Question

A 1.052g sample of an unknown compound, composed only of carbon and hydrogen, produced 3.19g of CO2 and 1.63g of h2o in a combustion analysis. What is the mass percent composition of hydrogen of the unknown compound?

Answer 1

Answer on Question 82370 in General Chemistry

.m $(C_xH_y) = 1.052$ g

.m(CO$_2$) = 3.19 g

.m $(H_2O) = 1.63$ g

.ω (H) = ?

Find the amount of substance of CO$_2$ n = $\frac{m}{Mr} = \frac{3.19}{44} = 0.0725$ mol

Mr(CO$_2$) = Ar (C) + 2 Ar(O) = 12 + 2 × 16 = 44

.n (C) = n(CO$_2$) = 0.0725 mol

.m(C) = n × Ar = 0.0725 × 12 = 0.87 g

Find the amount of substance of $H_2O$ n = $\frac{m}{Mr} = \frac{1.63}{18} = 0.091$

.n(H) = 2 n(H$_2$O) = 2 × 0.091 = 0.182

.m (H) = Ar × n = 0.182 × 1 = 0.182

Mr(H$_2$O) = 2 Ar(H) + Ar (O) = 2 × 1 + 16 = 18

Total weight of compound m = m (C) + m(H) = 0.87 + 0.182 = 1.052

There is no other element except carbon and hydrogen in the compound.

.ω (H) = $\frac{m(H)}{m(compound)} \times 100\% = \frac{0.182}{1.052} \times 100\% = 17.3\%$

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