Module 8b: Chemical Reactions

So from this module on,

we're going to be talking
about chemical reactions.

In this particular video,

we're going to introduce
what a chemical reaction is

and show you how
to balance them.

So chemical reactions
are the way that we--

you know, they're what we run
in a chemical laboratory.

They're what chemists,
most chemists, do every day.

They run chemical reactions.

They mix two
or more things together,

and they get a different product,

different chemical product,

out of that process.

And we have certain notation
that we use

when writing chemical reactions
so that we can all understand

what's going on
in the chemical reaction.

We always write
what we're putting

into the reaction on the left,

and we call those the reactants.

So in this particular reaction,

we're going to put in zinc,
which is actually that gray powder,

and iodine,

and remember, iodine
is one of those diatomics.

So it always is paired up
in twos.

So we will put--
and if you look at the purple,

it actually has little pairs
that are in that solid,

little pairs of iodine.

And so, zinc and iodine
are mixed together.

And the product of the reaction

between the two of them

is zinc iodide.

So we draw an arrow
to indicate that they reacted.

And then on the right side
of that arrow,

we draw what they became.

What is
the new chemical compounds

that came out of this reaction?

In this case, it's zinc iodide,
a brand new chemical react--

a brand new chemical compound,

where the two are linked,

and this is two ions,

zinc 2+ and I-.

Okay, so can you think
of a chemical reaction

that you encountered today?

What makes it
a chemical reaction?

What do you think?

So I drove in a car today.

So I know a chemical reaction
I encountered

was the combustion of fuel.

So the fuel in my car

reacted with oxygen in the air

in order to form CO2 and water.

And that went out the tailpipe,

and a whole lot of energy
that my car used to run.

The same kind
of combustion reaction,

a little different fuel source,

happened when I ate food.

Glucose in my body is burned up

and produces
a whole lot of energy,

and CO2 that I breathe out.

Some other types of reactions
we might see...

fireworks are reactions.

So this particular firework is--

the color is white.

So it is probably magnesium

reacting with oxygen
to make magnesium oxide.

And the different colors

of fireworks
are usually different metals.

So strontium
creates a red firework.

And that's pretty neat
to think about.

So in any chemical reaction,

one or more reactant

is converted to one
or more products.

So you have to have
different chemical species

on each side of the arrow.

That's what makes it
a chemical reaction.

So just boiling water
is not a chemical reaction,

because you have water
on one side,

and water on the other side.

So reactants to products.

Okay, so,
chemical reaction is a process

in which one or more substances
is changed

into one or more new substances.

We use a chemical equation

to write
down a chemical reaction.

So the chemical reaction
is what actually takes place.

The chemical equation
are the chemical symbols

that we use
to show the chemical reaction.

And we have a number of ways

of representing
a chemical reaction.

So this first way,

when you see the actual pictures
of the molecules,

that's kind of what you
should be thinking of in your brain.

So that's the picture
that you should think about

in your brain when you think
of a reaction.

So this is the reaction

of hydrogen and oxygen.

Both of those are diatomics,
right?

Remember, they always pair up,
hydrogen and oxygen,

and they form water.

So it takes two hydrogens,

two hydrogen molecules,

and one oxygen molecule

to form two water molecules.

So that's kind of what
you should be thinking of.

We could write this out in words.

That's the second line.

No, let me underline
that so I can show you.

So, we could write it out
in words.

This would be our written
in words.

The form that we most often use,
however,

is this last one,

where we wrote it out in symbols.

And this means that

two hydrogen molecules reacted,

that's the reacted,
with one oxygen molecule,

so we don't write the one,
it's implied,

to form,

so this arrow means to form,

or reacted.

And...

it formed two water molecules.

So, this two tells us how many,

and then this is the molecule
that formed.

Okay, and we know
that we always have reactants

on the left,
products on the right.

Now we want to look at how
to read chemical equations.

So when we see
a chemical equation,

you kind of know the picture
in your mind,

I want you to think about,
How can we read that?

So what do the coefficients mean?

By coefficients, I mean the numbers
in front of the atoms.

And this reaction is
that first reaction

that you saw in the firework,
the magnesium

reacting with oxygen

to form magnesium oxide.

And so what do those numbers,

the two in front
of the magnesium,

the implied one
in front of that O2,

and the two in front
of magnesium oxide,

what do those mean?

How can we read them?

Think about that for a minute.

Okay, so we can read them.

Hopefully, you said, Oh, well,
that's the number of molecules.

So we would say--

or the number of units.

