Question

A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula? Insert subscripts as needed. empirical formula: CH 0 The molecular mass of the compound is 132 amu. What is its molecular formula? Insert subscripts as needed. molecular formula: 44

Answer 1

Calculate atomic ratio by dividing % compossition by atomic mass

atomic ratio for C = 54.53 / 12 = 4.54

atomic ratio for H = 9.15 / 1 = 9.15

atomic ratio for O = 36.32 / 16 = 2.27

now calculate simplest ratio by dividing atomic ratio by smallest atomic ratio

simplest ratio for C = 4.54 /2.27 = 2

simplest ratio for H = 9.15/2.27 = 4

implest ratio for O = 2.27/2.27 = 1

Empirical formula for compound is C2H4O

Now Calculate empitical formula mass

Empirical formula mass = [(2 X 12) + ( 4 X 1) + ( 1 X 16 )] = 44

n = molecular formula mass / empirical formula mass

n = 132 / 44 = 3

molecular formula = n X empirical formula

molecular formula = 3 X C2H4O = C6H12O3

molecular formula = C6H12O3