Chapter 3, Problem 116AP

Interpretation Introduction

Interpretation:

The empirical formula and the molecular formula of the compound are to be determined with given molar mass.

Concept introduction:

The simplest atomic ratio of the mass of the elements present in a compound is known as the empirical formula.

The actual atomic ratio of the mass of the elements present in a compound is the molecular formula.

The mass of hydrogen in ${\text{H}}_2\text{O}$

is

$\text{mass of H = }\text{}\frac{\text{wt}\text{of}{\text{H}}_{\text{2}}\text{O×}\text{atomic}\text{mass}\text{of}\text{H}}{\text{molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}$

The mass of oxygen in ${\text{H}}_2\text{O}$

is

$\text{mass of O = }\text{}\frac{\text{wt}\text{of}{\text{H}}_{\text{2}}\text{O×}\text{atomic}\text{mass}\text{of}\text{O}}{\text{molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}$

The value of $n$ is calculated as

$n=\frac{\text{molar mass}}{\text{empirical formula mass}}$

The molecular formula of the compound is given as

$\text{Molecular formula }=\text{Empirical formula}\times n$

Answer to Problem 116AP

Solution: The empirical formula and the molecular formula of the compound are ${\text{CH}}_2\text{O}$ and ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$, respectively.

Explanation of Solution

Given information: The compound contains $40\%\text{ carbon}$.

The hydrogen to oxygen ratio in the compound is $2:1$.

The given compound contains $40\%\text{ carbon}$, which means that in a $100\text{ g}$ sample, the mass of carbon is $40\text{ g}$.

Therefore, the weight of the ${\text{H}}_2\text{O}$ in the compound will be $(100-40)=60\text{ g}$.

The amount of $\text{H}$ present in ${\text{60 g H}}_2\text{O}$ is calculated as

$18.0\text{ g}$ of water contains $\text{2}\text{.0 g}$ of hydrogen. Therefore, $\text{60}\text{.0 g}$ of water will contain:

$\text{mass of H = }\text{}\frac{\text{wt}\text{of}{\text{H}}_{\text{2}}\text{O×}\text{atomic}\text{mass}\text{of}\text{H}}{\text{molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}$

$\frac{60.0\text{g}\times 2.0\text{g/mol}}{18.0\text{g/mol}}=6.67\text{ g}$ of hydrogen.

The amount of oxygen present in ${\text{60 g H}}_2\text{O}$ is calculated as

$18.0\text{ g}$ of water contains $\text{16}\text{.0 g}$ of oxygen. Therefore, $\text{60}\text{.0 g}$ of water will contain:

$\text{mass of O = }\text{}\frac{\text{wt}\text{of}{\text{H}}_{\text{2}}\text{O×}\text{atomic}\text{mass}\text{of}\text{O}}{\text{molecular}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}$

$\frac{60.0\text{g}\times 16.0\text{g/mol}}{18.0\text{g/mol}}=53.33\text{ g}$ of oxygen.

The formula for computing the number of moles of an element is as follows:

$\text{Number}\text{of}\text{moles =}\frac{\text{Given mass}}{\text{Atomic mass}}$ …… (1)

Substitute the values of the given mass and the molar mass of the hydrogen atom in equation (1)

$\begin{array}{c}\text{Moles}\text{of}\text{H}=\frac{6.67\text{g}}{1.00\text{g/mol}}\\ =6.67\text{ mol}\end{array}$

Substitute the values of the given mass and the molar mass of the carbon atom in equation (1)

$\begin{array}{c}\text{Moles}\text{of}\text{C}=\frac{40.0\text{g}}{12.01\text{g/mol}}\\ =3.33\text{ mol}\end{array}$

Substitute the values of the given mass and the molar mass of the oxygen atom in equation (1)

$\begin{array}{c}\text{Moles}\text{of}\text{O}=\frac{53.33\text{g}}{16.00\text{g/mol}}\\ =3.33\text{ mol}\end{array}$

The simplest molar ratio between C, H, and O is $1:2:1$ (obtained by dividing the values of moles by the least number, i.e., $3.33$). Therefore, the empirical formula is ${\text{CH}}_2\text{O}$.

The empirical formula mass of the compound is calculated as

$\begin{array}{c}\text{Empirical}\text{formula}\text{mass}=\left(1\times \text{atomic}\text{mass of C}\right)+\left(\text{2}\times \text{atomic}\text{mass of H}\right)+\left(\text{1}\times \text{atomic}\text{mass}\text{of O}\right)\\ =\left(\text{1}\times \text{12}\text{.0}\text{g/mol}\right)+\left(\text{2}\times 1.0\text{g/mol}\right)+\left(1\times 16.0\text{g/mol}\right)\\ =30.0\text{ g/mol}\end{array}$

The value of $n$ is calculated as

$\begin{array}{c}n=\frac{\text{molar mass}}{\text{empirical formula mass}}\\ =\frac{178\text{g/mol}}{30\text{g/mol}}\\ =5.9\\ \cong 6\end{array}$

The molecular formula of the compound is given as

$\begin{array}{c}\text{Molecular formula }=\text{Empirical formula}\times n\\ ={\text{CH}}_2\text{O}\times 6\\ ={\text{C}}_6{\text{H}}_{12}{\text{O}}_6\end{array}$

Conclusion

The empirical formula of the compound is ${\text{CH}}_2\text{O}$.

The molecular formula of the compound is ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$.