Question

The phosphorus present in an organic compound of mass 30 g is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as$MgN{H}_{4}P{O}_4$, which on ignition is converted into $M{g}_2{P}_2{O}_7$. If the percentage composition of phosphorus present in the sample of the organic compound is 14 percent, find the approximate mass of $M{g}_2{P}_2{O}_7$ formed.

• A. 14 g
• B. 16 g
• C. 15 g
• D. 17 g

Answer 1

The correct option is C 15 g

The mass of the organic compound taken be= $Wg$Mass of $M{g}_2{P}_2{O}_7$ obtained $=W1g$

From stoichimetry,$M{g}_2{P}_2{O}_7=2P1$

$mol2mol$.$1\times$ molecular mass of $M{g}_2{P}_2{O}_7=2\times$ atomic mass of $P222$ $g2\times 31g=62g$

Then, mass of phosphorous in $W1$ g of $M{g}_2{P}_2{O}_7=W1\times \frac{62}{222}g$

Then, percentage of phosphorous in the compound = $W1\times \frac{62}{222}\times \frac{100}{W}=14$

Here, $W=30$ g So, from above equation weight of $W1$ can easily be calculated.