Can someone please help me set up this problem:
If you burned 6.10 *10^24 molecules of ethane (C2H6) what mass of ethane did you burn?
2006-12-04 12:05:47 · 2 answers · asked by Mia16 3 in Science & Mathematics ➔ Chemistry
1 mol = 6.02*10^23 molecules
x mol = 6.10*10^24 molecules
x = 10.1 mol
Up to this point, it does not matter if you have ethane or water or another compound. Now we calculate for ethane based on its molecular weight.
1 mol C2H6 = 30 g C2H6
10.1 mol C2H6 = z
z = 30 * 10.1 = 303 g C2H6
2006-12-05 07:12:14 · answer #1 · answered by Dr. J. 6 · 0⤊ 0⤋
1. calculate formula mass for C2H6
(2 x 12.01g/mole) + (6 x 1.01 g/mol) = 30.08 g/mol
2. dim analysis setup
? g C2H6 = [6.10x10^24 molecules] x
[(1 mole C2H6)/(6.02x10^23 molecules C2H6)] x
[(30.08 g C2H6)/(1mole C2H6)] = your answer
2006-12-04 20:14:11 · answer #2 · answered by rm 3 · 0⤊ 0⤋