Volumetric Analysis - Online Tutor, Practice Problems & Exam Prep

Concentration of solutions can be expressed through weight percent, volume percent, and weight volume percent. Weight percent is calculated as the mass of solute divided by the mass of the solution, multiplied by 100. Volume percent uses the volume of solute over the volume of solution, also multiplied by 100. Weight volume percent combines both, using the mass of solute in grams over the volume of solution in milliliters, multiplied by 100. Understanding these concepts allows for interconversion with molarity and molality, enhancing comprehension of solution chemistry.

Volumetric Analysis Concept 1

Now, in addition to molarity and molality, we can express the concentration of solutions in a few other ways. Here, we have three formulas: weight percent, volume percent, and weight volume percent. With weight percent, which is also called mass percent, we handle it as weight over weight or mass over mass. That will equal the mass of our solute divided by the mass of our solution times 100.

On the other hand, we have volume percent. Whereas weight or mass percent considers mass, volume percent will consider volume. So we'd have volume over volume here. In this case, it would be the volume of solute, usually in milliliters, divided by the volume of our solution, also in milliliters, and it would be times 100.

Then, we have weight volume percent, which is a mix of these two. So here, it's a weight over volume type of situation. In this scenario, we consider weight as the mass of our solute in grams divided by the volume of our solution, usually in milliliters, times 100.

Returning to weight percent, this typically is grams over grams as well. So, just remember, we have these three other ways of looking at the concentration of a given solution, and they can connect to molarity and molality through different types of manipulations. Pay attention as we cover different types of problems that really ask us to interconvert between these five different ideas.

Volumetric Analysis Example 1

Here in this example question, it says a 5.12-liter sample of a solution contains 0.30 grams of potassium sulfate. Determine the weight percent of potassium sulfate if the density of the solution is 1.30 grams per milliliter. We're also given the molecular weight of Potassium Sulfate as 174.26 grams per mole. Alright. So, we need to find the weight percent or mass percent. We'd say here that when it comes to this weight percent, it'll equal the mass of our solute, in this case, potassium sulfate. We already have that portion, 0.30 grams of potassium sulfate, and it's going to be divided by the mass of our solution, so grams of solution. And then we'd multiply that by 100. Now here, if we look at the information given to us, we have a 5.12-liter sample of the solution. So, we have the volume of the solution, and we're given the density of our solution. We can use those two together to help isolate the grams of our solution. So what we do here is we have a 5.12-liter solution. We're going to say here that 1 liter is equal to 1000 milliliters. You could also say 1 milliliter is equal to 10-3liters. Either way is fine. Since we have milliliters of solution now, we can bring in the density, which is 1.30 grams of solution divided by 1 ml of solution. Milliliters cancel out, and now we have grams of solution at the end, which comes out to 6656 grams of solution. We can take that and plug it in. When we plug this into our calculators, we get at the end 3.456×10-3%. If we look at the number of significant figures within the question, we have three significant figures, three significant figures, and three significant figures. So, again, we'd write 3.46×10-3%. So, this would be our final answer.

Parts per million of Lead

Alright. So here it says, when lead levels in blood exceed 0.80 parts per million, the level is considered dangerous. We're told that 0.80 parts per million means that 1,000,000 grams of blood contain 0.80 grams of lead. We're also told that given the density of blood is 1,060.0 kilograms per cubic meters, how many grams of lead would be found in 550 milliliters of blood with a lead level of 0.583 parts per million? All right. So this one is a big jumbled mess of numbers and expressions. Let's organize things out a little bit. First of all, they want us to figure out the grams of lead. So that's what we're looking for, grams of lead. Now, they're telling us a lot of different things. We're accustomed to saying that parts per million means milligrams per liter. But here we don't need the molarity of the solution. And plus they tell us here that in this case, we're going to say that 0.80 parts per million means 1,000,000 grams of blood connected to 0.80 grams of lead. So they're givin

Percent Composition

In the following example, we're told an 8.13% aluminum sulfate solution has a measured density of 1.235 grams per milliliter. Calculate the molar concentration of sulfate ions in the solution. Now realize here that we don't want the molarity of the aluminum sulfate solution. We just want the molarity of the sulfate ions themselves. So here we're looking for molarity of sulfate ions. So that would just be the moles of sulfate ions divided by the liters of solution. What information are we given? Well, we're told that we have 8.13% of aluminum sulfate solution. So that 8.13%. What does that really mean? Well, that means that we have 8.13 grams of aluminum sulfate for every 100 grams of solution. We're also told that the density of the solution is 1.235 grams of solution for every 1 milliliter of solution.

In addition, we're given the mass of aluminum sulfate, but with these types of calculations, we need that. We can write that down too. We'll incorporate it within our calculations. These are the two pieces of information that we're given, and with this, we have to figure out the molarity of sulfate ions. We're going to start out by bringing down the mass percent of aluminum sulfate solution. We have 8.13 grams of aluminum sulfate and that's over 100 grams of solution.

