Question

0.378g grams of a compound is made up of 0.273g of Mg and 0.105g nitrogen. Find its empirical formula.

Answer 1

The empirical formula of the compound is
`Mg_3N_2`.

Step-by-step explanation:

Mass of magnesium in compound = 0.273 g

Mass of nitrogen in the compound = 0.105 g

Moles of magnesium =
`(0.273 g)/(24.305 g/mol)=0.0112 mol`

Moles of nitrogen =
`(0.105 g)/(14.0067 g/mol)=0.00750 mol`

Form empirical formula divides the lowest value of moles of an element present with all the moles of elements present.

Magnesium
`=(0.0112 mol)/(0.00750 mol)=1.5`

Nitrogen
`=(0.00750mol)/(0.00750 mol)=1`

The empirical formula of a compound:

`Mg_(1.5)N_1=Mg_{(15)/(10)}N_1=Mg_(15)N_(10)=Mg_3N_2`

The empirical formula of the compound is
`Mg_3N_2`.