Question

A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer 1

The percentage composition of any constituent of the compound in a given compound is nothing but the percentage of the total mass of the given compound, that is being contributed by the constituent in question.

Formula used: $\dfrac{{{m_{total}}}}{{{m_{component}}}} \times 100$

Complete step by step answer:
In this case, the mass of the compound is 0.24 gm. Which contains 0.096 g of boron and 0.144 g of oxygen. Therefore, the percentage composition of boron and oxygen is
$\dfrac{{{m_{total}}}}{{{m_{component}}}} \times 100$ . Where is the total mass of the compound, and is the mass of boron and oxygen individually.
Therefore, for boron the percentage of composition is
$\dfrac{{0.24}}{{0.096}} \times 100$
=40
And the percentage of composition of oxygen is,
$\dfrac{{0.24}}{{0.144}} \times 100$
=60
Therefore, the total percentage of composition of the compound is the sum of the percentage of composition of the boron and oxygen as follows,
40+60=100
So, the percentage composition of the compound by weight is 100 gm.

Additional information:
In simpler terms, we can say that the percentage composition of any constituent is a way to determine the total contribution of the constituent to the total weight of the compound in a percentage form.
To calculate the percentage composition, we must follow the following steps:
1. We must find the molar mass of all elements in the given compound in gram per mole.
2. Then, we must find the molecular mass of the entire compound.
3. After this, we must divide the desired constituent’s molar mass by the entire molecular mass of the compound.
4. Following this, we must multiply the above number with the number of molecules of the constituent present in the compound.
5. Finally, multiply the above number with 100 to get the percentage composition of the constituent

If we were to put these steps in the form of a formula, we would get;
$\%$ Composition of ‘x’ = $\dfrac{{n \times {M_x}}}{{{M_{comp}}}} \times 100$
Where, n = number of molecules of ‘x’ in the given compound, ${M_x}$ = Molar mass of ‘x’, ${M_{comp}}$ =Molar mass of the compound

Note: To get the percentage composition of nitrogen in ammonium sulfate, we will first find the molar mass of the compound.
Therefore, the molar mass of ammonium sulfate ${\left( {N{H_4}} \right)_2}S{O_4}$ is:
$= 2\left[ {14 + 1\left( 4 \right)} \right] + 32 + \left( {16X4} \right)$
$= 132gmo{l^{ - 1}}$
It is important to note that there are 2 nitrogen atoms each belonging to one ammonium ion, $N{H_4}^ +$ .
Now, the percentage composition of nitrogen in ammonium sulfate is given by dividing the molar mass of the two nitrogen atoms by the molar mass of the compound and then multiply by 100.
Therefore, $\%$ composition of nitrogen $= \dfrac{{2 \times 14}}{{132}} = 21.2\%$ .This means that every 100g of ammonium sulphate contain 21.2g of nitrogen.