Question

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce 0.166 g of water and 0.403 g of carbon dioxide. What is the empirical formula for valproic acid? If the molar mass is 144 g>mol, what is the molecular formula?

Answer 1

Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g  /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.036412  / 15.999  = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,  

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$