Chapter 3, Problem 3.36QP

Interpretation Introduction

Interpretation: By using percent composition by mass concept the identification of unknown compound has to be explained.

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 explain: The identification of unknown compound by using percent composition by mass concept.

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

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{Where}\text{n}\text{is}\text{the}\text{number}\text{of}\text{atoms}\text{of}\text{the}\text{element}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$ Percent composition is fixed for each pure substance. So, with the help of this fixed value we can calculate the formula of unknown compound.

For example,

$\begin{array}{l}\text{The}\text{percent}\text{by}\text{mass}\text{of}\text{\%}\text{H}\text{=}\text{5}\text{.926\%}\\ \text{The}\text{percent}\text{by}\text{mass}\text{of}\text{\%}\text{O}\text{=}\text{94}\text{.06 \%}\\ \text{Molecular mass}\text{of}\text{unknown}\text{compound}\text{=}\text{34}\text{.02}\text{amu}\end{array}$

With the help of above details we can easily identify the unknown compound. The molecular formula of the unknown compound is calculated 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}$$\text{we have to find the "n'' value for both}\text{H}\text{and}\text{O}\text{atoms}$

The number of hydrogen atoms present in the unknown molecule is calculated as follows:

$\begin{array}{l}\text{percent}\text{by}\text{mass}\text{of}\text{H}\text{=}\frac{\text{n×atomic}\text{mass}\text{of}\text{H}\text{in}\text{amu}}{\text{molecularmass}\text{of}\text{unknown}\text{compound}}\\ \text{5}\text{.926}\text{\%}\text{=}\frac{\text{n×1}\text{.000794}\text{amu}}{\text{34}\text{.02}\text{amu}}\text{×}\text{100}\\ \text{5}\text{.926}\text{×34}\text{.02}\text{=}\text{n×1}\text{.00794×100}\\ \text{n}\text{=}\frac{\text{5}\text{.926}\text{×34}\text{.02}}{\text{1}\text{.00794}\text{×100}}\end{array}$

$\begin{array}{l}\text{n}\text{=}\frac{\text{201}\text{.60252}}{\text{100}\text{.794}}\\ \text{n}\text{=}\text{2}\text{.0001}\end{array}$

The number of hydrogen atoms present ion unknown compound is 2.

The number of oxygen atoms present in unknown the molecule is calculated as follows:

$\begin{array}{l}\text{percent}\text{by}\text{mass}\text{of}\text{O}\text{=}\frac{\text{n×atomic}\text{mass}\text{of}\text{O}\text{in}\text{amu}}{\text{molecular mass}\text{of}\text{unknown}\text{compound}}\\ \text{94}\text{.06}\text{\%}\text{=}\frac{\text{n×15}\text{.9994}\text{amu}}{\text{34}\text{.02}\text{amu}}\text{×}\text{100}\\ \text{94}\text{.06}\text{×34}\text{.02}\text{=}\text{n×15}\text{.9994×100}\\ \text{n}\text{=}\frac{\text{94}\text{.06}\text{×34}\text{.02}}{\text{15}\text{.9994}\text{×100}}\end{array}$

$\begin{array}{l}\text{n}\text{=}\frac{\text{3199}\text{.9212}}{\text{1599}\text{.94}}\\ \text{n}\text{=}\text{2}\text{.00000}\end{array}$

The number of oxygen atoms present ion unknown compound is 2.

$Molecularformulaoftheunknowncompoundis{H}_2{O}_2$

Conclusion

The identification of unknown compound was explained by percent composition by mass concept.