Posted by Anonymous on Wednesday, October 14, 2009 at 3:31pm.
Yes, but why do you want to do the problem piece-meal? Write all of it down, show me how you have solved it and let me take a look at it. We get it done one time instead of over four days.
I just wanted to make sure I'm on the right track
%Mg = [mass Mg/mass MgO]*100 = ??
%O = [mass O/mass MgO]*100 = xx
Now ??/atomic mass Mg = 0.03
and xx/atomic mass O = 0.02
And I'm stuck again... Where do I go from here?
OK. You have 62.97% Mg and 37.04% O.
Take a 100 g sample and you will have
62.97 g Mg
37.04 g O.
62.97/24.3 = ??
37.04/16 = ??
Then find the ratio of the two.
I don't know where the 0.03 and 0.02 came from. (By the way, I showed you the long way around. There is a little faster way of doing it but we'll just stay on course.)
Would it be 3:2 then?
62.97/24.32 = 2.59 moles Mg.
37.04/16 = 2.31 moles O.
Now find the ratio. The easiest way to do this is to divide the smaller number by itself; that way you know it will be 1.00 and we'll see what the other number is.
2.31/2.31 = 1.00 moles O
2.59/2.31 = 1.12 moles Mg
Now round to the nearest whole number.
1.00 rounds to 1 obviously.
1.12 rounds to 1 and the formula is MgO.
Here is the faster way of doing it. You had 0.17 g Mg and 0.10 g O.
There is no need to go through the percent step. We can work directly with the grams.
0.17/24.32 = 0.00699 moles Mg.
0.10/16 = 0.00625 moles Oxygen.
Now divide by the smaller number.
0.00625/0.00625 = 1.00
0.00699/0.00625 = 1.11
which rounds to 1:1 and the formula is MgO. You may wonder why these didn't come out to be EXACTLY 1:1 instead of the 1:1.11. That 1.11 is how much oxygen you determined, in the experiment, it took to form the oxide. That is experimental error. Done very carefully, with high precision balances, and ultra pure Mg metal, etc., it would have come out 1:1
Got it. I knew it should come out to be MgO but for some reason my mind slipped when it came to the actual calculations that would make it so.
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