Answer: C2H6
Explanation:
g g/mol AVG 0.375 g/mol
C 32 12 2.667 1 12 2 24
H 8 1 8 3 3 6 6
15 30
Obtain the weighted average of C and H by multiplying the grams of substance times its repective molar mass. The AVG column is this result. The number for C is 2.667, but we know we need a whole number for both C and H, so divide both averages by (1/2.667) 0.375 to convert C to a whole number. H also converts to a whole number with this ratio, which is further evidence we now have the correct ratio of C and H: they are 1:3 C:H, so the empirical formula is CH3. The actual formula could be a multiple of this, such as C2H6, C3H9, C4H12, etc. But we are given the molar mass if this compound, 30 g/mole, ! mole of CH3 would only be 15 g/mole. One mole of C2H6, however, is exactly 30 g/mole. That matches the value we were given, so the formula is C2H6.
