Question

What is the percentage composition of nitrogen for $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ ?

Answer 1

Hint :The molecular weight of the given compound is 211.62 gram. The atomic weight of nitrogen is 14.0067 gram, and there are two nitrogen atoms in the given compound. To know the percentage composition, check the amount of nitrogen in the given compound in percentage.

Complete Step By Step Answer:
The percentage composition of nitrogen in $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ is 13.237%.
Let us know how and why.
The molecular weight of the given compound $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ (Strontium nitrate) is 211.62 gram.
Atomic weight of any element is approximately the double of its own atomic number. As the atomic number of oxygen is 8, the atomic weight of oxygen is generally taken as 16. But, precise weight may change.
The atomic weight of nitrogen is 14.0067 gram. There are a total two nitrogen atoms in the given compound, thus the total atomic weight of nitrogen in the given compound is 28.0134 gram.
Now, as there is 28.0134 gram of nitrogen in 211.62 gram of Strontium nitrate ( $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ ), let us check how much nitrogen is there in 100 gram of Strontium nitrate ( $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ ) i.e. 100 percent of Strontium nitrate ( $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ ).
Therefore, $ 28.0134\times \dfrac{100}{211.62}=13.237% $ . Thus, there is 13.237% of nitrogen in 211.62 grams of Strontium nitrate ( $ Sr{{\left( N{{O}_{3}} \right)}_{2}} $ ).
Similarly, the percentage composition of strontium and oxygen can also be found out in this similar way.

Note :
In these types of questions, to know the approximate molecular weight of any compound, its atomic weight has to be known. Through the atomic weight of all the elements, we can find out the molecular weight of the compound.