Chapter 5, Problem 5.74QP

Interpretation Introduction

Interpretation: The percent composition by mass of chloroform $\left({\text{CHCl}}_{\text{3}}\right)$ has to be calculated.

Concept introduction: Percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

To determine: The percent composition by mass of chloroform $\left({\text{CHCl}}_{\text{3}}\right)$

Answer to Problem 5.74QP

The percent composition by mass of chloroform is $\text{\%}\text{C}\text{=}\text{10}\text{.06}\text{\%}\text{;}\text{\%}\text{H}\text{=}\text{0}\text{.85}\text{\%}\text{;}\text{\%}\text{Cl}\text{=}\text{89}\text{.0946}\text{\%}$ because of rounding the overall percentage is slightly more $\text{100}\text{.0046}\text{\%}$.

Explanation of Solution

Percent composition by mass: percent composition by mass of each element present in a compound is known as the compound’s percent composition by mass. The formula is expressed mathematically as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$ By using the above formula percent composition by mass of chloroform is calculated as follows:

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{CHCl}}_{\text{3}}\text{=}\text{mass}\text{of}\text{C}\text{in}\text{amu}\text{+}\text{mass}\text{of}\text{H}\text{in}\text{amu}\text{+}\left(\text{3×mass}\text{of}\text{Cl}\text{in}\text{amu}\right)\\ \text{=}\text{12}\text{.0107}\text{amu}\text{+}\text{1}\text{.0079}\text{amu}\text{+}\left(\text{3×35}\text{.453}\right)\text{amu}\\ \text{=}\left(\text{12}\text{.0107}\text{+}\text{1}\text{.0079}\text{+}\text{106}\text{.359}\right)\text{amu}\\ \text{=}\text{119}\text{.3776}\text{amu}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{C}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{C}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CHCl}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{12}\text{.0107}\text{amu}}{\text{119}\text{.3776}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{10}\text{.06}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{H}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{H}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CHCl}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{1}\text{.00794}\text{amu}}{\text{119}\text{.3776}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{0}\text{.844}\text{\%}\end{array}$ $\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}{\text{Cl}}_{\text{3}}\text{=}\frac{\text{3}\text{×atomic}\text{mass}\text{Cl}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CHCl}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{3×35}\text{.453}\text{amu}}{\text{119}\text{.3776}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{106}\text{.359}\text{amu}}{\text{119}\text{.3776}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{89}\text{.0946}\text{\%}\end{array}$

The percent composition by mass of chloroform is $\text{\%}\text{C}\text{=}\text{10}\text{.06}\text{\%}\text{;}\text{\%}\text{H}\text{=}\text{0}\text{.85}\text{\%}\text{;}\text{\%}\text{Cl}\text{=}\text{89}\text{.0946}\text{\%}$ because of rounding the overall percentage is slightly more $\text{100}\text{.0046}\text{\%}$.

Conclusion

The percent composition by mass of chloroform $\left({\text{CHCl}}_{\text{3}}\right)$ was calculated.

                      $\text{percent}\text{mass}\text{of}\text{\%}\text{C}\text{=}\text{10}\text{.06}\text{\%}$

                      $\text{percent}\text{mass}\text{of}\text{\%}\text{H}\text{=}\text{0}\text{.85}\text{\%}$

                      $\text{percent}\text{mass}\text{of}\text{\%}\text{Cl}\text{=}\text{89}\text{.0946}\text{\%}$