Introduction of Fundamental Concepts
This opening chapter sets the quantitative foundation for the rest of MCAT chemistry. The AAMC MCAT content outline expects you to use the mole concept and Avogadro's number, convert between mass, moles and molarity, determine empirical and molecular formulas from percent composition, identify the limiting and excess reactants in any equation, and compute theoretical, actual and percent yield. These are the numerical skills that recur throughout the Chemical & Physical Foundations section.
Moles and Avogadro's Number
The mole is the SI unit of amount of substance. One mole of any species contains Avogadro's number, NA = 6.022 × 1023 particles (atoms, molecules, ions or formula units).
Three faces of the mole
| 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 × NA (NA = 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.
Empirical and Molecular Formula
The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of atoms per molecule and is always a whole-number multiple of the empirical formula.
Percent composition gives the mass fraction of each element: % element = (mass of that element in 1 mol / molar mass) × 100.
Finding the empirical formula from percent composition
- Assume a 100 g sample, so each percent becomes grams.
- Convert each element's mass to moles (divide by its atomic mass).
- Divide every mole value by the smallest one to get the ratio.
- Multiply through to clear fractions (e.g. ×2 if a value is 1.5).
A compound is 40.0% C, 6.7% H, 53.3% O (by mass). Molar mass = 180 g/mol. Find both formulas.
Step 1. In 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
Step 2. Moles: C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.
Step 3. Divide by 3.33: C = 1, H = 2, O = 1 → empirical formula CH2O (empirical mass = 30 g/mol).
Step 4. Molecular multiple = 180 / 30 = 6, so molecular formula = C6H12O6 (glucose).
Molarity and Solution Concentration
Molarity (M) is the most common concentration unit on the MCAT: moles of solute per liter of solution.
M = moles of solute / liters of solution
- Rearrange to find moles (n = M × V) or the mass needed (mass = M × V × molar mass).
- Dilution: M1V1 = M2V2 (moles of solute are conserved when solvent is added).
- Related units: molality (mol solute / kg solvent) and mole fraction are temperature-independent, whereas molarity varies slightly with temperature because volume does.
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.
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 recrystallization.
- Impure starting materials.
Worked MCQs
Seven 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:
Molar mass of H2O = 18 g/mol, so 18 g = 1 mole. One mole contains exactly NA = 6.022 × 1023 molecules.
Q2. 2 g of H2 reacts with 32 g of O2 to form water (2H2 + O2 → 2H2O). The limiting reactant is:
Moles of H2 = 2/2 = 1; moles of O2 = 32/32 = 1. Divide by stoichiometric coefficients: H2 → 1/2 = 0.5; O2 → 1/1 = 1. The smaller ratio (0.5) is H2, so H2 is the limiting reactant: 1 mol H2 needs only 0.5 mol O2, so 0.5 mol O2 is left in excess.
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:
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:
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.
Q6. A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Its empirical formula is:
In 100 g: C = 40/12 = 3.33 mol, H = 6.7/1 = 6.7 mol, O = 53.3/16 = 3.33 mol. Dividing by the smallest (3.33) gives C:H:O = 1:2:1, so the empirical formula is CH2O.
Q7. How many grams of NaOH (M = 40 g/mol) are needed to prepare 500 mL of a 2.0 M solution?
moles = M × V = 2.0 mol/L × 0.500 L = 1.0 mol. Mass = moles × molar mass = 1.0 × 40 = 40 g.
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 × NA; molarity M = mol solute / L solution; dilution M1V1 = M2V2.
- Empirical formula = simplest whole-number atom ratio (from percent composition); molecular formula = whole-number multiple of it.
- 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.