Question

A compound contains only c, h, and n. it is 58.51% c and 7.37% h by mass. helium effuses through a porous frit 3.20 times as fast as the compound does. determine the empirical and molecular formulas of this compound. (type your answer using the format c2h2n for c2h2n.

Answer 1

Let us assume that there are 100 grams of substance present. The masses of the constituents will be:

C - 58.51 g
H - 7.37 g
N - 100 - 58.51 - 7.37 = 34.12 g

Next, we determine the moles of each substance by dividing by the atomic masses:

C - 58.51/12 = 4.88
H - 7.37/1 = 7.37
N - 34.12/14 = 2.44

Dividing all of the moles by the lowest number of moles, which is 2.44 of nitrogen, we get the ratio:

C - 2
H - 3
N - 1

The empircal formula becomes C₂H₃N

Next, we apply Graham's Law of diffusion, which mathematically is:

v₁/v₂ = √M₂ / √M₁

Let species 1 be helium, meaning v₁/v₂ = 3.2 and M₁ = 4

3.2 = √M₂ / √4

M₂ = 40.96

The mass of the empircal formula was:
12 * 2 + 1 * 3 + 14 = 41

The empircal formula is the molecular formula as their masses are approximately the same. This compound is C₂H₃N