Collin - 2-4-2006 at 13:03
So, I have balanced a chemical equation as required and got the one below.
This is baking soda when heated:
2NaHCO3 (s) --> Na2CO3(s) + H2O (g) + CO2
I have to find the mass of Na20 (the sodium compound).
What I have is the mass of baking soda before the reaction, the molar mass of both the Sodium compound and baking soda and number of moles of the
baking soda.
I used the formula: n = m/M (Number of moles = mass/Molar mass) to find the number of moles in baking soda.
I hope I did all of this correctly, though my question is, how do I find the mass of the Sodium compound? I only have the molar mass of it, and I do
need one more value to be able to use the formula....
Please help
Thanks..
12AX7 - 2-4-2006 at 13:55
Well, you know the Na isn't going anywhere, it's solid all the way through. So whatever you started with equals whatever you're left with. Add up
the atomic weights to find molecular weights then divide to get a dimensionless ratio of input to output (thus cutting out the extra step of finding
moles, unless moles are asked for, or your teacher just furrows his/her brow at the ease of ignoring moles in such problems 
). Now, you never have the compound Na2O, but you aren't asked for that, you're
asked for the Na2O *content* (I presume you made a typo with "Na[twenty]"!). Na2O only goes together one way, and for that matter you already have
sodium and oxygen in the compound to begin with, so it's pretty easy to see that all the sodium in the baking soda can just be tossed into the "Na2O"
formula and chugged against the weights of the other two compounds to find the percentage.
Tim
thanks
Collin - 2-4-2006 at 23:17
Okay, hey I appologize but I made a minor mistake, you are right, it wasnt Na20, nor Na2O! I have to find the mass of "Na2CO3" - (the sodium compound
that occurs from the heating of baking soda: 2NaHCO3)
2NaHCO3 (s) --> Na2CO3(s) + H2O (g) + CO2
Of "2NaHCO3" (baking soda) I have the mass and I found the molecular mass from theoretical values (84 g/mol). So I calculated the "number of Moles"
(N) with this formula:
n = m/M (Number of moles = mass/Molar mass)
Ok, now, I have the Molecular mass of Na2CO3 - But thats ALL! And to be able to use the above mentioned formula, I do need at least two values...
Is it possible to deduct the "number of Moles" (n) from baking soda (2NaHCO3) BEFORE the reaction (which I have all the "numbers" for)..
Hope this doesnt sound too confusing
regards
Odyssèus - 3-4-2006 at 05:16
If you dont have the molecular mass of something: Pull out your periodic table and look up the atomic masses of all the components, and add them
together multiplying each componet by the number in the subscript (i.e. 4 in my example).
So for example, if you want the molecular mass of MgSO4:
atomic mass of Mg + atomic mas of S + (atomic mass of O * 4)
?
Collin - 3-4-2006 at 06:21
But I do have the molecular mass of the Sodium compound I wrote. I just dont have the MASS (nor the number of Moles).
What I need to find, is the MASS!
Is it correct to just divide by 2 at the number of moles in "2NaHCO3" before the reaction, and then get the number of moles for the
sodium compound "Na2CO3(s)" which is created????
Can that be done? A friend of mine told me, but I dont completely trust him.. ;P
Odyssèus - 3-4-2006 at 08:00
Oh, doh.
You have twice the number of moles of NaHCO3, because it has a two in front of it and the Na2CO3 only has a 1 which is assumed. (2/1 or just 2, but if
you have some reaction for example with 2 moles of reactant and 3 moles of product it would be 2/3). So yes you can just divide the number of moles of
NaHCO3 by 2. To convert moles back to grams of a compound just multiply by the molecular mass of that compound.