A 7.85-g sample of compound with empirical formula C_5H_4 is dissolved in 301g o

Question

A 7.85-g sample of compound with empirical formula C_5H_4 is dissolved in 301g of benzene. The freezing point of the solution is 1.05 degree C below that of pure benzene (5.5 degree C). What are the molar mass and molecular formula of this compound? The elemental analysis of an organic solid extracted from gum arabic showed that it contained 40.0% C, 6.7% H, and 53.3% O. A solution of 0.650 g of the solid in 27.8 g of the solvent diphenyl gave a freezing-point depression of 1.56 degree C. Calculate the molar mass and molecular formula of the solid (k_f for diphenyl is 8.00 degree C/m). Coniferia, a sugar derivative found in conifers such as fir trees, has a composition of 56.13% C, 6.48% H, and 37.39% O by mass. A 2.216-g sample is dissolved in 48.68 g of H_2O and the solution is found to have a boiling point of 100.068 degree C. What is the molecular formula of coniferia?

Explanation / Answer

1) change in temperature = 1.05 C

KF for benzene = 5.12

We know that,

Delta T = Kf * molality of solute

-1.05 = -5.12 * m

m = 0.2050

We know m = (mass of solute/molar mass)*(1/kg solvent)

= (7.85/molar mass) *(1/0.301)

0.2050 = 26.08 / molar mass

Molar mass = 26.08/0.2050 = 127.22 g/mol

Emperical formula = C5H4

Emperical weight = 12*5+4*1 = 64 g/mol

Molar mass/ emperical weight = 2

So molecular formula = 2*C5H4 == C10H8

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