1. Convert the given masses of CO2 and H2O to moles.
- Moles of CO2 = 103.9 g / 44.01 g/mol = 2.36 mol
- Moles of H2O = 8.50 g / 18.02 g/mol = 0.472 mol
2. Determine the moles of carbon and hydrogen in the sample.
- Moles of C = Moles of CO2 = 2.36 mol
- Moles of H = 2 x Moles of H2O = 2 x 0.472 mol = 0.944 mol
3. Assume that the total mass of the sample is 59.5 g, which represents the sum of the masses of carbon, hydrogen, and oxygen.
- Mass of oxygen = Total mass - (Mass of carbon + Mass of hydrogen)
- Mass of oxygen = 59.5 g - (2.36 mol x 12.01 g/mol + 0.944 mol x 1.01 g/mol)
- Mass of oxygen = 59.5 g - (28.36 g + 0.952 g)
- Mass of oxygen = 59.5 g - 29.312 g
- Mass of oxygen = 30.188 g
4. Convert the masses of carbon, hydrogen, and oxygen to moles.
- Moles of C = 2.36 mol
- Moles of H = 0.944 mol
- Moles of O = 30.188 g / 16.00 g/mol = 1.887 mol
5. Divide the moles of each element by the smallest number of moles to obtain the simplest mole ratio.
- Moles of C / 1.887 mol = 2.36 mol / 1.887 mol ≈ 1.25
- Moles of H / 1.887 mol = 0.944 mol / 1.887 mol ≈ 0.5
- Moles of O / 1.887 mol = 1.887 mol / 1.887 mol = 1
6. Multiply the mole ratios by a suitable factor to obtain whole numbers. In this case, multiplying by 2 gives:
- Moles of C ≈ 2.5
- Moles of H ≈ 1
- Moles of O ≈ 2
Therefore, the empirical formula of the carbohydrate is approximately C2.5H1O2.
