Question

A much sought-after high explosive is composed of C, 20.7%; N, 24.1%; O, 55.2%. Determine its empirical formula from the following options:

A. C₃N₂O₆
B. C₂N₃O₅
C. CNO₂
D. CNO₃

Answer 1

Based on the percentage composition of the high explosive and conversion to moles, the empirical formula is identified as CNO₂ (Option C), which reflects a molar ratio of 1:1:2 for carbon, nitrogen, and oxygen respectively.

Step-by-step explanation:

To determine the empirical formula of the high explosive composed of C, 20.7%; N, 24.1%; O, 55.2%, we first need to convert these percentages to moles using the atomic weights of each element (C: 12.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol). Then, we calculate the simplest molar ratio for each element to get the empirical formula.

For Carbon (C):
20.7 g C × (1 mol C / 12.01 g) = 1.724 moles C

For Nitrogen (N):
24.1 g N × (1 mol N / 14.01 g) = 1.721 moles N

For Oxygen (O):
55.2 g O × (1 mol O / 16.00 g) = 3.450 moles O

The molar ratio of C: N: O is approximately 1:1:2, which corresponds closely to the empirical formula CNO₂.

Therefore, the correct empirical formula from the given options for the high explosive is C. CNO₂.