mole ratio from the equation
4. Calculate required amount

Example: What mass of sodium chloride is formed when $2.3$ g of sodium reacts with excess chlorine? $$2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}$$

• Moles of Na: $n = \frac{2.3}{23} = 0.1$ mol
• Ratio Na : NaCl = $2 : 2 = 1 : 1$, so moles of NaCl = $0.1$ mol
• Mass of NaCl: $m = 0.1 \times 58.5 = 5.85$ g

Titration Calculations

For acid–base titration: $C_1V_1 = C_2V_2$ (at equivalence point for monoprotic acids and bases).

If acid has $n$ acidic protons per molecule: $n \times C_{\text{acid}} V_{\text{acid}} = C_{\text{base}} V_{\text{base}}$

Example: $25.0$ cm³ of $0.10$ mol/dm³ NaOH neutralises $20.0$ cm³ of HCl. Find concentration of HCl. $$C_{\text{HCl}} = \frac{C_{\text{NaOH}} \times V_{\text{NaOH}}}{V_{\text{HCl}}} = \frac{0.10 \times 25.0}{20.0} = 0.125 \text{ mol/dm}^3$$

Percentage Composition: $$\text{% of element} = \frac{A_r \times \text{no. of atoms of element}}{M_r \text{ of compound}} \times 100%$$

⚡ NECO Common Mistakes:

• Forgetting to balance equations before calculating
• Mixing up cm³ and dm³ — always convert
• Using molecular mass when the question gives mass of an individual atom (e.g., calculating from isotope mass)
• Not identifying the limiting reagent in reactions with two reactants

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for NECO and JAMB Chemistry preparation.

Finding Empirical and Molecular Formula

Example: A compound contains 40% C, 6.7% H, and 53.3% O by mass. Its relative molecular mass is 180. Find the formula.

• Divide by atomic masses: C: $40/12 = 3.33$, H: $6.7/1 = 6.7$, O: $53.3/16 = 3.33$
• Divide by smallest: C: $1$, H: $2$, O: $1$ → empirical formula = CH₂O
• Empirical formula mass = $12 + 2 + 16 = 30$
• $n = 180/30 = 6$ → molecular formula = C₆H₁₂O₆

Yield Calculations

• Theoretical yield: Maximum mass of product predicted by stoichiometry
• Actual yield: Mass of product actually obtained
• Percentage yield: $\text{% yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100%$

Atom Economy (how efficiently a reaction uses atoms): $$\text{atom economy} = \frac{\text{molar mass of desired product}}{\text{total molar mass of all reactants}} \times 100%$$

Example: In the Haber process: N₂ + 3H₂ → 2NH₃ $$\text{atom economy} = \frac{2 \times 17}{28 + 6} \times 100% = \frac{34}{34} \times 100% = 100%$$

Ideal Gas Equation: $$PV = nRT$$ where $P$ = pressure (Pa), $V$ = volume (m³), $n$ = moles, $R = 8.314$ J/mol·K, $T$ = temperature (K).

Note: $R = 0.0821 \text{ L·atm/mol·K}$ when $V$ is in litres and $P$ in atm.

Gas Volume Calculations with Changes in Conditions:

$$PV = nRT \Rightarrow \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

Limiting Reagent Problems

Example: $10$ g of hydrogen and $64$ g of oxygen react to form water. $$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

• Moles H₂ = $10/2 = 5$ mol; requires $2.5$ mol O₂
• Moles O₂ available = $64/32 = 2$ mol → O₂ is limiting
• Water formed = $2 \times 2 = 4$ mol = $72$ g

Molarity from Percentage and Density:

If a solution is $w$% (w/w) and has density $d$ g/cm³: $$\text{Molarity} = \frac{d \times 1000 \times w}{M_r}$$

NECO/JAMB Patterns:

• Titration questions are very common in Section B
• Always write the balanced equation first
• Double-check your mole ratio from the balanced equation
• Watch for units — if concentration is in g/dm³, convert to mol/dm³ before titrating

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