A 24.3g sample of a compound XO2 contains 3.365 x 10^23 oxygen atoms. Determine the molar mass in g/mol of the element X. Show your work.

5 answers

# mols O atoms = (3.365E23/6.023E23) = ?
Then mols O2 = 1/2 that.
grams O2 = mols O2 x molar mass O2.
grams X = 24.3-g O2
molar mass X = gX/mols X = ?
Post your work if you get stuck.
Do you think this can be MnO2?
Thank you so much for this. This question confused me so much. Just one thing. Could you check my work for me?
3.365x10^23/6.022x10^23 atoms = 0.558784457 mol XO2
0.558784457 mol XO2/2 = 0.27939228mol O2
grams O2 = 0.27939228 mol O2*32.00 = 8.94055296gO2
grams X = 24.3g - 8.94055296gO2 = 15.35944704gX
15.35944704gX/0.279392228molX=54.9744862 g/mol
Oh and yes I do think that this could be MnO2.
I would argue with two things.
1. 0.559 is not mols XO2; rather it is mols O atoms. Then 0.559/2 = 0.279 or so = mols O2 molecules = mols XO2. What you have done is used moles O atoms to calculate mols XO2 (and your number for mols XO2 is 0.559 which isn't right), then used that number to solve for mols O2 molecules (which is right) but I think your steps are reversed. I think you must solve for mols O2 molecules first before you can get to mols XO2. Then X + O2 ==> XO2 and you can see that mols O2 = mols XO2 = mols X.
molar mass X = grams X/mols X

2. You have used far more significant figures than allowed. I never worried about that too much when I taught but many of the profs now really take that seriously. The 3.365 has 4 s.f. and the 24.3 has 3 so you are allowed only 3 s.f. in that part.
Alright I have one more question. Mol X is 0.279 right?