2.3.1.2.1: Strontium treats bone cancer pain- Solutions
- Page ID
- 188937
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The element strontium (Sr) is right underneath calcium (Ca) on the periodic table, which means that they tend to have similar chemical behavior and Sr can easily bind to bones in the same places Ca would. This can be good—Sr is used in Sensodyne toothpaste minimize tooth sensitivity and 89Sr (isotopic mass = 88.907 g/mol) is used to treat pain in bone cancer patients. The 90Sr isotopes released by the Chernobyl disaster, however, are extremely dangerous for this same reason.
1. Use the following information about the 3 naturally occurring isotopes of Sr to calculate its average atomic mass. Does your calculation match that on the periodic table?
|
Isotope |
Isotopic abundance (%) |
Isotopic mass (g/mol) |
|
86Sr |
? |
85.90926 |
|
87Sr |
7.00% |
86.90888 |
|
88Sr |
82.58% |
87.90561 |
I have the isotopic abundance for 2 of the 3 naturally occurring isotopes, so I can find the 3rd given that the percentages must add up to 100%.
Isotopic abundance of 86Sr is 100.00-7.00-82.58% = 10.44%. Wait, but you said that 89Sr and 90Sr exist, too? Why don’t I have to include them? The question asked you to focus on the 3 naturally occurring isotopes of Sr. 89Sr and 90Sr are highly radioactive elements that are man-made. They occur in such small quantities they don’t change the average atomic mass of Sr.
Now that I know the fractional isotopic abundances and isotopic masses for each naturally occurring isotope, I can find the average atomic mass using the following equation:
NEEDS equation
Average = 0.8256 x 87.90561 g/mol
0.0700 x 86.90888 g/mol
+ 0.1044 x 85.90926 g/mol
87.6 g/mol
This is the same value as the periodic table given that the 3 significant figures from 0.0700 x 86.90888 g/mol limit the outcome to 3 sig figs.
2. I would like to know how many moles of 89Sr are administered to a cancer patient who weighs 16 kg. A radiation dose of 1.5 MBq/kg of body mass using 89Sr will decrease pain in 80% of bone cancer patients without significant toxicity. Note that Bq are a unit of radiation and MBq is “mega” Bq. 89Sr is delivered as a drug called MetastronTM that delivers 37 MBq/mL. Each mL of Metastron contains 15 mg of the drug, which is 55% 89Sr by mass.
Soooo many conversions to do….
What I know:
Patient mass = 16 kg
Dose = 1.5 MBq/kg
Dose delivered by Metastron = 37 MBq/mL
Metastron contains 15 mg of drug/mL
Metastron is 55% 89Sr by mass
isotopic mass of 89Sr = 88.907 g/mol
Thought process
I am going to do a problem-solving bottom-up strategy, starting with recognizing that the answer I need is in terms of moles of 89Sr. I see a whole bunch of information above, but the two things involving 89Sr are the isotopic mass and the fact that Metastron is 55% 89Sr by mass. So, if I can figure out the mass of Metastron delivered, I will know the mass of 89Sr and I can convert that to moles. I see that there are 15 mg of Metastron/mL, so if I can figure out the volume delivered I can get the mass of Metastron. The dose of the medicine is 37 MBq/mL, so if I get the total radiation in MBq needed, I can get the volume. And lastly, I have the dose required (1.5 MBq/kg), so I can use this plus the mass to get the total radiation. Here is the process:
Step 1: Use the patient mass to get how much radiation we need.
16 kg x 1.5 MBq/1kg = 24 MBq
Step 2: Use the total radiation needed to find out what volume of the drug we need.
24 MBq (1 mL/37 MBq) = 0.65 mL
Step 3: Use the volume of the drug to find out the mass needed.
0.65 mL x 15 mg/mL = 975 mg
Step 4: Use the mass of Metastron to get the mass of 89Sr
0.975 g Metastron x 0.55 g 89Sr/g Metastron = 0.536 g 89Sr
Step 5: Convert the mass of 89Sr to moles of 89Sr
0.536 g 89Sr x 1 mol/88.907 g 89Sr = 6.0x10-3 mol 89Sr