3.12 Chemical formulas from experimental data

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Do curso por University of Kentucky

Chemistry

560 classificações

University of Kentucky

Chemistry

560 classificações

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

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Week 3

Now that we know the structure of an atom, we can explore how atoms combine to form either molecular or ionic compounds. Then we will learn the rules of nomenclature that ensure that a compound is named according to IUPAC rules. We will end this unit by looking at quantitative relationships for compounds including the molar mass of and mass percent of an element in a compound.

3.01 Elements vs. Compounds 14:40

3.02 Types of Chemical Bonds 7:25

3.03 Chemical Formulas 7:37

3.04 Formula of ionic compounds 12:46

3.04a Formula of the compound formed from aluminum and phosphorus 1:19

3.04b Formula of the compound formed from ammonium and phosphate 1:27

3.05 Nomenclature 6:26

3.06 Nomenclature of Ionic Compounds 10:42

3.06a IUPAC Name for K2SO4 0:57

3.06b IUPAC Name for NiCO3 1:54

3.06c Formula for manganese(II) flouride 0:52

3.06d IUPAC Name for AgNO3 1:04

3.06e Formula for chromium(III) oxide 1:00

3.07 Nomenclature of Molecular Compounds 8:41

3.07a IUPAC Name for PCl3 1:21

3.07b IUPAC Name for N2O1:01

3.07c Formula for boron triflouride 0:40

3.08 Nomenclature of Acids 5:54

3.09 Molar Mass of Compounds 11:41

3.09a Molecular Mass of Al2(Cr2O7)31:47

3.09b Mass of a sample of vitamin C 2:39

3.09c Moles of sucrose within a sugar cube 2:03

3.10 Mass percent composition 8:55

3.10a Mass percent of chromium 1:53

3.10b Mass of CaCl2 containing chlorine 2:29

3.11 Relationships from Chemical Formulas 11:39

3.11a Mass of H in Ca(H2PO4)2 3:30

3.11b Mass of NH3 2:13

3.12 Chemical formulas from experimental data 9:46

3.12a Empirical formula of a compound containing sulfur and oxygen 2:13

Conheça os instrutores

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this module, we're going to look at how we determine the chemical formula of

a compound from experimental data.

By the end of this module,

you should actually be able to use this data to find the empirical formula.

Some experimental techniques actually give us a mass percent of

elements in a compound.

And we can use this information to determine empirical formula for

the compound.

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

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

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

we can actually figure out what this formula is.

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

formula of a compound.

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

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

We will go through an actual example for each step.

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

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

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

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

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

then we determine the subscripts.

Let's look at an example with phenol.

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

6.43% hydrogen, and 17% oxygen by mass.

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

convert to mass.

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

So, the first step is actually very easy.

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

So we can actually say that we have 76.57

grams of carbon, 6.43 grams of hydrogen,

and 17.00 grams of oxygen.

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

these compounds to find the moles of each compound.

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

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

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

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

Now I can do the calculation and

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

So, 76.57 divided by 12.01 is

equal to 6.38 moles of carbon.

Then I move on to the hydrogen, and 6.43 grams of hydrogen

divided by 1.008 grams of hydrogen gives me 6.38 moles of hydrogen.

And last, I'll look at my oxygen,

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

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

I've converted from mass to moles.

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

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

1.06 because that's the smallest value.

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

the elements in the compound.

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

hydrogen 6, oxygen 1.

Now we already have our subscripts for

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

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

some value in order to get whole number subscripts.

For example, if we had C2H4.5, we

would need to actually multiply through by 2 so that

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

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

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

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

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

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

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

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

9.50% hydrogen, and 45.23% oxygen by mass.

If we assume a 100 gram sample, what is the mass of carbon?

Remember that when we assume a 100 gram sample, that makes our

life much easier because we can convert the percentage directly into the mass.

So we have 45.27 grams of carbon.

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

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

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

So in this step,

you need to calculate the moles of carbon that are present in the sample.

Now that we know the grams of carbon,

we can convert it to moles by dividing by its molar mass of 12.01 grams per mole.

And what we find is that we have 3.769 moles of carbon in the sample.

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

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

9.42 moles of hydrogen, and 2.827 moles of oxygen.

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

we need to divide by the smallest number of moles.

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

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

Now we need to determine the actual subscripts.

Now we need to figure out what we need to multiply through in order to get rid of

those decimal values and have only whole number subscripts.

Now figure out what you need to multiply through by in order to get to

the right formula.

If we multiply through by 3, you should have gotten the formula C4H10O3.

This gives us the smallest whole number ratio we can get between those three

elements in this compound based on the data given in the problem.

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

is a empirical formula.

It's not the molecular formula.

We would have to have some additional information such as

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

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

was determined from experimental data.

And we find that the formula is C3H3O.

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

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

where we have 3 times 12.01 for carbon,

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

And what we find is we get approximately 55 amu.

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

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

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

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

Remember that for molecular compounds,

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

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

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

plus 6 times 1.008, plus 2 times 16.

And when we add these values up, and

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

So we can see that our final formula for

our compound has a molecular mass which matches that given in the problem.

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