Question

What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine

Answer 1

Answer: $MgCl_2$

Explanation:

25.3% Mg

74.7% Cl

First step: change % to g

25.3g Mg

74.7g Cl

Second step: calculate g/mol of each compound. You can do this by using the atomic mass.

$25.3gMg(\frac{1mol}{24.30g})=1.04mol$

$74.7gCl(\frac{1mol}{35.45g} )=2.11mol$

Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

$Mg=\frac{1.04}{1.04} =1$

$Cl=\frac{2.11}{1.04}=2$

Fourth step: Write each compound with their respective number below.

This empirical formula should be: $MgCl_2$