Molarity Made Easy: How to Calculate Molarity and Make Solutions
- Welcome to Molarity Made Easy, brought to you by Ketzbook. In this video, we are going to learn how to do molarity calculations and how to make a solution of a certain molarity. But first, what is molarity? As the name suggests, molarity has something to do with moles, and we remember that the mole is a scientist's way of counting the number of particles there are.
Molarity is a concentration measurement that uses moles. It measures how concentrated a solution is based on the number of solute particles dissolved. The formula of molarity is M equals n over V, where M is molarity, n is the number of moles of solute, and V is the volume of the solution in liters. Let's go ahead and try a sample problem.
What is the molarity of a 125 mL solution containing .05 moles of hydrochloric acid. Before we start plugging numbers in our calculator, it's good to identify what we are solving for and the other variables in the question. The question starts out, "what is the molarity," which means we are solving for molarity. 125 mL is the volume, and .05 is the number of moles.
Next, write down the appropriate equation, which is M equals n over V. Because M is what we are solving for, we do not need to rearrange the equation. Substitute .05 moles in for n, and divide that by the volume. You may be tempted to write 125 mL in the denominator, but that would be incorrect because V should be the volume in liters, not mL. In order to convert 125 mL to liters, you can multiply it by a conversion factor fraction. Because one liter equals 1000 mL, we put 1000 mL on the bottom and one liter on the top.
The volume works out to be 0.125 L. However, because we are dealing with metric prefixes, we could have done the conversion simply by moving the decimal point according to the prefix. For milli, just move the decimal point three spots to the left. Back to the original problem, V equals 0.125 L, so put that in the denominator. Do the division, and the answer works out to be 0.40 moles per liter. In general, the units of molarity are moles per liter. Let's try another problem where we are solving for something different.
How many moles of sodium hydroxide are in 38 mL of 0.5 mol/L sodium hydroxide? The problem is asking us to calculate moles, and it tells us that the volume is 38 mL. However, for molarity we will need to convert that to liters, which we can do by simply moving the decimal point three places to the left. 0.50 mol/L is the molarity.
Next, write down the appropriate equation. M equals n over V. Because we are solving for n, we multiply both sides of the equation by V. On the right side, V over V cancels out, leaving n all by itself. We can rewrite the equation as n equals M times V. Plug in the values for M and V remembering that V has to be in liters. 0.5 mol/L times 0.038 liters equals 0.019. As for the units, liters on the top and bottom cancel, and the answer has the units of moles. Let's try another one.
How much 2.5 mol/L sulfuric acid should you use if you need 0.12 moles of sulfuric acid? This problem simply asks us "how much sulfuric acid," which is a little vague, but think about what you would do in a lab. How would you measure out a solution of sulfuric acid? By its volume, of course, so the problem is asking us to solve for volume. 2.5 mol/L is the molarity, and 0.12 moles is n. Next, write down the appropriate equation.
We use n equals M times V because it is closer to the form of the equation we need. It order to get V all by itself, divide both sides of the equation by M. On the right side, M over M cancels out, leaving V all by itself. We can rewrite the equation as V equals n over M. Finally, plug in the values for n and M. 0.12 moles divided by 2.5 mol/L equals 0.048. As for the units, moles on the top and bottom cancel, leaving us with liters as the units. Because the number is less than one, we convert it to mL by moving the decimal point three places to the right, giving us the final answer of 48 mL.
Now for our final problem, let's talk about how you would actually make a solution of a given molarity. How would you make 100 mL of a 0.4 molar solution of copper(II) sulfate? In this question, we notice something different. 0.40 M. That italicized M not only stands for molarity, but it also stands for the units of molarity, mol/L, and is pronounced "molar." So, a 0.4 molar solution is the same as 0.4 mol/L The volume of the solution that we need to make is 100 mL, which is the same as 0.1 liters. This problem is more complex than the previous ones, but let's go ahead and solve it like the previous problems first.
If we know V and M, then we can calculate n. M equals n over V rearranges to n equals MV. Substituting in for M and V give us 0.4 mol/L times 0.1 liters. Liters on the top and bottom cancel, and the answer is 0.04 moles. But, if we are going to make this in the lab, we need to need to know how much copper(II) sulfate to weigh out, so we need to convert the moles into grams. In order to do that, we need the molar mass, which is our conversion factor. The easiest thing to do would be to look at the chemical label, which almost always has the molar mass written on it; but let's suppose
we can't do that.
The Roman numeral II means that the copper has a +2 charge, and sulfate is SO4 two minus. Because the charges are the same, the compound is simply CuSO4. However, copper(II) sulfate is almost always found as the pentahydrate, which means it has 5 water molecules that have cocrystallized with the copper(II) sulfate. These water molecules aren't really doing anything; they are basically just hitchhiking with the copper(II) sulfate. However, these water molecules do affect the molar mass, which we will go ahead and calculate now. The formula contains one copper, with a molar mass of 63.55 g/mol, one sulfur which is 32.07 g/mol, 9 oxygens, so we multiply 9 times 16 which equals 144 g/mol, and 10 hydrogens, so we multiply 10 times 1.008 to get 10.08 g/mol. Add all of these together to get the molar mass of 249.7 g/mol. This is our conversion factor between moles and grams.
In order to calculate the mass of copper(II) sulfate needed we multiply 0.04 moles by a conversion factor fraction. Because we are staring with moles, one mole goes on the bottom. Because we are solving for grams, 249.7 grams goes on the top. Moles on the top and bottom cancel out. Because the one is on the bottom, multiply 0.04 times 249.7 to get 9.988 grams. So how to you actually make the solution?
First, get a 100 mL volumetric flask with a stopper. It should look something like this. Next, weigh out 9.988 grams of copper(II) sulfate pentahydrate, and carefully add all of that to the flask. You may need to use a funnel. Then, add distilled water to the flask until the water reaches the fill line. Put the stopper on the flask, and with your thumb on the stopper gently hold the flask upside down then right side up. Repeat this a few times.
Copper(II) sulfate tends to dissolve slowly unless you have a fine powder, so you may want to just set it down and let it dissolve on its own. Eventually, it will form a nice blue solution. Thanks for watching. If this video helped you at all, please give me a thumbs up. Feel free to also share any comments or questions you have below, subscribe to my channel, or check me out at ketzbook.com.