Calculate the percentage composition of glucose, C₆H₁₂O₆ (C = 12, H = 1, O = 16).

How to lay out the answer. Follow the frame, then write it in your own words.

1 · Define in one line. Percentage composition is the percentage by mass of each element present in a compound.
2 · Write the formula. % of an element = (mass of the element in 1 mole ÷ molar mass) × 100
3 · Find the molar mass. 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g
4 · Work out each element.
- % C = (72 ÷ 180) × 100 = 40.00%
- % H = (12 ÷ 180) × 100 = 6.67%
- % O = (96 ÷ 180) × 100 = 53.33%
5 · Check the total. 40.00 + 6.67 + 53.33 = 100%
6 · Real-life link. Glucose is the sugar in ORS and in our blood. Its carbon leaves the body as CO₂ when we breathe out.
7 · Close in one line. Therefore glucose contains 40% carbon, 6.67% hydrogen and 53.33% oxygen.

Scoring tip: Find the molar mass first, show each element on its own line and check that the total is 100%.