What is the percent of mass of GaBr3 in the solid mixture?

Question

A solid mixture weighs 0.6813 g. It contains gallium bromide (GaBr₃) and other inert impurities. When the solid mixture was dissolved in water and treated with excess silver nitrate (AgNO₃), 0.368 g of AgBr was precipitated. A balanced chemical equation describing the reaction is provided below. 

      GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

What is the percent of mass of GaBr₃ in the solid mixture? 

Answer 1

What this question boils down to is this: What is the mass of GaBr₃ in this solid mixture? (Forget what else is in there...who even cares. All we care about is GaBr₃). Now what's the deal with this excess AgNO₃? Well according to what they tell us, AgNO₃ reacts with GaBr₃, producing two products. And we know how many grams of one product is produced.

If we know how many grams of product were produced, then we can determine how many moles were produced (via the molar mass). If we know how many moles of a product we have, we can use the stoichiometric ratio to find out how many moles of reactant we have. So let's find out how many moles of the product were made.

We have 0.368g of AgBr. Referring to the periodic table, the molar mass of Ag is 107.9g/mol, and the molar mass of Br is 79.9g/mol. So to find the molar mass of AgBr (you may be way ahead of me here), we are going to add them together.

Molar Mass AgBr = 107.9 + 79.9 = 187.8g/mol

Now we can find number of moles AgBr...

0.368g AgBr x (1 mol/187.8 g) = 0.00196 mol AgBr

Now we ask ourselves, how can we relate moles of AgBr to moles of GaBr₃? The stoichiometric ratio. This equation gives us a 1 to 3 ratio of GaBr₃ to AgBr. So we set up a unit conversion...

0.00196 mol AgBr x (1 mol GaBr₃/3 mol AgBr) = 0.000653 mol GaBr₃

So we must have started with 0.000653 mol GaBr₃

Almost there peeps. So what is the question? Let's not lose sight of that... We want to know what percent of the original 0.6813g of solid is GaBr₃? So we must convert moles of GaBr₃ to grams of GaBr₃.

Again, the molar mass presents itself for the finding...I trust you.....I have faith in you.

Molar mass GaBr₃ = 69.7 + 3(79.9) = 309.4g/mol GaBr₃

Last unit conversion...

0.000653 mol GaBr₃ x (309.4g/1 mol GaBr₃) = 0.202g GaBr₃

So there were 0.202g GaBr₃ originally present in the solid mixture. Nothing else in there matters.

What percent of the total mass is this?

(0.202g/0.6813g) x 100 = 29.6%

So GaBr₃ accounts for 29.6% of the mass of the unknown solid.