Question

2.516 g of a compound containing carbon, hydrogen and oxygen (C_XH_YO_Z) is subjected to combustion analysis.

The results show that 3.082 g of CO₂ and 2.705 g of H₂O were produced.

What is the empirical formula for the compound?

If the molecular weight of the compound is 160.2 g/mol, what is the molecular formula of the compound?   

Answer 1

1) Determine the grams of carbon in 3.082 g CO₂ and the grams of hydrogen in 2.705 g H₂O.

carbon: 3.082 g x (12.011 g / 44.0098 g) = 0.8411 g

hydrogen: 2.705 g x (2.0158 g / 18.0152 g) = 0.302674 g

Determine the grams of oxygen in the sample by subtraction.

2.516 - (0.8411 g + 0.302674) = 1.372226 g

2) Convert grams of C , H and O to their respective amount of moles.

carbon: 0.8411 g / 12.011 g/mol = 0.070027 mol

hydrogen: 0.302674 g / 1.0079 g/mol = 0.3003 mol

Oxygen : 1.372226 g/15.9994 g/mol = 0.0876 mol

3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.

Carbon:     0.070027 mol / 0.070027 mol = 1

hydrogen: 0.3003 mol / 0.070027 mol = 4.2 = 4

oxygen:      mol / 0.070027 mol = 1.22 = 1

4) So, we can write Empirical formula as, C₁H₄O₁

5) Given that molecular formula is 160.2 g/mol

Therefore 160.2/Empirical formula weight = 160.2/32 = 5

Therefore now molecular formula will be C₅H₂₀O₅