Question

Nicotine, a component of tobacco, is composed of c, h, and n. A 2.625-mg sample of nicotine was combusted, producing 7.1210 mg of co2 and 2.042 mg of h2o. What is the empirical formula for nicotine?

Answer 1

The mass sample of nicotine combusted = 2.625 mg (given)

Mass of
`CO_2` produced = 7.1210 mg (given)

Mass of
`H_2O` produced = 2.042 mg (given)

Molar mass of
`CO_2` = 44 g/mol

Molar mass of
`H_2O` = 18 g/mol

Percentage of Carbon =
`(12 g)/(44 g/mol)* (7.120 mg of CO_2)/(2.625 mg sample)* 100 = 74.04`%

Percentage of hydrogen =
`(2 g)/(18 g/mol)* (2.04 mg of H_2O)/(2.625 mg sample)* 100 = 8.62`%

Now, for percentage of nitrogen =
`100 - (74.04+8.62) = 17.34`%

Calculating the moles of each element:

`Number of moles = (given mass)/(Molar mass)`

• For
  `C`

`(74.04 g)/(12 g/mol) = 6.17 mol`

• For
  `H`

`(8.62 g)/(1 g/mol) = 8.62 mol`

• For
  `N`

`(17.34 g)/(14 g/mol) = 1.24 mol`

Dividing with the smallest mole value to calculate the molar ratio of each element:

`C_{(6.17)/(1.24)} = C_(4.9) = C_5`

`H_{(8.62)/(1.24)} = H_(6.9) = H_7`

`N_{(1.24)/(1.24)} = N_1`

Hence, the empirical formula for nicotine is
`C_5H_7N`.