Question

A compound of magnesium contains $21.9\%$ magnesium, $27.8\%$ phosphorous and $50.3\%$ oxygen. What will be the simplest formula of the compound?
A. ${\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$
B. ${\text{MgP}}{{\text{O}}_{\text{3}}}$
C. ${\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{2}}}$
D. ${\text{Mg}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{4}}}$

Answer 1

To solve this we must know the atomic mass of magnesium, phosphorus and oxygen. From the atomic mass and the given mass percent calculate the number of moles of each element.

Formula Used: ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$

Complete answer:
We are given a compound of magnesium that contains $21.9\%$ magnesium, $27.8\%$ phosphorus and $50.3\%$ oxygen.
Thus, $100{\text{ g}}$ of compound contains $21.9{\text{ g}}$ magnesium, $27.8{\text{ g}}$ phosphorous and $50.3{\text{ g}}$ oxygen.
Calculate the number of moles of magnesium in $21.9{\text{ g}}$ magnesium as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $21.9{\text{ g}}$ for the mass of magnesium, $24{\text{ g/mol}}$ for the molar mass of magnesium and solve for the number of moles of magnesium. Thus,
${\text{Number of moles of magnesium}} = \dfrac{{{\text{21}}{\text{.9 g}}}}{{{\text{24 g/mol}}}}$
${\text{Number of moles of magnesium}} = 0.91{\text{ mol}} \approx {\text{1 mol}}$
Thus, the number of moles of magnesium in $21.9{\text{ g}}$ magnesium are $1{\text{ mol}}$.
Calculate the number of moles of phosphorus in $27.8{\text{ g}}$ phosphorous as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $27.8{\text{ g}}$ for the mass of phosphorus, $31{\text{ g/mol}}$ for the molar mass of phosphorus and solve for the number of moles of phosphorus. Thus,
${\text{Number of moles of phosphorus}} = \dfrac{{{\text{27}}{\text{.8 g}}}}{{{\text{31 g/mol}}}}$
${\text{Number of moles of phosphorus}} = 0.89{\text{ mol}} \approx {\text{1 mol}}$
Thus, the number of moles of phosphorus in $27.8{\text{ g}}$ phosphorus are $1{\text{ mol}}$.
Calculate the number of moles of oxygen in $50.3{\text{ g}}$ oxygen as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $50.3{\text{ g}}$ for the mass of oxygen, $16{\text{ g/mol}}$ for the molar mass of oxygen and solve for the number of moles of oxygen. Thus,
${\text{Number of moles of oxygen}} = \dfrac{{{\text{50}}{\text{.3 g}}}}{{{\text{16 g/mol}}}}$
${\text{Number of moles of oxygen}} = 3.14{\text{ mol}} \approx {\text{3 mol}}$
Thus, the number of moles of oxygen in $50.3{\text{ g}}$ oxygen are $3{\text{ mol}}$.
The number of moles of magnesium in $21.9{\text{ g}}$ magnesium are $1{\text{ mol}}$, the number of moles of phosphorus in $27.8{\text{ g}}$ phosphorous are $1{\text{ mol}}$ and the number of moles of oxygen in $50.3{\text{ g}}$ oxygen are $3{\text{ mol}}$.
Thus, the formula for the compound is ${\text{MgP}}{{\text{O}}_{\text{3}}}$.

**Thus, the correct option is (B) ${\text{MgP}}{{\text{O}}_{\text{3}}}$.

Note:**
The simple formula is written by writing the moles of each element as a subscript. When the number of moles is one no subscript is written. Percentage composition gives the simplest form of the compound. To find the molecular formula we need the molecular weight of the compound.