A 0.167-g sample of an unknown compound contains 0.00278 mol of the compound. Elemental analysis of the acid gives the following

Question

3 years ago

9

percentages by mass: 40.00% C; 6.71% H; 53.29% O.

1 answer:

ratelena [41] · Answer 1 · 2021-12-10T10:56:30+03:00

Answer:

Emperical formula of the acid is CH2O

Molecular formula is C2H4O2

Explanation:

C. H. O

% by mass. 40. 6.71 53.29

MW. 12. 1. 16

Mole 40/12. 6.71/1. 53.29/16

3.33. 6.71. 3.33

3.33/3.33 6.71/3.33 3.33/3.33

1. 2. 1

Emperical Formula (EF): CH2O

Molecular Formula (MF)

Molecular Weight (MW) of acid = mole/mass = 0.00278/0.167 = 60g/mol

MF = (EF)n = 60

(CH2O)n = 60

(12+2+16)n = 60

30n = 60

n = 60/30 = 2

Molecular formula = (EF)n = (CH2O)2 = C2H4O2