Question

A certain organic substance used as a solvent in many reactions contains carbon , hydrogen , ...
Weight % of hydrogen in the compound is 7.7 . The weight ratio C : O : S = 3 : 2 : 4 . What is the
least possible molar mass of the compound

Answer 1

Answer:

We need to find each of the elements mass percentage in simplest ratio then we will get the ratio of the entire compound.

So we can find the mass of carbon by.

=(100−7.7)∗3/(3+2+4)

=30.8 g

We will get the mass % of hydrogen as 7.7  so the simplest ratio  will be 6.

The mass % of the carbon as calculated 30.4 and the simplest ratio=2 .

Accordingly the mass % of oxygen=20.52  hence, the simplest ratio =1.

Similarly the mass % of the sulfur will be 41.04 so, the simplest ratio will be 1.

 

So by the above result we will get the empirical formulae of the compound as C2H6OS.

Hence, the molar mass of the compound will be accordingly 2*12+6*1+16+32

=78g.

Answer 2

The weight ratio of C , S and O = 3 : 2 : 4.

also given, percentage weight of hydrogen is 7.7%.so, rest 92.3% is for C , S and O.

percentage weight of C = 3/(3 + 2 + 4) × 92.3 = 92.3/3 = 30.76%

percentage weight of S = 2/9 × 92.3

= 184.6/9 = 20.51 %

percentage weight of O = 4/9 × 92.3

= 369.2/9 = 41.02 %

now, find percentage - atomic mass ratio :

C = 30.76/12 = 2.56

S = 20.51/32 = 0.64

O = 41.02/16 = 2.56

H = 7.7/1 = 7.7

now ratio of all above terms

C : S : O : H = 4 : 1 : 4 : 12

so, least possible molar formula of compound is $C_4SO_4H_{12}$

and molar mass = 12 × 4 + 32 × 1 + 16 × 4 + 12 × 1

= 48 + 32 + 64 + 12

= 156 g/ mol