Question

the percent composition by mass of a compound is 76.0% c, 12.8% H, and 11.2% O. the molar mass of this compound is 284.5 g/mol. what is the molecular formula of the compound?

Answer 1

C.... 76%........ 76/12 = 6.33 (carbon atomic mass = 12)

H....12.8%.......12.8/1 = 12.8 ( H atomic mass = 1 )

O....11.2%......11.2/16 = 0.7 ( Oxygen atomic mass = 16 )

The values are devided with the lowest value

C....6.3/0.7 = 9

H....12.8/0.7 = 18.2

O....0.7/0.7 = 1

From the above caluculation Empirical formula is C₉H₁₈O.

The empirical formula weight = 143.2

According to the question the molar mass is 284.5 g/mol

The formula units = molar mass/ empirical weight

= 284.5/143.2

= 1.98 (consider as 2)

The molecular formula = Empirical formula x 2

Molecular formula = (C₉H₁₈O) x 2

Molecular formula = C₁₈H₃₆O.