What ratio is used to carry out each conversion?
a.mol CH4 to grams CH4
b.L CH4(g) to mol CH4(g)(at STP)
c.molecules CH4 to mol CH4
I honestly don't kno what it's asking me
I got 1 mol CH4/ 16(g) CH4 for a.
1 mol/22.4 L for b.
and 6.02 x 10^23 molecules/1 mol
but I don't think this is right, is it?
Chemistry - DrBob222, Saturday, July 17, 2010 at 12:13am
1 mol CH4 x (16 g CH4/1 mole CH4) = ?g CH4
L CH4 x (1 mole CH4/22.4 L) = ? moles CH4
#molecules CH4 x (1 mole/6.022 x 10^23 molecules) = ? moles CH4
The fraction (ratio) in parentheses is the ratio the problem is asking for. You did b right. Note that in my solutions for
a). mole cancels to leave g as the unit.
b). L cancels to leave moles as the unit.
c). # molecules cancel to leave moles as the unit.