You have a small but seemingly pure sample of an unknown white powder containing

Question

You have a small but seemingly pure sample of an unknown white powder containing only C, H, and O. To analyze the compound, you determine its empirical formula by analysis for C and H, and its molar mass by freezing point depression. a) what is the empirical formula of a 0.01510 g sample gives 0.01102 g of H2O and 0.04484 g of CO2 on combustion? b) if you dissolve 1.000 g of the powder in 100.000 g of camphor, you find that the melting point of the solution is 177.05 C. Pure camphor melts at 179.75 C. The Kp for camphor is !40.0 deg/molal. What is the molar mass of the white powder? c) what is the molecular formula of the white powder ?

Explanation / Answer

mass of H in 0.01102g H2O
2.016/18.015*0.01102 = 0.0012332g

mass C in 0.04484g CO2
12.011/44.0096* 0.04484 = 0.01224g

Calculate mass O = 0.01510 - ( 0.0012332+0.01224) = 0.001629g

divide each by respective atomic mass
C = 0.01224g/12.011 = 0.0010191
H = 0.0012332g /1.008 = 0.0012234
O = 0.001629/15.999 = 0.000101819

Divide through by smallest:
C = 10.0
H = 12.0
O = 1.0
Empirical formula = C10H12O

Question b) you have dissolved 10g powder in 1kg camphor
melting point lowered by 179.75 - 177.05 = 2.7°C
Kf for camphor = 40°C/molal
therefore you have a 2.7/40 = 0.0675 molal solution.
10g = 0.0675mol
1mol = 10/0.0675 = 148.148g/mol

Empirical formula, C10H12O = 120+12+16 = 148g/mol
The empirical formula is the molecular formula.

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