Chapter 3, Problem 3.37QP

Tin (Sn) exists in Earth’s crust as SnO₂. Calculate the percent composition by mass of Sn and O in SnO₂.

Interpretation Introduction

Interpretation: The percent composition by mass of $\text{Sn}$ and $\text{O}$ in ${\text{SnO}}_{\text{2}}$ 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.

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

Answer to Problem 3.37QP

The percent composition by mass of $\text{\%}\text{Sn}\text{=}\text{78}\text{.7677}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{21}\text{.2322}\text{\%}$ because of rounding the overall percentage is slightly less $\text{99}\text{.9999}\text{\%}$.

Explanation of Solution

To determine: The percent composition by mass of $\text{Sn}$ and $\text{O}$ in ${\text{SnO}}_{\text{2}}$

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 is calculated as follows:

$\begin{array}{l}\text{Formula}\text{mass}\text{of}{\text{SnO}}_{\text{2}}\text{=}\text{mass}\text{of}\text{Sn}\text{in}\text{amu}\text{+}\left(\text{2×mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{118}\text{.710}\text{amu}\text{+}\left(\text{2×15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{118}\text{.710+}\text{31}\text{.9988}\right)\text{amu}\\ \text{=}\text{150}\text{.7088}\text{amu}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{Sn}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{Sn}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{SnO}}_{\text{2}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{118}\text{.710}\text{amu}}{\text{150}\text{.7088}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{78}\text{.7677}\text{\%}\end{array}$

 $\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{O}\text{=}\frac{\text{2}\text{×atomic}\text{mass}\text{O}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{SnO}}_{\text{2}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{2×15}\text{.9994}\text{amu}}{\text{150}\text{.7088}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{31}\text{.9988}\text{amu}}{\text{150}\text{.7088}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{21}\text{.2322}\text{\%}\end{array}$

The percent composition by mass of $\text{\%}\text{Sn}\text{=}\text{78}\text{.7677}\text{\%}\text{;}\text{\%}\text{O}\text{=}\text{21}\text{.2322}\text{\%}$ because of rounding the overall percentage is slightly less $\text{99}\text{.9999}\text{\%}$.

Conclusion

The percent composition by mass of ${\text{SnO}}_{\text{2}}$ was calculated.

$\text{percent}\text{mass}\text{of}\text{\%}\text{Sn}\text{=}78.7677\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{O}\text{=}21.2322\text{\%}$