Question 3.17: Lycopene (molar mass 536.88 g/mol) and cymene (134.22 g/mol)...

Collect and Organize The problem gives the percent composition and molar masses of two compounds. We want to determine both the empirical and molecular formulas of lycopene and cymene.

Analyze The data on percent composition lead directly to the empirical formula by using the procedure described in Section 3.6. The molar masses of lycopene and cymene in g/mol are numerically the same as their molecular masses in amu and can be used to determine the value of the multipliers needed to convert their empirical formulas into molecular formulas.

Solve Assuming a 100 g sample, we have 89.49 g of C and 10.51 g of H, which we convert to moles:

89.49  \sout{g  C} \times \frac{1  mol  C}{12.01  \sout{g  C}}=7.451  mol  C

10.51  \sout{g  H} \times \frac{1  mol  H}{1.008  \sout{g  H}}=10.43  mol  H

The mole ratio of C to H is 7.451:10.43, or 1:1.4. We can convert the fractional mole ratio to a ratio of whole numbers by multiplying by 5:

C:H= 5 × (1:1.4) = 5:7

which means the empirical formula is C_{5}H_{7}.
To determine the molecular formula, we first determine the molar mass of the empirical formula C_{5}H_{7}:

5  \sout{mol}(12.01  g/\sout{mol}) + 7  \sout{mol}(1.008  g/\sout{mol}) = 67.11  g

Next we divide the molar masses for lycopene and cymene by their respective empirical formula masses to determine the values of the multiplier n:

n=\frac{molar  mass  lycopene}{empirical  formula  mass}=\frac{536.88  \sout{g/mol}}{67.11  \sout{g/mol}}=8

n=\frac{molar  mass  cymene}{empirical  formula  mass}=\frac{134.22  \sout{g/mol}}{67.11  \sout{g/mol}}=2

The molecular formula of lycopene is

(C_{5}H_{7})_{n}=(C_{5}H_{7})_{8}=C_{40}H_{56}

and the molecular formula of cymene is

(C_{5}H_{7})_{n}=(C_{5}H_{7})_{2}=C_{10}H_{14}

Think About It The percent composition of a compound reveals only the empirical formula.
All compounds with the same empirical formula must have the same percent composition.