Example Calculation - 2

This worked example shows why the coefficients in the equation matter in an atom economy calculation.Question: hydrogen is made industrially by reacting methane with steam. Calculate the atom economy for making hydrogen. (A_r: H = 1.0, C = 12.0, O = 16.0)CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)Step 1: mass of desired product from the equation. Three moles of H₂ form, so use 3 × 2.0 = 6.0Step 2: total mass of reactants. CH₄ = 16.0 and H₂O = 18.0, so the total is 34.0Step 3: atom economy.atom economy = (6.0 ÷ 34.0) × 100 = 17.6%Check: the total mass of products must equal the total mass of reactants. CO = 28.0 and 3H₂ = 6.0 give 34.0, which matches.Interpreting the result: only about a sixth of the mass of the reactants ends up as hydrogen; the rest forms carbon monoxide. The atom economy is low, but the process becomes much more efficient if the carbon monoxide is used too, for example as a fuel or to make more hydrogen. If both products are wanted, all the mass ends up in useful products and the atom economy is effectively 100%.Common error: using the M_r of one H₂ (2.0) instead of 3 × 2.0. That would give 5.9%, which is wrong.