TL;DR:
Stoichiometry is a three-step bridge: grams to moles, moles to moles using the balanced equation, moles to grams.
The coefficients of a balanced equation are the mole ratios.
Burning 16 g of methane (about 1 mol) makes 1 mol of CO₂, which is 44 g.
A recipe says one cup of flour makes twelve pancakes. If you want thirty-six pancakes, you use three cups. Chemistry questions work the same way: given an amount of one substance, how much of another is involved? The balanced equation is the recipe.
The balanced equation is the recipe
Methane, the main part of natural gas, burns with oxygen: CH₄ + 2 O₂ → CO₂ + 2 H₂O. The coefficients say that 1 molecule of methane reacts with 2 of oxygen to make 1 of carbon dioxide and 2 of water. Since a mole is a counting unit (see the mole, explained with a dozen), the same numbers hold for moles.
Those coefficients give mole ratios, such as 1 mol CO₂ : 1 mol CH₄ or 2 mol H₂O : 1 mol CH₄. Balancing comes first. If the equation is not balanced, the ratios are wrong. See balancing chemical equations by counting atoms if you need that step.
The three-step bridge
Grams to moles: Divide the given mass by its molar mass.
Moles to moles: Multiply by the ratio from the balanced equation, wanted over given.
Moles to grams: Multiply by the molar mass of the wanted substance.
You can never go straight from grams of one substance to grams of another, because the coefficients count particles, not mass. The mole is the middle step.
Worked example: 16 g of methane burning
How much carbon dioxide forms when 16.0 g of methane burns completely, and how much oxygen is used? Use molar masses CH₄ = 16.0 g/mol (12.0 + 4 × 1.0), O₂ = 32.0 g/mol, CO₂ = 44.0 g/mol (12.0 + 2 × 16.0) and H₂O = 18.0 g/mol.
Grams to moles of methane:
16.0 g × (1 mol / 16.0 g) = 1.00 mol CH₄.Mole ratio, CO₂ over CH₄:
1.00 mol CH₄ × (1 mol CO₂ / 1 mol CH₄) = 1.00 mol CO₂.Moles to grams:
1.00 mol CO₂ × (44.0 g / 1 mol) = 44.0 g CO₂.
Notice that the grams unit cancels in step 1, the moles of methane cancel in step 2, and the moles of CO₂ cancel in step 3. Only grams of CO₂ remain, which is a built-in check that you arranged the fractions correctly.
Now the oxygen. The ratio is 2 mol O₂ : 1 mol CH₄, so 1.00 × 2 = 2.00 mol O₂, and 2.00 × 32.0 = 64.0 g O₂. Water: 2.00 mol H₂O × 18.0 = 36.0 g.
Check conservation of mass. Reactants: 16.0 + 64.0 = 80.0 g. Products: 44.0 + 36.0 = 80.0 g. They match, which is strong evidence the arithmetic is right.
A different amount scales the same way. With 8.0 g of methane (0.50 mol), you get 0.50 mol CO₂, which is 0.50 × 44.0 = 22 g. Half the methane gives half the carbon dioxide.
Combustion is not a home experiment: Methane and other fuel gases form explosive mixtures with air, and burning in poor ventilation makes carbon monoxide, which is poisonous. Do not try combustion reactions yourself. Do the arithmetic on paper.
Here is the same method on a different question, to see that it is a pattern. How many grams of oxygen are needed to burn 4.0 g of methane? Grams to moles: 4.0 ÷ 16.0 = 0.25 mol CH₄. Mole ratio: 0.25 × 2 = 0.50 mol O₂. Moles to grams: 0.50 × 32.0 = 16 g O₂. One quarter of the methane needs one quarter of the 64.0 g of oxygen from before, and 64.0 ÷ 4 = 16. It agrees.
The method works for any pair of substances in the equation, reactants or products. To find how much water forms from 4.0 g of methane, use the ratio 2 mol H₂O : 1 mol CH₄: 0.25 × 2 = 0.50 mol, and 0.50 × 18.0 = 9.0 g. The ratio changes with the substance, but the three steps stay the same.
How to check your answer
Mass balance. Total mass of reactants equals total mass of products.
Units cancel. Write the units on every fraction and make sure only the wanted one remains.
Ratio direction. The wanted substance goes on top in the mole ratio.
Significant figures. Match the least precise input, as in measurement uncertainty and significant figures.
Mistakes that cost marks
Using an unbalanced equation.
Using the ratio of grams, or of coefficients, as if they were masses.
Putting the mole ratio upside down.
Using the molar mass of the wrong substance in the first or last step.
Forgetting the second conversion back to grams.
Where you will use it next
The method extends to gases, solutions and yields. When more than one reactant is given, you need to find which one runs out first, which is the subject of limiting reagents.
Frequently asked questions
Can I skip the mole step if the ratio is 1 to 1?
No. Even with a 1 to 1 ratio, the molar masses differ, so equal masses are not equal moles. Methane and CO₂ at 16 g and 44 g show it.
What if the question gives volume of a solution?
Use molarity, moles per litre, to turn volume into moles, then continue with the same steps.
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