Chapter 10, Problem 166A

Cholesterol Heart disease is linked to high blood cholesterol levels. What is the percent composition of the elements in a molecule of cholesterol $\left({\text{C}}_{\text{27}}{\text{H}}_{\text{45}}\text{OH}\right)$ ?

Interpretation Introduction

Interpretation:

The percent composition of the elements in a molecule of cholesterol ( ${\text{C}}_{\text{27}}{\text{H}}_{\text{45}}\text{OH}$ ) needs to be determined.

Concept introduction:

Empirical formula is the simplest formula of any organic or inorganic compound that represents the simple ratio of all atoms present in the molecule. The empirical and molecular formula can be used to calculate the mass percent of each element given in the formula. The mathematical expression for mass percent of element can be shown as:

$\text{Mass percent = }\frac{\text{number of element ×molar mass of element }}{\text{molar mass of compound}}\times 100$

Answer to Problem 166A

$\begin{array}{l}\text{Mass percent of C =83}\text{.9\%}\\ \text{Mass percent of H =11}\text{.98\%}\\ \text{Mass percent of O = 4}\text{.14\%}\end{array}$

Explanation of Solution

Molecular formula = ${\text{C}}_{\text{27}}{\text{H}}_{\text{45}}\text{OH}$

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.007 g/mol

Molar mass of O = 15.99 g/mol

Molar mass of ${\text{C}}_{\text{27}}{\text{H}}_{\text{45}}\text{OH}$ = $\text{(27×12}\text{.01)+(46×1}\text{.007)+(1×15}\text{.99) = 324}\text{.27+46}\text{.322+15}\text{.99=386}\text{.582g/mol}$

$\text{Mass percent = }\frac{\text{number of element ×molar mass of element }}{\text{molar mass of compound}}\times 100\%$

Calculate mass percent of each element:

$\begin{array}{l}\text{Mass percent of C = }\frac{\text{27 × 12}\text{.01 g/mol }}{\text{386}\text{.582 g/mol }}\text{×100 \%=83}\text{.9\%}\\ \text{Mass percent of H = }\frac{\text{46 × 1}\text{.007 g/mol }}{\text{386}\text{.582 g/mol }}\text{×100 \%=11}\text{.98\%}\\ \text{Mass percent of O = }\frac{\text{1 × 15}\text{.99g/mol }}{\text{386}\text{.582 g/mol }}\text{×100\% = 4}\text{.14\%}\end{array}$

Conclusion

Thus,

$\begin{array}{l}\text{Mass percent of C =83}\text{.9\%}\\ \text{Mass percent of H =11}\text{.98\%}\\ \text{Mass percent of O = 4}\text{.14\%}\end{array}$