Math 1010
Molarity Project
Math Background:
The mathematics used in concentration calculations stem from the relationship between
amount (percentage), rate (percent) and base. Recall, EMBED Equation.DSMT4 ,
amount equals rate times base. So whether we are computing the tip on the cost of a
meal, the discount on an item on sale, or the amount of pure substance in a solution, you
would multiply the rate (percent) times Base. This basic formula can be “solved” for any
of the specific quantities,
or
.
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EMBED Equation.DSMT4
When doing calculations you must be aware of how units are behaving. The desired unit
must be the outcome of the computation.
Chemistry Background:
When referring to the concentration of a solution, we generally mean the amount of
solute contained in a certain amount of solution. By varying the amount of solute
dissolved in a solution the concentration changes. The strength of the concentration is
given in terms of a numerical value. The percentages will have units of measurement
using combinations of mass, volume or moles. For example, volume/volume percent,
weight (mass)/weight percent, weight/volume percent, or molarity.
Volume percent is used when two liquids are mixed.
EMBED Equation.DSMT4
Example:
If a solution has a (v/v%) concentration of 32.5%. How many mL of
solute are present in 350 mL solution?
Here you would use the formula amount of pure substance (volume of solute) equals
concentration times volume (solution). The concentration of 32.5% means 32.5 mL
solute contained in 100 mL solution.
EMBED Equation.DSMT4
Weight/volume percent is used when a mass in grams is dissolved in a liquid measured in
milliliters.
EMBED Equation.DSMT4
Example:
If a solution is 37.5% (m/v) glucose, how many mL of solution must
be administered to the patient if the doctor orders 9 grams of glucose?
The concentration tells us there is 37.5 grams glucose in 100 mL solution. This
is an equivalence relationship, a conversion factor that can be used to set up the
computation. The desired outcome is mL of solution and the doctor ordered 9
grams of glucose. The units of mass must cancel out, so the “conversion factor”
can be set up as
EMBED Equation.DSMT4
EMBED Equation.DSMT4
depending on what units are needed in the desired outcome.
Therefore,
EMBED Equation.DSMT4
You would administer 24 mL of the solution to the patient.
In the formula EMBED Equation.DSMT4 , the goal is to solve for V, so divide by the
concentration, hence
will give the volume to administer.
EMBED Equation.DSMT4
. To get the outcome, recall how fractions are divided.
EMBED Equation.DSMT4
We often express concentrations in terms of Molarity, which is the number of
moles of solute dissolved in a liter of solution.
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The units would be moles of solute per liter of solution.
Example:
A solution that contains 0.7 moles of solute per 5 L solution would have molarity:
EMBED Equation.DSMT4
Dilutions
In a lab, often you are working with stock solutions and need to dilute the concentration
in order to perform an experiment. The important concept to remember is that diluting a
substance does not change the amount of that substance. So the number of moles of the
substance before the dilution is the same as the number of moles after the dilution
(conservation of matter). Thinking about what that means, If you are making orange
juice and have one can of orange juice and add three cans of water, you have a solution
that is
now if you dilute it by adding another can of water
EMBED Equation.DSMT4
you now have
. The concentration decreased, but you still
EMBED Equation.DSMT4
have only one can of orange juice in the solution. The total amount of pure substance did
not change, a formula to reflect this concept is
. In words
EMBED Equation.DSMT4
initial concentration times initial volume equals final concentration times final volume.
Example: You are a intern in a lab and your supervisor has just asked you to prepare
500mL of 3M HCl for the experiment. The stockroom cabinet contains "Stock
Solutions". In it there is a 2.5L bottle labeled "11.6M HCl". The concentration of the
HCl is 11.6M. Please prepare a flask containing 500 ml of a 3 M solution, relabeling it
with its precise molarity. How do you prepare this solution?
The formula to use is
because the total number of moles will
EMBED Equation.DSMT4
be the same. How much stock do you use? Solve for EMBED Equation.DSMT4 .
