Question

Calculate the percentage composition of (a) ${\text{KN}}{{\text{O}}_3}$ (b) ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$.
$[{\text{At}}{\text{. wts}}{\text{. are K}} = 39,{\text{N}} = 14,{\text{O}} = 16,{\text{Ca}} = 40,{\text{P}} = 31]$

Answer 1

We need to calculate the percentage composition of each of the elements present in a given compound. The molar mass of each element can be calculated and using a unitary method we will calculate for 100 g.

Complete step by step answer:
Percentage composition is defined as the amount of solute present in the 100 g of substance. Let us first calculate the molar mass of ${\text{KN}}{{\text{O}}_3}$. The molar mass will be calculated by adding the mass of individual elements present multiplied with their respective number of atoms.
${\text{Molar mass of KN}}{{\text{O}}_3} = 39 + 14 + 3 \times 16 = 101$
In one mole of ${\text{KN}}{{\text{O}}_3}$ one mole atoms of potassium are present. The molar mass of K that is potassium is 39. Molar mass is the mass of 1 mole of an atom. Hence we can say that,
101 g of ${\text{KN}}{{\text{O}}_3}$ contains 39 g of ${\text{K}}$. For 1 g of ${\text{KN}}{{\text{O}}_3}$, $\dfrac{{{\text{39}}}}{{101}}$ grams of ${\text{K}}$ is present. In 100 g of ${\text{KN}}{{\text{O}}_3}$ the amount of potassium present will be: $\dfrac{{{\text{39}}}}{{101}} \times 100 = 38.61\%$
The percentage composition of ${\text{K}}$ is $38.61\%$.
In the same way we can do it for others.
101 g of ${\text{KN}}{{\text{O}}_3}$ contains 16 g of ${\text{O}}$, but there are 3 moles atom of oxygen so we will multiply the mass by 3. For 1 g of ${\text{KN}}{{\text{O}}_3}$, $\dfrac{{3 \times 16}}{{101}}$ grams of ${\text{O}}$ is present. In 100 g of ${\text{KN}}{{\text{O}}_3}$ the amount of oxygen present will be: $\dfrac{{3 \times 16}}{{101}} \times 100 = 47.52\%$
The percentage composition of ${\text{O}}$ is $47.52\%$
101 g of ${\text{KN}}{{\text{O}}_3}$ contains 14 g of ${\text{N}}$. For 1 g of ${\text{KN}}{{\text{O}}_3}$, $\dfrac{{14}}{{101}}$ grams of ${\text{N}}$ is present. In 100 g of ${\text{KN}}{{\text{O}}_3}$ the amount of nitrogen present will be: $\dfrac{{14}}{{101}} \times 100 = 13.86\%$
The percentage composition of ${\text{N}}$ is $13.86\%$.
For the ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ we will proceed in the same way,
${\text{Molar mass of C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2} = 3 \times 40 + (39 + 4 \times 16) \times 2 = 326$
310 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ contains $3 \times 40 = 120$ g of ${\text{Ca}}$. For 1 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$, $\dfrac{{120}}{{310}}$ grams of ${\text{N}}$ is present. In 100 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ the amount of calcium present will be: $\dfrac{{120}}{{310}} \times 100 = 38.71\%$
The percentage composition of ${\text{Ca}}$ is $38.71\%$.
310 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ contains $2 \times 31 = 62$ g of ${\text{P}}$. For 1 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$, $\dfrac{{62}}{{310}}$ grams of ${\text{P}}$ is present. In 100 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ the amount of phosphorous present will be: $\dfrac{{62}}{{310}} \times 100 = 20\%$
The percentage composition of ${\text{Ca}}$ is $20\%$.
310 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ contains $8 \times 16 = 128$ g of ${\text{O}}$. For 1 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$, $\dfrac{{128}}{{310}}$ grams of ${\text{O}}$ is present. In 100 g of ${\text{C}}{{\text{a}}_3}{{\text{(P}}{{\text{O}}_4})_2}$ the amount of oxygen present will be: $\dfrac{{128}}{{310}} \times 100 = 41.29\%$
The percentage composition of ${\text{O}}$ is $41.29\%$

Note:
The percentage composition can be defined in two ways that is percentage by mass and percentage by volume. If nothing is specified in question we will consider it percentage by mass. The total percentage of all the elements will always be 100.