I need help solving this problem

Question

A 4.46 g sample of lead oxide was reacting with carbon and 4.14 g of lead was obtained. Calculate the empirical formula of lead oxide.

Answer 1

Hello, Yesenia,

A tricky, but fun question. We are only given the starting and ending masses for the lead oxide reactant and lead product. The conservation of matter tells us that all of the 4.14 g of lead came from the lead oxide. So we can calculate the mass of oxygen in the lead oxide starting material:

4.46 g lead oxide - 4.41 g Pb = 0.32 grams of oxygen.

Hmmmm . . .

We need to convert the masses of oxygen and lead into the actual number of moles of each element. If we get the exact number of atoms of each, we should be able to find their ratio, and, therefore, the empirical formula.

Moles O = 0.32g/16 g/mole = 0.020 moles [we must be on the right track - look how this became a simple calculation of 0.32 over 16 = 0.020 mole!]

Moles Pb = 4.14/(207.2 g/mole Pb) = 0.020 moles Pb. [Another 0.02!. YES!]

Aha! The lead oxide brought 0.02 moles of Pb and 0.02 moles of O. That means there is one Pb for every one O:

0.02 mole Pb/0.02 mole O = 1:1 So the empirical formula is Pb₁O₁, or PbO. Empirical, since all we have is the ratio of the two elements. The formula could be Pb₂O₂, Pb₃O₃, etc., for all that we know. But since you asked a trustworthy Wyzant tutor, I'll share with you the closely guarded secret that, yes, in this case, the empirical formula is the actual formula.

Thanks for asking, and I hope this helps, It was fun.

Bob

Answer 2

Since you started with 4.46 g of lead oxide (made of lead and oxygen), and you ended up with 4.14 g of lead, then based on the law of conservation of matter, there must have been 0.32 g of oxygen present in the original compound (4.46 g - 4.14 g = 0.32 g).

Now we convert masses to moles as follows:

4.14 g Pb x 1 mol Pb/207 g = 0.02 moles Pb

0.32 g O x 1 mol O/16 g = 0.02 moles O

So, for the empirical formula, you have a 1:1 mole ratio of Pb and O

Empirical formula = PbO