Initial:
EMBED Equation.DSMT4
Final:
,
EMBED Equation.DSMT4
, EMBED Equation.DSMT4
EMBED Equation.DSMT4
Note 500 mL = 0.5 L
Therefore
EMBED Equation.DSMT4
becomes
EMBED Equation.DSMT4
(conservation of mass)
This simplifies as EMBED Equation.DSMT4 solving for final volume (divide both sides
by 11.6) notice moles cancel, this gives
. Therefore, take 130
EMBED Equation.DSMT4
mL of the 11.6 M stock solution and add enough water so that the total volume is 500
mL.
Lab Activity:
You are working as an intern in a biotechnology lab. Teddi, your supervisor, has just
asked you to perform the following tasks:
Please describe your complete procedure and the key quantities
you measure. Points are based on whether or not you explain your
procedure in sufficient detail for us to know what you did.
1. Prepare 500 mL of a 5% (w/v) solution of NaCl.
5mL solute/100 mL solution x 500 mL solution = 25 mL solute NaCl
In a 500 mL graduated cylinder pour 25 mL of the NaCl solute and add 475 mL
of water.
2. Prepare 30 g of a 25% (w/w) solution of NaCl.
100 mL solution/ 25 g NaCl x 30g NaCl = 120mL NaCl solution
To make a 25% solution of 30g NaCl you would need 120 mL of water to dilute.
3. Prepare 200 mL of a 0.8 M solution of CaCl2.
.001L= 1 mL 200 mL= 0.2L Molarity= moles of solute/volume of solution (L)
0.8M CaCl2= x molesCaCl2 /0.2L CaCl2
0.2L(0.8 moles CaCl2/L CaCl2)= (x moles CaCl2/0.2L CaCl2)0.2L
0.16 moles= x
In a graduated cylinder pour .16 moles of CaCl2 into a 0.2L solution.
4. Prepare 450 mL of a 35% stock solution using a 70% stock solution and a
20% stock solution.
c
v
a
!st
70
x
70x
2nd
20
450-x
20(450-x)
Mix
35
450
35(450)
70x+20(450-x)=35(450)
70x+9000-20x=15750
50x+9000=15750
50x=6750
x=135 mL
450-135=315 mL
Add 135 mL of the 70% stock solution into a beaker than pour 315 mL of the
20% stock solution into the same beaker to get 450 mL of a 35% solution.
5. Prepare 500 mL of 2.00 M buffer solution for an experiment. In the Stock
Solutions cabinet you will find a 2.50 liter bottle containing 5.6 M buffer
solution.
CiVi= CfVf
5.6 M= Ci
?= Vi
Cf=2.00 M
Vf=500 mL=0.5L
5.6 moles/L x ViL= 2.00moles/L x 0.5L
5.6molesVi/5.6 moles= 1mole/5.6 moles
Vi= 0.18L= 180mL
Add 180 mL of the 5.6M buffer solution into a graduated cylinder than add 320
mL of water to get a 500 mL solution of 2.00M buffer solution.
6.What volume of a 40% acidic solution contains 4.2 grams of acid?
100 mL/40g x 4.2g= 10.5 mL = 0.0105L
To get a 40% acidic solution put 4.2 grams of acid in a graduated cylinder and put
0.0105L of water to dilute.
Reflection:
What are you trying to do in this lab?
What technique or basic principle are you using or testing?
Describe any source of experimental error. How can they be avoided?
What problems did you encounter in completing this assignment?
How is this assignment related to other coursework?
In this lab I was trying to figure out the desired concentrations for different
solutions. I learned to see how much of a certain solute will give us the desired
concentration or percentage in a solution. The techniques used in computing the problems
all have something in common, I had to cancel out the units to get the desired unit at the
end. I had to find equivalent statements to see how different units are related. I had a hard
time figuring out how to explain the steps in using the math in order to apply the
information in a lab. I also made the mistake of not canceling out certain units and
changing liters into milliliters. To fix the problem I wrote out all the units and saw how
they cancelled out and avoided future errors. The most difficult part of the assignment
was understanding what the question was asking me to find. I also had a hard time
making correct equivalence statements that made sense. This assignment relates to many
different courses that I had to take and am currently studying. Most noticeably chemistry
and lab work, it also corresponds with most sciences such as physics, and many of the
formulas and applications learned can be applied to business, science, math, architecture,
etc.