Stoichiometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary of stoichiometry for NABTEB chemistry.

Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations.

Key Definitions:

• Relative atomic mass ($A_r$): Average mass of an atom relative to 1/12 of Carbon-12
• Relative molecular mass ($M_r$): Sum of $A_r$ of all atoms in a molecule
• Mole (mol): Amount of substance containing $6.02 \times 10^{23}$ particles (Avogadro’s number, $N_A$)
• Molar mass: Mass of one mole of a substance (g/mol) — numerically equal to $M_r$

Essential Formulas:

Quantity	Formula	Unit
Number of moles	$n = \frac{m}{M}$	mol
Number of particles	$n = \frac{N}{N_A}$	mol
Molar volume (gases at STP)	$V_m = 24 , \text{dm}^3/\text{mol}$	dm³/mol
Concentration	$C = \frac{n}{V}$	mol/dm³
Mass	$m = n \times M$	g

Where: $m$ = mass, $M$ = molar mass, $N$ = number of particles, $N_A = 6.02 \times 10^{23}$, $V$ = volume in dm³, $C$ = concentration.

Molar Volume of Gases: At standard temperature and pressure (STP: 0°C, 1 atm), one mole of any gas occupies 24 dm³. At room temperature and pressure (RTP: 25°C, 1 atm), one mole of gas occupies 24 dm³ (approximately).

⚡ NABTEB Exam Tip: Always check units — mass must be in grams, volume in dm³ (not cm³). To convert cm³ to dm³, divide by 1000.

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🟡 Standard — Regular Study (2d–2mo)

For NABTEB students who want thorough understanding of stoichiometric calculations.

Balancing Chemical Equations:

A balanced equation shows equal numbers of each atom on both sides.

Example — Burning methane: $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

• C: 1 each side ✓
• H: 4 each side ✓
• O: 4 each side (2×2 on left, 2+2×1 on right) ✓

Mole Calculations from Equations:

From the balanced equation: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

• 1 mole of $CH_4$ reacts with 2 moles of $O_2$
• Produces 1 mole of $CO_2$ and 2 moles of $H_2O$

Mole Ratio is the key to stoichiometry.

Example Calculation: Calculate the mass of $CO_2$ produced when 4 g of $CH_4$ burns completely.

Step 1: $M_r(CH_4) = 12 + 4 = 16$ g/mol $$n(CH_4) = \frac{4}{16} = 0.25 , \text{mol}$$

Step 2: Mole ratio $CH_4 : CO_2 = 1 : 1$ $$n(CO_2) = 0.25 , \text{mol}$$

Step 3: $M_r(CO_2) = 12 + 32 = 44$ g/mol $$m(CO_2) = 0.25 \times 44 = 11 , \text{g}$$

Concentration Calculations:

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

Where $C$ is in mol/dm³, $V$ is in dm³.

Example: Calculate the mass of NaOH in 250 cm³ of 0.5 mol/dm³ solution.

Step 1: Convert volume to dm³: $V = \frac{250}{1000} = 0.25 , \text{dm}^3$ Step 2: $n(NaOH) = C \times V = 0.5 \times 0.25 = 0.125 , \text{mol}$ Step 3: $M_r(NaOH) = 23 + 16 + 1 = 40$ g/mol Step 4: $m(NaOH) = 0.125 \times 40 = 5.0 , \text{g}$

Empirical and Molecular Formulas:

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

Example: A compound contains 40% C, 6.7% H, and 53.3% O by mass. Find empirical formula.

Step 1: Divide by atomic masses:

• C: $40/12 = 3.33$
• H: $6.7/1 = 6.7$
• O: $53.3/16 = 3.33$

Step 2: Divide by smallest (3.33):

• C: $3.33/3.33 = 1$
• H: $6.7/3.33 = 2$
• O: $3.33/3.33 = 1$

Empirical formula: $CH_2O$

If molecular mass is 180 g/mol, then molecular formula is $(CH_2O)n = 30n = 180$, so $n = 6$. Molecular formula: $C_6H{12}O_6$.

⚡ NABTEB Exam Tip: In limiting reagent problems, calculate the amount of product from EACH reactant separately. The reactant that produces the LEAST product is the limiting reagent and determines the yield.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage of stoichiometry for thorough NABTEB preparation.

