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A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this...

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A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compund to 50.0 mL benzene, C6H6 (d= 0.879 g/mL; Kf= 5.12 degrees celcius/m), lowers the freezing point from 5.53 to 1.37 degrees celcius. What is the molecular formula of this compound?

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The compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass.

Atom ratio C:H:N:O =  42.9/12:2.4/1.008:16.7/14:38.1/16 = 3:2:1:2

So, empirical formula of the compound is C3H2NO2.

Let its molecular formula be (C3H2NO2)n.

The molar mass can be obtained from freezing point lowering experiments.

We know lowering of freezing point is a colligative property, given by ∆Tf = Kf  × m  ----- (i)

Where, ∆Tf is the extent of lowering, Kf is the molal cryoscopic constant of the solvent, and m, molality = moles of solute dissolved per 1000 gram of solvent.

Here, 6.45 g=6.45/M moles of this compound is dissolved per 50.0 mL benzene = 50*0.879 g = 43.95 g benzene.

Molality of this solution is: 6.45 *1000/(43.95*M) = 146.7577/M

∆Tf = 5.53 – 1.37 = 4.16 ᵒC = 4.16 K.

Putting the values in eqn (i), we get

4.16 = 5.12*146.7577/M

Or, M = 5.12*146.7577/4.16 = 180.62

Now, from its molecular formula, (3*12+2*1.008+14+2*16)n = 180.62

Or, 84n = 180.62,

Or, n = 180.62/84 = 2.15 = 2 (because n must be a whole number).

Therefore, the molecular formula of the compound is (C3H2NO2)2 , or C6H4N2O4.

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