What we're going to do first is we're going to convert the grams of aluminum sulfate into moles of aluminum sulfate by using the molecular mass or molecular weight they were given within the question. We want to get rid of grams so it goes on the bottom. So we have 342.17 grams of aluminum sulfate on the bottom and then 1 mole of aluminum sulfate on top. This cancels out with this. The reason we have to figure out the moles of aluminum sulfate is that once we have the moles of the entire compound, we can do a mole to mole comparison to find the moles of just sulfate ions. At this point, we're going to say that for every 1 mole of the entire aluminum sulfate compound, there are exactly 3 moles of sulfate ions.

So that's 3 moles SO42- ions. The moles cancel out. At this point, we have moles of sulfate ion over grams of solution. We're not done yet because we need liters of solution on the bottom.

Now, to get rid of those grams of solution, we have the density of our solution. We're going to bring that down, wanting to cancel out these grams of solution here on the bottom. So we're going to have the grams of solution of the density on top and then, 1 milliliter of solution on the bottom. So grams of solution cancel out. Now I have milliliters of solution on the bottom. I want to cancel out those mL, so I'm going to put 1,000 mL of solution on top and then 1 liter of solution on the bottom. So that at the end, what do we have? We have exactly moles of sulfate on top, liters of solution on the bottom. That'll give us the molarity of sulfate ions.

Now, based on the information given, we have 3 significant figures here, so we'll go with 3 significant figures for our answer. That gives me 0.880 molar sulfate ions as our final answer.

Now that you've seen this approach, look to see if you can figure out how to set up example 2. This one is vastly different from the first one we just did. So if you can't approach it, don't worry. Just come back and see how I tackle this example 2. So attempt it on your own. If you're stuck, just come back and see how I approach that same question.

Percent Composition

Alright. So here it says, the density of a 33.8% solution of sodium acetate is 1.10 grams per milliliter. It says a reaction requires 68.8 grams of sodium acetate. What volume of the solution do you need if you want to use a 50 percent excess of sodium acetate? Alright. So this question is a bit different from what we've seen in the past. We're dealing here with an excess of a particular compound. But before we talk about that excess, let's see what they're asking us to find. They want us to find the volume of the solution. So here we're looking for milliliters as my default amount or units, milliliters of solution. Next, we're going to write down all the given information and from that, we'll solve for the milliliters of solution.

We're told that we have a 33.8% solution. So that means we have 33.8 grams of sodium acetate divided by 100 grams of solution. We're told that we have a density of solution that's 1.10 grams per milliliter. So that's 1.10 grams of solution per 1 milliliter of solution. And we can see here that the mL of solution I want to isolate, we can find them right here. We know that we're going to end with a density of solution in some way to get milliliters of solution at the end. Also, they're telling us that we required 68.8 grams of sodium acetate but that we want a 50% excess of that. So what exactly does that mean? Well, in a reaction, when we're talking about theoretical yield, theoretical yield represents 100%. What we want now is we want an additional 50%. So we want 150%. But when it comes to percentages, we don't use them in our calculations that way. What we're going to do is divide this by 100, so that's going to come out to 1.50. We're going to take that 1.50 and multiply it by the 68.8 grams of sodium acetate. That will give us our 50% excess that we need. So multiply it by 1.50, that gives me 103.2 g of sodium acetate.

Now, we know that we want to isolate mL of solution. We're going to start off by using first grams of sodium acetate because it's the easiest to manipulate because it's just one unit. So we have 103.2 grams of sodium acetate. These grams of sodium acetate can cancel out with these grams of sodium acetate. So we're going to put on the bottom Now we want to cancel out those grams of solution. So take the density, we take the 1.10 grams of solution, put them on the bottom. And then the 1 milliliter of solution on top. So grams of solution cancel out, so at the end I'll have mL of solution. This comes out to 277.569 mL of solution.

Here, this has 3 significant figures, 3 significant figures, 3 significant figures. 50%, that has one significant figure but we'll go with the other ones. So that gives me 278 mL of my solution. Okay. So that's my answer there. So we're accustomed to seeing these types of word problems. The new twist was looking at excess. Realize that when we're talking about theoretical yield, that's 100%. So any excess would just be that number added to the 100%. We would divide that new percentage by 100 to get its decimal form and then multiply it by the theoretical yield which in this was 68.8 grams of sodium acetate. Doing that helped us to isolate the volume of my solution in the end. Now we've seen this one. Try to attempt the practice question left here on the bottom. Again, don't worry if you get stuck. Just come back and see how I approach that practice problem left at the bottom of the page.

Determine the number of mmoles of sulfuric acid that would be found within a 120 g sample that is 79.9% sulfuric acid.

