 ## GOB Chemistry

Learn the toughest concepts covered in your GOB - General, Organic, and Biological Chemistry class with step-by-step video tutorials and practice problems.

Chemical Reactions & Quantities

# Empirical Formula

The empirical formula gives the relative number of atoms.

Empirical Formula
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concept

## Empirical Formula 2m
Play a video:
in this video, we're gonna take a look at the difference between our empirical formula and molecular formula. First of all, when it comes to the empirical formula of a compound, we're going to say it's related to the mass percentage of its constituent elements using the mold concept. Remember, the mold concept has to do with us going between grams moles and you did, such as molecules, ions or atoms. We're going to say here that when it comes to our molecular formula, this gives us our actual number of atoms in a compound, and the empirical formula gives us the relative number of atoms and represents the most simplified form. So by convention, we're going to say any formula that's contained whole numbers of each Adam in what's called the whole number ratio. Now, let's put this to practice and really understand what's the difference between molecular formula and empirical formula? Here we have the molecular formulas of different compounds. This is how many actual elements of each type that are present within that compound. The empirical formula can be thought of as it's reduced or simplified form. So if we take a look here we have C three h 603 realize that here 36 and three are all divisible by three. So if I divide this by three, what I'll have left is C H 20 That's C H 20 represents the reduced simplified form or the empirical formula. If we look at the next one, we have C 10 h 14 and to all of these numbers here are divisible by two. So when we divide everything by two, we're going to get C five h seven and that represents the empirical formula. And then finally we have C 12 h 22 0 11. They don't share any number in common where we can reduce it to a simplified form. So we actually have the same empirical formula as molecular formula and we'll see from time to time. That is true. So just remember that your molecular formula is the actual number of each of the elements within a compound, and the empirical formula is the reduced reduced form of that compound. Keep this in mind when doing a comparing off molecular formula versus empirical 2
example

## Empirical Formula Example 1 5m
Play a video:
we're not going to calculate the empirical formula forgiven compound now. Empirical formula itself can be calculated from the masses or percentages off elements within a compound. We're going to say this example question, it says determined the empirical formula off a compound that is 68.40% chromium and 31.60% oxygen. So step one, you're going to write down the symbols for each element in the question we're dealing with chromium and its symbol a CR and oxygen with its simple as just Oh, now, Step two. We're gonna write down the masses in grams of each element given notice here in this question that they're not giving us masses what they're giving us our percentages. So we're going to convert all the percentages into grams by assuming there are 100 g of the compound. And also, if you look at the percentages and you add them up, so 68.40% plus 31.60% you'll see that you get 100% of your total compound. This makes sense because if our compound is made up of a chromium and Onley oxygen together, they should add up to the complete compound. So we're going to convert 68.40% to 68. g of C R and 31 g of just up Step three. Convert all the masses into moles. Now, this requires some mole concept theory on your part. Remember, this is just a simple conversion. So we have 68 40 g of C. R. We want to get rid of grams of C. R. So we put grams of c are on the bottom. We put moles on top. We're gonna say one mole of CR. If you look at your periodic table, you'll see that one mole of CR weighs 51.996 grants. So grams here will cancel out. And here's the thing we're gonna say tow. Avoid rounding errors. Make sure the values have at least four decimal places. So plugging this into our calculator we get 1. moles of CR. So again, tow. Avoid rounding errors. It's best to have at least four decimal places. We converted grant of chromium into moles. Now let's do the same with grands of oxygen. 31.60 g of oxygen. We want to get rid of grands of oxygen, so place it on the bottom. We want to convert it into moles of oxygen. So put one mole of oxygen on top. Look at the periodic table. The mass of oxygen on the periodic table is 16. So here grams of oxygen cancel out. And when we do that, we're gonna get 1.9750 moles off oxygen. Next. This is important. Step four, You're going to divide each more answer by the smallest mole value in order to obtain whole numbers for each element. So here we have 1.3155 moles of CR and 19750 moles of O. The smaller number is 1.3155 So both will get divided by that number. No. 155 When we do that, we're gonna get one chromium Well, and then when we do that here, we're going to get 1.5 oxygen's. Okay, so we've run into an issue here. Step five. After dividing by the smallest number. If you get a value, that is 0.1 some number 0.1 and some number 0.9, then you can round to the nearest whole number. Unfortunately, we got 0.5 here. We can't just simply round that. So what do we do instead? Well, in these situations, if you can't round, we multiply by a factor to create whole numbers. So think to yourself 1.5. What number can I multiply that by To get a whole number. If we start out by multiplying by two, that's gonna give me 33 is the whole number, so we have three oxygen's. But here's the thing. If I multiply the number of oxygen's by two, then I'd have to multiply the number of chromium, also by two. So the empirical formula contains two chromium atoms and three oxygen atoms. So that would mean that the empirical formula here is CR two 03 So this would be my empirical formula for this question, and these are the steps you have to employ in order to get the empirical formula when given information on the percentages or mass of different elements within it,
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Problem

A compound that contains only carbon, hydrogen, and oxygen is composed of 48.64% C and 43.2% O by mass. What is the empirical formula of this compound?

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Problem

Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula of the compound?

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Problem

A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 1023 hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula? 