Question

what is the molecular and empirical formula for the given datas..
C is 75.92 percentage H is 6.32 percentage and N is 17.72 percentage..
The vapour density of the compound is 39.5
C is 12,H is 1,N is 14​

Answer 1

Answer:

Molecular formula ≡ C2H4Br2

Explanation:

As always with these problems, it is useful to assume 100⋅g of unknown compound....and thus find an empirical formula...

Moles of carbon=12.8⋅g12.011⋅g⋅mol−1=1.066⋅mol.

Moles of hydrogen=2.1⋅g1.00794⋅g⋅mol−1=2.084⋅mol.

Moles of bromine=85.1⋅g79.90⋅g⋅mol−1=1.066⋅mol.

And we divide thru by the LOWEST molar quantity to give an empirical formula of.....

CH2Br

But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.

And so 187.9⋅g⋅mol−1=n×(12.011+2×1.00794+79.9)⋅g⋅mol−1

And thus n=2, and the molecular formula is C2H4Br2