The previous lesson addressed mixture problems that had a monetary value. This lesson is on mixtures of chemicals. So far students have learned how to solve mixtures of chemicals when either water or other substance is added to dilute a solution, or a pure chemical (like 100% sulfuric acid) is added to a weaker solution to make it stronger. These problems lend themselves to being solved with one variable. With two variables such problems tend to be confusing to students.
In this lesson we look at mixtures of two chemicals of varying strengths to make a new mixture. For example: “A chemist has two solutions of sulfuric acid; one at 50% and the other at 75%. He mixes them to make 10 L of a solution at 60%. How many liters of each solution does he use?”
The principle behind such problems is that the amount of substance in the two mixtures is conserved—the amount of substance in the new mixture will be the same.
In addition to the two types of mixtures these lessons have addressed, there are also cost type problems that are similar to but not the same as the mixture/value problems addressed previously. For example: “Three chairs and a table cost $149. Four chairs and a table cost $177. How much does each item cost?” In these problems, the variables represent the cost so that the two equations for the problem would be:
I have typically worked in such problems among Warm-Ups rather than devoting a separate lesson to them, and that has worked out. Generally students find them fairly straightforward to set up and solve.
Students have had some problems similar to Problem 1 shown in class; if not, then I use the Warm-Up to provide guidance. Prompts: “Do we know the cost of each type of ticket?” “How do we represent the cost of 2 adult tickets?” Problem 2 reviews graphing of inequality; students find the slope by inspection as well as the y-intercept, to obtain the equation of the line. Problem 3 is a percent question that students have had previously. Problem 4 is a mixture problem in which pure water is added to dilute the solution. Students have had such problems previously. Problem 5 is a fractional equation which can be solved by taking the LCD for all terms, or using the cross multiplication method as shown.
Conservation of Matter. “If I have two marbles in one back, and five marbles in another, and I empty it out into a new bag, how many marbles are in that bag?” Students quickly answer “five”.
“You may be wondering what I’m getting at with such a simple question. It applies to mixtures of chemicals which is what we’ll be doing today. Yesterday we talked about mixtures that had a cost value. Today, we are talking about chemical mixture.”
I write on the board:
20 L of 25% alcohol solution + 10 L of 50% alcohol solution.
“I have two solutions of alcohol and water. How much alcohol is in the first solution?” This is similar to what they’ve had with mixture problems previously and they’ll tell me “5 liters”.
“And how much alcohol in the second solution?” Again, they tell me the answer: 5 Liters.
“And if I mix 20 l of solution with 10 L of solution, how much total solution do I have?” They tell me “30 Liters”
“And in this 30 liter solution, how much alcohol is there?”
Now there is silence. “You just told me how much alcohol was in each solution.” This generally gets them going and I soon hear “10 Liters”
“Yes, correct.” I pass out the illustration below for them to glue in their notebooks.
“The diagram shows what we just discussed, but there’s some new information. It shows the percent of alcohol in the new mixture: 33 1/3 %. Now although the percent of alcohol in the new mixture is different than the percentages in the two solutions, the amount of alcohol in the two solutions equals the same amount in the 30 liter solution—that is, 10 liters on the left hand side of the equation, and 10 liters on the right. Which leads to this statement for you to write in your notebooks.”
Material on left hand side of the equation = material on the right hand side of the equation.
“This is our prime directive, if you will, for the mixture problems we will be doing today. These are similar to what we did yesterday, except that instead of finding what the total cost is, we find out what the total amount of the material or chemical is on the right hand side of the equation.”
Examples. I work with the class on some examples, and then have them work independently, with guidance from me as needed.
1. A 35% sulfuric acid solution is mixed with a 65% sulfuric acid solution to obtain 20 ounces of a 41% solution. How many ounces of each solution are mixed?
“Like yesterday, the first equation is the ‘amount of solution’ equation. How much total solution is produced? Read the problem.” A few people will raise their hands: 20 ounces.
“We’ll let x = amount of 35% solution, and y= 65% solution. So what will the first equation be?” A few hands go up and I’m given the correct answer:
“That’s the easy equation. Now let’s work on the ‘amount of substance’ equation. How do I represent the amount of sulfuric acid in the first solution? What is our percent?” Someone tells me it’s 35 percent.
“OK, so we use 35 percent as a decimal. 35 percent of the amount of solution, which we are calling x. How do I write that?”
I hear 0.35x.
“Good, and now the amount of sulfuric acid in the 65% solution?”
I hear 0.65y.
“And now the final part. Yesterday, we calculated total cost. Today we calculate total amount of the substance—in this problem it’s sulfuric acid. What percent of sulfuric acid is in the total mixture?” 41%.
“And what is the total amount of that solution? Look at the problem.” I hear 20 ounces.
“Someone tell me the amount of sulfuric acid in that total mixture. How do you do it?”
I will usually hear the hesitant “I’m not sure, but…” and then 41% of 20, which is 8.2 The decimal in the answer throws students because they like whole numbers.
“Yes, 8.2 is correct, so let’s write our second equation.”
“And how do I get rid of the decimals?” I’m told to multiply each by 100 which we do, and I list the two equations:
“I strongly suggest you use substitution, but you can use elimination if that’s easier for you. Go ahead and solve it.”
They will get x = 16 and y = 4.
2. How many pounds of a 70% copper alloy must be melted with how many pounds of a 90% copper alloy to obtain 100 pounds of an 81% copper alloy?
Homework. Problems contain the two solution type discussed in this lesson, mixed in with problems that have dilution or addition of a pure substance, and using only one variable. Problems in future warm-ups will be varied so that they are not just the same problem with different numbers. But to start off, it’s best that students become comfortable with the general structure before giving them problems that vary. An example of such a problem is: How many ounces of a 75% acid solution must be added to 30 ounces of a 15% acid solution to produce a 50% acid solution?