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Solving Mixture Problems using Weighted Average

Solving Mixture Problems using Weighted Average

Author: Sophia Tutorial
Description:

Use weighted average to calculate the concentration of a mixed solution.

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Tutorial
what's covered
  1. Average/Mean
  2. Weighted Average
  3. Mixture Problems

1. Average/Mean

Before we begin discussing weighted average, it may be helpful to review how to calculate the simple average or mean of a data set. In order to find the average, we find the sum of all data values, and then divide by the number of data values in our set. This is illustrated in the example below:

left curly bracket 4 comma space 4 comma space 8 comma space 12 comma space 14 right curly bracket

4 plus 4 plus 8 plus 12 plus 14 equals 42
Add all values
42 divided by 5 equals 8.4
Divide sum by number of data values, 5
8.4
Our Solution


2. Weighted Average

Weighted average is different from simple average in that certain data values carry more weight, or influence the average more heavily. With simple average, every data point is equally represented in the calculation for the mean, but in weighted averages, some data points are multiplied by a number in its calculation.

Often times, we see weighted average in class grades. For example, tests and projects may be 50% of your grade, daily assignments are 30% of your grade, and participation is worth 20% of your grade. To calculate your final grade for the class, we multiply your test/project score by its weight of 50%, your assignment score by its weight of 30%, and your participation grade by its weight of 20%, all before adding these figures together. This is illustrated below:

Test/Project Grade: 83% Assignment Grade: 95% Participation Grade: 80%
Weight: 50% Weight: 30% Weight: 20%

0.83 times 0.5 equals 0.415
Weighted test/project grade
0.94 times 0.3 equals 0.282
Weighted assignment grade
0.80 times 0.2 equals 0.16
Weighted participation grade
fraction numerator 0.415 plus 0.282 plus 0.16 over denominator 0.5 plus 0.3 plus 0.2 end fraction equals x
Weighted portion in numberator, individual weights in denominator
fraction numerator 0.857 over denominator 1 end fraction equals x
Simplify numerator and denominator
0.857 equals x
Divide 0.857 by 1
85.7 percent sign
Our Solution

In general, data values in the denominator of the fraction are multiplied by its corresponding weight. The corresponding weights are then added together to form the denominator of the fraction. In this case, the weights summed to 100% because the weights of tests/projects, assignments, and participation need to reflect 100% of your grade in total.


3. Mixture Problems

Next, we will use the concept of weighted average to set up and solve a mixture problem. Consider the following example:

To prepare for a lab experiment, you mix two concentrations of HCl (hydrochloric acid): 30 mL of 15% HCl solution, and 20 mL of 40% HCl solution. What is the concentration of the mixed solution?

To solve this problem, we need to multiply the quantity of each solution by its concentration (or its weight to be included in our calculation). This will represent the numerator of our fraction for weighted average. As for the denominator, we divide by the total quantity of the solution (so we add the individual amounts to get a total amount of liquid). The solution is worked out below:

fraction numerator open parentheses 30 m L times 0.15 close parentheses plus open parentheses 20 m L times 0.4 close parentheses over denominator 30 m L plus 20 m L end fraction
Multiply the quantity by its weight in numerator. Add quantities in denominator
fraction numerator open parentheses 4.5 m L close parentheses plus open parentheses 8 m L close parentheses over denominator 50 m L end fraction

Simplify the numerator and denominator
fraction numerator 12.5 m L over denominator 50 m L end fraction

Add 4.5 to 8
0.25
Divide by 50mL
25 percent sign space m i x e d space s o l u t i o n
Our Solution

Some mixture problems will ask us to find specific amounts of each solution that must be mixed together in order to yield a certain amount of a specific concentration. Let's again use the HCl example. We have two kinds of solutions already made: a 15% solution, and a 40% solution. To prepare for a lab experiment, we are going to need 120 mL of a 33.33% solution. How many mL of each solution must be mixed in order to create this mixture?

Let's define variables for this situation:

  • x = mL of 15% solution
  • y = mL of 40% solution

We have an immediate relationship between x and y, that x + y = 120. How does this affect our equation for weighted average?

fraction numerator 0.15 x plus 0.4 y over denominator x plus y end fraction equals 0.3333

Weighted average equation
fraction numerator 0.15 x plus 0.4 y over denominator 120 end fraction equals 0.3333

x+y=120
0.15 x plus 0.4 y equals 40
Multiply by 120

We have simplified our equation, but it still has two variables. If this equation could be expressed using only 1 variable, we could solve for that variable, and then use substitution again to solve for the other variable. Let's return to the relationship x + y = 120. We can write this equivalently as y = 120 – x.

Now that we have an expression for y, we can write 120 – x into the original equation instead of y. The result will be a single-variable equation, and we'll be able to solve for x.

0.15 x plus 0.4 y equals 40

0.15 x plus 0.4 open parentheses 120 minus x close parentheses equals 40
y=120minusx
0.15 x plus 48 minus 0.4 x equals 40
Distribute 0.4 into (120minusx)
short dash 0.25 x plus 48 equals 40
Combine like terms
short dash 0.25 x equals short dash 8
Subtract 48
x equals 32
Divide by short dash 0.25

This tell us that we must use 32 mL of the 15% solution in our mixture. Since we know we need a total of 120 mL, we can deduce that we'll need 88 mL of the 40% solution. This combination will yield 120 mL of 33.33% solution.

summary
Recall that the average/mean is finding the sum of all data values, and then divide by the number of data values in our set. A weighted average gives you the average of a set of values that may carry different weights. The more weight that a value has, the more it is accounted for in the calculation. We can use the concept of weighted average to calculate mixture problems.

Formulas to Know
Weighted Average

fraction numerator open parentheses C subscript 1 close parentheses open parentheses Q subscript 1 close parentheses plus open parentheses C subscript 2 close parentheses open parentheses Q subscript 2 close parentheses over denominator Q subscript 1 plus Q subscript 2 end fraction equals C subscript 3