Explore
For Enterprise
Join for Free
Log In

Arts and Humanities
Business
Computer Science
Data Science
Information Technology
Health
Math and Logic
Personal Development
Physical Science and Engineering
Social Sciences
Language Learning
Degrees
Certificates

Explore all of Coursera

Browse
Search
For Enterprise
Log In
Join for Free

3.11 Relationships from Chemical Formulas

Loading...

Chemistry

University of Kentucky

4.7 (652 ratings) | 66K Students Enrolled

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.

Reviews

4.7 (652 ratings)

5 stars
489 ratings
4 stars
134 ratings
3 stars
14 ratings
2 stars
8 ratings
1 star
7 ratings

TM

Jun 21, 2017

This is a great course! Very well explained! It's managing to keep my attention on 100% and that's not an easy task for anybody. Thank you for sharing your knowledge with us!

M

Dec 30, 2016

I enjoyed every minute of this course and I have learnt so much. Thank you Coursera for this opportunity. I'll definitely recommend this course to my friends and colleagues.

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

Taught By

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

Transcript

In this module, we're going to

look at the relationships we find in chemical formulas.

By the end of this module, you should be able to write relationships based on

the subscripts and the formulas of compounds.

When I look at a molecule of water, what I see is that each molecule contains two

atoms of hydrogen and one atom of oxygen.

Much like if I disassemble a bike, I'm left with two wheels and one frame.

However, looking at individual molecules isn't always the most convenient way to

look at relationships in compounds.

Sometimes we need to look at it from much larger samples.

The same ratio that applies at the molecular and

atomic level, also applies when we have a lot of those molecules.

Let's look at what happens when we have one mole of water molecules.

Remember, this just means that we have 6.022 times 10 to

the 23rd molecules of water.

If I go from having 1 molecule of water to having 1 mole of water molecules,

it's just a counting number.

I'm just multiplying the number of molecules by Avogadro's number.

Therefore, if I have 1 mole of water,

I can say it contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Just like if I have a bike or

I have a mole of bikes, I have two moles of wheels and one mole of frames.

Once we've seen what makes up each molecule we can

then write relationships between moles of atoms and moles of the compound.

These relationships will be useful in calculations.

We've already seen that we can say 1 mol is equal to 6.022 times 10

to the 23rd atoms.

Remember, a mol is just a counting number, but

now we can also say that we have 1 mol of oxygen atoms for

every 1 mol of water, or 2 mols of hydrogen atoms for every mol of water.

And just like with our other conversion factors and

relationships, these don't always have to be used in the exact same way.

Depending on the information we're given and the information we are trying to

find we could invert these values to get to the units we need for our answer.

And it's perfectly okay to say that one mol of water contains one mol of

oxygen atoms.

Just as we said one mol of oxygen atoms is in one mol of water.

Remember, any relationship that we set up as fraction we can invert that as long as

we keep the right numbers and units together.

Let's look at an example of how we would use this information.

Here in this problem we're given grams of aluminum chloride and

we're looking to find how many chlorine atoms we have.

So, we know we're going to have to start with a 25.1 grams of

aluminum chloride because it's the only number we're given in the problem.

Now, I can't go from grams directly to atoms.

We have to go through mols.

And this will be a common theme both looking at compounds, but also when we

start looking at reactions, that the first thing we do is we convert to mols.

So in order to get from grams of aluminum chloride to moles of aluminum chloride,

I'm first going to have to know the molar mass.

So I see I have one aluminum, which has a mass of 27.00,

and I have three chlorines, each with a mass of 35.45.

Now I can find the molar mass by simply adding these values together.

So 27 plus 3 times 35.45 equals 133.35,

and remember this is a molar mass, so

it has units of grams per mol of aluminum chloride.

And notice that where I started my problem, I have grams in the numerator.

So I'm going to have to have grams in the denominator of the next step.

So I have 133.35 grams per mol.

Now my grams will cancel and this gets me to mols of aluminum chloride, but

now I need to make the jump from mols of aluminum chloride to mols chlorine atoms.

And I can do this using the subscript in the chemical formula.

So now I can say that for

every 1 mol of aluminum chloride I would get 3 mols of chlorine.

Remember I'm just breaking that aluminum chloride apart into it's pieces and

I get 3 chlorines for every 1 aluminum chloride.

Now, mols of aluminum chloride cancel out, and I'm left with mols of chlorine.

Because it says how many chlorine atoms,

we don't want to leave that in terms of mols.

We want to find out how, many atoms we actually have.

So now, I can use Avogadro's number.

Because mols is on the top in this step here,

I'm going to need mols on the bottom in the next step.

So one mol is 6.022 times 10 to the 23rd atoms of chlorine.

Note that whatever I put on the top is always going to

be equal to what is on the bottom.

In this case, it was 3 mols of chlorine to 1 mol of AlCl3.

