Find the molecular formula of a compound of boron with hydrogen if the mass of 1 litre of this gas equals the mass of 1 litre of nitrogen under same condition and the boron content in the substance is 78.2%. (B=11, N=14, O=16)

Find the molecular formula of a compound of boron with hydrogen if the

1.	Find the molecular formula of a compound of boron with hydrogen if the mass of 1 litre of this gas equals the mass of 1 litre of nitrogen under same condition and the boron content in the substance is 78.2%. (B=11, N=14, O=16)
Answer» Solution :100g of the compound contains 78.2g of B and 21.8g of H. Moles of `B= (78.2)/(11)=7.1` Moles of `H= (21.8)/(1)= 21.8` `therefore B:H= 1: 3` EMPIRICAL formula is `BH_(3)` `therefore` empirical formula WEIGHT `=11 +3=14` Now, since equal VOLUMES fo two gases contain the same number of molecules or moles if their temperature and pressure are the same, the number of moles of the gaseous compound is equal to the number of moles of nitrogen, both the gases occupying 1 litre of volume. Moles of the compound= moles of `N_(2)`. `("weight of the compound")/("mol. wt. of the compound")= ("weight of "N_(2))/("mol. wt. of"N_(2))` `because` weight of compound= weight of `N_(2)` (as given). Molecular weight of the compound= molecular weight of `N_(2)=28` We KNOW, `("molecular formula weight of the compound")/("empirical formula wt. of the compound")= (28)/(14)=2` `therefore` molecular formula is `(BH_(3))_(2)`, i.e., `B_(2)H_(6)`