Algebra/Chemistry Mixture Problems

©Copyright October 28, 2001. All rights reserved. This lesson is intended to be used by classroom teachers, not for publication. You have permission to make copies to use as classroom exercises. If you wish to make copies for any other purpose, please contact the author via email or snail mail at the address listed on the lesson plan page.

Example like Part 1 #1 of Guided Practice Worksheet

If Dr. M.J. Sanchez  mixes 75 ml of a 20% acid in water solution with 40 ml of acid, what is the quantity and concentration of the final solution?


First encourage the students to draw pictures of what is happening and then label it with the quantity and concentration information.
 
 
+
 =
75 ml of solution
+
40 ml of acid
=
115 ml of mixture
20% 100%
As you discovered in the previous exercises, percents cannot be added like quantities can.  However these percents can give us quantities of acid in the components of the mixture. These amounts are derived by multiplying the percent of acid by the quantity of the solution. This is shown in the next table.
 
 
+
 =
75 ml of solution
+
40 ml of acid
=
115 ml of mixture
20%
  
100%
    
15 ml of acid
+
40 ml of acid
=
55 ml of acid

Now that these amounts are derived, we can answer the question.

The final mixture (solution) is 55 ml of acid in a total of 115 ml. Therefore the concentration of the solution is derived by dividing the two (that is, 55/115). So the concentration, rounded to the nearest tenth of a percent,  is 47.8%.

Example like Part 2 #1 of Guided Practice Worksheet

How many gallons of a 15% sugar solution must be mixed with 5 gallons of a 40% sugar solution to make a 30% sugar solution?

Begin by creating a table/drawing of the situation and filling in the known information, as shown below.
 
 
+
 =
Quantity of Solution
  
+
5
=
  
Percents
15%
  
40%
  
30%
Quantity of Sugar
  
+
  
=
  

Now we must decide how to represent the unknown quantities. Since the question is asking us to find the number of gallons of 15% solution we will use to accomplish the required result, it will be beneficial to have our variable represent the amount of 15% sugar solution we are using.
Therefore:

x = # of gallons of 15% sugar solution
We place this variable in the appropriate place in our table/drawing and then perform all the mathematical operations which are applicable. See the table below for these results.
 
 
+
 =
Quantity of Solution
x gallons
+
5 gallons
=
x+5 gallons
Percents
15%
  
40%
  
30%
Quantity of Sugar
15% times x gallons
+
40% times 5 gallons
=
30% times (x+5) gallons

Since each element in the last row is in gallon units we can use this line to create an equation to solve for x. We need to change each percent value to its decimal equivalent and then multiply. The resulting equation would be:

0.15x + (0.40)(5) = 0.30(x + 5)
The steps to solving this equation are:
 
0.15x + 2 = 0.30x + 1.5
0.15x - 0.15x + 2 = 0.30x - 0.15x + 1.5
2 = 0.15x + 1.5
2 -1.5 = 0.15x + 1.5 - 1.5
0.5 = 0.15x
0.5/0.15 = 0.15x/0.15
3 1/3 = x
So, we will need to add 3 and 1/3 gallons of 15% sugar solution to 5 gallons of 40% sugar solution to produce a 30% sugar solution.

At some point the students should be able to create the algebraic equation without drawing the picture and without the table. That will take time and practice and each student will move to this point in their own time based on the maturity of the their problem solving techniques and their ability to reason abstractly.
 

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