Question

A compound gave a following data:
C= 57.82% O= 38.58% and the rest hydrogen. Its relative molecular mass is 166. Find its empirical formula and molecular formula. [C = 12, O= 16, H=1]​

Answer 1

Answer:

Answer

Correct option is

C

C

4

H

3

O

2

Molecular formula = 2 x V.D. = 2 x 83 = 166

% of Hydrogen =100−(57.82+38.58)=3.6

Element % Ratio Simplest Ratio Empirical formula

Carbon 57.82/12=4.82 4.82/2.41=2 C

4

H

3

O

2

Hydrogen 3.6/1=3.6 3.6/2.41=1.49

Oxygen 38.58/16=2.41 2.41/2.41=1

The empirical formula weight =12×4+3×1+16×2=83

Formula ratio =166/83=2

The molecular formula of compound =C

8

H

6

O

4

Explanation:

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Answer 2

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 C = 57.82 g

Mass of O = 38.58 g

Mass of H = (100-(57.82+38.58) = 3.6 g

Step 1 : convert given masses into moles.  

for C = 2 mole

for 0 = 1 MOLE

FOR H = 1.5 mole

Converting them into whole number ratios by multiplying by 2.

The ratio of C : O : H= 4: 2 : 3

Hence the empirical formula is c4o2 h3

hope it help  

and 2 multiply = c804h6