Lesson 4: Percentage Concentrations
The use of percentages is a common way of expressing the concentration of a solution. It is a straightforward approach that you have used earlier when dealing with the composition of compounds. There are, however, some differences. One is that the concentrations of solution are variable while the composition of compounds is constant. Another is that the percentages can be calculated using volumes as well as weights, or even both together.
One way of expressing concentrations, with which you might be familiar, is by volume percent. Another is by weight percent, which you learned about when you studied the percent composition of compounds in CH 104. Still another is a hybrid called weight/volume percent. In the pages of this section we will look at how to calculate each and the cases where each is generally used.
Volume Percent  Weight Percent  Weight/Volume Percent
Volume Percent
Volume percent is usually used when the solution is made by mixing two liquids.
For example, rubbing alcohol is generally 70% by volume isopropyl alcohol. That means that 100 ml of solution contains 70 ml of isopropyl alcohol. That also means that a liter (or 1000 ml) of this solution has 700 ml of isopropyl alcohol plus enough water to bring it up a total volume of 1 liter, or 1000 ml. 

Please note that the denominator is the volume of solution, not solvent. In all of the percentage concentration calculations, it will be the amount (whether volume or mass) of solution on the bottom.
One potentially confusing thing about volume percent stems from the fact that the volumes of liquids are not always additive. Sometimes the volumes change when two liquids are mixed together. For example, mixing 70 ml of isopropyl alcohol and 30 ml of water will not give you exactly 100 ml of solution.
Weight Percent
Another similar way of expressing the concentration of a solution is to express it in weight percent (or mass percent, if you prefer).

You've done this kind of calculation before (in the lesson on composition in CH104). You may already be prepared to answer the questions in exercise 2 in your workbook. If so, do that and check your answers at the bottom of the page. If not, read on.
As an example, let's consider a 12% by weight sodium chloride solution. Such a solution would have 12 grams of sodium chloride for every 100 grams of solution. To make such a solution, you could weigh out 12 grams of sodium chloride, and then add 88 grams of water, so that the total mass for the solution is 100 grams. Since mass (unlike volume) is conserved, the masses of the components of the solution, the solute and the solvent, will add up to the total mass of the solution. 
12 % NaCl solution = 
12 g NaCl 100 g solution 
12 g NaCl (12 g NaCl + 88 g water) 
= 12% NaCl solution 
To calculate the mass percent or weight percent of a solution, you must divide the mass of the solute by the mass of the solution (both the solute and the solvent together) and then multiply by 100 to change it into percent.
Examples (Ex. 1)
Your workbook has some examples of calculations involving weight percent in example 1. An explanation of those examples is given here.
Example 1a asks, "What is the weight percent of glucose in a solution made by dissolving 4.6 g of glucose in 145.2 g of water?" The way that I recommend you go about doing this is to look at what you need to find, look at what you are given, and determine what the relationship is. Let's start with what you are trying to find, the weight percent of glucose in the solution. What do we need in order to calculate that? We need to divide the weight of glucose by the weight of the solution. We have the weight of glucose, that is 4.6 g. What about the weight of the solution? That is not given, but we can figure it out by adding together the weight of glucose and water to get 149.8 g. Now we can calculate the weight percent of glucose as shown to get 3.1%. 

The next question is a little bit different. You are asked how you would prepare 400 g of a 2.50% solution of salt. You are given 400. g of solution (that is total) and you know that 2.50% of that is going to be salt. You need to find out how much salt you need and how much water you need. You can simply multiply 400. g by 2.50% to find out how much salt there is (shown in the top line), or you can set up the calculation shown on the next line. Either way, you find that you need 10.0 g of salt. Since you need a total mass of 400. g and 10. g of that is salt, the remaining 390. g would have to be water. So, to prepare this solution you would have to mix 10.0 g of salt with 390. g of water. 

If you have any questions about these calculations be sure to stop and go over them again or work with the instructor if you need to, so that you can get squared away on how to work these kinds of problems.
Practice (Ex. 2)
Take time now to answer the following questions (also given in exercise 2 in your workbook). The third question, you will note, has an extra twist to it. Take some time to do these now, get some help if you need it and then check your answers below.
a. 

b. 

c. 

Answers (Ex. 2)
The answer to "a" is 5.9% ethanol. If you got 6.2% probably what you did was to divide by the mass of the solvent (water); you should have added together the mass of the ethanol and the water to get the total mass of the solution.
Question "b" brings up a couple points. One is that the question asks "how would you prepare?" So the answer is not just 18.75g of glucose. Since the question asks you how to do something, you need to answer with how you would do it. You would dissolve 18.75 g of glucose in 231.25 g of water. The other point this brings up is a matter of significant digits. According to the number of digits given in the question it would be very legitimate to say that you would dissolve 19 g of glucose in 231 g of water. That would be a very appropriate thing to do. However, if you are trying to prepare a solution, try to prepare it as well as you can. Get as close to the 7.5% as you can. If you were to round this value to 19g and then dissolved it in 231 g of water, what would actually prepare would be 7.6% solution. In terms of significant digits, that is not a problem, you just have a slight variation in the last digit and that fits. But if you want to prepare a 7.5% solution instead of a 7.6% solution, then you have to be more careful than just two significant digits.
The answer to "c" is 1.96% potassium chloride. Depending on how you went about doing your calculations and rounding off you may have gotten 1.95% instead of 1.96%. Don't worry too much about that. One more point about the last question here is that in order to do it you needed to change from moles of KCl to grams of KCl. This is just the opposite of what you will need to do when calculating molarity a little later in this lesson.
Weight/Volume Percent
Another variation on percentage concentration is weight/volume percent or mass/volume percent. This variation measures the amount of solute in grams but measures the amount of solution in milliliters. An example would be a 5%(w/v) NaCl solution. It contains 5 g of NaCl for every 100. mL of solution.

Because of the different units in the numerator and denominator, this type of concentration is not a true percentage. It is used as a quick and easy concentration unit because volumes are often easier to measure than weights and because the density of dilute solutions is generally close to 1 g/mL. Thus, the volume of a solution in mL is very nearly numerically equal to the mass of the solution in grams. (The more concentrated the solution, the less accurate is this approximation.)
In the next section, we'll examine another way to quantify concentration. For now, make sure that you understand how to calculate the three different types of percentage concentration and then move on to the next section, "Molarity."