Hg[tex]Cl_{2}[/tex] is the empirical formula for a compound which contains 73.9%Hg and 26.1% Cl.
Explanation:
given that:
Hg = 73.9 % atomic mass of Hg= 200.59
Cl = 26.1 atomic mass of Cl= 35
let us assume that this percentage is in 100 grams.
number of moles = [tex]\frac{mass}{atomic mass of one mole}[/tex]
number of moles of Hg = [tex]\frac{73.9}{200.59}[/tex]
= 0.363 moles
number of moles of Cl = [tex]\frac{26.1}{35}[/tex]
= 0.735 moles
molar ratio of Cl and Hg = [tex]\frac{0.735}{0.363}[/tex]
= 2:1
In emperical formula first cation then anion is written.
We can see that Cl is twice the number of Hg. So the formula is
Hg[tex]Cl_{2}[/tex] is the empirical formula
