Question

What is the empirical formula of a compound composed of O and Mn in equal weight ratio?
(a) $ MnO $
(b) $ Mn{O_2} $
(c) $ M{n_2}{O_3} $
(d) $ M{n_2}{O_7} $

Answer 1

Hint :The molar mass of oxygen (O) is $ 16g/mol $ and the molar mass of Manganese (Mn) is $ 54.93g/mol $ , approximately taken as $ 55g/mol $ . We know that the number of moles (n) of any substance present in the sample is the ratio of the given weight (m) of the substance by its molar mass (M).
 $ Number\,of\,moles\,(n)\, = \,\dfrac{{Given\,weight\,(m)}}{{Molar\,Mass\,(M)}} $

Complete Step By Step Answer:
According to the question, the oxygen and Manganese are present in equal weight ratio in the compound. So,
Assume the mass of oxygen (O) and Manganese (Mn) in the compound will be equal to ‘X’ each.
The number of moles (n) of O will be $ \dfrac{X}{{16}} $
The number of moles (n) of Mn will be $ \dfrac{X}{{55}} $
Now, Determine the molar ratio of oxygen and Manganese.
 $ \dfrac{{Number\,of\,moles\,of\,Manganese\,({n_{Mn}})}}{{Number\,of\,moles\,of\,Oxygen\,({n_o})}}\, = \dfrac{X}{{55}} \times \dfrac{{16}}{X} $
∴ Molar ratio = $ \dfrac{{16}}{{55}} $
Therefore, the empirical formula will be $ M{n_{16}}{O_{55}} $
Now, Convert the empirical formula obtained in simplest ratio
We know that both are present in equal percentages in the compound, consider it to be $ 50\; $ and $ 50\; $ each
For Mn= $ \dfrac{{50}}{{55}} = 0.909 $
For O = $ \dfrac{{50}}{{16}} = 3.125 $
So... calculating the simplest ratio by dividing it with the smallest value.
For Mn = $ \dfrac{{0.909}}{{0.909}} = 1 $
For O = $ \dfrac{{3.125}}{{0.909}} = 3.5 $
Simplest ratio = $ 1:3.5\; $
Multiply the above ratio by $ 2 $ to convert it into a whole number.
Hence, simplest ratio will be = $ 2:7\; $
Therefore, the empirical formula= $ M{n_2}{O_7} $
The correct answer is option (D).

Additional Information:
Don’t be confused between the empirical formula and molecular formula. Molecular formula indicates the actual number of each different atom present in the molecule whereas the empirical formula gives the simplest ratio of the number of different atoms present in the molecule. So, basically the empirical formula is the simplest chemical formula of molecular formula.

Note :
Always remember to convert the molar ratio into the simplest form. The Empirical formula helps to determine the molecular formula as well as molecular mass for practical purposes. Some common molecular mass like oxygen = $ 16g/mol $ and nitrogen = $ 14g/mol $ should be remembered.