Chapter 3, Problem 79E

Express the composition of each of the following compounds as the mass percents of its elements.

a. formaldehyde, CH₂O

b. glucose, C₆H₁₂O₆

c. acetic acid, HC₂H₃O₂

Interpretation Introduction

Interpretation: The percentage composition (by mass) of each element in the given compounds ${\text{CH}}_{\text{2}}\text{O},{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ and ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

To determine: The percentage composition (by mass) of each element in formaldehyde $\left({\text{CH}}_{\text{2}}\text{O}\right)$ compound.

Explanation of Solution

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In the compound ${\text{CH}}_{\text{2}}\text{O}$, only one carbon atom is present. Hence, mass of carbon in ${\text{CH}}_{\text{2}}\text{O}$ is $\text{12}\text{.01}\text{g}/\text{mol}$.

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{15}\text{.999}\text{g}/\text{mol}$. In the compound ${\text{CH}}_{\text{2}}\text{O}$, only one oxygen atom is present. Hence, mass of oxygen in ${\text{CH}}_{\text{2}}\text{O}$ is $\text{15}\text{.999}\text{g}/\text{mol}$.

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In the compound ${\text{CH}}_{\text{2}}\text{O}$, two hydrogen atoms are present. Hence, mass of hydrogen in ${\text{CH}}_{\text{2}}\text{O}$ is,

$\text{2}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{2}\text{.016}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it.

Hence, molar mass of ${\text{CH}}_{\text{2}}\text{O}$ is calculated as,

$\text{12}\text{.01}\text{g/mol}+\text{15}\text{.999}\text{g}/\text{mol}+\text{2}\text{.016}\text{g}/\text{mol}=\text{30}\text{.025}\text{g/mol}$

Formula

The mass percent of element is calculated as,

$\text{Mass\%}\text{of}\text{element}=\frac{\text{Mass}\text{of}\text{element}}{\text{Molar}\text{mass}\text{of}\text{compound}}$                                        (1)

Substitute the values of mass of carbon and molar mass of ${\text{CH}}_{\text{2}}\text{O}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{2}}\text{O}}\\ =\left(\frac{\text{12}\text{.01}\text{g/mol}}{\text{30}\text{.025}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{40.00\%}\end{array}$

Substitute the values of mass of hydrogen and molar mass of ${\text{CH}}_{\text{2}}\text{O}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{2}}\text{O}}\\ =\left(\frac{\text{2}\text{.016}\text{g/mol}}{\text{30}\text{.025}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{6.714\%}\end{array}$

Substitute the values of mass of oxygen and molar mass of ${\text{CH}}_{\text{2}}\text{O}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{O}\text{atom}=\frac{\text{Mass}\text{of}\text{O}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{2}}\text{O}}\\ =\left(\frac{\text{15}\text{.999}\text{g}/\text{mol}}{\text{30}\text{.025}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{53.28\%}\end{array}$

The percentage composition (by mass) of carbon, hydrogen and oxygen is $\underset{\_}{40.00\%}$, $\underset{\_}{6.714\%}$ and $\underset{\_}{53.28\%}$ respectively.

Conclusion

The percentage composition of any element is calculated by dividing the total mass of that element with molar mass of compound.

Interpretation Introduction

Interpretation: The percentage composition (by mass) of each element in the given compounds ${\text{CH}}_{\text{2}}\text{O},{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ and ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

To determine: The percentage composition (by mass) of each element in glucose $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)$ compound.

Explanation of Solution

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$, six carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is,

$\text{6}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{72}\text{.06}\text{g}/\text{mol}$

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{15}\text{.999}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$, six oxygen atoms are present. Hence, mass of oxygen in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is,

$\text{6}\times \text{15}\text{.999}\text{g}/\text{mol}=\text{95}\text{.994}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$, twelve hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is,

$\text{12}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{12}\text{.096}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it.

Hence, molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is calculated as,

$\text{72}\text{.06}\text{g}/\text{mol}+\text{95}\text{.994}\text{g}/\text{mol}+\text{12}\text{.096}\text{g}/\text{mol}=\text{180}\text{.15}\text{g/mol}$

Formula

The mass percent of element is calculated as,

$\text{Mass\%}\text{of}\text{element}=\frac{\text{Mass}\text{of}\text{element}}{\text{Molar}\text{mass}\text{of}\text{compound}}$                                        (1)

Substitute the values of mass of carbon and molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ =\left(\frac{\text{72}\text{.06}\text{g/mol}}{\text{180}\text{.15}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{40.00\%}\end{array}$

Substitute the values of mass of hydrogen and molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ =\left(\frac{\text{12}\text{.096}\text{g/mol}}{\text{180}\text{.15}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{6.714\%}\end{array}$

Substitute the values of mass of oxygen and molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{O}\text{atom}=\frac{\text{Mass}\text{of}\text{O}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ =\left(\frac{\text{95}\text{.994}\text{g/mol}}{\text{180}\text{.15}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{53.28\%}\end{array}$

Conclusion

The percentage composition of any element is calculated by dividing the total mass of that element with molar mass of compound.

Interpretation Introduction

Interpretation: The percentage composition (by mass) of each element in the given compounds ${\text{CH}}_{\text{2}}\text{O},{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ and ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction: The atomic mass is defined as the sum of number of protons and number of neutrons.

Molar mass of a substance is defined as the mass of the substance in gram of one mole of that compound.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

To determine: The percentage composition (by mass) of each element in acetic acid $\left({\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)$ compound.

Explanation of Solution

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In the compound ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$, two carbon atoms are present. Hence, mass of carbon in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is,

$\text{2}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{24}\text{.02}\text{g}/\text{mol}$

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{15}\text{.999}\text{g}/\text{mol}$. In the compound ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$, two oxygen atoms are present. Hence, mass of oxygen in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is,

$\text{2}\times \text{15}\text{.999}\text{g}/\text{mol}=\text{31}\text{.998}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In the compound ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$, four hydrogen atoms are present. Hence, mass of hydrogen in ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is,

$\text{4}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{4}\text{.032}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it. Hence, molar mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is calculated as,

$\text{24}\text{.02}\text{g}/\text{mol}+\text{31}\text{.998}\text{g}/\text{mol}+\text{4}\text{.032}\text{g}/\text{mol}=\text{60}\text{.05}\text{g/mol}$

Formula

The mass percent of element is calculated as,

$\text{Mass\%}\text{of}\text{element}=\frac{\text{Mass}\text{of}\text{element}}{\text{Molar}\text{mass}\text{of}\text{compound}}$                                        (1)

Substitute the values of mass of carbon and molar mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{24}\text{.02}\text{g/mol}}{\text{60}\text{.05}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{40.00\%}\end{array}$

Substitute the values of mass of hydrogen and molar mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{4}\text{.032}\text{g/mol}}{\text{60}\text{.05}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{6.72\%}\end{array}$

Substitute the values of mass of nitrogen and molar mass of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{O}\text{atom}=\frac{\text{Mass}\text{of}\text{O}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}\\ =\left(\frac{\text{31}\text{.998}\text{g/mol}}{\text{60}\text{.05}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{53.28\%}\end{array}$

Conclusion

The percentage composition of any element is calculated by dividing the total mass of that element with molar mass of compound.