Question

The bitter-tasting compound quinine is a component of tonic water and is used as a protection against malaria. It contains only C, H, N and O. When a sample of mass 0.487 g was burned, 1.321 g of carbon dioxide, 0.325 g of water, and 0.0421 g of nitrogen were produced. The molar mass of quinine is 324 g/mol. Determine the empirical and molecular formulas of quinine. (Type your answer using the format CO2 for CO2 and use the order CHNO)
empirical
........
molecular
..........

Answer 1

Answer:

Empirical formula = $\mathbf{C_{10}H_{12}N_{2}}$

Molecular formula = $\mathbf{C_{20}H_{24}N_{4}}$

Explanation:

From the given information:

we need to estimate the mass of carbon C in 1.321 g of $CO_2$

before that, the number of moles of C is:

$C = 1.321 \ g \times \dfrac{1 \ mol \ CO_2}{44/010 \ g \ of \ CO_2}$

c = 0.03002 mol

we know that:

number of moles = mass/ molar mass

mass = number of moles × molar mass

mass = 0.03002  × 12.011g of C

mass of C = 0.3606 g

Similarly; for hydrogen

the number of moles of H = $0.325 g \ of \ H_2O \times \dfrac{ 1\ mol \ H_2O}{18.02g \ of \ H_2O}\times \dfrac{2 \ mole \ of H }{1 \ mol \ H_2O }$

the number of moles of H = $0.325 g \ of \ H_2O \times \dfrac{2 \ mole \ of H}{18.02g \ of \ H_2O}$

the number of moles of H = 0.0361 mol of H

mass of H = $0.0361 \ mol \ of \ H \times \dfrac{1.008 g \ of \ H}{ 1 \ mol \ of \ H}$

mass of H  = 0.0364 g

The mass of N will therefore be the difference the sample burnt with the mass of carbon and hydrogen.

i.e

mass of N = 0.487 g - 0.3606 of C - 0.0364 g of H

mass of N = 0.0900 g

however, the number of moles of nitrogen  = mass/ molar mass

the number of moles = 0.0900 g /14.007 g

the number of moles of nitrogen = 0.00643 mol

Thus, the formula is:  $\mathsf{C_{0.03002}H_{0.0361}N_{0.00643}}$

If we divide by the smallest number (0.00643); we have:

$\mathsf{C_{\dfrac{0.03002}{0.006432}}H_{\dfrac{0.0361}{0.00643}}N_{\dfrac{0.00643}{0.00643}}}$

= $\mathsf{C_{4.7}H_{5.7}N}$

Thus, multiplying the subscript by 2.1,  we have:

 $\mathsf{C_{4.7 \times 2.1}H_{5.7 \times 2.1}N_{1\times 2.1}}$

Thus, the empirical formula = $\mathbf{C_{10}H_{12}N_{2}}$

The mass of the empirical formula is:

= (10 × 12.010 u) + (12 × 1.008 u) + ( 2 × 14.007 u)

= 160.21 u

Thus, because the molecular mass 324 g/mol is double the value of the empirical formula, the molecular mass is definitely double the empirical formula;

i.e

Molecular formula = $\mathbf{C_{10\times 2}H_{12\times 2}N_{2\times 2}}$

Molecular formula = $\mathbf{C_{20}H_{24}N_{4}}$