Question

You are searching for an inexpensive synthetic route to make anthraquinone to use it as the active component in a new laxative formulation. Reaction of benzene and phthalic anhydride yielded a yellow solid thought to be anthraquinone. Combustion analysis of 0.520 g of the yellow solid yielded 1.54 g of carbon dioxide and 0.180 g of water; (all carbon dioxide and water formed by combustion were isolated and weighed). In another experiment, the molar mass of the yellow solid was found to be approximately 205 g/mol. Use the analysis information above to determine the molecular formula of the yellow solid.

Answer 1

% by mass of C = 12 * mass of CO2 * 100 / (44 * mass of yellow solid) = 12 * 1.54 * 100 / (44 * 0.520)

% by mass of C = 80.8 %

% by mass of H = 2 * mass of H2O * 100 / (18 * mass of yellow solid) = 2 * 0.180 * 100 / (18 * 0.520)

% by mass of H = 3.85 %

Now, % by mass of oxygen = 100 - 3.85 - 80.8 = 15.4 %

Element mass Moles division by small number simple whole number

C             80.8    80.0/12=6.73    6.72/0.962=6.99                   7

H           3.85    3.85/1 = 3.85      3.85/0.962 = 4.00                 4

O           15.4   15.4/16=0.962     0.962/0.92 = 1                       1

Therefore, emperical formula = C₇H₄O

Emperical formula mass = 7(12) + 4(1) + 1 (16) = 104

Molecular mass = 205 g/mol

n = molecular mass / emperical formula mass

n = 205 / 104

n = 2 (approx.)

Therefore,

Molecular formula = (emperical formula)n = (C₇H₄O)₂ = C₁₄H₈O₂