Question

an unknown molecule is found to consist of 24.2% carbon by mass, 4.0% hydrogen by mass and the remaining mass is due to chlorine. what is the molecular formula of the molecule given that the molar mass is found to be approximately 150?

Answer 1

Answer: C3 H6 Cl3

Explanation:

Molecular formula = (Empirical formula)n

where n= num of moles

Then, let's calculate empirical formula

The molecule consists of Carbon (24.2%), Hydrogen (4.0%) and Chlorine (??)

The abundance of Chlorine is = 100% - ( 24.2% + 4.0%)  

                                                  = 100% - 28.4%

                                                  = 71.8%

Therefore, we have, Carbon 24.2%, Hydrogen 4.0% and Chlorine 71.8%

                                                             C                   H                Cl

                                                          24.2                4.0              71.8

divide each by mass its num             12                    1                 35.5

                                                             2                    4                 2

Divide through by the smallest          2                     2                 2

(i.e. 2)                                                    1                      2                  1

Therefore, the empirical formula= CH2Cl

Molecular formular =  (CH2Cl)n

But n = ??

         (CH2Cl)n = 150;                                C=12, H=1, Cl=35.5

          (12 + (2 x 1) + 35.5 )n = 150

           (49.5)n = 150

            n=3

Therefore, Molecular fromula = (CH2Cl)3

                                             = C3 H6 Cl3

           

Answer 2

Answer:

C3 H6 Cl 3

Explanation:

C -24.2%

H -  4.0%

Cl - (100-24.2 - 4.0)=73.8 %

We can take 100g of the substance, then we have

C -24.2 g

H -  4.0 g

Cl - 73.8 g

Find the moles of these elements

C -24.2 g/12.0 g/mol =2.0 mol

H -  4.0 g/1.0 g/mol = 4. 0 mol

Cl - 73.8 g/ 35.5 g/mol = 2.1 mol

Ratio of these elements gives simplest formula of the substance

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

CH2Cl

Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol

Real molar mass = 150  g/mol

real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3

So, Real formula should be C3 H6 Cl 3.