Question

How do you calculate percent composition of a compound?

Answer 1

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, for hydrogen it is $1\,g\,mo{l^{ - 1}}$ in one mole of it. Similarly, carbon is $12\,g\,mo{l^{ - 1}}$ and oxygen is $16\,g\,mo{l^{ - 1}}$.

Complete step-by-step answer:
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 $100$ to convert it in percentage.

If we write it in mathematical form like to represent a formula it will be like this,
$Percent\,composition\,of\,Carbon\, = \left( {\dfrac{{Mass\,of\,carbon\,in\,compound}}{{Total\,mass\,of\,compound}}} \right)\, \times 100$
Now let’s take an example for your better understanding in which we will take this formula in consideration. Suppose we have given ethyl alcohol, we also called it as ethanol. Ethanol has a molecular formula as ${C_2}{H_5}OH$ in which there are two carbons, six hydrogen and one oxygen. So we know that hydrogen has a molar mass approx. $1\,g\,mo{l^{ - 1}}$ carbon has molar mass of $12\,g\,mo{l^{ - 1}}$ and oxygen has molar mass of $16\,g\,mo{l^{ - 1}}$ . If we have to find out the percent composition of hydrogen it means we have to find out the percent of hydrogen which will present in ethyl alcohol.

Let’s start with hydrogen- $Percent\,composition\,of\,Hydrogen = \left( {\dfrac{{Mass\,of\,hydrogen\,in\,compound}}{{Total\,mass\,of\,compound}}} \right)\, \times 100$
As we have a total of six hydrogen then the mass of hydrogen will be $(6 \times 1\,g\,mo{l^{ - 1}})$ and the total mass of the compound will be the sum of mass of all atoms in it.

Total mass of compound= $(12\,g\,mo{l^{ - 1}} \times 2\,carbons) + (16\,g\,mo{l^{ - 1}} \times 1\,oxygen) + (1\,g\,mo{l^{ - 1}} \times \,6\,\,hydrogen\,)$
= $(24 + 16 + 6)\,g\,mo{l^{ - 1}}$ = $(46)\,g\,mo{l^{ - 1}}$
$Percent\,composition\,of\,Hydrogen = \left( {\dfrac{{6\, \times \,1\,g\,mo{l^{ - 1}}}}{{46\,g\,mo{l^{ - 1}}}}} \right)\, \times 100 = \,0.13043\, \times 100 = \,13.04\%$

Similarly, for calculating the percent composition of carbon we will take two carbon masses divided with the total mass of the compound.
$Percent\,composition\,of\,Carbon\, = \left( {\dfrac{{12 \times 2\,g\,mo{l^{ - 1}}}}{{46\,g\,mo{l^{ - 1}}}}} \right)\, \times 100 = \,52.17\%$

Here as we see that we take the mass of two carbons present in the ethyl alcohol divided by the total mass of compound multiplied with $100$ to convert it in percentage. So, $24$ is the mass of two carbon atoms which is divided with $46\,g\,mo{l^{ - 1}}$ , we will get $52.17\%$ of carbon. Now, what $52.17\%$ means is that this amount of carbon is present in the total one mole of compound.

Now let’s calculate for one atom of oxygen which is our alcoholic oxygen (oxygen from alcohol group) here for it we will take $Percent\,composition\,of\,Oxygen = \left( {\dfrac{{Mass\,of\,oxygen\,in\,compound}}{{Total\,mass\,of\,compound}}} \right)\, \times 100$
$Percent\,composition\,of\,Oxygen = \left( {\dfrac{{16\,g\,mo{l^{ - 1}}}}{{46\,g\,mo{l^{ - 1}}}}} \right)\, \times 100 = \,34.78\%$

So, it is calculated that in one mole of ethyl alcohol we have $34.78\% \,oxygen$ , $\,52.17\% \,\,carbon\,$ and $13.04\% \,\,hydrogen$ .

Note: For finding the percent composition of elements in a compound you have to multiply at last by number $100$ . We get a fraction when we just divide the mass of the element with the total mass of the compound. So in many questions it was asked about the mass fraction thus you have to keep in mind that the difference between both the terms is a multiplication of $100$.