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I take the MCAT on Tuesday, and I have a last minute question.
If both compounds undergo complete combustion, 1 mole of which will require the least amount of oxygen?
Propanol (H3C-CH2-CH2-OH) or Glucose (C6H12O6)?
*There was a similar question to this on AAMC 9, which I got correct, but under a different logic. According to AAMC, the compound with the lowest carbon-to-oxygen ratio will undergo combustion more easily, which in this case would be glucose (1:1 ratio as opposed to propanol's 3:1 ratio). However, glucose is much bigger, so my instinct is telling me that it would require more oxygen. Also:
C3H8O + 4.5O2 --> 4H20 + 3CO2
C6H12O6 + 6O2 --> 6H20 + 6CO2
I'm sure there are probably other factors involved to go along with AAMC's explanation, but could someone please clarify? Thanks.
If both compounds undergo complete combustion, 1 mole of which will require the least amount of oxygen?
Propanol (H3C-CH2-CH2-OH) or Glucose (C6H12O6)?
*There was a similar question to this on AAMC 9, which I got correct, but under a different logic. According to AAMC, the compound with the lowest carbon-to-oxygen ratio will undergo combustion more easily, which in this case would be glucose (1:1 ratio as opposed to propanol's 3:1 ratio). However, glucose is much bigger, so my instinct is telling me that it would require more oxygen. Also:
C3H8O + 4.5O2 --> 4H20 + 3CO2
C6H12O6 + 6O2 --> 6H20 + 6CO2
I'm sure there are probably other factors involved to go along with AAMC's explanation, but could someone please clarify? Thanks.