Question

A 2.077-g sample of an element, which has an atomic mass between 40 and 55, reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide and identify the element.

Short answer

The formula of the oxide is${\mathrm{CaO}}_2$(Calcium peroxide)and the element is identified asCalcium($\mathrm{Ca}$).

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

Molar mass$=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{number}\mathrm{of}\mathrm{moles}}$

Determination of x

Here, $\mathrm{E}+{\mathrm{O}}_2\to {\mathrm{E}}_{\mathrm{x}}{\mathrm{O}}_{\mathrm{y}}$

${\mathrm{mO}}_2=3.708\mathrm{g}-2.077\mathrm{g}$

$=1.631\mathrm{g}$

${\mathrm{nO}}_2=1.631\mathrm{g}\times \frac{1\mathrm{mol}}{32.0\mathrm{g}}$

$=0.0510\mathrm{mol}$

Coefficient of ${\mathrm{O}}_2$when element’s molar mass is 44.0 g/mol

$2.077\mathrm{g}\times \frac{1\mathrm{mol}\mathrm{E}}{40.0\mathrm{g}}\times \frac{\mathrm{x}\mathrm{mol}{\text{O}}_2}{1\mathrm{mol}\mathrm{E}}=0.0510\mathrm{mol}{\text{O}}_2$

$\mathrm{x}=\frac{\left(0.0510\mathrm{mol}\right)\left(40.0\mathrm{g}\right)}{2.077\mathrm{g}}=0.982\mathrm{mol}$

Coefficient of ${\mathrm{O}}_2$when element’s molar mass is 55.0 g/mol

$2.077\mathrm{g}\times \frac{1\mathrm{mol}\mathrm{E}}{55.0\mathrm{g}}\times \frac{\mathrm{x}\mathrm{mol}{\text{O}}_2}{1\mathrm{mol}\mathrm{E}}=0.0510\mathrm{mol}{\text{O}}_2$

$\mathrm{x}=\frac{\left(0.0510\mathrm{mol}\right)\left(55.0\mathrm{g}\right)}{2.077\mathrm{g}}=1.35\mathrm{mol}$

Determination of the element and its oxide formula

As we know, for 1 mol of the element E, there is 1 mol of $O_{\mathit{2}}$

$\mathrm{E}+{\mathrm{O}}_2\to {\mathrm{EO}}_2$

As$0.98⩽\mathrm{x}⩽1.35$

$\mathrm{x}\approx 1\mathrm{mol}$

$\text{0.0510}\mathrm{mol}{\text{O}}_2\times \frac{1\mathrm{mol}\mathrm{E}{\text{O}}_2}{1\mathrm{mol}{\text{O}}_2}\times \frac{\text{molar}\mathrm{mass}{\text{EO}}_2}{1\mathrm{mol}\mathrm{E}{\text{O}}_2}=3.708\mathrm{g}$

$\text{molar}\mathrm{mass}{\text{EO}}_2=\frac{3.708\mathrm{g}}{\text{0.0510}\mathrm{mol}}$

$=72.7\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

$\text{molar}\mathrm{mass}\text{E}=72.7\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.-32.0\raisebox{1ex}{${\displaystyle \mathrm{g}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathrm{mol}}$}\right.$

role="math" localid="1650350835087" $=40.7\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

$\text{Molar}\mathrm{mass}\text{E}=\text{Molar}\mathrm{mass}{\text{EO}}_2-\text{Molar}\mathrm{mass}{\text{O}}_2$

Therefore the element is Calcium ($\mathrm{Ca}$) and the compound is Calcium peroxide (${\mathrm{CaO}}_2$).