Question

Find the molecular formula for a compound that is 40.01% C, 53.26 % O and 6.73% H. The molecular mass of the compound is 240 g/mol. Which ingredient is this compound?

Answer 1

Answer: C8H16O3 Octopyranose, a type of sugar

Explanation: When given the organic compounds that have C H O you can take the percentages and change them to grams and divide by molecular weight to get the moles.

40.01g/ 12.0g/mol = 3.33

53.26 g O/15.999g/mol = 3.33

6.73g H/ 1.008g/mol = 6.68

So divide all by 3.33 so C is 1, O is 1 and H is 2

CH2O which would only be 30 g. So if you multiple CH2O each element by 8 you get C8H16O8, and that has an atomic mass of 240g/mol. This is Octopyranose. (ose endings indicate a type of sugar.