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Determine the empirical and molecular formulas of each of the following substances. For example, butane has an empirical formula of C2H5 (lowest whole-number ratio) and a molecular formula of C4H10, where the molecular formula corresponds to the molar mass of 58.12 g/mol.

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol.

Answer :

Answer:

Empirical formula of ibuprofen : [tex]C_{13}H_{18}O_2[/tex]

Molecular formula of ibuprofen : [tex]C_{13}H_{18}O_2[/tex]

Explanation:

Let the empirical formula of ibuprofen be = [tex]C_xH_yO_z[/tex]

Considering 100 gram of ibuprofen with 75.69% C, 8.80% H, and 15.51% O.

Mass of carbon = [tex]100\times \frac{75.69}{100}=75.69 g[/tex]

Moles of carbon = [tex]\frac{75.69 g}{12 g/mol}=6.3075 mol[/tex]

Mass of hydrogen = [tex]100\times \frac{8.80 }{100}=8.80 g[/tex]

Moles of hydrogen = [tex]\frac{8.80 g}{1 g/mol}=8.80 mol[/tex]

Mass of oxygen = [tex]100\times \frac{15.51}{100}=15.51 g[/tex]

Moles of oxygen = [tex]\frac{15.51 g}{16 g/mol}=0.9694 mol[/tex]

For empirical formula, divide the lowest value of moles from each moles of element present in the compound.

Carbon = [tex]x=\frac{6.3075 mol}{0.9694 mol}=6.5[/tex]

Hydrogen = [tex]y=\frac{8.80 mol}{0.9694 mol}=9[/tex]

Oxygen =  [tex]z=\frac{0.9694 mol}{0.9694 mol}=1[/tex]

Empirical formula of ibuprofen :

[tex]C_xH_yO_z=C_{6.5}H_9O_1=C_{13}H_{18}O_2[/tex]

Molecular mass = n × empirical mass

Empirical mass of ibuprofen= [tex]13\times 12 g/mol+18\times 1g/mol+2\times 16 g/mol=206 g/mol[/tex]

Molecular mass of ibuprofen = 206 g/mol

206 g/mol = n × 206 g/mol

n = 1

Molecular formula = [tex]C_{13\times n}H_{18\times n}O_{2\times n}[/tex]

Molecular formula of ibuprofen :

= [tex]C_{13\times 1}H_{18\times 1}O_{2\times 1}=C_{13}H_{18}O_2[/tex]

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