Calculating Percent Composition and Empirical Formulas

September 26, 2019

Howdy! Mr. Causey here coming to you from beautiful Atascocita, Texas in my virtual studios and today we’re going to talk about percent compositions and percentages. This is a very important, empirical lesson that chemists use to determine the composition of unknown compounds. Get out your calculators and get out your periodic tables and let’s get started. In
this lesson, you’re going to talk about percent by mass, percent composition, empirical formulas, molecular formulas, and much, much more. You’ll need your periodic table, you’ll need your calculator and you must know the periodic table. You must know how to find molar mass and you need to be able to write formulas and of course all of this can be found in some of my videos on my video channel. And you need to know how to do percent. So are you ready? Let’s get started! One of the first things we want to talk about is compounds and some things that we need to know about compounds is that they have a definite composition and a definite ratio and this has been determined by the law of definite composition and the law of multiple proportions. These two laws, determined in the eighteenth-century, help us to understand that compounds are always going to be the same. So that means that water, H2O, is always H2O. It’s always going to be 2 moles of Hydrogen and 1 mole of Oxygen and with that knowledge we can use it to help us write chemical formulas and figure out the percent composition and the percentage composition of different compounds. A percentage composition is the composition of a compound given in percentages, of course that’s using percent, and then we can determine our empirical formulas and we can even determine molecular formulas using this information and you’re going to want to listen to the whole thing because I do two examples at the end dealing with finding the empirical formula and then finding a molecular formula but before we do that let’s look at percent and percentages. Percent by mass, we’re going to take the mass of an element divided by the mass of the compound and then multiply it by 100 and that’s how you find a percent. So if I take grams of Hydrogen and divide it by the grams of Ammonia and then multiply it by 100, that gives me the percent of Hydrogen. On a percentage, it’s just taking the mass of the compound multiplied by the percent. Grams Ammonia multiplied by the percent Hydrogen and don’t forget to change the percent to a decimal and that will give us the grams Hydrogen. Let’s do a practice problem. What percent of HNO3 is Hydrogen? Well, first we need to determine the molar masses and then we need to realize that HNO3 is the whole, it’s the main part, and the Hydrogen is the part of the whole. And if you know about fractions, we can take the part divided by the whole, multiply by 100 and that is our percent. so let’s look at that problem, let’s go to the magic blackboard. HNO3’s molar mass, well we have 1 mole of Hydrogen, we have 1 mole of Nitrogen and 3 moles of Oxygen and we can get all these values off of the periodic table. That adds up to 63.02 grams of Nitric acid and of course there’s one 1 mole of Hydrogen and that’s 1.01 grams. So there’s our two values, let’s take them divide them. We take the part, right here the part, divide it by the whole, times 100 and it equals a percent. Now finding percentage, how much Iron can be obtained from 125.0 grams of Fe2O3 which is Iron 3 Oxide, commonly known as rust? So we need to determine the molar masses, we then need to find the percent Iron and then calculate the percentage. Three steps. Let’s go to the magic blackboard. First thing is taking Iron 3 Oxide and determining the mole amounts. There are 2 moles of iron and 3 moles of oxygen. Add those together and there we go, there’s our molar mass of Iron 3 Oxide and we want to compare it to the moles Iron and there’s 2 moles of Iron in there so we want to make sure that we use that number. And we take the part, divide it by the whole, times 100 and it’s 69.94 percent Iron. So take the total amount of Iron 3 Oxide that we had, 125 grams, multiply it by the percent and now notice we have to change it to a decimal. That gives us 87.43 grams of Iron and so in 125 grams of Iron 3 Oxide, there are 87.43 grams of Iron. And I think our sig figs are right there but if they aren’t you can check them. I believe we have four sig figs. Let’s recap! I’ve told you or answered the question, what is percentage composition? I’ve shown you how to find percent by mass and I’ve shown you how to find a percentage. Alright, now let’s go ahead and use this information to find empirical formulas and molecular formulas. Find the empirical formula of a compound that has the following following composition: 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. Alright, one of the first things you want to do is step one, assume that we have 100.0 grams of the compound. The reason I’ve chosen 100 grams is because it makes the math easier, it’s easier to work with 100 grams or multiples of 10 than is anything else. So that would give us 40 grams of carbon, 6.70 grams of hydrogen, and 53.3 grams of oxygen. Now, let’s convert the masses to moles using the periodic table. We can go and use the molar mass to convert these 2 moles. So to the blackboard! First thing we do is multiply 40 grams of carbon by its molar mass ratio and we get 3.33 moles of carbon. We do the same thing to hydrogen and we get 6.63 moles of hydrogen. And then we take oxygen, do the same thing, and we get 3.33 moles of oxygen. So now we have our molar values and we can use this to do step number three. In step 3, we want to write a ratio using the smallest value as the denominator and if you remember, on the blackboard, 3.33 is the smallest of the values. So let’s write our ratios. Again, we knew that there were 3.33 moles of carbon, we divide that by 3.33 and we get 1 mole of carbon. Do the same thing with hydrogen and we get 2 moles of hydrogen and we get one mole of oxygen. Well if we look at that, that means we get 1 carbon, 2 hydrogen and 1 oxygen and our empirical formula is CH2O but that’s the empirical formula and not necessarily the molecular formula. Let’s take a look at the molecular formula. If the sample in the preceding problem has a molar mass of 60.06 grams, find the molecular formula. We have three steps in this calculation. Step 1: Determine the mass of the empirical formula, the empirical formula of course being CH2O. To the blackboard! We get CH2O, 1 mole of carbon, 2 moles of hydrogen, and 1 mole of Oxygen and it all adds up to 30.03 grams of that compound or at least the empirical formula. Step 2: Divide the actual molar mass by the empirical formula. Now remember the actual molar mass is 60.06 grams. 60.06 grams divided by 30.03 grams gives us ratio of 2 I really don’t mean a ratio but a value of 2 which then tells us that everything in the empirical formula needs to be multiplied by two and that should give us C2H4O2 and we see that is our molecular formula. aAlright, let’s recap. We looked at how to find the empirical formula, we looked at how to find the molecular formula, and if you have any questions, send an email to mrcausey and you can check it out by going to or and check out PowerPoint videos and much much more. Also, you can subscribe to my YouTube channel, studies have shown that it increases your IQ.

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