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

Answer:

The empirical formula of the compound is $Mg_3N_2$.

Explanation:

Mass of magnesium in compound = 0.273 g

Mass of nitrogen in the compound = 0.105 g

Moles of magnesium =$\frac{0.273 g}{24.305 g/mol}=0.0112 mol$

Moles of nitrogen = $\frac{0.105 g}{14.0067 g/mol}=0.00750 mol$

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

Magnesium $=\frac{0.0112 mol}{0.00750 mol}=1.5$

Nitrogen $=\frac{0.00750mol}{0.00750 mol}=1$

The empirical formula of a compound:

$Mg_{1.5}N_1=Mg_{\frac{15}{10}}N_1=Mg_{15}N_{10}=Mg_3N_2$

The empirical formula of the compound is $Mg_3N_2$.