Question

Question #0ccb5

Answer 1

The empirical formula and molecular formulas are `C_2H_4Cl_2`

Explanation:

First change the percentages to masses as if there are 100 grams then each 1 % = 1 gram.

`4.07 % H = 4.07 g H`

`24. 27% C = 24.27 g C`

`71.65% Cl = 71.65 g Cl`

Second change the grams to moles by dividing by the g/mole

`4.07/ 1 = 4`moles H

`24.27/12 = 2` moles C

`71.65/35.4 = 2` moles Cl

Third change the ratio of moles to a whole number ratio ( done)

The empirical formula is `C_2 H_4 Cl_2`

The molecular weight of this formula is
4 x H = 2 x 1 = 4

2 x C = 2 x 12 = 24

2 x Cl = 2 x 35.4 = 70.8

adding these together gives 98.8 approximately the same as
the molecular mass of the compound 98.96

So the empirical formula is the same as the molecular formula
Cl Cl
| |
H- C- C - H
| |
H H

The bonding works well for this formula.

Answer 2

Dividing the given percentages of the constituent elements in the compound by the respective atomic mass and taking their ratio we get the ratio of number of atoms of constituent elements in the molecule of the given compound as

Ratio of number of atoms

`C:H:Cl=24.27/12:4.07/1:71.65/35.5`

`=2.02:4.07:2.02~~1:2:1`

So empirical formula `CH_2Cl`
Let its molecular formula be `(CH_2Cl)_n`

Given the molar mass of the compound `98.96g"/"mol`

So `(12+2xx1+35.5)xxn=98.96`

`=>49n=98.96`

`=>n~~2`

Hence its molecular formula is `(CH_2Cl)_2=C_2H_4Cl_2`