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The molecular formula of the given compound is [tex]$\mathrm{H}_{2} \mathrm{CO}_{3}$[/tex] also known as Carbonic acid.
What is empirical formula and molecular formula?
The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.
The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.
Given that,
H = 3.3%
C = 19.3%
O = 77.4%
No. of moles of H = 3.3/1
No. of moles of H = 3.3
No. of moles of C = 19.3 / 12
No. of moles of C = 1.60
No. of moles of O = 77.4/16
No. of moles of O = 4.83
Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83
Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60
The ratio of the atoms of C, H and O = 2 : 1 : 3
So, the empirical formula is [tex]$\mathrm{H}_{2} \mathrm{CO}_{2}$[/tex]
Let the molecular formula is [tex]$\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$[/tex]
Then, molar mass [tex]$=(2 \times 1+1 \times 12+3 \times 16) n\\[/tex]
Molar mass = 62n
As the question, 62 n = 60
n = 0.96 or n = 1 (rounded off to nearest ones)
So, the molecular formula is [tex]$\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$[/tex] i.e., the compound is Carbonic acid.
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