a. methyl butyrate (component of apple taste and smell):

C 58.80%, H 9.87%, O 31.33%

b, vanillin (responsible for the taste and smell of vanilla):

C 63.15%, H 5.30%, O 31.55%

explain, please!

thank you!

### 2 Answers

Generally: divide the mass percent by the molar masses to get the formula.

a: C – 58.8/12 = 4.9, H – 9.87/1 = 9.87, O – 31.33/16 = 1.96

Then divide by the smallest amount to get a ratio

C – 4.9/1.96 = 2.5, H – 9.87/1.96 = 5.0, O – 1.0

Then multiply to get whole numbers (here with 2) and you get C5H10O2

If you do the same for vanillin (try it out!), you’ll get C8H8O3.

The ratio of the subscripts in the empirical formula are equal to the mole ratio of each element! I let the total mass of the sample = 100 grams, so the % = mass of each element. Moles = mass ÷ molar mass 1) methyl butyrate (component of apple taste and smell): C58.80 H9.87 O31.33. Moles of C = 58.80 ÷ 12= 4.9 Moles of H = 9.87 ÷ 1 = 9.87 Moles of O = 31.33 ÷ 16= 1.96 Divide each by the smallest number of moles. C = 4.9 ÷ 1.96 = 2.5 H = 9.87 ÷ 1.96 = 5.04 O = 1 C2.5 H5O1 double each subscript to have whole numbers C5 H10O2 Use the same procedure below 2)glucose (a source of energy and metabolic intermediate): C40.00 H6.71 O53.29. Answer = C1H2O1