From Grams and Moles to Theoretical Yield

Measured reactant masses become the maximum mass of product through two linked calculations: converting mass to moles using molar mass, then using the balanced equation's stoichiometric ratios to find the product moles. The path from grams and moles to theoretical yield is a straight division and multiplication chain, but the most common failure is using the wrong molar mass. For a reaction like 2H₂ + O₂ → 2H₂O, start with 4.032 g of hydrogen gas (H₂) and 32.00 g of oxygen gas (O₂). Molar mass of H₂ is 2.016 g/mol, so moles H₂ = 4.032 / 2.016 = 2.000 mol. Molar mass of O₂ is 32.00 g/mol, so moles O₂ = 32.00 / 32.00 = 1.000 mol. The balanced equation shows that 2 mol H₂ react with 1 mol O₂, so hydrogen is the limiting reactant: 2.000 mol H₂ requires exactly 1.000 mol O₂, which you have. The theoretical yield of water is then 2.000 mol H₂ × (2 mol H₂O / 2 mol H₂) × 18.015 g/mol = 36.03 g H₂O. This assumes 100% reaction efficiency, which never happens in a lab.

Take the combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O. You have 8.035 g of methane (molar mass 16.042 g/mol, from C=12.011 + 4×1.008) and 32.00 g of oxygen gas. Moles CH₄ = 8.035 / 16.042 = 0.5009 mol. Moles O₂ = 32.00 / 32.00 = 1.000 mol. The stoichiometric ratio is 1 mol CH₄ to 2 mol O₂, so the required O₂ for 0.5009 mol CH₄ is 1.0018 mol. You have 1.000 mol O₂, so oxygen is the limiting reactant. Theoretical yield of CO₂: 1.000 mol O₂ × (1 mol CO₂ / 2 mol O₂) × 44.01 g/mol = 22.01 g CO₂. The failure case: if you mistakenly treat methane as the limiting reactant, you would calculate 0.5009 mol CH₄ × (1 mol CO₂ / 1 mol CH₄) × 44.01 g/mol = 22.05 g CO₂, a small error from rounding, but with different masses the error can be large.

You start with measured masses in grams. Convert each to moles using the formula n = m / M. Identify the limiting reactant by comparing the available moles to the required stoichiometric ratio. Then use the theoretical yield calculator to multiply the limiting reactant moles by the stoichiometric ratio from the balanced equation and the molar mass of the product. The governing formula is: Theoretical yield (g) = (moles of limiting reactant) × (coefficient ratio) × (molar mass of product). The IUPAC defines those coefficients as exact mole ratios, so you must have a correctly balanced equation. An unbalanced equation produces wrong stoichiometric ratios and a useless yield number.

Application to Ammonia Synthesis

For the Haber process: N₂ + 3H₂ → 2NH₃. With 28.01 g N₂ (molar mass 28.013 g/mol) and 6.048 g H₂ (molar mass 2.016 g/mol). Moles N₂ = 28.01 / 28.013 = 1.000 mol. Moles H₂ = 6.048 / 2.016 = 3.000 mol. The ratio is 1:3, exactly matched. Neither reactant is limiting. Theoretical yield of NH₃: 1.000 mol N₂ × (2 mol NH₃ / 1 mol N₂) × 17.031 g/mol = 34.06 g NH₃. The failure mode here is misreading the periodic table: using atomic mass of nitrogen (14.007) instead of molecular mass of N₂ (28.013) gives 2.000 mol N₂ and an overestimated yield of 68.12 g NH₃.

The conversion from grams to moles depends on substance-specific molar masses. Molar mass of water is 18.015 g/mol, not 18.0. The 0.015 g/mol gap comes from the weighted average of isotopes and matters in precise lab work. For sodium chloride, molar mass is 58.44 g/mol. For glucose, it is 180.156 g/mol. The Avogadro constant is exactly 6.02214076 × 10²³ mol⁻¹, though rounding to 6.022 × 10²³ is acceptable for most problems. Theoretical yield assumes complete conversion of the limiting reactant, but actual yield is always less due to side reactions, incomplete reactions, or losses during separation. If you have a hydrate like CuSO₄·5H₂O, you must include the water mass in the molar mass, or the mole calculation will be wrong.