Chapter 2, Problem 80PS

Calculate the mass percent of each element in the following compounds:

(a) C₈H₁₀N₂O₂, caffeine

(b) C₁₀H₂₀O menthol

(c) CoCl₂   6 H₂O

Interpretation Introduction

Interpretation: The mass percent of each element in the given molecules are needed to be calculated.

Concept introduction:

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Equation for finding molar mass from mass of a molecule and number moles is,

    $Molarmass=\frac{Massingrams}{Numberofmoles}$ 

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.

Answer to Problem 80PS

The mass percent of $\text{C, H, N and O}$ in $C_8{H}_{10}{N}_2{O}_2$ is $14.42\%$, $5.19\%$, $49.47\%$ and $16.47\%$ respectively.

Explanation of Solution

Molar mass of $C_8{H}_{10}{N}_2{O}_2$ is $\text{194}\text{.19g/mol}$ .

Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

Therefore,

The mass percent of nitrogen in $C_8{H}_{10}{N}_2{O}_2$ is,

    $\frac{2\times 28.014g}{194.19g}\times 100\%=28.85\%$

The mass percent of hydrogen in $C_8{H}_{10}{N}_2{O}_2$ is,

    $\frac{10\times 1.008g}{194.19g}\times 100\%=5.19\%$

The mass percent of carbon in $C_8{H}_{10}{N}_2{O}_2$ is,

    $\frac{8\times 12.01g}{194.19g}\times 100\%=49.47\%$

The mass percent of oxygen in $C_8{H}_{10}{N}_2{O}_2$ is,

    $\frac{2\times 16g}{194.19g}\times 100\%=16.47\%$

Interpretation Introduction

Interpretation: The mass percent of each element in the given molecules are needed to be calculated.

Concept introduction:

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Equation for finding molar mass from mass of a molecule and number moles is,

    $Molarmass=\frac{Massingrams}{Numberofmoles}$ 

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.

Answer to Problem 80PS

The mass percent of $\text{C, H and O}$ in $C_{10}{H}_{20}O$ is $76.85\%$, $12.9\%$ and $10.23\%$ respectively.

Explanation of Solution

Molar mass of menthol ($C_{10}{H}_{20}O$) is $\text{156}\text{.27g/mol}\text{.}$

Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

The mass percent of carbon in $C_{10}{H}_{20}O$ is,

    $\frac{10\times 12.01g}{156.27g}\times 100\%=76.85\%$

The mass percent of hydrogen in $C_{10}{H}_{20}O$ is,

    $\frac{20\times 1.008g}{156.27g}\times 100\%=12.9\%$

The mass percent of oxygen in $C_{10}{H}_{20}O$ is,

    $\frac{1\times 15.99g}{156.27g}\times 100\%=10.23\%$

Interpretation Introduction

Interpretation: The mass percent of each element in the given molecules are needed to be calculated.

Concept introduction:

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Equation for finding molar mass from mass of a molecule and number moles is,

    $Molarmass=\frac{Massingrams}{Numberofmoles}$ 

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.

Answer to Problem 80PS

The mass percent of $\text{Co, H, O and Cl}$ in $CoC{l}_2.6H_{2}O$ is $24.77\%$,$5.084\%$ and $29.79\%$ respectively.

Explanation of Solution

Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

The mass percent of Cobalt in $CoC{l}_2.6H_{2}O$ is,

    $\frac{1\times 58.933g}{237.923g}\times 100\%=24.77\%$

The mass percent of hydrogen in $CoC{l}_2.6H_{2}O$ is,

    $\frac{12\times 1.008g}{237.923g}\times 100\%=5.084\%$

The mass percent of oxygen in $CoC{l}_2.6H_{2}O$ is,

    $\frac{6\times 15.99g}{237.923g}\times 100\%=40.32\%$

The mass percent of chlorine in $CoC{l}_2.6H_{2}O$ is,

    $\frac{2\times 35.45g}{237.923g}\times 100\%=29.79\%$