Question

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.

Answer 1

# 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?

Answer 2

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

Answer 3

Oh and yes I do think that this could be MnO2.

Answer 4

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.

Answer 5

Alright I have one more question. Mol X is 0.279 right?