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Given :
A compound has a molar mass of 129 g/mol .
Empirical formula of compound is C₂H₅N .
To Find :
The molecular formula of the compound.
Solution :
Empirical mass of compound :

Now, n-factor is :

Multiplying each atom in the formula by 3 , we get :
Molecular Formula, C₆H₁₅N₃
1 mol = 6.022 x 10²³ atoms
In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.
So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃
But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