Question

An unknown compound with a molar mass of 155.06 g/mol consists of 46.47% c, 7.80% h, and 45.72% cl. find the molecular formula for the compound. c6h12cl2 chcl c9h18cl3 c6h12cl

Answer 1

The emperical formula of compound is calculated as:

Atomic ratio for Carbon,
`C`:

`atomic ratio = (46)/(12) = 3.833`

Atomic ratio for Hydrogen,
`H`:

`atomic ratio = (7.8)/(1) = 7.8`

Atomic ratio for Chlorine,
`Cl`:

`atomic ratio = (45.72)/(35.5) = 1.29`

The simplest ratio is:

For Carbon =
`(3.833)/(1.29) = 2.97 \simeq 3`

For Hydrogen =
`(7.8)/(1.29) = 6.05 \simeq 6`

For Chlorine =
`(\1.29)/(1.29) = 1`

So, the emperical formula of the compound is
`C_3H_6Cl`.

For determining the molecular formula:

`n = (Molar Mass)/(Emperical Formula Mass)`

Molar Mass =
`155.06 g/mol`

Emperical Formula Mass =
`C_3H_6Cl = 12* 3+1* 6+35.5 = 77.5 g/mol`

Substituting the values,

`n = (155.06 g /mol)/(77.5 g /mol) = 2`

Since,
`Molecular Formula = (Emperical Formula)_(n)` so, the molecular formula is:

`(C_3H_6Cl)_(2) = C_6H_(12)Cl_2`

Answer 2

Answer:- Molecular formula of the compound is
`C_6H_1_2Cl_2` .

Solution:- From given information:

C = 46.47%

H = 7.80%

Cl = 45.72%

First of all we find out the empirical formula from given percentages. We divide the given percentages by their respective atomic masses to calculate the moles:

`C=(46.47)/(12.01)` = 3.87

`H=(7.80)/(1.01)` = 7.72

`Cl=(45.72)/(35.45)` = 1.29

Now, we divide the moles of each by the least one of them. Least one is Cl as it's moles are least as compared to the moles of C and H. So, let's divide the moles of each by 1.29.

`C=(3.87)/(1.29)` = 3

`H=(7.72)/(1.29)` = 6

`Cl=(1.29)/(1.29)` = 1

So, the empirical formula of the compound is
`C_3H_6Cl` .

Empirical formula mass = 3(12.01) + 6(1.01) + 1(35.45)

= 36.03 + 6.06 + 35.45

= 77.54

To calculate the number of formula units we divide molar mass by empirical formula mass.

number of empirical formula units =
`(155.06)/(77.54)`

= 2

So, the molecular formula would be two times of empirical formula that is,
`C_6H_1_2Cl_2` .