Question

Calculate the molecular formula of the compound which contains 12% Mg and 88% Br by weight, if the molecular weight of compound is 184 g.

• A. $M{g}_{\frac{3}{2}}Br$
• B. $MgB{r}_2$
• C. $MgBr$
• D. $M{g}_{2}Br$

Answer 1

The correct option is B $MgB{r}_2$

Let us assume that the total weight of the compound = 100 g

Mass of Br present in the compound = 88 g
$\therefore$ Moles of Br present in the compound,
= $\frac{88}{80}=1.1mol$

Mass of Mg present in the compound = 12 g
$\therefore$ Moles of Mg present in the compound,
= $\frac{12}{24}=\frac{1}{2}mol$

The simplest whole number ratio of Mg:Br = 1 : 2
Hence, the empirical formula becomes $MgB{r}_2$
The empirical formular mass $=24+2\times 80=184$
The given molecular weight of the compound = 184 g

Empirical formula mass = molecular formula mass.
Hence, the molecular formula is the same as the empirical formula.

So, the molecular formula of the compound is $MgB{r}_2$.