Question

A compound containing carbon, hydrogen,and oxygen gave the following analytical data:C-40.0percent and H-6.67 percent calaulate the
molecular formula of the compounds if it's molecular mass is 180.

Answer 1

Answer:

Given that

molecular mass of compound=180

Elements

i) carbon (C)

%=40

Atomic mass=12

moles=40/12=3.34

(ii) Hydrogen (H)

%= 6.67

Atomic mass = 1

moles = 6.67 / 1 = 6.67

(iii) oxygen (O)

%= 100-40--6.67=53.33

Atomic mass = 16

moles=53.33 /16=3.34

Now,

Atomic ratio (Molar ratio) of three elements

C 33.4÷3.34=1

H. 6.67÷3.34=2

0. 33.4÷3.34=1

simplest whole number ratio

C : H : 0 = 1 : 2 : 1

so,

emptical formula =CH2O

emptical formula mass= 12X1+1X2 +16X1

= 12+2+16

= 30

n =molecular mass/emptical formula mass

= 180/30=6

So,

molecular formula = n (emptical formula )

= 6(CH2O)

=C6H1206