Question

If you have 34.85 grams of Lead (II) Nitrate, how many atoms of oxygen will there be?

Answer 1

Well, what is the molar quantity of this mass of lead nitrate...`Pb(NO_3)_2`?

Explanation:

I make it `(34.85*g)/(331.20 *g*mol^-1)=0.1052*mol`.

But given the formula of `Pb(NO_3)_2`, if there are `0.1052*mol` of salt...there are `6xx0.1052*mol` oxygen atoms....i.e. `0.6313*mol` `"oxygen atoms"`... What is the mass of these oxygen atoms....? What is the number of oxygen atoms?

How many moles of nitrogen atoms? And how many moles of lead atoms? And when we sum the masses of these molar quantities would we get `34.85*g`? Why?

Answer 2

`"34.85 g Pb"("NO"_3)_2"` contains `3.802xx10^23` `"O atoms"`.

Explanation:

We need to calculate the moles `"Pb(NO"_3)_2"`. Then calculate the number of moles of oxygen atoms. One mole of anything is `6.022xx10^23` anythings. So we will multiply the moles oxygen by `6.022xx10^23` `"atoms/mol"`.

Moles lead(II) nitrate

To calculate moles `"Pb"("NO"_3)_2"`, divide the given mass by its molar mass `("331.208 g/mol")`.

`(34.85color(red)cancel(color(black)("g")) "Pb"("NO"_3)_2)/(331.208 color(red)cancel(color(black)("g"))/"mol")="0.10522 mol Pb"("NO"_3)_2"`

Moles oxgen

One mole of `"Pb"("NO"_3)_2"` contains `6` moles of oxygen atoms. `(2xx3)`.

Calculate mol `"O"` atoms by multiplying the calculated mol `"Pb"("NO"_3)_2"` by `"6 mol O atoms"`.

`0.10522color(red)cancel(color(black)("mol Pb"("NO"_3)_2))xx(6" mol O atoms")/(1color(red)cancel(color(black)("mol Pb"("NO"_3)_2)))="0.63132 mol O atoms"`

Atoms oxygen

Calculate the number of `"O"` atoms by multiplying mol `"O"` atoms by `6.022xx10^23` `"atoms/mol"`.

`0.63132color(red)cancel(color(black)("mol O atoms"))xx(6.022xx10^23"O atoms")/(1color(red)cancel(color(black)("mol O atoms")))=3.802xx10^23` `"O atoms"` (rounded to four significant figures)