A compound containing only C, H and O (0.215 g) on Liebig combustion gave 0.473 g of CO2 and 0.1548 g of H2O. Oxygen is obtained by difference. Using H = 1, C = 12, O = 16, the empirical formula of the compound is

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A compound containing only C, H and O (0.215 g) on Liebig combustion gave 0.473 g of $CO X_{2}$ and 0.1548 g of $H X_{2} O$ . Oxygen is obtained by difference. Using H = 1, C = 12, O = 16, the empirical formula of the compound is

A $C X_{3} H X_{4} O$
B $C X_{4} H X_{8} O$
C $C X_{5} H X_{8} O X_{2}$
D $C X_{5} H X_{10} O X_{2}$

Answer: C

Mass of C =

\frac{12}{44} \times 0.473 = 0.1290

g, mass of H =

\frac{2}{18} \times 0.1548 = 0.0172

g, so mass of O = 0.215 - 0.1290 - 0.0172 = 0.0688 g. Moles: C = 0.01075, H = 0.0172, O = 0.00430. Dividing by the smallest gives C : H : O = 2.50 : 4.00 : 1, and doubling to remove the 2.5 gives

C X_{5} H X_{8} O X_{2}

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A compound containing only C, H and O (0.215 g) on Liebig combustion gave 0.473 g of COX2​ and 0.1548 g of HX2​O. Oxygen is obtained by difference. Using H = 1, C = 12, O = 16, the empirical formula of the compound is

Answer: C

More questions on Estimation of elements: Lassaigne, Kjeldahl, Carius

A compound containing only C, H and O (0.215 g) on Liebig combustion gave 0.473 g of $CO X_{2}$ and 0.1548 g of $H X_{2} O$ . Oxygen is obtained by difference. Using H = 1, C = 12, O = 16, the empirical formula of the compound is