Answer:
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of compound = 0.519 g
Mass of CO₂ = 1.24 g
Mass of H₂O = 0.255 g
Empirical formula = ?
Solution:
%age of C,H,O
C = 1.24 g/0.519 × 12/44 ×100 = 65.5%
H = 0.255 g/0.519 × 2.016/18 ×100 = 5.6%
O = 100 - (65.5+5.6)
O = 28.9%
Number of gram atoms of H = 5.6 / 1.01 = 5.5
Number of gram atoms of O = 28.9 / 16 = 1.81
Number of gram atoms of C = 65.5 / 12 = 5.5
Atomic ratio:
C : H : O
5.5/1.81 : 5.5/1.81 : 1.81/1.81
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
