I recently performed an experiment in which magnesium oxide formed.
I have to find out the simplest whole number ratio of Mg atoms to O atoms.
How do I go about doing this?
* Chemistry - DrBob222, Tuesday, October 13, 2009 at 11:19pm
Mass MgO - mass Mg initially = mass oxygen added.
%Mg = [mass Mg/mass MgO]*100 = ??
%O = [mass O/mass MgO]*100 = xx
Now ??/atomic mass Mg =
and xx/atomic mass O =
and go from there.
Ok, so:
Mass MgO (0.27g) - Mass Mg initially (0.17g) = 0.1g (Mass oxygen added)
Does that look alright so far?
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
Next
%Mg = [mass Mg/mass MgO]*100 = ??
.17/.27= 62.97%
%O = [mass O/mass MgO]*100 = xx
.10/.27=37.04%
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.
Thank you!
Yes, that looks correct so far. You have calculated the mass of oxygen added, which is the difference between the mass of magnesium oxide formed and the mass of magnesium initially.
Now, let's move on to calculating the percentages of magnesium and oxygen in magnesium oxide.
To calculate the %Mg:
%Mg = (mass Mg / mass MgO) * 100
In this case, the mass of magnesium is 0.17g, and the mass of magnesium oxide is 0.27g. Plugging these values into the formula:
%Mg = (0.17g / 0.27g) * 100
Calculating this equation will give you the percentage of magnesium in magnesium oxide.
Similarly, to calculate the %O:
%O = (mass O / mass MgO) * 100
The mass of oxygen added is 0.1g (as you calculated earlier), and the mass of magnesium oxide is 0.27g. Plugging these values into the formula:
%O = (0.1g / 0.27g) * 100
Calculating this equation will give you the percentage of oxygen in magnesium oxide.
Once you have the percentages of magnesium and oxygen, you can proceed to find the simplest whole number ratio of Mg atoms to O atoms by dividing the percentages by their respective atomic masses:
Mg atoms = %Mg / atomic mass of Mg
O atoms = %O / atomic mass of O
Using these ratios, you can determine the simplest whole number ratio of Mg atoms to O atoms in magnesium oxide.