Chapter 24, Problem 24.67E

Interpretation Introduction

Interpretation:

The ketone present in the patient’s blood is to be stated. The number of ketone molecules in $1.0\text{mL}$ of patient’s blood is to be calculated.

Concept introduction:

Ketones are organic compounds containing carbonyl group. Both diabetic and non-diabetic human body contains some amount of ketone in it. Liver turns fat into ketones. High concentration of ketone in blood can be harmful.

Answer to Problem 24.67E

Acetone is present in the patient’s blood.

The number of ketone molecules in $1.0\text{mL}$ of patient’s blood is $5.18\times {10}^{15}$.

Explanation of Solution

It is given that the molecular weight of ketone is $58.1\text{g}/\text{mol}$.

The starting member of ketone family is acetone.

The molecular formula of acetone is ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}$.

The molar mass of hydrogen is $1\text{g}/\text{mol}$.

The molar mass of oxygen is $16\text{g}/\text{mol}$.

The molar mass of carbon is $12\text{g}/\text{mol}$.

Therefore, the molecular weight of acetone is calculated as shown below.

$\begin{array}{rll}\text{Molecular weight of acetone} & =& \left(3\times \text{C}+1\times \text{O}+6\times \text{H}\right)\text{g}/\text{mol}\\ & =& \left(3\times 12+1\times 16+6\times 1\right)\text{g}/\text{mol}\\ & =& \left(36+16+6\right)\text{g}/\text{mol}\\ & =& 58\text{g}/\text{mol}\end{array}$

The molecular weight of ketone present in patient’s blood is approximately equal to that of acetone. Therefore, acetone is present in the patient’s blood.

It is given that the concentration of ketone in $100\text{mL}$ of patient’s blood is $0.050\text{mg}$.

Therefore, the concentration of ketone in $1.0\text{mL}$ of patient’s blood is $\frac{0.050}{100}\text{mg}$.

The conversion of $0.050\text{mg}$ to $\text{g}$ is shown below.

$0.050\text{mg}=0.050\times {10}^{-3}\text{g}$

Therefore, the concentration of ketone in $100\text{mL}$ of patient’s blood is $0.050\times {10}^{-3}\text{g}$.

Therefore, the concentration of ketone in $1.0\text{mL}$ of patient’s blood is $\frac{0.050\times {10}^{-3}}{100}\text{g}$.

The number of acetone molecules in $1.0\text{mL}$ of patient’s blood is calculated by the formula shown below.

$\text{Number}\text{of}\text{ketone}\text{molecules}=\left(\frac{\begin{array}{l}\text{Concentration}\text{of}\text{ketone}\text{in}\\ \text{1}\text{.0}\text{mL}\text{of}\text{blood}\end{array}}{\text{Molecular}\text{weight}\text{of}\text{ketone}}\right)\times {\text{N}}_{\text{A}}$…(1)

Where,

• ${\text{N}}_{\text{A}}$ is Avogadro’s constant.

The concentration of ketone in $1.0\text{mL}$ of patient’s blood is $\frac{0.050\times {10}^{-3}}{100}\text{g}$.

The molecular weight of ketone is $58.1\text{g}/\text{mol}$.

The value of ${\text{N}}_{\text{A}}$ is $6.022\times {10}^{23}{\text{mol}}^{\text{-1}}$.

Substitute the value of concentration of ketone in $1.0\text{mL}$ blood, molecular weight of ketone and ${\text{N}}_{\text{A}}$ in equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{ketone}\text{molecules}& =& \left(\frac{\begin{array}{l}\text{Concentration}\text{of}\text{ketone}\text{in}\\ \text{1}\text{.0}\text{mL}\text{of}\text{blood}\end{array}}{\text{Molecular}\text{weight}\text{of}\text{ketone}}\right)\times {\text{N}}_{\text{A}}\\ & =& \left(\frac{\frac{0.050\times {10}^{-3}}{100}\text{g}}{58.1\text{g}/\text{mol}}\right)\times 6.022\times {10}^{23}{\text{mol}}^{\text{-1}}\\ & =& 5.18\times {10}^{-8}\times {10}^{23}\\ & =& 5.18\times {10}^{15}\end{array}$

Therefore, the number of ketone molecules in $1.0\text{mL}$ of patient’s blood is $5.18\times {10}^{15}$.

Conclusion

Acetone is present in the patient’s blood and the number of ketone molecules in $1.0\text{mL}$ of patient’s blood is $5.18\times {10}^{15}$.