Chapter 2, Problem 140QRT

Interpretation Introduction

Interpretation:

Four potassium compounds of oxygen with mass percent of potassium as in $\text{83}\text{.0\%}$ compound I,  $\text{55}\text{.0\%}$ in compound II, $\text{44}\text{.9\%}$ in compound III and $\text{71}\text{.0\%}$ in compound IV.  The molar mass of one compound is $\text{110}\text{.2}\text{g/mol}$.  The chemical formula of each compound has to be calculated.

Explanation of Solution

Given,

The molar mass of one compound is $\text{110}\text{.2}\text{g/mol}$.

Molar mass of potassium is $\text{39}\text{.0983}\text{g}$ and oxygen is $\text{15}\text{.9994}\text{g}$.

Mass percent of potassium in compound I is $\text{83}\text{.0\%}$.

Mass percent of potassium in compound II is $\text{55}\text{.0\%}$.

Mass percent of potassium in compound III is $\text{44}\text{.9\%}$.

Mass percent of potassium in compound IV is $\text{71}\text{.0\%}$.

The general formula of the compound with potassium and oxygen will be ${\text{K}}_{\text{x}}{\text{O}}_{\text{y}}$.

For compound I:

The mass percent of potassium is $\text{83}\text{.0\%}$.

The mass percent of oxygen is $\text{100}\text{-}\text{83}\text{.0}\text{=}\text{17\%}\text{O}$.

$\text{100}\text{g}$ of sample has $\text{83}\text{.0}\text{g}\text{K}$ and $\text{17}\text{.0}\text{g}\text{O}$.

Number of moles of potassium is:

    $\text{83}\text{.0}\text{g}\text{K}\text{×}\frac{\text{1}\text{mol}\text{K}}{\text{39}\text{.0983}\text{g}\text{K}}\text{=}\text{2}\text{.12}\text{mol}\text{K}$.

Number of moles of oxygen is:

    $\text{17}\text{.0}\text{g}\text{O}\text{×}\frac{\text{1}\text{mol}\text{O}}{\text{15}\text{.9994}\text{g}\text{O}}\text{=}\text{1}\text{.06}\text{mol}\text{O}$.

Mole ratio is $\text{2}\text{.12}\text{mol}\text{K}\text{:}\text{1}\text{.06}\text{mol}\text{O}$

Simplify by dividing $\text{1}\text{.06}\text{mol}$.

Then the ratio will be $\text{2}\text{mol}\text{K}\text{:}\text{1}\text{mol}\text{O}$

The empirical formula is ${\text{K}}_2\text{O}$.

The empirical formula mass of ${\text{K}}_2\text{O}$ is

  $\text{(2×39}\text{.0983}\text{g/mol}\text{K)}\text{+}\text{(15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{94}\text{.1960}\text{g/mol}$.

For compound II:

The mass percent of potassium is $\text{55}\text{.0\%}$.

The mass percent of oxygen is $\text{100}\text{-}55.\text{0}\text{=}45.0\text{\%}\text{O}$.

$\text{100}\text{g}$ of sample has $\text{55}\text{.0}\text{g}\text{K}$ and $\text{45}\text{.0}\text{g}\text{O}$.

Number of moles of potassium is:

    $\text{55}\text{.0}\text{g}\text{K}\text{×}\frac{\text{1}\text{mol}\text{K}}{\text{39}\text{.0983}\text{g}\text{K}}\text{=}\text{1}\text{.41}\text{mol}\text{K}$.

Number of moles of oxygen is:

    $\text{45}\text{.0}\text{g}\text{O}\text{×}\frac{\text{1}\text{mol}\text{O}}{\text{15}\text{.9994}\text{g}\text{O}}\text{=}\text{2}\text{.81}\text{mol}\text{O}$.

Mole ratio is $\text{1}\text{.41}\text{mol}\text{K}\text{:}\text{2}\text{.81}\text{mol}\text{O}$

Simplify by dividing $\text{1}\text{.41}\text{mol}$.

Then the ratio will be $\text{1}\text{mol}\text{K}\text{:}\text{2}\text{mol}\text{O}$

The empirical formula is ${\text{KO}}_2$.

The empirical formula mass of ${\text{KO}}_2$ is

  $\text{(39}\text{.0983}\text{g/mol}\text{K)}\text{+}\text{(2}\times \text{15}\text{.9994}\text{g/mol}\text{O)}\text{=71}\text{.0971}\text{g/mol}$.

For compound III:

The mass percent of potassium is $\text{44}\text{.90\%}$.

The mass percent of oxygen is $\text{100}\text{-}\text{44}\text{.9=}\text{55}\text{.1\%}\text{O}$.

$\text{100}\text{g}$ of sample has $\text{44}\text{.9}\text{g}\text{K}$ and $\text{55}\text{.1}\text{g}\text{O}$.

Number of moles of potassium is:

    $\text{44}\text{.9}\text{g}\text{K}\text{×}\frac{\text{1}\text{mol}\text{K}}{\text{39}\text{.0983}\text{g}\text{K}}\text{=}\text{1}\text{.15}\text{mol}\text{K}$.

Number of moles of oxygen is:

    $\text{55}\text{.1}\text{g}\text{O}\text{×}\frac{\text{1}\text{mol}\text{O}}{\text{15}\text{.9994}\text{g}\text{O}}\text{=}\text{3}\text{.44}\text{mol}\text{O}$.

Mole ratio is $\text{1}\text{.15mol}\text{K}\text{:}\text{3}\text{.44}\text{mol}\text{O}$

Simplify by dividing $\text{1}\text{.15}\text{mol}$.

Then the ratio will be $\text{1}\text{mol}\text{K}\text{:}3\text{mol}\text{O}$

The empirical formula is ${\text{KO}}_3$.

The empirical formula mass of ${\text{KO}}_3$ is

  $\text{(39}\text{.0983}\text{g/mol}\text{K)}\text{+}\text{(3}\times \text{15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{87}\text{.0965}\text{g/mol}$.

For compound IV:

The mass percent of potassium is $\text{71}\text{.0\%}$.

The mass percent of oxygen is $\text{100}\text{-}\text{71}\text{.0}\text{=}\text{29\%}\text{O}$.

$\text{100}\text{g}$ of sample has $\text{71}\text{.0}\text{g}\text{K}$ and $\text{29}\text{.0}\text{g}\text{O}$.

Number of moles of potassium is:

    $\text{71}\text{.0}\text{g}\text{K}\text{×}\frac{\text{1}\text{mol}\text{K}}{\text{39}\text{.0983}\text{g}\text{K}}\text{=1}\text{.82mol}\text{K}$.

Number of moles of oxygen is:

    $\text{29}\text{.0}\text{g}\text{O}\text{×}\frac{\text{1}\text{mol}\text{O}}{\text{15}\text{.9994}\text{g}\text{O}}\text{=}\text{1}\text{.81}\text{mol}\text{O}$.

Mole ratio is $\text{1}\text{.82}\text{mol}\text{K}\text{:}\text{1}\text{.81}\text{mol}\text{O}$

Simplify by dividing $\text{1}\text{.81}\text{mol}$.

Then the ratio will be $\text{1}\text{mol}\text{K}\text{:}\text{1}\text{mol}\text{O}$

The empirical formula is $\text{KO}$.

The empirical formula mass of $\text{KO}$ is

  $\text{(39}\text{.0983}\text{g/mol}\text{K)}\text{+}\text{(15}\text{.9994}\text{g/mol}\text{O)}\text{=}\text{55}\text{.098}\text{g/mol}$.