Chapter 1, Problem 54QRT

Interpretation Introduction

Interpretation:

The element chromium forms three different oxides (contains only chromium and oxygen).  The percentage of chromium (number of grams of chromium in $\text{100}\text{g}$ oxide) in these compounds is $\text{52}\text{.0}\text{\%,}\text{68}\text{.4\%}\text{and 76}\text{.5\%}$.  Does this data confirms the law of multiple proportions has to be identified and if so the reason has to be given.

Concept Introduction:

Law of multiple proportions:

Compounds are made up of atoms of different elements.  Some compounds are made up of same elements.  The law of multiple proportions states that, when two elements are combined to form more than one compound, the mass of one element is fixed and the mass of the other element will be a ratio of whole number.

For example, carbon monoxide ($\text{CO}$) and carbon dioxide (${\text{CO}}_{\text{2}}$) are formed with same elements carbon ($\text{C}$) and oxygen ($\text{O}$).  In both compounds, the composition of carbon is same and oxygen is varied.

Explanation of Solution

The given three different proportions of chromium in $\text{100}\text{g}$ of oxide are $\text{52}\text{.0}\text{\%,}\text{68}\text{.4\%}\text{and 76}\text{.5\%}$.

The mass of the oxygen can be calculated by subtracting the chromium from $\text{100}\text{g}$ of oxide.

In compound 1:

  $\begin{array}{l}\text{=}\text{100}\text{.0}\text{g}\text{-52}\text{.0}\text{g}\text{of}\text{Cr}\\ \\ \text{=}\text{48}\text{.0}\text{g}\text{of}\text{O}\end{array}$

In compound 2:

  $\begin{array}{l}\text{=}\text{100}\text{.0}\text{g}\text{-68}\text{.4}\text{g}\text{of}\text{Cr}\\ \\ \text{=}\text{31}\text{.6}\text{g}\text{of}\text{O}\end{array}$

In compound 3:

  $\begin{array}{l}\text{=}\text{100}\text{.0}\text{g}\text{-76}\text{.5}\text{g}\text{of}\text{Cr}\\ \\ \text{=}\text{23}\text{.5}\text{g}\text{of}\text{O}\end{array}$

The oxygen mass is made same in all compounds as in compound 1.

  $\begin{array}{l}\text{100}\text{.0}\text{g}\text{of}\text{compound}\text{1}\text{has}\text{48}\text{.0}\text{g}\text{O}\left(\frac{\text{48}\text{.0}\text{g}}{\text{48}\text{.0}\text{g}}\right)\text{=}\text{1}\text{.00}\\ \\ \text{100}\text{.0}\text{g}\text{of}\text{compound}\text{2}\text{has}\text{31}\text{.6}\text{g}\text{O}\left(\frac{\text{48}\text{.0}\text{g}}{\text{31}\text{.6}\text{g}}\right)\text{=}\text{1}\text{.52}\\ \\ \text{100}\text{.0}\text{g}\text{of}\text{compound}\text{3}\text{has}\text{23}\text{.5}\text{g}\text{O}\left(\frac{\text{48}\text{.0}\text{g}}{\text{23}\text{.5}\text{g}}\right)\text{=}\text{2}\text{.04}\end{array}$

From the above oxygen mass, the mass of chromium in each sample is determined.

In compound 1:

The mass of oxygen is $\text{48}\text{.0}\text{g}$ and the mass of chromium is $\text{52}\text{.0}\text{g}$.

In compound 2:

  $\text{(1}\text{.52)×100}\text{.0}\text{g}\text{of}\text{compound}\text{2}\text{has}\text{(1}\text{.52)×31}\text{.6}\text{g}\text{O}\text{and}\text{(1}\text{.52)×68}\text{.4}\text{g}\text{Cr}$

So, $\text{151}\text{.9}\text{g}$ of compound has $\text{48}\text{.0}\text{g}$ of oxygen and $\text{104}\text{g}$ of chromium.

In compound 3:

  $\text{(2}\text{.04)×100}\text{.0}\text{g}\text{of}\text{compound}\text{has}\text{(2}\text{.04)×23}\text{.5}\text{g}\text{O}\text{and}\text{(2}\text{.04)×76}\text{.5}\text{g}\text{Cr}$

So, $\text{204}\text{g}$ of compound has $\text{48}\text{.0}\text{g}$ of oxygen and $\text{156}\text{g}$ of chromium.

The ratios of the mass of chromium from three compounds are taken.

  $\begin{array}{l}\frac{\text{Cr}\text{(g)}\text{in}\text{compound}\text{2}}{\text{Cr}\text{(g)}\text{in}\text{compound}\text{1}}\text{=}\frac{\text{104}\text{g}}{\text{52}\text{.0}\text{g}}\text{=}\text{2}\text{.00=}\frac{\text{2}}{\text{1}}\\ \\ \frac{\text{Cr}\text{(g)}\text{in}\text{compound}\text{3}}{\text{Cr}\text{(g)}\text{in}\text{compound}\text{1}}\text{=}\frac{\text{156}\text{.3}\text{g}}{\text{52}\text{.0}\text{g}}\text{=}\text{3}\text{.01=}\frac{\text{3}}{\text{1}}\\ \\ \frac{\text{Cr}\text{(g)}\text{in}\text{compound}\text{3}}{\text{Cr}\text{(g)}\text{in}\text{compound}\text{2}}\text{=}\frac{\text{156}\text{.3}\text{g}}{\text{104}\text{g}}\text{=}\text{1}\text{.50=}\frac{\text{3}}{\text{2}}\end{array}$

As the ratios are whole numbers, the given data confirms the law of multiple proportions.