Basic Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. The word comes from the Greek stoicheion (element) and metron (measure) — literally, measuring elements.

A balanced chemical equation tells you exactly how much of each substance is involved. The coefficients in front of each formula give you the ratio of moles — and from moles, you can calculate masses, volumes, and number of particles.

Every stoichiometry problem follows the same core pattern: use the balanced equation to build a mole ratio (also called a stoichiometric factor), then use that ratio to convert between substances.

The coefficients in a balanced equation directly give you the mole ratios between any two substances in the reaction. Consider the synthesis of ammonia:

3 H₂ + N₂ → 2 NH₃

From this equation, you can derive several stoichiometric factors:

• 3 mol H₂ : 1 mol N₂
• 3 mol H₂ : 2 mol NH₃
• 1 mol N₂ : 2 mol NH₃

Each ratio can be flipped depending on what you're solving for. If you know the moles of one substance, multiply by the appropriate mole ratio to find the moles of another.

Example: How many moles of I₂ react with 0.429 mol Al?

2 Al + 3 I₂ → 2 AlI₃

The mole ratio is 3 mol I₂ per 2 mol Al:

0.429 mol Al × (3 mol I₂ / 2 mol Al) = 0.644 mol I₂

In the lab, you measure masses — not moles. A mass-to-mass stoichiometry problem adds two conversion steps around the mole ratio:

1. Mass → Moles: Divide the given mass by the molar mass of that substance
2. Moles → Moles: Apply the mole ratio from the balanced equation
3. Moles → Mass: Multiply by the molar mass of the target substance

The pattern is always: grams → moles → mole ratio → moles → grams.

Example: What mass of NaOH is needed to produce 16 g of Mg(OH)₂?

MgCl₂ + 2NaOH → Mg(OH)₂ + 2NaCl

Step 1: 16 g Mg(OH)₂ ÷ 58.32 g/mol = 0.274 mol Mg(OH)₂

Step 2: 0.274 mol Mg(OH)₂ × (2 mol NaOH / 1 mol Mg(OH)₂) = 0.549 mol NaOH

Step 3: 0.549 mol NaOH × 40.00 g/mol = 22.0 g NaOH

When two or more reactants are mixed in amounts that don't match the exact mole ratio from the balanced equation, one will run out first. The reactant that is completely consumed is the limiting reagent — it limits how much product can form. The other reactant is present in excess.

To identify the limiting reagent:

1. Convert each reactant's mass to moles
2. Divide each by its coefficient in the balanced equation
3. The reactant with the smallest value is the limiting reagent

Alternative method: Calculate how much product each reactant could produce independently. The reactant that yields the lesser amount of product is limiting.

Example: Which is limiting when 3 mol H₂ and 2 mol Cl₂ react?

H₂ + Cl₂ → 2 HCl

The stoichiometric ratio is 1:1. You have 3:2. Since Cl₂ has fewer moles relative to its coefficient (2/1 = 2) compared to H₂ (3/1 = 3), Cl₂ is the limiting reagent. It will be completely consumed, producing 4 mol HCl, with 1 mol H₂ left over.

The theoretical yield is the maximum amount of product that could be formed from a given amount of limiting reagent, calculated using stoichiometry. It assumes the reaction goes to completion with no losses.

In practice, you almost always get less product than the theoretical yield. Reasons include:

• Side reactions that produce unwanted products
• Incomplete reactions that don't go to 100% completion
• Mechanical losses during collection, transfer, or purification

The actual yield is the amount you actually obtain in the lab. The percent yield compares the two:

Percent yield = (actual yield / theoretical yield) × 100%

Both yields must be in the same units (grams, moles, etc.) for this formula to work.

Example: Reacting 1.274 g CuSO₄ with excess Zn gives 0.392 g Cu. The theoretical yield of Cu is 0.508 g. Percent yield = (0.392 / 0.508) × 100% = 77.2%

Once you identify the limiting reagent, you can calculate exactly how much of the excess reagent remains unreacted:

1. Use the moles of the limiting reagent to calculate how many moles of the excess reagent are consumed (via the mole ratio)
2. Subtract the consumed amount from the original amount of the excess reagent

Example: If 3 mol H₂ and 2 mol Cl₂ react (1:1 ratio), Cl₂ is limiting. It consumes 2 mol H₂. Remaining H₂ = 3 − 2 = 1 mol H₂ unreacted.

To convert this to grams, multiply by the molar mass: 1 mol H₂ × 2.016 g/mol = 2.016 g H₂ remaining.

At standard temperature and pressure (STP: 0 °C and 1 atm), one mole of any ideal gas occupies 22.4 L. This is called the standard molar volume.

This fact gives you an additional conversion factor for stoichiometry problems involving gases at STP:

1 mol gas = 22.4 L at STP

You can now convert between volume and moles without needing the ideal gas law (PV = nRT) — as long as the gas is at STP.

Example: What volume of O₂ at STP is needed to react with 2.7 L of propane (C₃H₈) at the same conditions?

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

When gases are at the same temperature and pressure, their volume ratios equal their mole ratios (Avogadro's law). Since the coefficient ratio is 1:5, you need 5 × 2.7 L = 13.5 L of O₂.

For gases not at STP, use the ideal gas law: n = PV/RT to convert between volume and moles, then proceed with the normal stoichiometric mole ratio.

Many real stoichiometry problems combine several of the skills above into a single calculation. A typical multi-step problem might ask you to:

• Start with a mass of one reactant
• Identify the limiting reagent
• Calculate the theoretical yield in grams
• Determine the percent yield from an actual yield
• Find how much excess reagent remains

The key to solving these is to work one step at a time. Don't try to do everything in one equation. Follow the chain:

Given quantity → moles → mole ratio → moles → target quantity

At each step, check your units and make sure you're using the right mole ratio from the balanced equation. Dimensional analysis is your best tool for keeping track of conversions.

Every stoichiometry problem follows the same core workflow:

1. Write and balance the equation. No calculation is valid without a balanced equation.
2. Convert the given quantity to moles (using molar mass, molarity, or molar volume at STP).
3. Use the mole ratio from the balanced equation to find moles of the target substance.
4. Convert moles of target to the requested unit (grams, liters, particles).
5. For limiting-reagent problems: calculate moles of product from each reactant separately. The reactant that gives the smaller amount of product is the limiting reagent.

Common mistakes: using an unbalanced equation, skipping the grams-to-moles conversion and plugging grams directly into the mole ratio, confusing limiting and excess reagents, and forgetting that percent yield = (actual/theoretical) × 100%. A quick check: if your percent yield exceeds 100%, re-examine your theoretical yield calculation.