3.09 Molar Mass of Compounds

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From the course by University of Kentucky

Chemistry

599 ratings

University of Kentucky

Chemistry

599 ratings

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.

From the lesson

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

Meet the Instructors

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

In this module, we're going to learn how to calculate the molar mass of compounds.

Our objective is to calculate both the molecular mass and

the molar mass of compounds.

So when we first started talking about elements,

we talked about the atomic mass with the mass of a single atom of an element.

And we looked at this in units of amu, or atomic mass units.

We're going to do something similar for looking at compounds, and

we're going to look at individual molecules, and we're looking at

this as the molecular mass, where it's the mass of a single unit of a compound.

We also see it called formula mass or

molecular weight and a few other things, but

they all essentially mean the same thing, the mass of a single unit of a compound.

We will also look at this in terms of amu, or atomic mass units.

And just like with atomic mass, we get the relevant values from the periodic table.

When I look at exam, at an example such as carbon monoxide,

we see we have one carbon atom and one oxygen atom.

We see that a carbon atom has a mass of 12.01 amu and

oxygen has a mass of 16.00 amu.

We simply add these values together and

get a total mass of 28.01 amu for each unit of carbon monoxide.

Now we're going to look at an example with magnesium chloride,

which is frequently used to melt ice on roads.

Magnesium chloride has a subscript other than 1.

With our carbon monoxide example, we had one atom of carbon and one atom of oxygen.

Now we have two atoms of chlorine in the unit, so

we have to take that into consideration when we're determining the molecular mass.

So to solve this problem, we take 1 times the atomic mass of magnesium plus 2

times the atomic mass of chlorine.

We still get these numbers directly from the periodic table.

For magnesium, the value is 24.31 amu, and for chlorine, it's 35.45 amu.

We can simply do the math and add up our results, and find that the molecular mass

of magnesium chloride is 95.21 amu, or atomic mass units.

Another example.

Here we have three elements in H2SO4, or sulfuric acid.

We have two hydrogens, so we have 2 times 1.008,

which is the atomic mass of each hydrogen,

1 times the atomic mass of sulfur, and 4 times the atomic mass of oxygen.

We add them up and we find the molecular mass of H2SO4 is 98.09 amu.

Note that we had to round up from the 98.086 so

that we had the correct number of significant figures.

Since we were adding values,

we had to look at the values with the least number of decimal places.

Now we can find the molecular mass of a compound with the formula C4H9OH.

Note the molecular mass is 74.12,

which we found from looking at four carbons, ten hydrogens, and one oxygen.

However, note that it doesn't tell us anything about the structure.

All of the compounds listed below represent C4H9OH.

We have carbons at the end and at the intersections of each of these.

And there are enough hydrogens to make four bonds around each carbon.

And so, if we look at this for

all four of these, they all have exactly the same molecular mass.

So while it's an important tool, and especially important when we're doing,

measuring out samples, it doesn't necessarily tell us

any more information about the identity of that particular compound.

Let's look at an example to find the molecular mass of ammonium dichromate.

The molecular mass of ammonium dichromate is 252.08 amu.

Remember from when we were talking about polyatomic ions and we had the number

outside parentheses that it belonged to everything inside the parentheses.

So in this unit, we have two nitrogen atoms, we have eight hydrogen atoms,

because we have two ammoniums, each which contain four hydrogens.

We have two chromium atoms and seven oxygen atoms.

So we can look up the atomic mass for each of those elements,

multiply it by the appropriate value, and then sum up our result to find 252.08 amu.

Now we want to switch from molecular mass,

where we're looking at a single molecule of a substance, to the molar mass,

where we're looking at one mole of the substance.

In this case, we're looking at water.

If I look at the molecular mass of water, what I find is it has a mass of 18.02 amu.

That's for one single molecule of water.

When I look at a bowl of water, where I have 6.022 times 10 to the 23rd molecules

of water, it now has a mass of 18.02 grams per mole.