The Mole Concept:

One mole of any substance contains $6.02 \times 10^{23}$ particles (Avogadro’s number, $N_A$).

This includes:

• Atoms in an elemental sample
• Molecules in a molecular sample
• Ions in an ionic compound
• Electrons in a discharge tube
• Formula units in a giant lattice structure

Molar Mass vs Atomic Mass:

• Atomic mass ($A_r$): Relative value (no units), e.g., $A_r(Cl) = 35.5$
• Molar mass ($M$): Absolute mass of one mole, e.g., $M(Cl) = 35.5$ g/mol

They are numerically equal but differ in units.

Gas Calculations Using the Ideal Gas Equation:

For ideal gases: $PV = nRT$

Where:

• $P$ = pressure (Pa or Nm⁻²)
• $V$ = volume (m³)
• $n$ = number of moles
• $R$ = gas constant = 8.31 J/mol·K
• $T$ = temperature (K)

Important conversions:

• 1 atm = 101,325 Pa = 101.325 kPa
• 0°C = 273 K
• 1 dm³ = 10⁻³ m³

Example: Calculate the volume of 0.5 mol of gas at 27°C and 100 kPa.

$T = 27 + 273 = 300$ K $$V = \frac{nRT}{P} = \frac{0.5 \times 8.31 \times 300}{100 \times 10^3} = \frac{1246.5}{100000} = 0.0125 , \text{m}^3 = 12.5 , \text{dm}^3$$

Yield Calculations:

Theoretical Yield: Maximum amount of product predicted by stoichiometry.

Actual Yield: Amount actually obtained in the experiment.

Percentage Yield: $$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100%$$

Atom Economy (Modern Concept):

$$\text{Atom economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100%$$

Higher atom economy = greener process.

Solution Stoichiometry — Titrations:

For acid-base titrations: $$C_1V_1 = C_2V_2$$ (when reacting in 1:1 ratio)

For reactions with different mole ratios, use mole ratios: $$C_1V_1/n_1 = C_2V_2/n_2$$

Where $n_1$ and $n_2$ are the stoichiometric coefficients.

Example Titration Calculation:

25.0 cm³ of 0.10 mol/dm³ HCl is titrated with NaOH. The titre is 20.0 cm³. Find concentration of NaOH.

$$C_{NaOH} = \frac{C_{HCl} \times V_{HCl}}{V_{NaOH}} = \frac{0.10 \times 25.0}{20.0} = 0.125 , \text{mol/dm}^3$$

Hydrated Salts:

Some salts contain water of crystallisation:

• $CuSO_4 \cdot 5H_2O$: Copper(II) sulphate pentahydrate
• $Na_2CO_3 \cdot 10H_2O$: Sodium carbonate decahydrate

When calculating moles, you must account for the water: $$M(CuSO_4 \cdot 5H_2O) = 63.5 + 32 + (4 \times 16) + 5 \times 18 = 250 , \text{g/mol}$$

Volumetric Analysis — Redox Titrations:

Using potassium manganate(VII): $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Or iodide-thiosulphate: $$I_2 + 2e^- \rightarrow 2I^-$$ $$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$$

Mass Spectrometry:

In a mass spectrometer:

• Sample is vaporised and ionised
• Positive ions are accelerated through magnetic field
• Lighter ions deflect more than heavier ions
• Mass/charge ratio ($m/z$) determines deflection

Relative atomic mass is calculated as the weighted average of isotopic masses using their abundance.

⚡ NABTEB Quick Reference:

• $n = m/M$ (moles from mass)
• $n = CV$ (moles from concentration and volume)
• $n = V/24$ (moles of gas at STP in dm³)
• $PV = nRT$ (ideal gas equation)
• $C_1V_1 = C_2V_2$ (titration — equal molar ratios)
• Percentage yield = $\frac{\text{actual}}{\text{theoretical}} \times 100$
• Atom economy = $\frac{\text{desired product mass}}{\text{reactants mass}} \times 100$
• $A_r = \frac{\sum(\text{isotope mass} \times \text{abundance})}{\text{sum of abundances}}$
• $N_A = 6.02 \times 10^{23}$ mol⁻¹
• Gas molar volume = 24 dm³/mol at STP