Chapter 3, Problem 3.51QP

Interpretation Introduction

Interpretation: The percent composition by mass of ${\text{CaSiO}}_{\text{3}}$ 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 ${\text{CaSiO}}_{\text{3}}$

Answer to Problem 3.51QP

$\text{\%}\text{Ca}\text{=}34.5019\text{\%}\text{;}\text{\%}\text{Si}\text{=}24.1779\text{\%}\text{;}\text{\%}\text{O}\text{=}41.3202\text{\%}$

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 is calculated as follows:

$\begin{array}{l}\text{Formula}\text{mass}\text{of}{\text{CaSiO}}_{\text{3}}\text{=}\text{mass}\text{of}\text{Ca}\text{in}\text{amu}\text{+}\text{mass}\text{of}\text{Si}\text{in}\text{amu}\text{+}\left(\text{3×mass}\text{of}\text{O}\text{in}\text{amu}\right)\\ \text{=}\text{40}\text{.078}\text{amu}\text{+}\text{28}\text{.0855}\text{amu}\text{+}\left(\text{3×15}\text{.9994}\text{amu}\right)\\ \text{=}\left(\text{40}\text{.078}\text{+}\text{28}\text{.0855}\text{+}\text{47}\text{.9982}\right)\text{amu}\\ \text{=}\text{116}\text{.1617}\text{amu}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{Ca}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{Ca}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CaSiO}}_{\text{3}}\text{in}\text{amu}}\text{ ×}\text{100\%}\\ \text{=}\frac{\text{40}\text{.078}\text{amu}}{\text{116}\text{.1617}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{34}\text{.5019}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{Si}\text{=}\frac{\text{1}\text{×atomic}\text{mass}\text{Si}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CaSiO}}_{\text{3}}\text{in}\text{amu}}\text{ ×}\text{100\%}\\ \text{=}\frac{\text{28}\text{.0855}\text{amu}}{\text{116}\text{.1617}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{24}\text{.1779}\text{\%}\end{array}$

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{\%}\text{O}\text{=}\frac{\text{3}\text{×atomic}\text{mass}\text{O}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}{\text{CaSiO}}_{\text{3}}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{3×15}\text{.9994}\text{amu}}{\text{116}\text{.1617}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{47}\text{.9982}\text{amu}}{\text{116}\text{.1617}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{41}\text{.3202}\text{\%}\end{array}$

The percent composition by mass of $\text{\%}\text{Ca}\text{=}34.5019\text{\%}\text{;}\text{\%}\text{Si}\text{=}24.1779\text{\%}\text{;}\text{\%}\text{O}\text{=}41.3202\text{\%}$ this will give total percentage is $\text{100}\text{\%}$

Conclusion

The percent composition by mass of ${\text{CaSiO}}_{\text{3}}$ was calculated.

$\text{percent}\text{mass}\text{of}\text{\%}\text{Ca}\text{=}\text{34}\text{.5019}\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{Si}\text{=}\text{24}\text{.1779}\text{\%}$

$\text{percent}\text{mass}\text{of}\text{\%}\text{O}\text{=}\text{41}\text{.3202}\text{\%}$