A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of carbon dioxide and 0.930g of water were produced. Find the empirical formula of the compound.

C₂ H₄ O

Explanation:

1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)

atomic mass of C: 12.0107 g/mol molar mass of CO₂: 44.01 g/mol Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂ Solve for x:

x = (12.0107 g of C / 44.01 g of CO₂) * 2.190 g of CO₂ = 0.59767 g of C

2) Mass of hydrogen (H) in 0.930 g of water (H₂O)

atomic mass of H: 1.00784 g/mol molar mass of H₂O: 18.01528 g/mol proportion: 2 * 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O Solve for x:

x = (2 * 1.00784 g of H / 18.01528 g of H₂O) * 0.930 g of H₂O = 0.10406 g of H

3) Mass of oxygen (O) in 1.0857 g of pure sample

Mass of O = mass of pure sample - mass of C - mass of H Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

4) Mole calculations

Divide the mass in grams of each element by its atomic mass:

C: 0.59767 g / 12.0107 g/mol = 0.04976 mol H: 0.10406 g / 1.00784 g/mol = 0.10325 mol O: 0.3840 g / 15.999 g/mol = 0.02400 mol

5) Divide every amount by the smallest value (to find the mole ratios)

C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2 H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4 O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

C₂ H₄ O ← answer