Note that the number stays the same, but that the units are different,

depending on whether I'm looking at the mass of a single molecule or

the mass of a mole of molecules, or 6.022 times 10 to the 23rd molecules.

Once we calculate the molar mass, we can use it to do calculations.

Much like we use many other conversion factors, we can use this relationship to

understand the connection between moles of a sample and the mass of a sample.

Now, unlike conversion factors like 12 inches in a foot or

5,280 feet in a mile, which are fixed regardless of what we're measuring,

the molar mass is going to vary depending on the identity of the substance.

For water, the relationship is 18.02 grams per mole.

For other substances, the value will be different.

What we always see is that the molar mass has units of grams per mole.

Just like our other conversion factors,

we can invert this value depending on the calculation we're trying to do.

So if we're looking at water, we know water has a molar mass

of 18.02 grams per mole, but if I'm trying to get rid of grams and

I need that on the bottom of my calculation, I can also write this

relationship as 1 mole over 18.02 grams.

Both of these are correct, and the one we use in our calculation depends on

what we know and what we're trying to find.

Let's look at an example.

How many moles are in 34.2 grams of water?

In this problem, we're given 34.2 grams of water as our starting point.

Since we're dealing with water, we've already calculated the molar mass, but

if you didn't know that, you could use the periodic table to

find the individual masses and find the molar mass as a compound.

So we have 18.02 grams per mole.

Now I need to remember that I'm trying to get rid of grams and

convert to moles, so I need to have grams on the bottom.

So I have 18.02 grams per mole of water.

Now that I've seen that my units cancel out correctly and

that the units remaining are moles of water, I can do the actual calculation.

So I have 34.2 divided by 18.02,

and I find that I have 1.90 moles of H20.

Note that I've rounded this to three significant figures because the number

given in the problem has the least number of significant figures, which is 34.2.

Our value of molar mass will be considered when determining the number of

significant for our answer.

Typically, we will have four or five digits in our molar mass.

So usually, it does not affect the significant figures, but

we must look at it and use that in our determination.

Let's look at another example where we don't know the molar mass before we start.

How many grams of C3H8, which is the chemical formula for

propane, are present in 4.32 moles of C3H8?

Now, I know that I'm starting with moles, my 4.32 moles of C3H8,

and I see that I'm going to need to find the molar mass.

So I can take that I have 3 carbons times the mass of carbon,

plus 8 times my mass of hydrogen.

And I can add those values up and find the molar mass of my compound.

So, 3 times 12.01.

Plus 8 times 1.008.

And I end up with 44.09 grams per mole, and this is for propane.

So now, when I look at my problem, I see I have 4.32 moles.

I'm trying to get to grams.

Since moles is on the top and

I want to get rid of that, it's going to have to go on the bottom of the next step.

So I'm going to leave the moles of propane on the bottom, and

I'm going to put the 44.09 grams on the top.

Remember, the number always belongs with the grams, so

I'm going to have to leave it, the right units with the right number.

Now, moles cancels with moles.

I see that my units are working out correctly.

When I solve for my answer, I'm going to have grams of C3H8.

So now, I can simply do the math where I have 4.32

moles of propane times 44.09 grams per mole.

And I end up with 190 grams of propane.

Here we need three sig figs in our answer because 4.32 has three sig figs, so

I put a decimal point at the end to indicate that this 0 is significant.

Alternatively, I could write this in scientific notation, so I have the same

numerical value and it's clear the number of significant figures is three.

Let's look at an example for you to try.

How many moles of caffeine are found in a 50 gram sample?

The formula for caffeine is given, as is the molar mass,

although you could find that if you needed to.

The correct answer is 0.257 moles.

Note that we have a 50 gram sample of our caffeine.

We know the molar mass is 194.19 grams per mole of caffeine.

And I put the grams on the bottom because I was trying to get rid of grams, so

I need them to cancel out, and I want to find moles.

This will allow me to calculate the moles and find that it's 0.257 moles.

Because 50 has three sig figs in it, I need three sig figs in my answer.

In the next module, we're going to look at how we find the percent composition of

elements in a compound.

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