Question

For analysis of a calcium sample, it is desirable to obtain a precipitate weighingabout 0.50g. What size sample should be taken from a mixture which isapproximately 56% CaO?

Answer 1

Ans. Atomic mass of Calcium = 40.078 g/mol

            Atomic mass of Oxygen = 15.999 g/mol

Molecular mass of CaO = 56.077 g/mol

From molecular formula, 1 mol CaO contains 1 mol Ca-atom. Or, out of total 59.077 gram CaO, the proportion of Ca-atoms alone is 40.078 gram.

Therefore,        

40.078 gram Ca is equivalent to 56.077 gram CaO

Or,        1 gram              -           -           (56.077 / 40.078) gram

Or,        0.5 gram           -           -           (56.077 / 40.078) x 0.5 gram

                                                            = 0.69959 gram

That is, if you take 0.69959 (= 0.7 gram, approx.) CaO, there is 0.5 gram calcium in it.

Now,

Given,

            % availability of CaO in the sample = 56%

That is-

56 gram CaO is present in 100 gram sample

Or,        1 gram -           -           -   (100 /56) gram

Or,        0.7 g     -           -           -    (100 /56) x 0.7 gram

                                                     = 1.25 gram

Thus, the required amount of sample to get 0.5 g precipitate of Ca = 1.25 gram