0:14

Some experimental techniques actually give us a mass percent of

Â elements in a compound.

Â And we can use this information to determine empirical formula for

Â the compound.

Â Remember that an empirical formula is kind of like a reduced fraction.

Â It's the most simple ratio between the elements in a compound.

Â So if we have experimental data that shows us the mass percent,

Â we can actually figure out what this formula is.

Â 0:41

This is an overview of the steps we're going to take to determine the empirical

Â formula of a compound.

Â Because we're frequently starting with mass percent, the first thing we

Â usually have to do is assume 100 gram sample and convert our percents to mass.

Â We will go through an actual example for each step.

Â For now, we're just looking at the general overview.

Â 1:00

Second, we're going to convert the mass of an element to moles of an element.

Â So once we find the masses in step one, we're going to convert those to moles.

Â Then, we divide by the smallest number of moles because we're trying to

Â get the smallest whole number ratio between elements in a compound, and

Â then we determine the subscripts.

Â 1:22

Let's look at an example with phenol.

Â It's a disinfectant which has the composition of 76.57% carbon,

Â 6.43% hydrogen, and 17% oxygen by mass.

Â The first thing we need to do is we're going to assume a 100 gram sample and

Â convert to mass.

Â Then we're going to convert from mass to moles, divide, and determine subscripts.

Â So, the first step is actually very easy.

Â Because we're assuming 100%, we can simply turn those percents into masses.

Â 2:00

So we can actually say that we have 76.57

Â grams of carbon, 6.43 grams of hydrogen,

Â and 17.00 grams of oxygen.

Â Now that we have these in terms of grams, we can use the molar mass of each of

Â these compounds to find the moles of each compound.

Â Note that I'm putting the 12.01 grams on the bottom because I need grams of

Â carbon to cancel with grams of carbon, and I'm going to be left with moles of carbon.

Â For hydrogen, I do the same thing, 1.08 grams per mole of hydrogen.

Â And then finally with the oxygen of 16.00 grams per mole of oxygen.

Â 2:51

Now I can do the calculation and

Â find the actual moles that I would have in a 100 gram sample.

Â So, 76.57 divided by 12.01 is

Â equal to 6.38 moles of carbon.

Â 3:26

And last, I'll look at my oxygen,

Â 17 divided by 16, and I get 1.06 moles of oxygen.

Â So now, I've assumed my 100 gram sample and converted to mass.

Â I've converted from mass to moles.

Â Now I need to divide, and I'm going to divide by the least number of moles.

Â So I'm going to divide each of these numbers by

Â 1.06 because that's the smallest value.

Â Remember I'm just trying to find the smallest whole number ratio between

Â the elements in the compound.

Â So I have 6.38 divided by 1.06, and I end up with carbon 6,

Â hydrogen 6, oxygen 1.

Â Now we already have our subscripts for

Â that empirical formula because what we ended up with were nice whole numbers.

Â If we don't, then we have to look at multiplying throughout by

Â some value in order to get whole number subscripts.

Â For example, if we had C2H4.5, we

Â would need to actually multiply through by 2 so that

Â I can get rid of this fraction because we can't have a decimal in our subscript.

Â So if I multiply through by 2, I end up with C4H9.

Â Now I have my smallest, whole number ratio between the elements in the compound.

Â Generally, you will only multiply by a number such as 2, 3, or 4.

Â If you find yourself continuing to multiply to get to the smallest whole

Â number ratio, check your math and make sure you have the correct number of moles.

Â 5:12

Now let's let you try one of these problems.

Â Diethylene glycol, used as an antifreeze, has the composition 45.27% carbon,

Â 9.50% hydrogen, and 45.23% oxygen by mass.

Â 5:47

Now, if we do this for all elements, what we see is that we get 45.27 grams

Â of carbon, 9.50 grams of hydrogen, and 45.23 grams of oxygen.

Â So now we have the masses of each of our elements, now we need to convert to moles.

Â 6:28

Now, we can do the same thing for the hydrogen and

Â oxygen, and what we find is that we have 3.769 moles of carbon,

Â 9.42 moles of hydrogen, and 2.827 moles of oxygen.

Â Now, to find the lowest whole number ratio between those elements,

Â we need to divide by the smallest number of moles.

Â When we do that, we find that we get 1.33 for

Â carbon, two, 3.28 for hydrogen, and 1 for oxygen.

Â Now we need to determine the actual subscripts.

Â 7:40

Now, remember that the formula we find based on experimental data

Â is a empirical formula.

Â It's not the molecular formula.

Â We would have to have some additional information such as

Â the approximate molar mass to find the actual formula for the compound.

Â 7:55

So let's look at an example where we have the empirical formula that

Â was determined from experimental data.

Â And we find that the formula is C3H3O.

Â And we know that the molecular mass of the actual compound is approximately 110 amu.

Â So what I need to find is what I would to call the empirical formula mass,

Â where we have 3 times 12.01 for carbon,

Â plus 3 times 1.008 for hydrogen, and 1 times 16.00 for oxygen.

Â And what we find is we get approximately 55 amu.

Â Now that we know the empirical formula mass, we can simply see what the factor

Â is between the actual molecular mass and the empirical formula mass.

Â So I can take 110 divided by 55 and get a factor of 2.

Â What that tells me is that my actual chemical formula will be C6H6O2.

Â Remember that for molecular compounds,

Â we don't simplify down those subscripts because this is the actual number of

Â each type of atom that is present in a single molecule of the compound.

Â If we want to check our work, we can go back and say 6 times 12.01,

Â plus 6 times 1.008, plus 2 times 16.

Â And when we add these values up, and

Â what we find is that the molecular mass for this compound is 110.1 amu.

Â