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, $C$:

$atomic ratio = \frac{46}{12} = 3.833$

Atomic ratio for Hydrogen, $H$:

$atomic ratio = \frac{7.8}{1} = 7.8$

Atomic ratio for Chlorine, $Cl$:

$atomic ratio = \frac{45.72}{35.5} = 1.29$

The simplest ratio is:

For Carbon = $\frac{3.833}{1.29} = 2.97 \simeq 3$

For Hydrogen = $\frac{7.8}{1.29} = 6.05 \simeq 6$

For Chlorine = $\frac{\1.29}{1.29} = 1$

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

For determining the molecular formula:

$n = \frac{Molar Mass}{Emperical Formula Mass}$

Molar Mass = $155.06 g/mol$

Emperical Formula Mass = $C_3H_6Cl = 12\times 3+1\times 6+35.5 = 77.5 g/mol$

Substituting the values,

$n = \frac{155.06 g /mol}{77.5 g /mol} = 2$

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:

$C=\frac{46.47}{12.01}$  = 3.87

$H=\frac{7.80}{1.01}$  = 7.72

$Cl=\frac{45.72}{35.45}$  = 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.

$C=\frac{3.87}{1.29}$ = 3

$H=\frac{7.72}{1.29}$ = 6

$Cl=\frac{1.29}{1.29}$ = 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 = $\frac{155.06}{77.54}$

= 2

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