Answer:
Molecular formula => C₁₂H₁₂O₃
Explanation:
From the question given above, the following data were obtained:
Molar mass of compound = 205gmol¯¹
Mass of Carbon (C) = 3.758 g
Mass of Hydrogen (H) = 0.316 g
Mass of Oxygen (O) = 1.251 g
Molecular formula =?
We'll begin by determining the empirical formula of the compound. This can be obtained as follow:
C = 3.758 g
H = 0.316 g
O = 1.251 g
Divide by their molar masses
C = 3.758 /12 = 0.313
H = 0.316 /1 = 0.316
O = 1.251 /16 = 0.078
Divide by the smallest.
C = 0.313 / 0.078 = 4
H = 0.316 / 0.078 = 4
O = 0.078 / 0.078 = 1
Thus, the empirical formula of the compound is C₄H₄O
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Molar mass of compound = 205gmol¯¹
Empirical formula => C₄H₄O
Molecular formula =>?
[C₄H₄O]ₙ = 205
[(12×4) + (4×1) +16]n = 205
[48 + 4 +16]n = 205
68n = 205
Divide both side by n
n = 205 / 68
n = 3
Molecular formula => [C₄H₄O]ₙ
Molecular formula => [C₄H₄O]₃
Molecular formula => C₁₂H₁₂O₃
