Question

Molecular Formula from Elemental Analysis and Molecular Mass Determination

An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 8.00 mg of this compound is burned, 20.52 mg of CO₂ and 2.40 mg of H₂O are produced.
The freezing point of camphor is lowered by 26.4°C when 3.052 g of the compound is dissolved in 19.25 g of camphor (K_f = 40.0°C kg/mol).

What is the molecular formula of the unidentified compound?

Choose appropriate coefficients in the molecular formula below.
C H O

Answer 1

% of Carbon (C) = (12/44)*(mass of CO2*100/mass of compoun)

% of C = (12/44)*(20.52*100/8) = 69.955%

% of H = (2/18)*(mass of H2O*100/mass of compound)

% of H = 2*2.40*100/18*8 = 3.333%

% of O = 100 - (69.955 + 3.333) = 26.712%

Moles of C = 69.955/12 = 5.83

Moles of H = 3.333/1 = 3.333

Moles of O = 26.712/16 = 1.67

Mole ratio;

C : H : O = 5.83 : 3.33 : 1.67

Simplest ratio;

C : H : O = (5.83/1.67) : (3.33/1.67) : (1.67/1.67)

C : H : O = 3.5 : 2 : 1

C : H : O = 7 : 4 : 2

Empirical formula,

C7H4O2

Empirical formula mass = 120 g/mol

Delta Tf = m * Kf

m = 26.4/40 = 0.66

Mass of solvent = 19.25 g = 0.01925 Kg

Molality (m) = moles of solute/mass of solvent in Kg

0.66 = n / 0.01925

n = 0.012705 moles

Moles = mass/molar mass

Molar mass = 3.052/0.012705 = 240.22 g/mol

i = 240/120 = 2

Molecular formula = (C7H4O2)2 = C14H8O4 ...Answer