Question 3.24: Taxol, known generically as paclitaxel, is a molecular compo...
Taxol, known generically as paclitaxel, is a molecular compound used in the treatment of cancer. It was originally isolated from the bark of the Pacific yew tree (Figure 3.39) and subsequently has been synthesized in the laboratory.
a. The yew tree provides about 100 mg of Taxol for every 1.00 kg of bark. If 1 tree has about 3 kg of bark and 9000 trees were needed to isolate enough Taxol for its first clinical trial, how many grams of Taxol were used for those initial tests?
b. One of the laboratory syntheses of Taxol begins with camphor (78.89\% C; 10.59\% H; \mathscr{M}= 152.23 g/mol) and requires 23 different chemical reactions in sequence to convert it into Taxol (66.11\% C; 6.02\% H; 1.64\% N; \mathscr{M} = 853.88 g/mol). The overall yield of the process to produce 1 mole of Taxol, starting with 1 mole camphor, is 0.10%. (i) What are the molecular formulas of camphor and Taxol? (ii) How much camphor must one use to make the amount of Taxol provided by 9000 trees?
Collect and Organize We are given information about the amount of Taxol in bark, the percent composition and molar mass of camphor and Taxol, and the yield of the laboratory synthesis.
Using dimensional analysis and other methods from this chapter, we can answer the questions posed.
Analyze The percent compositions given for both substances do not add to 100%. We may assume that both compounds contain oxygen.
Solve
a. We begin by calculating the amount of Taxol in 9000 trees:
9000 \sout{trees} \times \frac{3 \sout{kg bark}}{1 \sout{tree}} \times \frac{0.100 g Taxol}{1.00 \sout{kg bark}}= 3 \times 10^{3} g Taxol
b. (i) The percent carbon in camphor added to the percent hydrogen is less than 100%:
78.89\% + 10.59\% =5 89.48\%
Because the reported percentages do not add to 100%, the camphor most likely contains oxygen, the percent of which we cannot find in a combustion analysis. We can assume the percent oxygen in camphor is 100\% -89.48\% =10.52\%. We can now determine the empirical formula of camphor by considering that 100.00 g of camphor would contain 78.89 g of C, 10.59 g of H, and 10.52 g of O:
78.89 \sout{g C} \times \frac{1 mol C}{12.01 \sout{g C}} =6.569 mol C
10.59 \sout{g H} \times \frac{1 mol H}{1.008 \sout{g H}} =10.51 mol H
10.52 \sout{g O} \times \frac{1 mol O}{16.00 \sout{g O}} =0.6575 mol O
Dividing by the smallest number of moles:
\frac{6.569 mol C}{0.6575}=10 mol C
\frac{10.51 mol H}{0.6575}=16 mol H
\frac{0.6575 mol O}{0.6575}=1 mol O
The empirical formula of camphor is C_{10}H_{16}O. The corresponding mass is [(10 × 12.01 g/mol)+(16 × 1.008 g/mol)+(16.00 g/mol)] = 153.23 g/mol, which matches the molar mass of camphor. Therefore the molecular formula of camphor is the same as its empirical formula.
Carrying out the same procedure for Taxol:
66.11\% + 6.02\% + 1.64\% = 73.77\%
100\% -73.77\% = 26.23\% oxygen
Calculating the empirical formula:
66.11 \sout{g C} \times \frac{1 mol C}{12.01 \sout{g C}} =5.505 mol C
6.02 \sout{g H} \times \frac{1 mol H}{1.008 \sout{g H}} =5.97 mol H
1.64 \sout{g N} \times \frac{1 mol N}{14.01 \sout{g N}} =0.117 mol N
26.23 \sout{g O} \times \frac{1 mol O}{16.00 \sout{g O}} =1.639 mol O
Dividing by the smallest number of moles:
\frac{5.505 mol C}{0.117}=47 mol C
\frac{5.97 mol H}{0.117}=51 mol H
\frac{0.117 mol N}{0.117}=1 mol N
\frac{1.639 mol O}{0.117}=14 mol O
The empirical formula of Taxol is C_{47}H_{51}NO_{14}, which has a mass of [(47 × 12.01 g/mol)+(51 × 1.008 g/mol)+ (1 × 14.01 g/mol)+ (14 × 16.00 g/mol)]=853.88 g/mol. This value is the same as the molar mass of Taxol, so C_{47}H_{51}NO_{14} is also the molecular formula.
(ii) We need 3 × 10³ g of Taxol from a process that has a yield of 0.10%. The molar ratio in the process is 1 mole of camphor produces 1 mole of Taxol.
3 × 10^{3} g Taxol = 0.10\% of the theoretical yield
The theoretical yield for the process is
3 × 10^{3} g = 0.0010x x=3 × 10^{6} g
Starting with that amount of Taxol as the theoretical yield:
3 \times 10^{6} g Taxol \times \frac{1 \sout{mol Taxol}}{853.88 \sout{g Taxol}} \times \frac{1 \sout{mol camphor}}{1 mol Taxol} \times \frac{152.23 g camphor}{1 mol camphor} =5 \times 10^{5} g camphor
Thus, we need 500 kg of camphor to synthesize as much Taxol as there is in 9000 Pacific yew trees.
Think About It Camphor is quite inexpensive, but this example clearly illustrates how much material can be required to make a valuable product if the process has a low overall yield.