Chapter 3, Problem 3.78QE

(a)

Interpretation Introduction

Interpretation:

The percentage by mass of each element in ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ has to be calculated.

Explanation of Solution

First the molar mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ has to be calculated.

    $\begin{array}{l}\text{6(C) 6×12}\text{.01u=72}\text{.06u}\\ \text{12(H)12×1}\text{.01u=12}\text{.12u}\\ \text{--------------------------}\\ {\text{Molecular mass C}}_{\text{6}}{\text{H}}_{\text{12}}\text{ = 84}\text{.18u}\end{array}$.

Thus one mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ has a mass of $\text{84}\text{.18}\text{g}$ that comes from $\text{72}\text{.06g}$ carbon, and $\text{12}\text{.12g}$ hydrogen.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%C: }\frac{\text{72}\text{.06gC}}{\text{84}{\text{.18g C}}_{\text{6}}{\text{H}}_{\text{12}}}\text{×100\%=85}\text{.6\%C}\\ \text{\%H: }\frac{\text{12}\text{.12 gH}}{\text{84}{\text{.18g C}}_{\text{6}}{\text{H}}_{\text{12}}}\text{×100\% = 14}\text{.4\%H}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage by mass of each element in ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$ has to be calculated.

Explanation of Solution

First the molar mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$ has to be calculated.

    $\begin{array}{l}\text{5(C) 5×12}\text{.01u=60}\text{.05u}\\ \text{12(H) 12×1}\text{.01u=12}\text{.12u}\\ \text{1(O)}\text{1}\text{×}\text{15}\text{.999u}\text{=}\text{15}\text{.999}\text{u}\\ \text{--------------------------}\\ {\text{Molecular mass C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O = 88}\text{.169u}\end{array}$.

Thus one mole of ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$ has a mass of $\text{88}\text{.169}\text{g}$ that comes from $\text{60}\text{.05g}$ carbon, $\text{12}\text{.12g}$ hydrogen and $\text{15}\text{.999}\text{g}$ oxygen.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%C: }\frac{\text{60}\text{.05gC}}{\text{88}{\text{.169g C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}}\text{×100\%=68}\text{.11\%C}\\ \text{\%H: }\frac{\text{12}\text{.12 gH}}{\text{88}{\text{.169g C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}}\text{×100\% = 13}\text{.75\%H}\\ \text{\%O: }\frac{\text{15}\text{.999 gO}}{\text{88}{\text{.169g C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}}\text{×100\% = 18}\text{.15\%O}\end{array}$

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

    $\begin{array}{l}\text{1(Ni) 1×58}\text{.69u=58}\text{.69u}\\ \text{2(Cl) 2×35}\text{.45u=70}\text{.9u}\\ \text{--------------------------}\\ {\text{Molecular mass NiCl}}_{\text{2}}\text{ = 129}\text{.59u}\end{array}$.

Thus one mole of ${\text{NiCl}}_{\text{2}}$ has a mass of $\text{129}\text{.59}\text{g}$ that comes from $\text{58}\text{.69g}$ nickel and $\text{70}\text{.9}\text{g}$ chlorine.  These values are used to express the mass composition of the compound as percentages.

    $\begin{array}{l}\text{\%Ni: }\frac{\text{58}\text{.69gNi}}{\text{129}{\text{.59g NiCl}}_{\text{2}}}\text{×100\%=45}\text{.29\%Ni}\\ \text{\%Cl: }\frac{\text{70}\text{.9 gCl}}{\text{129}{\text{.59g NiCl}}_{\text{2}}}\text{×100\% = 54}\text{.71\%Cl}\end{array}$