The mole and stoichiometry

Why this matters

Reactions happen atom by atom, but you weigh chemicals by the gram. The moleA counting unit for particles: one mole is 6.022 × 10^23 of them. is the bridge between the two. With it, a lab technician knows how many grams of salt make a buffer, a process engineer knows how much product a batch should yield, and a chemist can tell whether a 60% yield is a problem or normal. This is the single most used calculation in introductory chemistry.

The mole

A mole is a counting unit, like a dozen, but much bigger:

1 mol = 6.022 × 10^23 particles   (Avogadro's number, N_A)

It is chosen so that the molar massThe mass of one mole of a substance in grams per mole, found by adding the atomic masses in its formula. of an element in grams per mole equals its atomic mass from the periodic table. Carbon's atomic mass is 12.011 u, so one mole of carbon atoms weighs 12.011 g.

Molar mass of a compound

Add up the atomic masses of every atom in the formula.

H2O   = 2 × 1.008 + 15.999           = 18.015 g/mol
CO2   = 12.011 + 2 × 15.999          = 44.009 g/mol
NaCl  = 22.990 + 35.45               = 58.44 g/mol
C6H12O6 (glucose) = 6 × 12.011 + 12 × 1.008 + 6 × 15.999 = 180.156 g/mol

The three conversions

moles     = grams / molar mass
grams     = moles × molar mass
particles = moles × 6.022 × 10^23

Example: how many moles are in 9.0 g of water? 9.0 / 18.015 ≈ 0.50 mol, which is about 3.0 × 10²³ molecules.

Stoichiometry: the road map

StoichiometryUsing a balanced equation's mole ratio to calculate amounts of reactants and products. uses the balanced equation's coefficients as a mole ratio. You cannot go straight from grams of A to grams of B; you always go through moles:

grams A  ->  moles A  ->  (mole ratio)  ->  moles B  ->  grams B
     ÷ molar mass A      × coef B / coef A        × molar mass B

Worked example 1: water from hydrogen

2 H2 + O2 -> 2 H2O

How much water can 4.00 g of hydrogen make (with plenty of oxygen)?

1. Moles of H₂: 4.00 g / 2.016 g/mol = 1.984 mol
2. Mole ratio H₂O : H₂ = 2 : 2 = 1, so 1.984 mol of H₂O
3. Grams of H₂O: 1.984 × 18.015 ≈ 35.7 g

Worked example 2: carbon dioxide from methane

CH4 + 2 O2 -> CO2 + 2 H2O

A furnace burns 1.00 kg (1000 g) of methane. How much CO₂ is released?

1. Moles of CH₄: 1000 / 16.043 ≈ 62.33 mol
2. Ratio CO₂ : CH₄ = 1 : 1, so 62.33 mol CO₂
3. Grams: 62.33 × 44.009 ≈ 2743 g ≈ 2.74 kg

Burning a kilogram of methane releases more than two and a half kilograms of CO₂, because each carbon picks up two oxygen atoms from the air.

Worked example 3: ammonia from nitrogen

N2 + 3 H2 -> 2 NH3

From 28.0 g of N₂: 28.0 / 28.014 ≈ 0.9995 mol N₂; ratio 2 : 1 gives 1.999 mol NH₃; × 17.031 g/mol ≈ 34.0 g of ammonia.

Limiting reactant and percent yield

Real reactions rarely have the exact ratio of reactants. The limiting reactantThe reactant that runs out first and so sets the maximum amount of product. is the one that runs out first; it sets the maximum product. To find it, calculate the product each reactant could make on its own; the smaller answer wins.

Example: 4.0 mol H₂ and 1.5 mol O₂. Hydrogen alone could make 4.0 mol H₂O; oxygen alone could make 3.0 mol (ratio 2 : 1). Oxygen is limiting: 3.0 mol of water at most.

The amount you calculate is the theoretical yield. What you actually collect is usually less (side reactions, losses in transfer, incomplete reaction):

percent yield = actual / theoretical × 100%

If the theoretical yield is 35.7 g and you collect 30.4 g, the percent yieldActual yield divided by theoretical yield, times 100%. is 30.4 / 35.7 × 100 ≈ 85%.

Common mistakes

• Using grams in the mole ratio. Coefficients compare moles, never grams.
• Forgetting the subscript multiplier in molar mass (two hydrogens in H₂O).
• Using an unbalanced equation. Balance first, always.
• Ignoring the limiting reactant when both amounts are given.
• Percent yields over 100%. Usually a wet or impure product, not a miracle.

Quick recap

• 1 mol = 6.022 × 10²³ particles; molar mass in g/mol comes from the periodic table.
• Grams ↔ moles with molar mass; moles of A ↔ moles of B with the mole ratio.
• The limiting reactant sets the theoretical yield; percent yield compares reality to it.

Try it

Open the Do tab: you will complete a Python function that predicts the theoretical yield of a reaction, then take the quiz.