# 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

CeCl3

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

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Molar Mass (g)

AgCl - 143.32

Ce - 140.12

Cl - 35.453

**Answer:**CeCl3

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

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