Question

in a compound of magnesium and nitrogen 54g of magnesium combined with21g of nitrogen.determine the emphirical formulae of te compound

Answer 1

Moles of Magnesium $= \dfrac{54}{24} = 2.25$

Moles of Nitrogen $= \dfrac{21}{14} = 1.5$

Relative number of moles of atom,

For Magnesium $= \dfrac{2.25}{1.5} = 1.5$

For Nitrogen $= \dfrac{1.5}{1.5} = 1$

Converting numbers into whole numbers.

Hence, Empirical Formula will be $= Mg_{3}N_2$