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Question: Determine empirical formula of a compound having Na - 58.97 % and O - 41.03 %. a.) \(N{a_2}O\) b...

Determine empirical formula of a compound having Na - 58.97 % and O - 41.03 %.
a.) Na2ON{a_2}O
b.) Na2O2N{a_2}{O_2}
c.) NaO
d.) NaO2Na{O_2}

Explanation

Solution

The empirical formula gives the minimum number of atoms of each element that must combine to give the specified compound. By dividing the given percentage with molar mass, we will get the approximate number of atoms of each element in the molecule which will give us the empirical formula of the molecule.

Complete step by step solution:
The empirical formula gives the minimum number of atoms of each element that must combine to give the specified compound.
We have been given in question that
Na - 58.97 %
and O - 41.03 %
We know that molar mass of Sodium (Na) = 23
And molar mass of Oxygen (O) = 16
So, by dividing the given percentage with molar mass, we will get the approximate number of atoms of each element in the molecule.
So, for Na = 58.9723\dfrac{{58.97}}{{23}}
Na = 2.56
For ‘O’ = 41.0316\dfrac{{41.03}}{{16}}
O = 2.56
The number of atoms of ‘Na’ and ‘O’ is the same.
So, the empirical formula = NaO

Thus, the option c.) is the correct answer.

Note: It must be noted that empirical formulas in simple terms can be defined as the simple positive integer ratio of atoms present in a compound. The term that we commonly use for molecular formula is a different one. It shows the number of atoms of each type present in a molecule. The empirical formula of two compounds can be the same even if the molecular formula is different.