Introduction of Fundamental Concepts

This opening chapter sets the quantitative foundation for the rest of MDCAT chemistry. The PMDC 2026 syllabus expects you to use the mole concept and Avogadro's number, identify the limiting and excess reactants in any equation, and compute theoretical, actual and percent yield. Mastering these three skills cracks ~3-4 numerical MCQs on the paper.

PMC Table of Specifications. Three subtopics — moles & Avogadro's number, limiting and excess reactants, and yield calculations. Practise short numerical MCQs — about three quarters of items on this chapter are calculations.

Moles and Avogadro's Number

The mole is the SI unit of amount of substance. One mole of any species contains Avogadro's number, N_A = 6.022 × 10²³ particles (atoms, molecules, ions or formula units).

Three faces of the mole

Mole conversions — the only formula triangle you need
Convert	From	To	Formula
Mass ↔ moles	g	mol	n = mass / M (M = molar mass in g/mol)
Moles ↔ particles	mol	atoms / molecules / ions	N = n × N_A (N_A = 6.022 × 10²³)
Moles ↔ gas volume (STP)	mol	L of gas at STP	V = n × 22.4 L (273.15 K, 1 atm)
Moles ↔ gas volume (any T, P)	mol	L	PV = nRT (R = 0.0821 L·atm/mol·K)
Concentration	mol of solute, L of solution	molarity	M = n / V

Calculating molar mass

Sum the atomic masses of all atoms in the formula. Examples: H₂O = 2(1) + 16 = 18 g/mol; CO₂ = 12 + 2(16) = 44 g/mol; H₂SO₄ = 2(1) + 32 + 4(16) = 98 g/mol.

Worked stoichiometry example

How many grams of CO₂ are produced when 24 g of carbon is burned completely in oxygen? (C + O₂ → CO₂)

Step 1. moles of C = 24 / 12 = 2 mol.

Step 2. Stoichiometric ratio C : CO₂ = 1 : 1, so moles of CO₂ = 2 mol.

Step 3. Mass of CO₂ = 2 × 44 = 88 g.

Memory aid. "Mass ÷ M = moles, moles × N_A = particles, moles × 22.4 = litres at STP." Internalise this triangle — you will use it in every numerical MCQ.

Limiting and Excess Reactants

When two reactants are mixed in non-stoichiometric proportions, one of them runs out first and stops the reaction. That reactant is the limiting reactant (LR); the other(s), present in surplus, are excess reactants.

Identifying the limiting reactant

Convert each reactant's mass to moles.
Divide each by its stoichiometric coefficient in the balanced equation.
The smallest ratio belongs to the limiting reactant.
Use the moles of LR (and the equation) to compute moles/mass of products formed and excess reactant left.

Why it matters

The theoretical yield of the product is dictated entirely by the limiting reactant.
Once the LR is consumed, the reaction stops irrespective of how much excess reactant remains.
In industry, the cheaper or environmentally friendlier reactant is normally added in excess to drive expensive reactants to completion.

Common trap. The reactant in smaller mass is not always the limiting reactant. You must compare moles ÷ coefficient, not raw mass. A heavy reactant with a high stoichiometric requirement can still be limiting.

Yield

In practice, no reaction gives 100% of the predicted product because of side reactions, reversible equilibria, mechanical losses, or impure reagents. Three terms quantify this:

Theoretical yield: Maximum amount of product calculable from the limiting reactant assuming the reaction goes to completion. Always determined by stoichiometry.
Actual yield: The mass of product actually obtained in the laboratory. An experimental quantity, always less than (or equal to) the theoretical yield.
Percent yield: % yield = (actual yield / theoretical yield) × 100. A measure of how efficient a reaction or procedure is.

Reasons for percent yield < 100%

Side reactions producing by-products.
Reversible reactions that never reach 100% conversion.
Loss during transfer, filtration or recrystallisation.
Impure starting materials.

Worked MCQs

Five MCQs that capture the high-yield testing patterns for stoichiometry. Read every explanation — the deeper concept lives there.

Q1. The number of molecules in 18 g of water at STP is approximately:

3.011 × 10²³
6.022 × 10²³
1.204 × 10²⁴
22.4

Molar mass of H₂O = 18 g/mol, so 18 g = 1 mole. One mole contains exactly N_A = 6.022 × 10²³ molecules.

Q2. 4 g of H₂ reacts with 32 g of O₂ to form water (2H₂ + O₂ → 2H₂O). The limiting reactant is:

H₂
O₂
H₂O
Both run out simultaneously

Moles of H₂ = 4/2 = 2; moles of O₂ = 32/32 = 1. Dividing by stoichiometric coefficients: H₂ → 2/2 = 1; O₂ → 1/1 = 1. Numerically equal — but reading the question as written (2 mol H₂ needs 1 mol O₂; we have exactly that), H₂ finishes first by virtue of being the species fully consumed in producing 2 mol H₂O. The "smallest mole-to-coefficient ratio" identifies the limiting reactant; the lighter species H₂ is consumed completely.

Q3. A reaction has a theoretical yield of 50 g and an actual yield of 40 g. The percent yield is:

% yield = (actual / theoretical) × 100 = (40 / 50) × 100 = 80%. A typical lab synthesis is in the 60-90% range.

Q4. The volume of 0.5 mole of an ideal gas at STP is:

22.4 L
11.2 L
5.6 L
44.8 L

At STP, 1 mol = 22.4 L, so 0.5 mol = 11.2 L. Equivalent to PV = nRT with P = 1 atm and T = 273.15 K.

Q5. The molar mass of CaCO₃ (Ca = 40, C = 12, O = 16) is:

56 g/mol
68 g/mol
100 g/mol
164 g/mol

M(CaCO₃) = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol. This is why one mole of limestone is exactly 100 g — useful in titration calculations.

Quick Recap

1 mole = 6.022 × 10²³ particles = molar mass in grams = 22.4 L of any gas at STP.
moles = mass / molar mass; particles = moles × N_A.
Limiting reactant = smallest mole-to-coefficient ratio; it dictates theoretical yield.
Excess reactant remains after the reaction stops.
Theoretical yield = stoichiometric maximum; actual yield = experimentally obtained.
Percent yield = (actual / theoretical) × 100.
Yield is reduced by side reactions, reversibility, transfer losses and impurities.

Test yourself. Take a timed practice test or browse the topic-wise MCQs to lock these concepts in.