Question

Using the Mole: Empirical Formulas, Molecular Formulas, % Composition 2. Find the empirical formulas for the following compounds: a. e. 46.2% Mg, 7.69% H, 46.2% O

Answer 1

ANSWER

the empirical formula is MgH8O3

EXPLANATION

Given that;

The % composition of Mg is 46.2%

The % composition of H is 7.69%

The % composition composition of O is 46.2%

Assume the mass of the sample is 100g

To find the empirical formula, follow the steps below

Step 1; Find the mass of the elements

For Magnesium

$\begin{gathered} \text{ Mg }=\text{ }\frac{46.2}{100}\times100 \\ \text{ mass of Mg }=\text{ 46.2 g} \end{gathered}$

For hydrogen

$\begin{gathered} \text{ mass of H }=\text{ }\frac{7.69}{100}\times\text{ 100} \\ \text{ mass of H }=7.69\text{ grams} \end{gathered}$

For oxygen

$\begin{gathered} \text{ Mass of O}=\text{ }\frac{46.2}{100}\times100 \\ \text{ Mass of O }=\text{ 46.2 grams} \end{gathered}$

Step 2; Find the molar mass of the elements

The molar mass of oxygen is 15.999 g/mol

The molar mass of hydrogen is 1.00 g/mol

The molar mass of magnesium is 21.904.305 g/mol

Find the mole of the element

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

For Mg

$\begin{gathered} \text{ mole }=\text{ }\frac{46.2}{24.305} \\ \text{ mole }=1.900\text{ mole} \end{gathered}$

For Hydrogen

$\begin{gathered} \text{ mole }=\frac{7.69}{1} \\ \text{ mole }=\text{ 7.69 mol} \end{gathered}$

For Oxygen

$\begin{gathered} \text{ Mole}=\text{ }\frac{46.2}{15.999} \\ \text{ mole}=\text{ 2.889 mol} \end{gathered}$

Step 3; find the mole ratio

In the above calculations, Mg has the least number of moles. Therefore to find the mole ratio divide the moles by the smallest moles

$\begin{gathered} \text{ For mg} \\ \text{ mole ratio}=\frac{1.900}{1.999} \\ mole\text{ ration }=\text{ 1} \\ \\ \text{ For H} \\ Mole\text{ ratio }=\frac{7.69}{1} \\ \text{ molenration }=\text{ 7.69} \\ \\ fOR \\ \text{ MOLE RATION }=\frac{42.6}{15.999} \\ 2.64\text{ mol} \end{gathered}$

Therefore, the empirical formula is MgH8O3