Question

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 63.57% carbon, 6.000% hydrogen, 9.267% nitrogen, 21.17% oxygen by mass. Which of the following compounds could the powder be?
a.) C11H15NO2 = 3,4-methylenedioxymethamphetamine (MDMA), illicit drug

b.) C3H6NO3 = hexamethylene triperoxide diamine (HMTD), commonly used explosive

c.) C21H23NO5 = heroin, illicit drug

d.) C8H9NO2 = acetaminophen, analgesic

e.) C7H5N3O6 = 2,4,6-trinitrotoluene (TNT), common used explosive

f.) C17H19NO3 = morphine, analgesic

g.)C10H15N = methamphetamine, stimulant

h.) C4H5N2O = caffeine, stimulant

Answer 1

Answer:

Option d: C₈H₉NO₂ = acetaminophen, analgesic

Explanation:

% composition of compound is:

63.57 g of C

6 g of H

9.267 g of N

21.17 g of O

First of all we divide each by the molar mass of the element

63.57 g / 12 gmol = 5.29 mol of C

6 g of H / 1 g/mol = 6 mol H

9.267 g of N / 14 g/mol =  0.662 mol of N

21.17 g of O / 16 g/mol = 1.32 mol of O

We divide each by the lowest value, in this case 0.662

5.29 / 0.662 = 8

6 / 0.662 = 9

0.662 / 0.662 = 1

1.32 / 0.662 = 2

Molecular formula of the compound is C₈H₉NO₂