Question

H, 0.97%; Na, 22.09%; O, 46.13%; S, 30.81% What is the empirical formula?

Answer 1

Let us assume that the total mass is 100 g

So, given percentage= the mass of each element

Here in this problem,

H= 0.97 g

Na= 22.09 g

O= 46.13 g

S= 30.81 g

Now we have to convert the mass of each element to moles using the molar mass given in the periodic table

1.01 g H = 1 mole of H

So, 0.97 g H = 0.97 g H x 1 mole of H/ 1.01 g of H = 0.96 mole H

Similarly, 22.09 g Na = 22.09 g Na x 1 mole of Na/ 22.98 g Na = 0.96 mole Na

46.13 g O = 46.13 g O x 1 mole of O/ 16 g O = 2.88 mole O

30.81 g S = 30.81 g S x 1 mole of S/ 32 g S = 0.96 mole S

Now we have to divide mole value by the smallest number of moles calculated

From above calculation it is evident that the smallest number of moles is 0.96

0.96 mole H/ 0.96 = 1

0.96 mole Na/ 0.96 = 1

2.88 mole O/ 0.96 = 3

0.96 mole of S/ 0.96 = 1

These obtained numbers are the mole ratio of the elements and is represented by subscripts in the emperical formula

So, the empirical formula is NaHSO₃

Note: If decimal numbers are obtained after the divison of mole numbers by smallest obtained mole number then round off to the nearest whole number