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How to Solve Stoichiometry Problems: Step-by-Step Chemistry Guide

Master mole ratios, grams-to-grams conversion bridges, theoretical yield, and limiting reactant calculations.

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Sparky Academic TeamCurriculum Specialist
9 min readUpdated 2026-10-17
Quick Summary / TL;DR

Stoichiometry uses the coefficients of a balanced chemical equation to convert between amounts of reactants and products. Always follow the golden 3-step mole bridge: (1) Convert Grams A to Moles A using molar mass, (2) Convert Moles A to Moles B using the balanced mole ratio, and (3) Convert Moles B to Grams B using molar mass!

Core Equation / Rule:
Grams A \xrightarrow{÷ MM_A} Moles A \xrightarrow{× coeff Bcoeff A} Moles B \xrightarrow{× MM_B} Grams B
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🌉 Grams to Moles, Mole to Mole, Moles to Grams: Cross the Stoichiometry Bridge!
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The 4-Step Solving Process

1

Ensure the Chemical Equation is Fully Balanced

N_2 + 3H_2 \longrightarrow 2NH_3

You cannot perform stoichiometry without correct coefficients from a balanced equation.

💡 Pro Tip: The large numbers in front (1, 3, 2) are your conversion factors (mole ratios)!
2

Step 1 of Bridge: Convert Given Grams to Moles

Moles A = Grams AMolar Mass A

Divide given mass of substance A by its molar mass from the periodic table.

💡 Pro Tip: You can NEVER jump straight from grams of A to grams of B—you MUST pass through the mole kingdom!
3

Step 2 of Bridge: Apply the Mole Ratio (B / A)

Moles B = Moles A × (Coefficient of BCoefficient of A)

Multiply by the fraction: (coefficient of unknown B) / (coefficient of known A).

💡 Pro Tip: Always put 'what you want' on top and 'what you have' on bottom so units cancel.
4

Step 3 of Bridge: Convert Moles B to Grams

Grams B = Moles B × Molar Mass B

Multiply moles of substance B by its molar mass to get the final theoretical mass.

💡 Pro Tip: Check significant figures based on the initial given value in the problem.

What is Stoichiometry?

Stoichiometry (pronounced stoy-kee-AH-meh-tree) is the math of chemistry. It allows you to predict:

  • •Exactly how many grams of product a chemical reaction will make.
  • •Exactly how many grams of reactant you need to mix together so nothing is wasted.

Think of a recipe: If 2 slices of bread + 1 slice of cheese makes 1 sandwich, stoichiometry lets you calculate how many sandwiches you can make from 20 slices of bread.


The Master 3-Step Stoichiometry Bridge

In chemistry, you cannot directly compare grams of one substance to grams of another because different molecules have different atomic weights. You must convert through MOLES!

Grams A \xrightarrow{÷ Molar Mass A} Moles A \xrightarrow{× Coeff BCoeff A} Moles B \xrightarrow{× Molar Mass B} Grams B
  1. Grams to Moles: Divide by molar mass of A.
  2. Mole to Mole Ratio: Multiply by (Coefficients from balanced equation: Want BHave A).
  3. Moles to Grams: Multiply by molar mass of B.

Worked Example 1: Full Grams-to-Grams Problem

Problem:

In the synthesis of ammonia gas (NH_3):

N_2 + 3H_2 \longrightarrow 2NH_3

How many grams of ammonia (NH_3) are produced when 28.02 grams of nitrogen gas (N_2) reacts completely with excess hydrogen?

Step 1: Molar masses from periodic table

  • •Molar mass of N_2 = 2 × 14.01 = 28.02 g/mol
  • •Molar mass of NH_3 = 14.01 + (3 × 1.01) = 17.04 g/mol

Step 2: Bridge Step 1 (Grams N_2 to Moles N_2)

Moles of N_2 = 28.02 g28.02 g/mol = 1.00 mol N_2

Step 3: Bridge Step 2 (Mole Ratio: N_2 → NH_3)

Look at the balanced coefficients: 1 N_2 produces 2 NH_3:

Moles of NH_3 = 1.00 mol N_2 × (2 mol NH_31 mol N_2) = 2.00 mol NH_3

Step 4: Bridge Step 3 (Moles NH_3 to Grams NH_3)

Grams of NH_3 = 2.00 mol × 17.04 g/mol = \mathbf{34.08 grams}

Final Answer: You will produce 34.08 g of ammonia gas.


Interactive Practice Check
For the reaction: $2H_2 + O_2 → 2H_2O$. Step 1: If you react 4 moles of $H_2$ with excess oxygen, how many moles of $H_2O$ will be produced? (Mole ratio is $2 mol H_2O : 2 mol H_2$).

Worked Example 2: How to Solve Limiting Reactant Problems

In real life, reactants rarely match the exact mole ratio. One reactant will run out first—this is the Limiting Reactant. The limiting reactant decides the maximum amount of product that can be formed (Theoretical Yield).

Problem:

You react 2.0 moles of Mg with 3.0 moles of HCl:

Mg + 2HCl \longrightarrow MgCl_2 + H_2

Which is the limiting reactant, and how many moles of H_2 gas will be formed?

Step 1: Calculate product yield from EACH reactant separately

  • •From Mg:
2.0 mol Mg × (1 mol H_21 mol Mg) = 2.0 mol H_2
  • •From HCl:
3.0 mol HCl × (1 mol H_22 mol HCl) = 1.5 mol H_2

Step 2: Compare results

  • •HCl produces only 1.5 mol H_2, while Mg could produce 2.0 mol H_2.
  • •The smaller number wins!
  • •Limiting Reactant: HCl (runs out first).
  • •Excess Reactant: Mg (some remains unreacted).
  • •Theoretical Yield of H_2: 1.5 moles.

Percent Yield Formula

On chemistry lab reports, you compare what you actually measured on the lab scale to what stoichiometry predicted on paper:

Percent Yield = (Actual Yield (from lab)Theoretical Yield (from math)) × 100%
⚠️WATCH OUT

Never Use Coefficients When Calculating Molar Mass!

In 2NH_3, the molar mass of NH_3 is 17.04 g/mol, NOT 34.08 g/mol. The 2 is accounted for strictly in the mole ratio step, never inside the molar mass calculation!

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