Question

5. (12 pts) A compound contains only carbon, hydrogen, and oxygen. If complete combustion of 70.86 g of this compound produces 103.9 g of CO2 and 42.52 g of H20, what is the molecular formula of this compound if it has a molar mass of 120.104 g/mol? (Carry each step to 3 digits after the decimal)

Answer 1

first calculate % of each element from given data

44 g CO2 contain 12 g C

103.9 g CO2 contain 103.9 x 12 / 44 = 28.34 g C

% C = (28.34 / 70.86) x 100

% C = 39.99

18 g H2O contain 2 g H

42.52 g H2O contain 42.52 x 2 / 18 = 4.72 g H

% H = (4.72 / 70.86) x 100

% H = 6.66

% O = 100 - 39.99 - 6.66 = 53.35

% elemen atomic ratio simplest ratio C = 39.99 39.99 / 12 = 3.33 3.33/3.33 = 1.0 H = 6.66 6.66/1 = 6.66 6.66 /3.33 = 2.0 0 = 53.35 53.35 / 16 = 3.33 3.33/3.33 = 1.0 empirical formula = CH2O empirical formula mss = 30 g/mol molar mass = 120.104 g/mol n = molar mass / empirical formula mass = 120.104/30 = 4.00 molecular formula = (empirical formula)n = (CH20)4 molecular formula = C4H804