• A. 0.960 mmol 
• B. 980 mmol
• C. 0.33 mmol
• D. 1.7 mmol

Dilution

So we're going to say here a dilution involves the addition of water to a concentrated solution. We're going to say typically a given volume of a concentrated solution. Talking about dilutions, we're going to use the dilution equation which is M₁V₁ equals M₂V₂. Here, M₁ represents concentrated molarity. This is basically our concentration of molarity before water has been added. V₁ represents our initial volume, so before any additional water has been added. M₂ represents our diluted concentration and V₂ represents our final volume after we've added the water.

Now we're going to say typically that M₁, which is more concentrated, will be the larger concentration. V₂ represents your final volume, which equals your initial volume which is V₁ plus volume of added water. So just remember, anytime we're adding water to any type of solution, that represents a dilution which means that we may use this formula in some way or another. Knowing this will help guide us to the exact way of solving the following examples that are given at the bottom. You can take a look at example 1, and attempt it on your own. But if you get stuck, just go on to the next video and see how I approach answering the example question given below.

Dilutions

Here it's asking how many grams of 53.1 weight percent sodium chloride should be diluted to 2.50 liters to make 0.15 molar NaCl. Now in this question, we're not dealing with m1v1 = m2v2 because we only have one volume and one molarity involved within the question. What this question itself is asking, it's asking us for the grams of solution. So how many grams of solution are needed based on what they've told us. Now with this information, we're going to put all the given information on the other side and organize it to help isolate our grams of solution. So we have 53.1 weight percent sodium chloride solution. So that means we have 53.1 grams of sodium chloride over one hundred grams of solution. There goes the grams of solution we know we need to isolate at the very end of all our calculations. We're told that we have 2.50 liters. Really mean? Well, molarity is moles over liters, so 0.15 molar really means that I have 0.15 moles of sodium chloride per 1 liter of solution. So we kno

If 920 mL of water is added to 78.0 mL of a 1.28 M HBrO₄ solution what is the resulting molarity? 

• A. 0.110 M
• B. 15.1 M
• C. 0.100 M
• D. 0.119 M

Volumetric Analysis Example 6

Here we're told that a student prepared a stock solution by dissolving 25 grams of Sodium hydroxide in enough water to make a 150 ml of solution. The student took 20 ml of the stock solution and diluted it with enough water to make 250 ml of solution. And then finally, taking 75 ml of that solution and dissolving it in water to make 500 ml of solution. Now, what is the concentration of sodium hydroxide for this final solution? We're also given the molecular weight of sodium hydroxide as being 40 grams per mole. Now this question here represents a serial dilution. Basically, we make our original concentration, which is our stock solution, and we're going to go through a series of dilutions of it to get to our final molarity at the end. So, our first step is to determine what is our stock solution concentration. We have 25 grams of sodium hydroxide. For every one mole of sodium hydroxide, it is 40 grams of sodium hydroxide. So this adds up to 0.625 moles of sodium hydroxide. We're doing this because we're going to determine what our initial molarity is. Our initial or \( M_1 \) is the moles of sodium hydroxide which 0.625 moles divided by the liters of solution. Initially, we have 150 ml's, so dividing that by 1000 gives us 0.150 liters. So this comes out to be 4.167 M molar. Alright. So that's \( M_1 \). Now, they are telling me that I am taking 20 ml's of \( M_1 \) and diluting it to 250 ml's. Remember, dilution is M_1 V_1 = M_2 V_2 . Our initial molarity is again 4.167 M molar. We took 20 ml of it. We don't know what our new molarity is just yet, but we do know that we're diluting it to 250 ml. Divide both sides here by 250 ml. And we're going to have \( M_2 \). \( M_2 \) here will equal 0.33336 M. We're not done because now they're saying we're taking 75 ml's of that, so we're going to say M_2 V_2 = M_3 V_3 . So we're going to say here 0.33336 M times we took 75 ml's. We don't know where our last molarity is, but we do know we're diluting all of this up to 500 ml's. Divide both sides here by 500 ml's. And when we do that, we'll get \( M_3 \) which will represent the final concentration of sodium hydroxide as being 0.0500 M . If we look, the number of significant figures, the lowest number of significant figures that we see within our question is this 500, which has only 1 significant figure. So, we can just say that this is 0.05 M as my final answer. So, this would be the final concentration of Sodium Hydroxide through these series of dilutions.

Dilutions

The density of 63.7 weight percent sodium hydroxide is 0.915 grams per milliliter. How many milliliters of water should be diluted to 850 ml to create 0.425 molar sodium hydroxide? Alright. So here they're asking us how many ml of water. It's really asking how many milliliters of solution should be diluted. That's what we're looking for. What we're going to do now is to isolate all the given information and see how we can isolate those ml of solution. We're told here that we have 63.7 weight percent sodium hydroxide. So that means we have 63.7 grams of sodium hydroxide for every 100 grams of solution. We're told that the density of the solution is 0.915, so that's 0.915 grams of solution per 1 milliliter of solution.

