Derive the molecular formula of a substance consisting of carbon, hydrogen and bromine, the mass fractions

Derive the molecular formula of a substance consisting of carbon, hydrogen and bromine, the mass fractions of which are 40%, 6.67% and 53.33%, respectively. The oxygen density of the substance vapors is 4.69.

Given:

w% (C) = 40%

w% (H) = 6.7%

w% (Br) = 53.3%

D (O2) = 4.69

Find:

Formula -?

Solution:

1) Find the molecular weight of a substance by its oxygen density:

D (O2) = Mr (things): Mr (O2)

Mr (things) = D (O2) * Mr (O2) = 4.69 * 32 = 150

2) Knowing the mass fraction of carbon, we find its molecular weight and the number of atoms:

150 – 100%

x – 40%,

then x = 40% * 150: 100% = 60

N (C) = 60: Mr (C) = 60: 12 = 5 atoms.

3) Knowing the mass fraction of bromine, we find its molecular weight and the number of atoms:

150 – 100%

x – 53.3%,

then x = 53.3% * 150: 100% = 79.95

N (Br) = 79.95: Mr (Br) = 79.95: 80 = ~ 1 atom.

4) Then the molecular weight of hydrogen is 150 – 60 – 79 = 11.

N (H) = 11: Mr (H) = 11: 1 = 11 atoms.

We got a substance with the formula C5H11Br – bromopentane.

Answer: C5H11Br – bromopentane.