So we might say two atoms...

of magnesium...

react with one molecule...

of O2...

to form...

two molecules...

of magnesium oxide.

And, you might see
that this was--

we might draw this in a picture,

so we'd have two atoms
of magnesium,

because those are just circles,

and then one molecule of oxygen.

And we would get two...

molecules...

of magnesium oxide.

And so that might be one way
we'd think of it.

There is another way
that we can actually write this.

We can think of it in terms
of this was an individual unit.

So I was thinking of it
in terms of individual units.

I can also write this
in terms of moles.

And this is actually more useful
because in a lab,

we can, remember,
see a mole of stuff.

We can see a mole of magnesium,

we can see a mole of oxygen.

We can't see two atoms
of magnesium,

or one molecule of oxygen.

So if we write this
in terms of moles,

then we can say two moles...

of magnesium...

plus one mole of oxygen...

forms...

two moles...

of magnesium oxide.

And that would be another way
to read it.

So we can read it in terms
of individual units,

or in terms of moles.

The one way
that we can't read it,

we cannot say two grams
of magnesium

plus one gram of oxygen

forms...

two grams of magnesium oxide.

That way does not work;

do not use grams.

It only applies
to individual units or moles,

[indistinct] not mass.

This brings us
to the Law of Conservation of Matter.

This law was observed by--
first observed

by Lavoisier in the 1700s

and early 1800s.

And he observed that mass

is neither created or destroyed

in chemical reactions.

And so we're going to prove that

to ourselves here
with this reaction.

So we have carbon monoxide

and lead oxide reacting

to give us carbon dioxide
plus lead.

And we know that one way
that we can write this

is using the mole fraction.

So we have a one
in front of each of those...

those compounds.

And so we can write this
as one mole of CO

plus one mole of lead oxide

gives us one mole of CO2

plus one mole of lead.

And we can calculate the mass
of each of those.

And so let's go ahead
and do that.

We want to compare the weight
of the total weight

of the reactants
with the total weight in grams.

So, actually mass,
that should say,

the mass of the products,

and see what we get.

Go ahead and try it
and then come back,

we'll do it on the--
we'll do it all together.

Okay, so the first thing
we need to do,

in order to get this,
is get the molar mass

of each of these compounds.

So one mole of CO, let's see,
is going to weigh...

we have one carbon, 12.01;

we have one oxygen, 16.00.

The one mole of CO

is going to weigh 28.01 grams.

And then one mole of lead oxide.

Lead is a heavy metal.

So we have 207.2

for lead,

and 16, for oxygen.

And so,

one mole of lead oxide here

is going to weigh

223.2 grams.

And then, let's see,
one mole of CO2

is going to weigh...

we have one carbon, that's 12.01,

and we have two oxygens,
that's 32.00.

And so we get 44.01 grams

for one mole of CO2,

and then lead by itself is 207.2.

Okay, so now we want to see,

how does the weight
in grams of the reactants

compare to the products?

So let's go down here

and see reactants.

We're going to have 28.01

plus 223.2.

Let's add those up,

we get...

251.2 grams for the reactants.

And the products....

What are we expecting?

Hopefully,
you're expecting 251.2, right?

Because matter is neither

created nor destroyed.

Add these up to--

I'm ignoring that last digit
because of the sigfigs.

251.2. Yay!

Mass is conserved.

So the weight of the reactants

equals the weight of the mass
of the products

in grams.

So we do have the matter
is conserved,

the Law of Conservation of Matter
is upheld.

So that's good information.

Okay, so, knowing

that the Law of Conservation
of Matter exists

this means that

we have to balance equations.

It means that if there is--

if there are two aluminum atoms
on the reactant side,

there has to be two aluminum atoms
on the product side.

If there are six oxygens
on the reactant side,

there has to be six oxygens
on the product side.

And so, because of the Law
of Conservation of Matter,

we have to balance equations.

So we are going to learn now how

to balance chemical equations.

Here is a chemical reaction,

we have a picture of it.

Oftentimes, when you're trying
to balance a reaction,

you'll get this reaction
without coefficients,

and you have to figure
out what the coefficients are.

Okay, so we know

we're creating Al2Br6.

So in order to balance this,

we would need two aluminums here,

and three bromine molecules,

because we have--
they come in pairs.

You see,
this two tells us they come in pairs.

We can see that
from the picture, too.

So to get six of them,
we only need three of those.

And then,
that would give us one molecule.

And I think
I have these written in,

so I'm going to
actually erase these

so that it doesn't
pop in on top of them, yep.