What else are we given? We're told that we have 850 ml to create 0.425 molar sodium hydroxide. So we have 850 ml. And then remember, molarity is moles over liters, so that's just 0.425 moles of sodium hydroxide per 1 liter. We're going to start out with 850 ml first because remember that's just one unit, easier to deal with. The others have a mixture of 2 units involved, harder to control. So we have 850 ml's. This 850 ml's is attached to this molarity. If we could change those mls into liters and multiply it by the molarity, we can isolate the moles of sodium hydroxide. So we're going to change mls into liters. Remember that 1 liter is 1,000 milliliters. Now that we have liters of solution, we can cancel out the liters from the molarity given. And that'll give us moles of NaOH at the end.

But here, we got to keep going. So far, we've used this portion of the given information. What do we have left? We have left the mass or weight percent and we're left with density. We know that we need to isolate ml's of solution which are right here. Next, I see that I have moles of NaOH. So if I could change them into grams, I can cancel out these grams of NaOH. So we're going to say here for every 1 mole of NaOH, what is the mass of NaOH? Well, it's made up of 1 sodium, 1 oxygen, and 1 hydrogen. And based on the atomic masses from the periodic table, each of them is 22.98977 grams, 15.99994 grams, and 1.00794 grams. Add them all up together, gives us 39.99765 grams. So we have grams of NaOH, so bring down the weight percent, 63.7 grams of NaOH here on the bottom, 100 grams of solution on top.

Finally, I can cancel out the grams of solution by using the density of the solution given. So we put 0.915 grams of solution here on the bottom and then, 1 ml of solution here on top. So when we punch all that in, that gives me 24.79 ml of solution. And here, we have 3 significant figures, 3 significant figures, 4 significant figures, and 3 significant figures. So we can round that just to 24.8 ml's of solution would need to be diluted to 850 ml's in order to create 0.425 molar NaOH solution.

Again, the wording of some of these problems can be a bit challenging, but the best approach like I've been saying is to write down what you're looking for first, write down all the given information, try to isolate the units that you need to get the desired answer at the end. Doing that helps us to navigate through all the wording, all the crazy jumbled numbers in order to get our final answer. So hopefully, you guys were able to follow along in terms of these typical types of dilution questions, and we'll continue onward into talking more about chemical reactions as well as stoichiometric calculations that'll become a focal point in discussing concentrations of solutions.

Here’s what students ask on this topic:

What is the difference between molarity and molality?

Molarity (M) is defined as the number of moles of solute per liter of solution. It is expressed as:

$M=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{liters} \mathrm{of} \mathrm{solution}}$

Molality (m) is defined as the number of moles of solute per kilogram of solvent. It is expressed as:

$m=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{kilograms} \mathrm{of} \mathrm{solvent}}$

The key difference is that molarity depends on the volume of the solution, which can change with temperature, while molality depends on the mass of the solvent, which remains constant regardless of temperature changes.

How do you calculate weight percent in a solution?

Weight percent (also known as mass percent) is calculated by dividing the mass of the solute by the mass of the solution and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 grams of solution, the weight percent would be:

$\frac{5}{100} \times 100=5\%$

What is volume percent and how is it calculated?

Volume percent is a way to express the concentration of a solution, calculated by dividing the volume of the solute by the volume of the solution and then multiplying by 100. The formula is:

$\mathrm{Volume} \mathrm{percent}=\frac{\mathrm{volume} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 10 mL of solute in 100 mL of solution, the volume percent would be:

$\frac{10}{100} \times 100=10\%$

How do you convert between molarity and weight percent?

To convert between molarity (M) and weight percent, you need the density of the solution and the molar mass of the solute. The steps are:

1. Calculate the mass of the solute using molarity and volume of the solution.

2. Determine the total mass of the solution using its density.

3. Use the weight percent formula:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have a 1 M solution of NaCl (molar mass = 58.44 g/mol) and the density of the solution is 1.05 g/mL, for 1 L of solution:

Mass of NaCl = 1 mol × 58.44 g/mol = 58.44 g

Mass of solution = 1 L × 1.05 g/mL × 1000 mL/L = 1050 g

Weight percent = (58.44 g / 1050 g) × 100 ≈ 5.56%

What is weight volume percent and how is it used?

Weight volume percent (w/v%) is a concentration measure that combines the mass of the solute and the volume of the solution. It is calculated by dividing the mass of the solute (in grams) by the volume of the solution (in milliliters) and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{volume} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 mL of solution, the weight volume percent would be:

$\frac{5}{100} \times 100=5\%$

This measure is commonly used in medical and biological contexts, such as preparing solutions for intravenous fluids.