Question

A binary compound created by the reaction of an unknown element E and hydrogen contains $\mathbf{91}\mathbf{.}\mathbf{27}\mathbf{}\mathit{P}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}\mathbf{}\mathbf{E}$and$\mathbf{8}\mathbf{.}\mathbf{73}\mathbf{}\mathit{P}\mathit{e}\mathit{r}\mathit{c}\mathit{e}\mathit{n}\mathit{t}$by mass. If the formula of the compound is${\mathbf{E}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{6}}$, calculate the atomic mass of E.

Short answer

The atomic mass of E is$28.08\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$ if the formula of the compound is ${\mathrm{E}}_3{\mathrm{H}}_6$.

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 isequal to the total mass of the component atoms. It may be stated mathematically as follows:
$\mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{Nunber}\mathrm{of}\mathrm{moles}}$

Determining the molar mass of a substance

A substance's molar mass (g) is the mass of one mole of that substance.

A compound's number of moles $\left({\mathrm{n}}_{\mathrm{Compounds}}\right)$is computed as follows:

$\left({\mathrm{n}}_{\mathrm{Compound}}\right)=\frac{\mathrm{given}\mathrm{mass}\mathrm{compound}}{\mathrm{molar}\mathrm{mass}\mathrm{compound}}$

E is has a mass percentage of $91.27\mathrm{Percent}$, hydrogen has a mass percentage of $8.73\mathrm{Percent}$, and the compound's formula is ${\mathrm{E}}_3{\mathrm{H}}_8$. Hydrogen's molar mass $\left({\mathrm{M}}_{\mathrm{H}}\right)$is $1.008\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$. Assume that the total mass of the binary compound ${\mathrm{E}}_3{\mathrm{H}}_8\left({\mathrm{m}}_{{\mathrm{E}}_3{\mathrm{H}}_8}\right)$is 100 g, that the mass of the unknown element $\mathrm{E}\left({\mathrm{m}}_{\mathrm{E}}\right)$ is 91.27 g, and that the mass of hydrogen is 8.73 g. Because the compound's formula is ${\mathrm{E}}_3{\mathrm{H}}_8$.

There are 1 moles ${\mathrm{E}}_3{\mathrm{H}}_8$of and 3 moles of E and 8 moles of hydrogen. Mass of H is 8.73 g and a molar mass of $1.008\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$. As a result, the number of moles of hydrogen is calculated as follows:

$\begin{array}{rcl}{\mathrm{n}}_{\mathrm{Compound}}& =& \frac{8.73\mathrm{g}}{1.008{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 8.66\mathrm{mol}\end{array}$

Determining the Atomic mass of E

Let’s consider x be the number of moles of E,

$\begin{array}{rcl}3:8& =& \mathrm{x}:8.66\\ \mathrm{x}& =& 3.25\mathrm{moles}\mathrm{E}\end{array}$

The Atomic mass of E is mentioned as follows,

$\begin{array}{rcl}\mathrm{Atomic}\mathrm{mass}& =& \frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{compound}}{\mathrm{Number}\mathrm{of}\mathrm{poles}}\\ & =& \frac{91.27\mathrm{g}}{3.25\mathrm{mol}}\\ & =& 28.08\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

Therefore the atomic mass is$28.08\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$.