Question

A 27.129 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 48.328 mg of carbon dioxide and 19.783 mg of water. In another experiment, 34.687 mg of the compound is reacted with excess oxygen to produce 14.992 mg of sulfur dioxide. Add subscripts below to correctly identify the empirical formula of this compound (use this order of elements: CHSO).

Answer 1

Answer on Question 76941 in General Chemistry

.m₁ (sample) = 27.129 g

.m(CO₂) = 48.328 g

.m(H₂O) = 19.783 g

.m₂ (sample) = 34.687 g

.m(SO₂) = 14.992 g

Formula = ?

Find the amount of substance of CO₂

.n(CO₂) = \frac{m(CO_2)}{Mr(CO_2)} = \frac{48.328}{44} = 1.1 mol

Mr(CO₂) = Ar (C) + 2 Ar(O) = 12 + 2 × 16 = 44

.n(C) = n(CO₂) = 1.1 mol

.m(C) = n × Ar(C) = 1.1 × 12 = 13.2 g

Find the amount of substance of H₂O

.n(H₂O) = \frac{m(H_2O)}{Mr(H_2O)} = \frac{19.783}{18} = 1.1 mol

Mr(H₂O) = 2Ar(H) + Ar(O) = 2 + 16 = 18

.n(H) = 2n(H₂O) = 2.2 mol

.m(H) = n(H) × Ar(H) = 2.2 × 1 = 2.2 g

By the proportion we find the mass of SO₂ which is formed during combustion of 27.127 g of substance

34.687 g of substance gives 14.992 g of SO₂

27.129 g of substance gives x g of SO₂

.x = m(SO₂) = \frac{27.129 \times 14.992}{34.687} = 11.725 g

Find the amount of substance of SO₂

.n = \frac{m(SO_2)}{Mr} = \frac{11.725}{64} = 0.18 mol

$$\begin{array}{l} \mathrm{Mr}(SO_2) = \mathrm{Ar}(S) + 2\mathrm{Ar}(O) = 32 + 2 \times 16 = 64 \\ \cdot n(S) = n(SO_2) = 0.18 \text{ mol} \\ \cdot m(S) = N \times \mathrm{Ar} = 32 \times 0.18 = 5.76 \text{ g} \end{array}$$

Find the mass of oxygen

$$\begin{array}{l} \cdot m(O) = m(\text{substance}) - m(C) - m(H) - m(S) = 27.129 - 13.2 - 2.2 - 5.76 = 5.969 \text{ g} \\ \text{Find the amount of substance of } O \\ \cdot n = \frac{m(O)}{Ar(O)} = \frac{5.969}{16} = 0.37 \text{ mol} \\ \cdot n(C) : n(H) : n(S) : n(O) = 1.1 : 2.2 : 0.18 : 0.37 = 6 : 12 : 1 : 2 \\ \end{array}$$

The formula is $C_6H_{12}SO_2$

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