Question 6.6: How many grams of nitrogen are present in a 0.10-g sample of......
How many grams of nitrogen are present in a 0.10-g sample of caffeine, the stimulant in coffee and tea? The formula of caffeine is C_{8}H_{10}N_{4}O_{2}.
Step 1: There is an important difference between this problem and the preceding one; here we are dealing with not one but two substances, caffeine and nitrogen. The given quantity is grams of caffeine (substance A), and we are asked to find the grams of nitrogen (substance B). This is a “grams of A” to “grams of B” problem.
\quad\quad\quad\quad 0.10 g C_{8}H_{10}N_{4}O_{2} = ? g N
Step 2: The appropriate set of conversions for a “grams of A” to “grams of B” problem, from Figure 6.7, is
\boxed{Grams of A}\xrightarrow[mass]{Molar}\boxed{Moles of A}\xrightarrow[subscripts]{Formula}\boxed{Moles of B}\xrightarrow[mass]{Molar}\boxed{Grams of B}The conversion factor setup is
0.10 \cancel{g c_{8}H_{10}N_{4}O_{2}}\times(\frac{1 \cancel{mole C_{8}H_{10}N_{4}O_{2}}}{194.26 \cancel{g C_{8}H_{10}N_{4}O_{2}}} )\times (\frac{4 \cancel{moles N}}{1 \cancel{mole C_{8}H_{10}N_{4}O_{2}}})\times (\frac{14.01 g N}{1 \cancel{mole N}} )
The number 194.26 that is used in the first conversion factor is the formula mass for caffeine. The conversion from “moles of A” to “moles of B” (the second conversion factor) is made by using the information contained in the formula C_{8}H_{10}N_{4}O_{2}. One mole of caffeine contains 4 moles of nitrogen. The number 14.01 in the final conversion factor is the molar mass of nitrogen.
Step 3: Collecting the numbers from the various conversion factors and doing the arithmetic give us our answer.
\quad\quad\quad\quad (\frac{0.10\times 1\times 4\times 14.01}{194.26\times 1\times 1}) g N=0.029 g N
