Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

In a 100 g sample of the compound, there are 63.57 g of carbon, 6 g of hydrogen, 9.267 g of nitrogen, and 21.17 g of oxygen. First, convert these masses into moles (n) using the formula n = m/M, where M is the molar mass from the periodic table. For carbon: 63.57 g C -> 63.57 g C / 12.01 g/mol = 5.29 moles C. For hydrogen: 6 g H -> 6 g H / 1.008 g/mol = 5.95 moles H. For nitrogen: 9.267 g N -> 9.267 g N / 14.01 g/mol = 0.6615 moles N. For oxygen: 21.17 g O -> 21.17 g O / 16.00 g/mol = 1.32 moles O. Thus, the mole ratio looks like this: C 5.29 H 5.95 N 0.6615 O 1.32. Now, divide each value by the smallest number (1.32): C 4 H 4.5 N 0.5 O 1. To eliminate fractions, multiply all values by 2, yielding C8H9N1O2. Now, all numbers are integers! Hence, the empirical formula is C8H9NO2. Although the empirical formula isn't always the same as the molecular formula, in this instance, it corresponds to acetaminophen.

The full question is as follows: "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?" The options are: a.) 3,4-methylenedioxymethamphetamine (MDMA), illicit drug; b.) hexamethylene triperoxide diamine (HMTD), commonly used explosive; c.) heroin, illicit drug; d.) acetaminophen, analgesic; e.) 2,4,6-trinitrotoluene (TNT), common explosive; f.) morphine, analgesic; g.) methamphetamine, stimulant; h.) caffeine, stimulant. The potential identity of the powder is acetaminophen, known as an analgesic, which has the chemical formula C8H9NO2. The percentages provided are treated as mass in a 100 g sample: 63.57 g C, 6.000 g H, 9.267 g N, and 21.17 g O. To derive the empirical formula, we'll perform some steps: first, convert the mass values into moles. Then, determine the mole ratios of each element using the smallest mole value to find the subscripts. The resulting ratio is C: H: N: O = 8: 9: 1: 2, leading to the empirical formula C8H9NO2. Thus, the powder could indeed be identified as acetaminophen.