We will solve this problem by using the three steps of dimensional analysis (Section 2.8) and Figure 6.7.
Step 1: The given quantity is 0.250 g of C_{6}H_{8}O_{6}, and the desired quantity is molecules of C_{6}H_{8}O_{6}.
\quad\quad\quad\quad0.250 g C_{6}H_{8}O_{6} = ? molecules C_{6}H_{8}O_{6}
In terms of Figure 6.7, this is a “grams of A” to “particles of A” problem. We are given grams of a substance, A, and desire to find molecules (particles) of that same substance.
Step 2: Figure 6.7 gives us the pathway we need to solve this problem. Starting with “grams of A,” we convert to “moles of A” and finally reach “particles of A.” The arrows between the boxes along our path give the type of conversion factor needed for each step.
\quad\quad\quad\quad \boxed{Grams  of  A} \xrightarrow[mass]{Molar} \boxed{Moles  of  A} \xrightarrow[number]{Avogadro’s} \boxed{Particles  of  A}
Using dimensional analysis, the setup for this sequence of conversion factors is
\quad\quad\quad\quad 0.250  \cancel{g  C_{6}H_{8}O_{6}}\times (\frac{1  moles  C_{6}H_{8}O_{6}}{176.14  \cancel{g  C_{6}H_{8}O_{6}}})\times (\frac{6.02\times 10^{23}  molecules  C_{6}H_{8}O_{6}}{1  \cancel{moles  C_{6}H_{8}O_{6}}})
\quad\quad\quad\quad\quad\quad\quadg C_{6}H_{8}O_{6} → moles C_{6}H_{8}O_{6} → molecules C_{6}H_{8}O_{6}
The number 176.14 that is used in the first conversion factor is the formula mass of C_{6}H_{8}O_{6}. It was not given in the problem but had to be calculated by using atomic masses and the method for calculating formula masses shown in Example 6.1.
Step 3: The solution to the problem, obtained by doing the arithmetic, is
\quad\quad\quad\quad\quad \frac{0.250\times 1\times 6.02\times 10^{23}}{176.14\times 1}  molecules  C_{6}H_{8}O_{6}\\\quad\quad\quad\quad\quad\quad\quad = 8.54\times 10^{20}  molecules  C_{6}H_{8}O_{6}