Question

A sample of a “suboxide” of cesium gives up $\mathbf{1}\mathbf{.}\mathbf{6907}\mathbf{\%}$ of its mass as gaseous oxygen when gently heated, leaving pure cesium behind. Determine the empirical formula of this binary compound

Short answer

The empirical formula of the binary compound is ${\text{Cs}}_7\text{O}$

Step-by-step solution

Step 1:Calculation of number of moles of each element

Empirical formula is the simplest whole number ratio of atoms.Suboxide of cesium has two elements, i.e., cesium and oxygen. So we need to find the number of moles of cesium and oxygen.

Let us assume the suboxide of cesium is of 100 g.

Mass of oxygen,

${\text{m}}_{\text{O}}=1.6907\text{g}$

Molar mass of oxygen,

${\text{M}}_{\text{O}}=16.00\text{g}$

Number of moles of oxygen,

${\text{n}}_{\text{O}}=\frac{{\text{m}}_{\text{O}}}{{\text{M}}_{\text{O}}}=\frac{1.6907}{16.00}=0.1057{\text{gmol}}^{-1}$

Mass of cesium,

$\begin{array}{rcl}{\text{m}}_{\text{Cs}}& =& 100-1.6907\\ & =& 98.3093\text{g}\end{array}$

Molar mass of cesium,

${\text{M}}_{\text{Cs}}=132.91{\text{gmol}}^{-1}$

Number of moles of cesium,

${\text{n}}_{\text{Cs}}=\frac{{\text{m}}_{\text{Cs}}}{{\text{M}}_{\text{Cs}}}=\frac{98.3093}{132.91}=0.7397{\text{gmol}}^{-1}$

Step 2: Simplification of ratios

The smallest value is of oxygen. So let us divide both the number of moles by this value.

$\text{O =}\frac{0.1057}{0.1057}=1$

$\text{Cs =}\frac{0.7397}{0.1057}=6.998~7.00$

Empirical formula

Since the ratio of oxygen and cesium is 1:7, the empirical formula of the compound is ${\text{Cs}}_7\text{O}$