**Solve the following problem:**

Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.

#### Solution

**Given: **Atomic mass of M = 56, Percentage of M = 70.0%

**To find:** The empirical formula of the compound

**Calculation:** %M = 70.0%

Hence, %O = 30.0%, Atomic mass of O = 16 u

Moles of M = `"% of M"/"Atomic mass of M"=70.0/56` = 1.25 mol

Moles of O = `"% of O"/"Atomic mass of O"=30.0/16` = 1.875 mol

Hence the ratio of number of moles of M : O is

`1.25/1.25` = 1 and `1.875/1.25` = 1.5

Convert the ratio into a whole number by multiplying by the suitable coefficient, i.e., 2.

Therefore, the ratio of the number of moles of M : O is 2 : 3.

Hence, the empirical formula is M_{2}O_{3}.

Empirical formula of the compound =** M _{2}O_{3}**