The Student Room Group

MCQ from OCR A-level chemistry

Q1) 0.541g of an element X is reacted with oxygen to form 0.790g of the oxide X2O3.
What is the element X ?
A - Al
B - Cr
C- Ga
D - Sc

Can some one please do the working out for this question step by step. Thankyou !
I dont know how to do it other than just by trial and error.
First write out the equation and balance:

4X + 3O2 -> 2X2O3

So starting with aluminium, its Ar = 27
so moles of aluminium would be = 0.541 / 27 = 0.02mol
If it was aluminium then X2O3 is Al2O3 and the Mr = 27 x 2 + 16 x 3 = 102
Moles of Al2O3 would be = 0.79 / 102 = 0.0077

The ratio of X : X2O3 should be 2:1 (simplified from the balanced equation from earlier)
That means that if 0.02mol aluminium reacted then 0.04mol of the oxide should form, which it doesn’t so the answer isn’t A. Then just repeat that using the mass numbers from the periodic table for the other elements given and you’ll eventually (hopefully, unless I’m getting it wrong lol) find one that gives the right ratio. I dont know any other way of doing it but it’s multiple choice so maybe they want trial and error?
Reply 2
Original post by Teribblestudent
I dont know how to do it other than just by trial and error.
First write out the equation and balance:

4X + 3O2 -> 2X2O3

So starting with aluminium, its Ar = 27
so moles of aluminium would be = 0.541 / 27 = 0.02mol
If it was aluminium then X2O3 is Al2O3 and the Mr = 27 x 2 + 16 x 3 = 102
Moles of Al2O3 would be = 0.79 / 102 = 0.0077

The ratio of X : X2O3 should be 2:1 (simplified from the balanced equation from earlier)
That means that if 0.02mol aluminium reacted then 0.04mol of the oxide should form, which it doesn’t so the answer isn’t A. Then just repeat that using the mass numbers from the periodic table for the other elements given and you’ll eventually (hopefully, unless I’m getting it wrong lol) find one that gives the right ratio. I dont know any other way of doing it but it’s multiple choice so maybe they want trial and error?


Thankyou ! i worked it out the answer is C
You can do it a faster method than trial and error,
First u can work out the mass of oxygen:
0.79-0.541=0.249g
Then u can find the moles of oxygen which is the mass divided by the molecular mass:
0.249/16=0.0155625
Then u can work out the moles if X in X2O3 by the ratio X : O is 2:3 so u divide 0.0155625 by 3 and multiply this by 2 to get the moles of X in the compound, which is 0.010375
And then work out the mr of X by dividing the mass over the moles:
0.541/0.010375= 52.1 so the answer is chromium
Reply 4
Original post by Whizzkid03
You can do it a faster method than trial and error,
First u can work out the mass of oxygen:
0.79-0.541=0.249g
Then u can find the moles of oxygen which is the mass divided by the molecular mass:
0.249/16=0.0155625
Then u can work out the moles if X in X2O3 by the ratio X : O is 2:3 so u divide 0.0155625 by 3 and multiply this by 2 to get the moles of X in the compound, which is 0.010375
And then work out the mr of X by dividing the mass over the moles:
0.541/0.010375= 52.1 so the answer is chromium

how comes you divide it by 16 instead of 32 since its O2 in the equation?
Original post by F.A.
how comes you divide it by 16 instead of 32 since its O2 in the equation?

Moles of oxygen atoms in oxide rather than oxygen molecules reacting.
Reply 6
Original post by Old man1234
Moles of oxygen atoms in oxide rather than oxygen molecules reacting.

Ohh makes sense thank you.
Original post by F.A.
Ohh makes sense thank you.

But you would get same answer if calculated moles of O2 reacting and used the 4:3 ratio.

Quick Reply

Latest