Chapter 3, Problem 3.77QE

(a)

Interpretation Introduction

Interpretation:

The percentage by mass of each element in ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ has to be calculated.

Explanation of Solution

First the molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ has to be calculated.

    $\begin{array}{l}\text{4(C) 4×12}\text{.01u=48}\text{.04u}\\ \text{8(H) 8×1}\text{.01u=8}\text{.08u}\\ \text{--------------------------}\\ {\text{Molecular mass C}}_{\text{4}}{\text{H}}_{\text{8}}\text{ = 56}\text{.12u}\end{array}$.

Thus one mole of ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$ has a mass of $\text{56}\text{.12}\text{g}$ that comes from $\text{48}\text{.04g}$ carbon, and $\text{8}\text{.08g}$ hydrogen.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%C: }\frac{\text{48}\text{.04gC}}{\text{56}{\text{.12g C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}}\text{×100\%=85}\text{.6\%C}\\ \text{\%H: }\frac{\text{8}\text{.08 gH}}{\text{56}{\text{.12g C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}}\text{×100\% = 14}\text{.4\%H}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage by mass of each element in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}$ has to be calculated.

Explanation of Solution

First the molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}$ has to be calculated.

    $\begin{array}{l}\text{3(C) 3×12}\text{.01u=36}\text{.03u}\\ \text{4(H) 4×1}\text{.01u=4}\text{.04u}\\ \text{2(N)}\text{2}\text{×}\text{14}\text{.01u}\text{=}\text{28}\text{.02}\text{u}\\ \text{--------------------------}\\ {\text{Molecular mass C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}\text{ = 68}\text{.09u}\end{array}$.

Thus one mole of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}$ has a mass of $\text{68}\text{.09}\text{g}$ that comes from $\text{36}\text{.03g}$ carbon, $\text{4}\text{.04g}$ hydrogen and $\text{28}\text{.02}\text{g}$ nitrogen.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%C: }\frac{\text{36}\text{.03gC}}{\text{68}{\text{.09g C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}}\text{×100\%=52}\text{.9\%C}\\ \text{\%H: }\frac{\text{4}\text{.04 gH}}{\text{68}{\text{.09g C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}}\text{×100\% = 5}\text{.93\%H}\\ \text{\%N: }\frac{\text{28}\text{.02 gN}}{\text{68}{\text{.09g C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}}\text{×100\% = 41}\text{.15\%N}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The percentage by mass of each element in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ has to be calculated.

Explanation of Solution

First the molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ has to be calculated.

    $\begin{array}{l}\text{2(Fe) 2×55}\text{.845u=111}\text{.69u}\\ \text{3(O) 4×15}\text{.999u=63}\text{.996u}\\ \text{--------------------------}\\ {\text{Molecular mass Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 175}\text{.69u}\end{array}$.

Thus one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ has a mass of $\text{175}\text{.69}\text{g}$ that comes from $\text{111}\text{.69g}$ iron and $\text{63}\text{.996}\text{g}$ oxygen.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%Fe: }\frac{\text{111}\text{.69gFe}}{\text{175}{\text{.69g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×100\%=63}\text{.57\%Fe}\\ \text{\%O: }\frac{\text{63}\text{.996 gO}}{\text{175}{\text{.69g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×100\% = 36}\text{.42\%O}\end{array}$