Question

An ionic compound is found to contain 10.9% magnesium, 31.8% chlorine and 57.3% oxygen. What is the empirical formula for this compound?

Answer 1

Mg 10.9g/24.31=0.018/0.018=1
Cl 31.8g/35.45=0.897/0.018=49.83=50
O 57.3g/16.00=3.581/0.018=198.94=199

empirical formula=MgCl50O199

I think it's right

Answer 2

Answer:

The empirical formula is MgClO₈.

Explanation:

To find the empirical formula, which is the minimum integer ratio between the atoms in a compound, we have to follow some steps.

Step 1: Divide each percentage by the atomic mass of the element

$Mg: \frac{10.9}{24.3} =0.449$

$Cl: \frac{31.8}{35.5} =0.896$

$O: \frac{57.3}{16.0} =3.58$

Step 2: We divide all the numbers by the smallest

In this case, the smallest number is 0.449

$Mg: \frac{0.449}{0.449} =1.00$

$Cl: \frac{0.896}{0.449} =2.00$

$O: \frac{3.58}{0.449} =7.97\approx 8$

Step 3: We incorporate these numbers as the atomicities in the empirical formula

The empirical formula is MgClO₈.