Answer:
EMPIRICAL FORMULA:
C₂H₃O₂
Explanation:
Mass of compound = 4.647 g
Mass of CO₂ = 8.635 g
Mass of H₂O = 1.767 g
Empirical formula = ?
Solution:
Percentage of C = 8.635/ 4.647 × 12/44 ×100
Percentage of C = 1.86× 12/44 ×100 = 50.7
Percentage of H = 1.767/4.647 × 2/ 18 × 100
Percentage of H = 0.38 × 2/ 18 × 100
Percentage of H = 4.18
Percentage of O = 100 - (50.7+4.18)
Percentage of O = 100 - 54.88
Percentage of O = 45.12
Number of grams atom:
Number of grams atom of C = 50.7/ 12 = 4.23
Number of grams atom of H = 4.18 /1 = 4.18
Number of grams atom of O = 45.12/ 16 = 2.82
Atomic ratio:
C : H : O
4.23/2.82 : 4.18/2.82 : 2.82/2.82
1 : 1.5 : 1
C : H : O (1 : 1.5 : 1)
C : H : O 2(1 : 1.5 : 1)
C : H : O (2 : 3 : 2)
EMPIRICAL FORMULA:
C₂H₃O₂
