Question

if $0.27g~$ of an organic compound gave on combustion $0.396g~\left( C{{O}_{2}} \right)$ and $0.216g~\left( {{H}_{2}}O \right)$, $0.36g~$ of the same substance gave $48.88mL~({{N}_{2}})$ at $290K$ and $740mm$ pressure. Calculate the percentage composition of the compound

Answer 1

We know that the percent composition is the percentage of elements present in the compounds. So you have to calculate that percentage of each component. Here take one mole of compound as standard it means that all elements are present in molar mass quantities

Complete step by step solution:
A composition is defined as the mixture of several elements in a defined ratio or proportion. The moles of elements and the atomic masses of the elements are used for calculation.
We know that for a compound which is made up of any element to have some mass, so for finding the percent composition of a compound we have to divide the mass of that atom with the total mass of the compound. The fraction is then multiplied with to convert it in percentage. If we write it in mathematical form like to represent a formula it will be like this,
$$ and $$
We know that, ideal gas equation is given as $PV=nRT$
Substituting all the values in above equation,
$\Rightarrow \left[ \left( \dfrac{740}{760} \right)\times \left( 48.88\times {{10}^{-3}} \right) \right]=n\times 0.082\times 290$
We get $n=0.002$
To calculate the mass of ${{N}_{2}}=0.002\times 28=0.056$
Thus, percentage of nitrogen $=\dfrac{14}{28}\times \dfrac{0.056}{0.36}\times 100=7.78$

Note: Percent composition is the amount of each element by mass present in a compound. It helps in determining the stoichiometry of each element present in a chemical composition. The percent composition is the percentage of the individual atoms which makes a compound. The percentage is obtained by the ratio of the mass of an atom and the total mass of the compound multiplied by $100$