Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

\[1.6g\] of an organic compound given \[2.6g\] of magnesium pyrophosphate. The percentage of phosphorus in the compound is:
  A.45.38%
  B..38%
  C.37.76%
  D.19.02%

Answer
VerifiedVerified
511.8k+ views
Hint: To answer these types of questions, first we need to have basic understanding of the quantitative analysis of the phosphorus compound. When a certain mass of any phosphorus presented organic chemical compound is heated with the fumes of nitric acid, the phosphorus compound gets oxidized to phosphoric acid. Then, the mass of the phosphoric acid obtained gives the percentage of used phosphorus compounds.

Complete step by step answer:
To calculate the percentage of phosphorus in an organic compound let’s first write down the value of all the given quantities.
As we know that, mass of phosphorus = 31 u
Now, we need to derive the chemical formula of magnesium pyrophosphate i.e. $M{g_2}{P_2}{O_7}$
Then, refer to the formula to understand that 1 mole of $M{g_2}{P_2}{O_7}$ contains 2 g atoms of phosphorus.
So, mass of phosphorus in $M{g_2}{P_2}{O_7}$ becomes $2 \times 31 = 62u$
Also, we need to find out the total mass of $M{g_2}{P_2}{O_7}$= \[222.5u\]
And the weight of $M{g_2}{P_2}{O_7}$ is given in the question as \[2.6g\].
Also, the weight of the organic compound is mentioned to be \[1.6g\].
Now, we will calculate the total percentage of phosphorus
= $\dfrac{{Mass Of Phosphorus}}{{MassofM{g_2}{P_2}{O_7}\;}} = \dfrac{{WeightofM{g_2}{P_2}{O_7}}}{{Weight Of Organic Compound}}$ $ \times 100$
Now, we will place all the values to do the further calculation.
$ \Rightarrow \dfrac{{62}}{{222.5}} \times \dfrac{{2.6}}{{1.6}} \times 100 = 45.38\% $
Thus, the percentage of phosphorus in the organic compound is \[45.38\% \]
Hence, option A is the correct answer.

Note:
We must know that the chemical element phosphorus is present in two different forms i.e., white phosphorus and red phosphorus. Phosphorous is one of the highly reactive elements that are not available for free on Earth.