There is a relationship between those two values.

Just like there's a relationship between the number of atoms,

Avogadro's number of atoms, and one mol.

Now, my mols of chlorine cancel, now I can actually solve my problem.

Now, I can multiply 25.1 times 3 times 6.022 times 10 to the 23rd,

and divide by 133.35.

And when I do, I get 3.40 times 10 to

the 23rd, and this is atoms of chlorine.

So even though we have multiple steps in the problem the thing that kept me

moving was looking at the units.

I converted to mols from grams, then I used my mol ratio between the mols of

the compound and the mols of the chlorine atoms in that compound.

And then I used Avogadro's number,

atoms per mol, to find the number of atoms of chlorine.

Now let's look at a slightly different problem.

If there are 4.7 times 10 to the 25th atoms of phosphorus in a sample of

potassium phosphate, then what is the mass of potassium phosphate?

Notice that the formula for the potassium phosphate is not given.

So that's going to be that first thing we have to figure out.

So what is the formula for potassium phosphate?

What we find is that it's K3PO4.

Remember that potassium in an ionic compound has a plus 1 charge.

Phosphate is a polyatomic ion that has a charge of 3 minus.

In order for there to be a balance of charge I have to have K3PO4.

Now we know the formula for potassium phosphate but

now we need to find the molar mass of potassium phosphate.

So just like I've done before I'm going to take 3 times the mass of

the potassium which is 39.10 times 1,

times the mass of the phosphorous which is 30.97.

And then 4 times the mass of the oxygen's and when I add that up

together I'm going to have the molar mass of my potassium phosphate.

I find that I have 212.27 grams per mole of potassium phosphate.

Now, we can use both the formula and the molar mass to figure out the mass of

potassium phosphate that contains 4.7 times 10 to the 25th atoms.

This is a little different than the example we just did, but

remember, we want to look at our units and look at what we're given.

Because we are given 4.7 times 10 to the 25th atoms, and

that's the only number in the problem, that's where we need to start.

And just like on the previous problem, our first goal is to convert to mols,

because then we can look at the relationship between the mols of

potassium in the compound and the mols of the compound.

What we find is that we get 1.7 times 10 to the 4th grams.

So for this problem we want to start with our number of atoms,

4.7 times 10 to the 25th atoms.

And the first thing I'm going to use is Avogadro's number, or

6.022 times 10 to the 23rd atoms, per mol.

And we're here, we're talking about our phosphorus.

Now, I want to use my ratio between mols of phosphorus and mols of the compound.

And the formula was K3PO4, so I know that for

every one mol of phosphorus, I had to have had one mol of K3PO4.

And now I can use my molar mass to get from moles of phos,

potassium phosphate, to grams of potassium phosphate.

And again, what I'm doing is looking at my units.

Because what I see is that atoms cancel with atoms,

mols of phosphorus cancel with mols of phosphorus,

moles of potassium phosphate cancel with mols of potassium phosphate.

Now the units I'm left with are grams of the potassium phosphate, so

now I just need to do my calculation.

So I take 4.7 times 10 to the 25th, times 212.27,

and divide it by 6.022 times 10 to the 23rd.

Because all of my ratios here were with a value of one, so

they're not effecting the mathematical answer.

When I do that I get 16567 grams, but I notice that I

have five digits here and I can only have two significant figures in my answer.

So I need to round this and I'm going to write it in scientific notation.

So there's no question of how many numbers I have in my answer.

So I write this as 1.7 times 10 to the 4th grams of potassium phosphate.

And that mass of potassium phosphate contains 4.7 times 10 to

the 25th atoms of phosphorus.

So as we've gone through doing a variety of calculations,

we've talked about different ways we can relate values to one another.

We've looked at things like our prefix meanings from gram to kilogram,

for example.

We've also done milligram and nanogram.

We can use the density to help us get between grams and millilitres or

centimeters cubed, remember that 1 millilitre equals 1 centimeter cubed.

We've used mols and Avogadro's number to find out how many of something we had.

And we've used molar mass to get between grams and mols.

Now, we have another tool that we can add to our toolbox, which is the,

relating the number of molecules to the number of atoms.

And this can be in terms of individuals molecules, or

in terms of mols of molecules and mols of atoms.

Next we're going to look at how we determine chemical formulas from

the experimental data.

Explore our Catalog

Join for free and get personalized recommendations, updates and offers.

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.

© 2019 Coursera Inc. All rights reserved.

Download on the App Store

Get it on Google Play

Coursera
About
Leadership
Careers
Catalog
Certificates
MasterTrack™ Certificates
Degrees
For Enterprise
For Government

Community
Learners
Partners
Developers
Beta Testers
Translators

Connect
Blog
Facebook
LinkedIn
Twitter
Instagram
YouTube
Tech Blog

More
Terms
Privacy
Help
Accessibility
Press
Contact
Directory
Affiliates