Question

calculate the % composition of elements in the following compounds a) KMnO4 b) [Ca3(PO4)2]​

Answer 1

Percentage Composition of an element in a substance means the measure of amount of mass present in that substance.

There are two compounds given in the question, We will solve it one by one with explanation.

$\sf (a) \quad \bold{KMnO_{4}}$

Here , There are two elements in this given substance namely K (Potassium) , Mn (Manganese) and O (Oxygen).

We know the mass of each element so lets find the mass of the compound. which is the sum  of each element.

⇒    Mass of substance = Mass of K + Mass  of Mn + 4 * (Mass of O)

⇒    Mass of substance = 39 + 55 + 4 * 16

⇒    Mass of substance = 158 g/mol

Mass of each element:

• Potassium (K) = 39 g/mol
• Manganese (Mn) = 55 g/mol
• Oxygen (O) = 16 g/mol

Percentage composition can be given as:

         $\qquad \boxed{\sf \dfrac{Mass \;of\; element}{Mass \;of \;Substance} \times 100}$

For Potassium (K):

⇒    % Composition = (39 / 158)  * 100

⇒    % Composition = 0.246 * 100

⇒    % Composition = 25 %

For Manganese (Mn):

⇒    % Composition = (55 / 158) * 100

⇒    % Composition = 0.348 * 100

⇒    % Composition = 35 %

For Oxygen (O):

⇒    % Composition = (4*16 / 158) * 100

⇒    % Composition = (64 / 158) * 100

⇒    % Composition = 0.405 * 100

⇒    % Composition = 40 %

Now, Lets move on to the other compound.

$\sf (b)\qquad \bold{Ca_{3}(PO_{4})_{2}}$

This compound contains the following elements:

• 3 Calcium (Ca)
• 2 Phosphorus (P)
• 8 Oxygen (O)

Now the mass of each element is:

• Calcium (Ca) = 40 g/mol
• Phosphorus (P) = 124 g/mol
• Oxygen (O) = 16 g/mol

Similarly as in question (A) lets find the mass of the compound.

⇒    Mass of substance = 3 * (Ca) + 2 * (P) + 8 * (O)

⇒    Mass of substance = 3 * 40 + 2 * 31 + 8 * 16

⇒    Mass of substance = 120 + 62 + 128

⇒    Mass of substance = 310 g/mol

Now, Lets find the percentage composition as we did in question (a),

For Calcium (Ca):

⇒    % Composition = (3*40 / 310) * 100

⇒    % Composition = (120 / 310) * 100

⇒    % Composition = 0.387 * 100

⇒    % Composition = 39 %

For Phosphorus (P):

⇒    % Composition = (2*31 / 310) * 100

⇒    % Composition = (2 / 10) * 100

⇒    % Composition = 0.20 * 100

⇒    % Composition = 20 %

For Oxygen (O):

⇒    % Composition = (8*16 / 310) * 100

⇒    % Composition = (128 / 310) * 100

⇒    % Composition = 0.412 * 100

⇒    % Composition = 41 %