Chapter 2, Problem 136QRT

Interpretation Introduction

Interpretation:

A $\text{10}\text{.0}\text{g}$ sample of stainless steel is treated to convert the metal oxides: $\text{10}\text{.3}\text{g}$ ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$, $\text{2}\text{.71}\text{g}$ ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$, $\text{1}\text{.14}\text{g}$ $\text{NiO}$.  The mass percent of each metal in the sample has to be calculated.

Explanation of Solution

Given the amount of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{10}\text{.3}\text{g}$.

Molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is:

    $\text{(2×55}\text{.854}\text{g/mol}\text{Fe)}\text{}\text{+}\text{(2×15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{159}\text{.688}\text{g/mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

Number of grams of $\text{Fe}$ is:

    $\text{10}\text{.3}\text{g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\left(\frac{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{159}\text{.688}\text{g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\text{×}\left(\frac{\text{2}\text{mol}\text{Fe}}{\text{1}\text{mol}\text{Fe}}\right)\text{×}\left(\frac{\text{55}\text{.854}\text{g}\text{Fe}}{\text{1}\text{mole}\text{Fe}}\right)\text{=}\text{7}\text{.16}\text{g}\text{Fe}$.

Percentage of iron present in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

    $\text{\%}\text{Fe}\text{=}\frac{\text{7}\text{.16}\text{g}\text{Fe}}{\text{10}\text{.0}\text{g}\text{sample}}\text{×}\text{100\%}\text{=}\text{71}\text{.6\%}\text{Fe}$

Given the amount of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{2}\text{.71}\text{g}$.

Molar mass of ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$ is:

    $\text{(2×51}\text{.9961}\text{g/mol}\text{Cr)}\text{}\text{+}\text{(2×15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{151}\text{.9904}\text{g/mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$.

Number of grams of $\text{Cr}$ is:

    $\text{2}\text{.71}\text{g}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\left(\frac{\text{1}\text{mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{151}\text{.9904}\text{g}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\text{×}\left(\frac{\text{2}\text{mol}\text{Cr}}{\text{1}\text{mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\text{×}\left(\frac{\text{51}\text{.9961}\text{g}\text{Cr}}{\text{1}\text{mole}\text{Cr}}\right)\text{=}\text{1}\text{.85}\text{g}\text{Cr}$.

Percentage of chromium present in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}$.

    $\text{\%}\text{Cr=}\frac{\text{1}\text{.85}\text{g}\text{Cr}}{\text{10}\text{.0}\text{g}\text{sample}}\text{×}\text{100\%}\text{=}\text{18}\text{.5\%}\text{Cr}$.

Given the amount of $\text{NiO}$ is $\text{1}\text{.14}\text{g}$.

Molar mass of $\text{NiO}$ is:

    $\text{(58}\text{.6934}\text{g/mol}\text{Ni)}\text{}\text{+}\text{(15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{74}\text{.6928}\text{g/mol}\text{NiO}$.

Number of grams of $\text{Ni}$ is:

    $\text{1}\text{.14}\text{g}\text{NiO}\text{×}\left(\frac{\text{1}\text{mol}\text{NiO}}{\text{74}\text{.6928}\text{g}\text{NiO}}\right)\text{×}\left(\frac{\text{1}\text{mol}\text{Ni}}{\text{1}\text{mol}\text{NiO}}\right)\text{×}\left(\frac{\text{58}\text{.6934}\text{g}\text{Ni}}{\text{1}\text{mole}\text{Ni}}\right)\text{=}\text{0}\text{.896}\text{g}\text{Ni}$.

Percentage of chromium present in $\text{NiO}$.

    $\text{\%}\text{Ni}\text{=}\frac{\text{0}\text{.896}\text{g}\text{Ni}}{\text{10}\text{.0}\text{g}\text{sample}}\text{×}\text{100\%}\text{=}\text{8}\text{.96\%}\text{}\text{Ni}$.