A researcher has attempted to synthesize the molecule (C8H9NO2)shown below, which has the potential to be an active pharmaceutical agent to treat chronic pain.

Question

A researcher has attempted to synthesize the molecule (C8H9NO2)shown below, which has the potential to be an active pharmaceutical agent to treat chronic pain. While analyzing their product, they performed an elemental analysis in which 0.2524 g of their sample was combusted in excess oxygen to produce 0.6107 g of CO2, 0.1458 g of H2O, and 0.1064 g of NO2. Assume thatthe sample contains only C, H, N, and O, and that all carbon in the sample was converted to CO2, all hydrogen was converted to water, and all nitrogen was converted to NO2. Determine the empirical formula of the sample and identify whether the researcher has successfully synthesized their desired product. When you reach the stage of determining the empirical formula, you may round numbers that are very close to integer values (e.g. 6.95 can be rounded to 7, but 3.3 cannot be rounded to 3).

Answer 1

mols C = 0.6107 g CO₂ x 1 mol CO₂/44 g x 1 mol C/mol CO₂ = 0.01388 mols C

mols H = 0.1458 g H₂O x 1 mol H₂O/18 g x 2 mol H/mol H₂O = 0.0162 mols H

mols N = 0.1064 g NO₂ x 1 mol NO₂/46.0 g x 1 mol N/mol NO₂ = 0.00231mols N

Now to find mols of oxygen (O), we need to first find grams of O. We do this by subtracting grams of other elements from 0.2524 g (original sample) to get grams O, then convert to moles.

g C = 0.01388 mols C x 12 g/mol = 0.1666 g C

g H = 0.0162 mols H x 1 g/mol = 0.0162 g H

g N = 0.00231 mols N x 14 g/mol = 0.03234 g N

g O = 0.2524 g - 0.1666 g - 0.0162 g - 0.03234 g = 0.03726 g O

mols O = 0.03726 g O x 1 mol / 16 g = 0.00233 mols O

SUMMARY OF MOLES:

0.01388 mols C

0.0162 mols H

0.00231mols N

0.00233 mols O

Divide all by smallest value (0.00231) in an attempt to whole numbers...

6.00 mols C

7.0 mols H

1.0 mols N

1.0 mols O

Empirical formula = C₆H₇NO

The researcher HAS NOT synthesized the desired product