Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
