A compound gave the following data :
C=57.82%,O=38.58% and the rest hydrogen. It's vapour density is 83. Find its empirical and molecular formula.
Answers
Answered by
23
Given:
Vapour density (V. D) = 83
As molecular weight = 2×V.D
Molecular weight = 83 × 2 = 166g
C = 57.82%
O = 38.58%
H = 100 - (57.82 + 38.58) = 3.60%
As number of moles = Given mass /molar mass
Therefore,
No. Of moles of C = 57.82/12 = 4.81
No.of moles of O = 38.58/16 = 2.42
No.of moles of H = 3.60/1 = 3.60
Taking the approx, we get the empirical formula as C4H3O2.
Now,
Molecular formula =molecular weight /empirical formula mass
=166/83
=2
Hence,
Empirical formula(C4H3O2) = 12 × 4 + 1× 3 + 16 × 2
= 48 +3 +32
= 83g
And,
Molecular formula = 2(C4H3O2)
= C8H6O4
Answered by
12
Answer:
It's empirical formula is C4H3O2.
It's molecular formula is C8O4H6.
Attachments:
Similar questions