Chapter 5, Problem 73E

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers:

a. C₃H₄O₂ (acrylic acid, from which acrylic plastics are made)

b. C₄H₆O₂ (methyl acrylate, from which Plexiglas is made)

c. C₃H₃N (acrylonitrile, from which Orion is made)

Interpretation Introduction

Interpretation: The percentage composition (by mass) of each element in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ compound 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$, three carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is,

$\text{3}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{36}\text{.03}\text{g}/\text{mol}$

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{16}\text{.00}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$, two oxygen atoms are present. Hence, mass of oxygen in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is,

$\text{2}\times \text{16}\text{.00}\text{g}/\text{mol}=\text{32}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$, four hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is calculated as,

$\text{36}\text{.03}\text{g/mol}+\text{4}\text{.032}\text{g}/\text{mol}+\text{32}\text{g}/\text{mol}=\text{72}\text{.053}\text{g/mol}$

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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{36}\text{.03}\text{g/mol}}{\text{72}\text{.053}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{50.00\%}\end{array}$

Substitute the values of mass of hydrogen and molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{4}\text{.032}\text{g/mol}}{\text{72}\text{.053}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{5.59\%}\end{array}$

Substitute the values of mass of oxygen and molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{O}\text{atom}=\frac{\text{Mass}\text{of}\text{O}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{32}\text{g/mol}}{\text{72}\text{.053}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{44.41\%}\end{array}$

The percentage composition (by mass) of carbon, hydrogen and oxygen is $\underset{\_}{50.00\%}$, $\underset{\_}{5.59\%}$ and $\underset{\_}{44.40\%}$ 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ compound 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 ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ compound.

Explanation of Solution

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$, four carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is,

$\text{4}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{48}\text{.04}\text{g}/\text{mol}$

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{16}\text{.00}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$, two oxygen atoms are present. Hence, mass of oxygen in ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is,

$\text{2}\times \text{16}\text{.00}\text{g}/\text{mol}=\text{32}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$, six hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is,

$\text{6}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{6}\text{.048}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it. Hence, molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is calculated as,

$\text{48}\text{.04}\text{g}/\text{mol}+\text{6}\text{.048}\text{g}/\text{mol}+\text{32}\text{g}/\text{mol}=\text{86}\text{.088}\text{g/mol}$

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 ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{48}\text{.04}\text{g/mol}}{\text{86}\text{.088}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{55.80\%}\end{array}$

Substitute the values of mass of hydrogen and molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{6}\text{.048}\text{g/mol}}{\text{86}\text{.088}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{7.02\%}\end{array}$

Substitute the values of mass of oxygen and molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ in equation (1).

$\begin{array}{c}\text{Mass\%}\text{of}\text{O}\text{atom}=\frac{\text{Mass}\text{of}\text{O}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{32}\text{g/mol}}{\text{86}\text{.088}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{37.17\%}\end{array}$

The percentage composition (by mass) of carbon, hydrogen and oxygen is $\underset{\_}{55.80\%}$, $\underset{\_}{7.02\%}$ and $\underset{\_}{37.17\%}$ 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ and ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ compound 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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ compound.

Explanation of Solution

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

$\text{3}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{36}\text{.03}\text{g}/\text{mol}$

The atomic weight of nitrogen $\left(\text{N}\right)$ is $\text{14}\text{.006}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$, only one nitrogen atom is present. Hence, mass of nitrogen in ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is $\text{14}\text{.006}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$, three hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is,

$\text{3}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{3}\text{.024}\text{g}/\text{mol}$

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

$\text{36}\text{.03}\text{g}/\text{mol}+\text{3}\text{.024}\text{g}/\text{mol}+\text{14}\text{.006}\text{g}/\text{mol}=\text{53}\text{.06}\text{g/mol}$

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 ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}\\ =\left(\frac{\text{36}\text{.03}\text{g/mol}}{\text{53}\text{.06}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{67.90\%}\end{array}$

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

$\begin{array}{c}\text{Mass\%}\text{of}\text{H}\text{atom}=\frac{\text{Mass}\text{of}\text{H}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}\\ =\left(\frac{\text{3}\text{.024}\text{g/mol}}{\text{53}\text{.06}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{5.69\%}\end{array}$

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

$\begin{array}{c}\text{Mass\%}\text{of}\text{N}\text{atom}=\frac{\text{Mass}\text{of}\text{N}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}\\ =\left(\frac{\text{14}\text{.006}\text{g/mol}}{\text{53}\text{.06}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{26.39\%}\end{array}$

The percentage composition (by mass) of carbon, hydrogen and nitrogen is $\underset{\_}{67.90\%}$, $\underset{\_}{5.69\%}$ and $\underset{\_}{26.39\%}$ respectively.

Conclusion

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