Question

Morphine has the formula C17H19NO3. It is a base and accepts one proton per molecule. It is isolated from opium. A 0.682g sample of opium is found to require 8.92 mL of a 0.0116 M solution of sulfuric acid for neutralization. Assuming that morphine is the only base present in opium, calculate the mass (in grams) of morphine in the sample of opium.
Reaction equation: 2 C17H19NO3 + H2SO4 --> Product

Answer 1

Answer: The mass of morphine is 0.059 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of sulfuric acid solution = 0.0116 M

Volume of solution = 8.92 mL

Putting values in above equation, we get:

$0.0166M=\frac{\text{Moles of sulfuric acid}\times 1000}{8.92}\\\\\text{Moles of sulfuric acid}=\frac{0.0116\times 8.92}{1000}=1.035\times 10^{-4}mol$

The given chemical equation follows:

$2C_{17}H_{19}NO_3+H_2SO_4\rightarrow \text{Product}$

By Stoichiometry of the reaction:

1 moles of sulfuric acid reacts with 2 moles of morphine

So, $1.035\times 10^{-4}mol$ of sulfuric acid will react with = $\frac{2}{1}\times 1.035\times 10^{-4}=2.07\times 10^{-4}mol$

To calculate the mass of morphine for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of morphine = 285.34 g/mol

Moles of morphine = $2.07\times 10^{-4}mol$

Putting values in above equation, we get:

$2.07\times 10^{-4}mol=\frac{\text{Mass of morphine}}{285.34}\\\\\text{Mass of morphine}=(2.07\times 10^{-4}mol\times 285.34g/mol)=0.059g$

Hence, the mass of morphine is 0.059 grams