Answer:
B) C3H3O and C6H6O2
Explanation:
Given data:
Molar mass of compound = 100 g/mol
Percentage of hydrogen = 5.45%
Percentage of carbon = 65.45%
Percentage of oxygen = 29.09%
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 5.45 / 1.01 = 5.4
Number of gram atoms of O = 29.09/ 16 = 1.8
Number of gram atoms of C = 65.45 / 12 = 5.5
Atomic ratio:
C : H : O
5.5/1.8 : 5.4/1.8 : 1.8/1.8
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 = 12×3 + 1.01 ×3 + 16×1 = 55.03
n = 100 / 5503
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₃H₃O)
Molecular formula = C₆H₆O₂
