When a 0.3167 gram sample of an unknown compound, (containing only carbon, hydro

Question

When a 0.3167 gram sample of an unknown compound, (containing only carbon, hydrogen and nitrogen), is burned, 0.7733 gram of carbon dioxide and 0.2109 gram of water are produced. When a 2.1078 gram sample of the same unknown compound is dissolved in 23.2370 grams of benzene, (C_6H_6), the mixture is found to freeze at 0.985degreeC. Given that benzene has a freezing point depression constant of 5.065degreeC per molal, a boiling point elevation constant of 2.61degreeC per molal, a freezing point of 5.455degreeC and a normal boiling point of 80.2degreeC, determine the molecular formula of the unknown compound.

Explanation / Answer

we know that

moles = mass / molar mass

so

moles of C02 = 0.7733 / 44 = 0.017575

so

moles of C = 0.017575

now

moles of H20 = 0.2109 / 18 = 0.01171666

so

moles of H = 2 x 0.01171666 = 0.02343333

now

mass = moles x atomic mass

so

mass of C = 0.017575 x 12 = 0.2109 g

now

mass of H = 0.02343333 x 1 = 0.02343333 g

so

mass of N = 0.3167 - ( 0.2109 + 0.02343333)

mass of N = 0.0823667

moles of N = 0.0823667 / 14 = 5.8833 x 10-3

now

consider the ratio of moles of C , H , N

C:H:N = 0.017575 : 0.02343333 : 5.8833 x 10-3

C:H:N = 3 : 4 : 1

so

the empirical formula is C3H4N

we know that

depression in freezing point is given by

dTf = Kf x m

now

dTf = freezing point of benzene - freezing point of solution

so

dTf = 5.455 - 0.985

dTf= 4.47

now

dTf = kf x m

4.47 = 5.065 x m

m = 0.882527

so

molality of the solution (m) = 0.882527

now

molality = moles of solute x 1000 / mass of benzene (g)

0.882527 = moles of solute x 1000 / 23.237

moles of solute = 0.020507283

now

moles = mass / molar mass

so

0.020507283 = 2.1078 / molar mass

molar mass = 102.783 g/ mol

let the molecular formula be (C3H4N)n

n = molar mass / empirical mass

n = 102.783 / 54

n = 1.9

so

the molecular formula is C6H8N2

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