Download Emmett Swearson Term project This problem for the term project is a

Emmett Swearson
Term project
This problem for the term project is a mixing solutions problem. It involves mixing two
different solutions each containing different concentrations of pure drug within them in order to
create a new solution with a percent strength between the percent strengths of the two original
solutions that were mixed together.
We are given a 32.5% solution and a 6.5% solution. Our goal is to create a 1 liter solution
with a strength of 9.5% using a mixture of the 32.5% and 6.5% solutions.
1. First, we interpret the strength of the 32.5% solution. To interpret the strength we simply
put the percent strength in the numerator as grams. The denominator of the fraction in
interpreting percent strength is 100 mL.
32.5%
32.5 g
100mL
2. To interpret the strength of 6.5% we do the same thing that we did in problem #1. We put
the percent strength in the numerator as grams while we put 100 mL in the denominator.
6.5%
6.5 g
100mL
3. To interpret the strength of a 9.5% we do the same thing that we did in the previous two
questions. We take 9.5% and put it in the numerator as grams. We put 100 mL in the
denominator of the fraction.
9.5%
9.5 g
100mL
4. In order to find the amount of 32.5% and 6.5% solution to mix together to get 1L of
9.5% we need to use Shane’s formula for mixing solutions to find the volume of 32.5%
solution needed in the 1L 9.5% solution, which is our goal.
F= 1000 (0.095-0.065) = 115.4 mL of 32.5% solution
(0.325-0.065)
Once we have figured out the amount of stronger solution we can figure out the volume
of weaker strength solution need by subtracting the volume needed of the strongest
strength solution from the volume of the 1L goal solution.
Volume of 6.5% solution need= 1000mL- 115.4mL
=884.6mL of 6.5% solution.
Now that we have done the arithmetic to find out the volume of each solution needed for
the intended goal we can mix them. To create a 1 Liter 9.5% solution we need to mix
884.6 mL of 6.5% solution with 115.4 mL of 32.5% solution.
5. We used Shane’s formula to figure out how many mL of 32.5% solution we needed in
order to create our goal of 1000 mL of a 9.5% solution. Our result was that we need
115.4 mL of 32.5% solution.
6. In order to find how many grams of pure drug are in the 115.4 mL of 32.5% solution we
multiply the volume (115.4 mL) by the decimal form of 32.5%. To find the decimal form
of 32.5% we divide it by 100.
32.5/100= 0.325
Next, multiply 115.4 mL by 0.325 to find the amount of pure drug in the solution.
115.4 x 0.325= 37.5 g
There is 37.5 grams of pure drug in the volume of needed 32.5% solution.
7. In order to find the volume of 6.5% solution needed in the intended 1 L, 9.5% solution
we have to subtract the volume of 32.5% solution needed from the goal of 1000 mL to
get the difference which is the answer for the volume of 6.5% solution needed to achieve
our intended solution. So we subtract 115.4 mL from 1000 mL to find the volume of
6.5% solution needed.
1000-115.4= 884.6
We need 884.6 mL of 6.5% solution.
8. To find the amount of pure drug in the 884.6 mL of 6.5% solution, we multiply the
volume of 6.5% solution by the decimal form of 6.5%. To find the decimal form of the
percent we need to divide 6.5% by 100.
6.5%/100=0.065 in decimal form.
Next, multiply 884.6 mL by 0.065
884.6 x 0.065= 57.5 g
So, there is 57.5 grams of pure drug in the needed 884.6 mL of 6.5% solution needed to
create a 1000 mL, 9.5% solution.
9. To calculate the sum in grams of pure drug in both 32.5% solution and 6.5% we add
together our answer from questions #6 and #8. So, we add 37.5 to 57.5 to get the total
amount of pure drug in the final solution.
37.5+57.5= 95
So, there is 95 grams of pure drug in our intended 1000 mL of 9.5% solution.
10. To directly compute the amount of pure drug within a 9.5% 1 liter solution we simply
convert liters to milliliters and multiply it by the decimal form of 9.5%. First lets convert
liters to millileters by moving the decimal 3 places to the left converting 1 L to 1000 mL.
1L
1000 mL
Next we find the decimal form of 9.5%. To do this we divide 9.5% by 100.
9.5%/100=0.095
Finally, to directly compute the amount of pure drug within the 1 L 9.5% solution we
multiply 1000 mL by 0.095.
1000 x 0.095= 95 g
Results from our calculation show that there is 95 grams of pure drug in a 1 L 9.5%
solution.
Reflection
The problem for this semesters math project was a mixing solutions problem. We took
two solutions, each with different strengths and mixed them together to create a new
solution with a specific strength and volume. After we correctly calculated the volume of
each concentrated solution, we mix them together to get our final solution. If we correctly
solved the equation, questions nine and ten will be equal to each other. In question nine
we are asked to find the amount of pure drug (in grams) we added from each different
solution to make the final solution. In question ten, we are asked to find the amount of
pure drug (in grams) in the final solution. If we add the amount of pure drug from the
two solutions we used to make the final solution are added together and equal the amount
of pure drug we should have in our final solution then we have done our arithmetic
correctly and we have correctly solved the problem. I learned from this project a new way
of checking my answers by adding the amounts of drugs together then cross referencing
them to the amount of pure drug in my intended solution. Medical math is more than just
a passing grade. While I personally am not going into the medical field, this class has
developed my math skills tremendously. I feel like I can complete problems in my head
much faster than I could before and I feel much more confident in my math skills.
Group projects in math class can be helpful. Sometimes having someone else besides the
teacher explain how to correctly complete problems can make more sense to me.
Sometimes if I don’t understand how to do a certain problem or get stuck I will ask
another student and often times figure out the problem with just a little help. The only
downside to group projects is that sometimes on group member will get left behind and
be to afraid to ask the group questions so that group member can fully understand how to
complete the problem. If one group member gets left behind it completely ruins the
whole point of group work. I have really learned a lot from this medical math class and
feel like my math skills are much stronger and efficient.