Question

In a sample of calcium phosphate. $ C{a_3}{(P{O_4})_{2.}}0.432 $ mole of phosphorus is present. What amount of calcium phosphate is present in the sample if the sample is $ 100\,\% $ pure? $ [Ca = 40,\,P = 31,\,\,O = 16] $

Answer 1

Hint :Before solving the problem let us first get some idea about Molar mass. In chemistry, the molar mass of a chemical compound is the mass of a sample divided by the amount of material in that sample, measured in moles. In grams, it is the mass of $ 1 $ mole of the substance, or $ 6.022 \times {10^{23}} $ particles. The molar mass of a substance is a bulk attribute, not a molecular property.

Complete Step By Step Answer:
Calcium phosphate refers to a group of materials and minerals that contain calcium ions $ \left( {C{a^{2 + }}} \right) $ and inorganic phosphate anions. Oxide and hydroxide are also present in some so-called calcium phosphates. Calcium phosphates are white solids with nutritional value that can be found in a wide range of living creatures, including bone mineral and tooth enamel. It resides in colloidal form in micelles bound to casein protein with magnesium, zinc, and citrate–collectively known as colloidal calcium phosphate–in milk $ (CCP) $ .
Calcium phosphate has the following formula: $ C{a_3}{(P{O_4})_2} $
There are $ \;3 $ moles of calcium atoms, $ 2 $ moles of phosphorus atoms, and $ 8 $ moles of oxygen atoms in the $ C{a_3}{(P{O_4})_2} $ molecule.
To proceed, we must determine the moles of $ C{a_3}{(P{O_4})_2} $ .
As, $ 2 $ moles of phosphorus present in $ 1 $ mole of $ C{a_3}{(P{O_4})_2} $ .
So, $ 0.432 $ moles of phosphorus present in $ \dfrac{{0.432}}{2} = 0.216 $ mole of $ C{a_3}{(P{O_4})_2} $
Now we have to calculate the mass of $ C{a_3}{(P{O_4})_2} $
Molar mass of $ C{a_3}{(P{O_4})_2} $ $ = (3 \times 40) + (2 \times 31) + (8 \times 16) = 310\,g/mol $
Mass of $ C{a_3}{(P{O_4})_2} $ $ = $ Moles of $ C{a_3}{(P{O_4})_2} $ $ \times $ Molar mass of $ C{a_3}{(P{O_4})_2} $
Mass of $ C{a_3}{(P{O_4})_2} $ $ = $ $ (0.216\,moles) \times (310\,g/mole) = 66.96g $
 $ \therefore $ The amount of calcium phosphate is present in the sample is $ 66.96\,g $

Note :
In order to solve this problem there are some very important points which we should keep on our fingertips. The concept behind Molar mass and mole calculation is one of the most important parts of this problem which we should focus on.