Question 3.EP.8: The explosive known as RDX contains 16.22% carbon,2.72% hydr...
The explosive known as RDX contains 16.22% carbon, 2.72% hydrogen, 37.84% nitrogen, and 43.22% oxygen by mass. Determine the empirical formula of the compound.
Strategy
The empirical formula is based on the mole ratios among the elements in the compound, and the data given are in terms of mass. Molar mass provides the link between mass and moles, as usual. We can begin by choosing a convenient mass of the compound—usually 100 g—and use the percentages given to find the mass of each element in that sample. Then we will convert those masses into numbers of moles of each element. The ratios between those numbers of moles must be the same for any sample of the compound. Finally, we will need to convert those ratios into whole numbers to write the empirical formula.
Consider a 100-g sample of RDX. From the percentages given, that sample will contain 16.22g C, 2.72g H, 37.84g N, and 43.22g O. We will convert each of these masses into moles:
16.22g \ C\times \frac{1 \ mol \ C}{12.011 \ g \ C}=1.350 \ mol \ C \ in \ 100 \ g \ RDX
2.72g \ H\times \frac{1 \ mol \ H}{1.0079 \ g \ H}=2.70 \ mol \ H \ in \ 100 \ g \ RDX
37.84g \ N\times \frac{1 \ mol \ N}{14.0067 \ g \ N}=2.702 \ mol \ N \ in \ 100 \ g \ RDX
43.22g \ O\times \frac{1 \ mol \ O}{15.9994 \ g \ O}=2.701 \ mol \ O \ in \ 100 \ g \ RDX
These numbers tell us the ratios of C:H:N:O in the compound. We might be tempted to write a formula of C_{1.35}H_{2.70}N_{2.70}O_{2.70}. But we know that the correct empirical formula requires whole numbers of each element. In this case, it may be easy to see the correct ratios, but we need to approach such a problem systematically. The usual strategy is to divide all of the numbers of moles we obtained above by the smallest number of moles. This assures that the smallest resulting number will always be one. In this case, we will divide all four numbers of moles by 1.350 because it is the smallest of the four.
\frac{1.350 \ mol \ C}{1.350}=1
\frac{2.70 \ mol \ H}{1.350}=2
\frac{2.702 \ mol \ N}{1.350}=2.001\approx 2
\frac{2.701 \ mol \ O}{1.350}=2.001\approx 2
The result is a small whole number ratio: 1 mole C:2 moles of H:2 moles N:2 moles O. So the empirical formula is CH_{2}N_{2}O_{2}. The mass percentage data allow us to determine only the empirical formula, and not the molecular formula, because we do not have any information about how large or small a molecule of the compound might be.
Analyze Your Answer
We know that the coefficients in a chemical formula must be whole numbers. The fact that all four of the coefficients we found are so close to integers means that our proposed formula is plausible.
Discussion
In working with this type of problem, it is a good idea to use accurate molar masses for the elements and to carry as many significant figures as possible throughout the calculation. This ensures that no large rounding errors occur and helps make it easy to decide whether the final coefficients obtained are whole numbers. The steps we used above will produce whole numbers for all of the elements only if at least one of the subscripts in the empirical formula is a one. Otherwise, the result for one or more elements will still not be an integer. In nearly all such cases, though, such nonintegral results will be readily identified as small rational fractions (e.g., 1.5 or 2.33). We can obtain integral values by multiplying all of the coefficients by an appropriate integer.