Chapter 10, Problem 164A

${\text{Vitamin D}}_3$ Your body's ability to absorb calcium is aided by vitamin Chemical analysis of vitamin yields the data shown in Figure 10.22. What are the empirical and molecular formulas for vitamin ${\text{D}}_3$ ?

Interpretation Introduction

Interpretation:

The empirical and molecular formula for vitamin D₃ should be predicted.

Concept introduction:

The representation of atoms of a compound in simple whole number ratio is known as empirical formula.

Chemical formula or molecular formula represents the type of atoms and number of atoms present in a molecule using numerical subscripts and atomic symbol.

The rules for determining empirical formula is given as follows:

• Determine the mass of elements in given compound.
• Calculate number of moles using molar mass of each compound or element.
• Divide each number of moles by smallest number of mole value calculated in second step.
• Round the value calculated in step 3 to nearest whole number.

Answer to Problem 164A

The empirical and molecular formula for vitamin D₃ is ${\text{C}}_{\text{27}}{\text{H}}_{\text{44}}\text{O}$.

Explanation of Solution

Assuming the percentage is given by mass that means 84.31 g, 4.16 g, and 11.53 g of carbon, oxygen, and hydrogen are present in the vitamin D₃.

1. Mass of carbon = 84.31 g

Mass of hydrogen = 11.53 g

Mass of oxygen = 4.16 g

Now, calculating the moles of carbon, hydrogen, and oxygen using their molar mass as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{84.31\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=7.02\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{11}\text{.53}\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=11.4\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{4.16\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=0.26\text{mol}\end{array}$

2. Now, dividing each mole value by smallest mole value as follows:

$\begin{array}{c}{\text{n}}_{\text{c}}=\frac{7.02}{0.26}\\ =27\\ {\text{n}}_{\text{H}}=\frac{11.4}{0.26}\\ =44\\ {\text{n}}_{\text{O}}=\frac{0.26}{0.26}\\ =1\end{array}$

3. Thus, the empirical formula of vitamin D₃ is ${\text{C}}_{\text{27}}{\text{H}}_{\text{44}}\text{O}$.

Now, n that is the ratio of given molar mass and empirical formula mass is calculated as follows:

$\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}$

The empirical formula mass is calculated as follows:

$\begin{array}{c}\text{Empirical}\text{}\text{formula}\text{mass}=\left(27\times \text{C}+44\times \text{H}+1\times \text{O}\right)\text{g/mol}\\ =\left(27\times 12.0107+44\times 1.008+1\times 15.999\right)\text{g/mol}\\ \text{Empirical}\text{}\text{formula}\text{mass}=384.6399\text{g/mol}\end{array}$

Calculation of n as follows:

$\begin{array}{c}\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}\\ =\frac{\text{384}\text{g/mol}}{\text{384}\text{.6399}\text{g/mol}}\\ =0.99\\ \text{n}=1\end{array}$

Now, the number obtained will be multiplied to empirical formula as follows:

$\begin{array}{c}\text{Molecular}\text{formula}={\text{C}}_{\text{27}}{\text{H}}_{\text{44}}\text{O}\times \text{1}\\ ={\text{C}}_{\text{27}}{\text{H}}_{\text{44}}\text{O}\end{array}$