Question

an organic compound on analysis was found to contain 71.7 % of chlorine ,4.04% of hydrogen and the rest is carbon. if its molecular weight is 99 .then calculate the molecular formula

Answer 1

Cl = 71.7 % , moles = 71.7/35.5 = 2.02

H = 4.04 %, moes = 4.04/1 = 4.04

C = 24.26 %, moles = 24.26/12 = 2.02

mole ratio C:H:Cl = 2:4:2 = 1:2:1

Empirical formula = CH2Cl.

E.F wt = 12+2+35.5 = 49.5

M.wt = 99 (given)

n = M.wt/E.F wt = 99/49.5 = 2

Therefore MF = (E.F)n = (CH2Cl)2

= C2H4Cl2 , dichloro ethane.

Answer 2

Answer: Thus molecular formula is $C_2H_4Cl_2$

Explanation:  Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Cl = 71.7 g

Mass of H = 4.04 g

Mass of C = 100-(71.7+4.040)= 24.26 g

Step 1 : convert given masses into moles.

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{71.7g}{35.5g/mole}=2.02mole$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.04g}{1g/mole}=4.04moles$

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{24.26g}{12g/mole}=2.02moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cl = $\frac{2.02}{2.02}=1$

For H = $\frac{4.04}{2.02}=2$

For C= $\frac{2.02}{2.02}=1$

The ratio of Cl : H : C= 1 : 2 : 1

Hence the empirical formula is $C_{1}H_{2}Cl_1$  

The empirical weight of $C_{1}H_{2}Cl_1$  = 1(12) + 2(1) + 1(35.5)= 49.5 g

The molecular weight = 99 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}$

$n=\frac{99g/mole}{49.5g/eq}=2$

Thus molecular formula is  $C_{1}H_{2}Cl_1\times 2=C_2H_4Cl_2$