A chemistry problem, empirical formula?

A 0.660g sample of a coumpound containing only Cerium and Chloride is dissolved in water and excess aqueous silver nitrate is added to precipitate 1.15g of silver Chloride. Calculate the empirical formula of the compound.

Molar Mass (g)
AgCl - 143.32
Ce - 140.12
Cl - 35.453


moles AgCl = 1.15g / 143.32 g/mol
moles AgCl = 0.008024 moles

gram Cl = 0.008024 mol * 35.453 g/mol
gram Cl = 0.284 grams

gram Ce = 0.660g - 0.284g
gram Ce = 0.376 grams

moles Ce = 0.376g / 140.12g/mol
moles Ce = 0.002683 moles

ratio of Ce and Cl,
Ce = 0.002683 mol / 0.002683 mol = 1
Cl = 0.008024 mol / 0.002683 mol = 3

hence, empirical formula is CeCl3
Since you have 1.15 g of AgCl, you can use the ratio of chlorine in AgCl to calculate exactly how much chlorine you started with. Subtract that from 0.660g and now you also have how many grams of Ce you started with.

This now becomes a regular empirical formula problem. Divide each of those 2 gram quantitites by the respective molecular mass of Cl and Ce. This gives you the molar ratio of each. Take the smaller of these 2 resulting numbers and divide both molar ratios by this number to get whole numbers.

Now you have the empirical formula

The answers post by the user, for information only, FunQA.com does not guarantee the right.

More Questions and Answers:
  • What is the oxidation number of sulfur in sodium thiosulphate?
  • I have the molar mass, sample mass and # of moles of a substance.how do i find the #of molecules/formula units
  • What is clonodine used for?
  • Help with a few easy chemistry problems?
  • How can I make sodium acetate? What do you do with the baking soda and the vinegar?
  • Does anyone know why hair doesn't decompose?
  • Can Carboanion be a leaving group?
  • What is 111111111x111111111=?
  • I have a chemistry question. Can someone please explain it to me?