Chapter 3, Problem 78E

Arrange the following substances in order of increasing mass percent of carbon.

a. caffeine. C₈H₁₀N₄O₂

b. sucrose, C₁₂H₂₂O₁₁

c. ethanol, C₂H₅OH

Interpretation Introduction

Interpretation: The compounds ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ are to be arranged in increasing order of mass percent of carbon.

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 arrangement of compounds ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in order of increasing mass percent of carbon.

Answer to Problem 78E

The increasing order of mass percent of carbon for the compounds ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is,

${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}<{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}<{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$

Explanation of Solution

To determine: The percentage composition (by mass) of carbon in the compound ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$.

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{15}\text{.999}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$, two oxygen atoms are present. Hence, mass of oxygen in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ 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 a compound ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$, ten hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is,

$\text{10}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{10}\text{.08}\text{g}/\text{mol}$

The atomic weight of nitrogen $(\text{N})$ atom is $\text{14}\text{.006}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$, four nitrogen atoms are present. Hence, mass of nitrogen in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is,

$\text{4}\times \text{14}\text{.006}\text{g}/\text{mol}=\text{56}\text{.024}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it. Hence, molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is calculated as,

$\left(\text{96}\text{.08}+\text{31}\text{.998}+\text{10}\text{.08}+\text{56}\text{.024}\right)\text{g/mol}=\text{194}\text{.182}\text{g/mol}$

Formula

The mass percent of element is calculated as,

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

Substitute the values of mass of carbon and molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}}\\ =\left(\frac{\text{96}\text{.08}\text{g}/\text{mol}}{\text{194}\text{.182}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{49.47\%}\end{array}$

The percentage composition (by mass) of carbon in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is $\underset{\_}{49.47\%}$.

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In the compound ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$, eight carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is,

$\text{8}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{96}\text{.08}\text{g}/\text{mol}$

To determine: The percentage composition (by mass) of carbon in compound ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$.

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$, twelve carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is,

$\text{12}\times \text{12}\text{.01}\text{g}/\text{mol}=\text{144}\text{.12}\text{g}/\text{mol}$

The atomic weight of oxygen $\left(\text{O}\right)$ is $\text{15}\text{.999}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$, eleven oxygen atoms are present. Hence, mass of oxygen in ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is,

$\text{11}\times \text{15}\text{.999}\text{g}/\text{mol}=\text{175}\text{.989}\text{g}/\text{mol}$

The atomic weight of hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$, twenty-two hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is,

$\text{22}\times \text{1}\text{.008}\text{g}/\text{mol}=\text{22}\text{.176}\text{g}/\text{mol}$

The molar mass is the sum of mass of individual atoms present in it. Hence, molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is calculated as,

$\left(\text{144}\text{.12}+\text{175}\text{.989}+\text{22}\text{.176}\right)\text{g/mol}=\text{342}\text{.285}\text{g/mol}$

Formula

The mass percent of element is calculated as,

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

Substitute the values of mass of carbon and molar mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}}\\ =\left(\frac{\text{144}\text{.12}\text{g}/\text{mol}}{\text{342}\text{.28}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{42.10\%}\end{array}$

To determine: The percentage composition (by mass) of carbon in compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$.

The atomic weight of carbon $\left(\text{C}\right)$ is $\text{12}\text{.01}\text{g}/\text{mol}$. In a compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$, two carbon atoms are present. Hence, mass of carbon in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ 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 a compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$, only one oxygen atom is present. Hence, mass of oxygen in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ 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 a compound ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$, six hydrogen atoms are present. Hence, mass of hydrogen in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ 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 ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is calculated as,

$\left(\text{24}\text{.02}+\text{15}\text{.999}+\text{6}\text{.048}\right)\text{g/mol}=\text{46}\text{.067}\text{g/mol}$

Formula

The mass percent of element is calculated as,

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

Substitute the values of mass of carbon and molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in the above equation.

$\begin{array}{c}\text{Mass\%}\text{of}\text{C}\text{atom}=\frac{\text{Mass}\text{of}\text{C}\text{atom}}{\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\\ =\left(\frac{\text{24}\text{.02}\text{g}/\text{mol}}{\text{46}\text{.067}\text{g/mol}}\right)\times \text{100}\\ =\underset{\_}{52.14\%}\end{array}$

To determine: The increasing order of mass percent of carbon for the compounds ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$.

The mass percent of carbon atom in ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}$ is $\text{49}\text{.47\%}$.

The mass percent of carbon atom in ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ is $\text{42}\text{.10\%}$.

The mass percent of carbon atom in ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $\text{52}\text{.14\%}$.

Hence, the increasing order of mass percent of carbon for the compounds ${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}},{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$ and ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is,

${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}<{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}<{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$

Conclusion

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