Okay,

so that is
a balanced chemical equation.

So now I'm going to show you
a method

that you can use

to balance more complicated
chemical equations,

and that you can always use
to balance equations.

Before we get to that method,

we're going to look at--
this is just another example

of a chemical reaction,

and how it would be written.

You can see
that in this reaction,

there are two iron atoms

that must react
with three chlorine molecules

to give us
two iron chloride units.

And so we can see here that

on the reactant side,

we have two irons

and six chlorines.

And on the product side,

we have two irons,

and we have two--
because we have two iron chlorides,

and each
of those have three chlorines,

we have six chlorines.

So this reaction is balanced.

Also note that in reactions,

we often indicate
what phase things are in.

And by phase,
I mean what state of matter.

And you'll see that this--
that's what this is.

So this means
that we react solid iron

with gaseous chlorine

to give us a solid product.

And so that's what
those little things in parentheses.

So you'll see s equals solid,

l equals liquid,

and g equals gas.

And then there's one more
that you'll find

that we'll encounter,

aq equals

aqueous solution.

So a solution where water

is the solvent.

Okay.

Okay, so let's start balancing
those chemical equations.

The first thing
that you're going to do

is write the correct formulas

for the reactants
on the left side,

the correct formulas for the products
on the right side.

Sometimes you'll get
a reaction in words,

and you will
have to write the formulas

from the names.

So if we said,
methane reacts with oxygen

to form carbon dioxide
and water--

Methane isn't one
I would expect you to know

on a test, however.

But the others you should know.

You would write this:

Methane
and oxygen are the reactants,

and CO2
and H2O are the products.

Okay, that's the first step.

Second step:
Change the numbers

in front of the formulas--

we call those coefficients,

--to make the number of atoms
of each element the same

on both sides of the equation.

DO NOT CHANGE the subscripts!

So this.

So if you want to put a two
in front of the methane,

you just put a two there,

don't change it to C2H8,

don't mess with the subscripts.

Once you've written the formulas,
they stay.

That's little thinker buddy
getting mad

if you mess
with the subscripts.

Okay.

Start by balancing the elements
that appear

in only one reactant
and one product.

So here, that would mean

I don't want to start
with oxygen.

Do you see how oxygen is
in both of the products?

That just makes things
more complicated.

So I want to make sure that I start
with either the carbon

or the hydrogen;

it wouldn't matter which one
I started with,

but one of those.

Okay.

Here,
I have one carbon on the left

and one carbon on the right.

I don't have to balance them,
they're already balanced.

So I'm going to leave them
the way they are.

Okay, so now,

I'm going to look
at the hydrogens.

I have four hydrogens on the left

and two hydrogens on the right.

How can I balance this?

Well, I can have two waters.

So I now have a reaction
with the two

in front of the waters.

Okay.

Then,
once I've balanced everything

that only appears once,
then I can go through

and balance the elements
that appear

in two or more reactants
or products.

So here we have this reaction

that we started to balance.

Now we can look at the oxygen.

So we have two oxygen
on the left,

two oxygen from CO2,

plus two oxygen from water.

We have four oxygens
on the right.

Okay, so we need to put a two
in front of O2

in order to balance this.

Go.... Okay.

Then at the end, we want to check

to make sure that--
double check that it's all balanced.

Once you think it's balanced,
double check it.

So we have our reaction
we think is balanced.

Now let's go through
and check it.

Okay, we have one carbon
and one carbon, yes.

Four hydrogens on the left,

four hydrogens on the right, yes.

And four oxygens on the left,

and four oxygens on the right.

Yep, it is balanced;
we are finished.

That is our answer.

Okay.

Okay, let's try this example.

I'm going to show you how I usually go
through these problems.

I'm going to rewrite this
to give myself

a little more space here,

because it's kind of squished
together.

And...

I always write

the number of atoms

that I have on each side

before I start.

So I say,
Okay, I have gold and chlorine,

and I have gold and chlorine
in this reaction.

Over on the reactant side,

I have one gold
and two chlorine right now.

And over on the product side,

I have one gold
and three chlorine.

So this gives me--
and I keep a running tally

so I know what I need
to balance still

and what I don't.

So this, just by looking at this,
I can see,

okay, gold's already balanced,

I need to balance chlorine.

Okay, how can I balance chlorine.

I have three and two,

I could put a two here
and a three here

to give me six on each side.

So I then have six chlorine
on each side.

However,
that also changes the number

of gold that I have.

So I have two gold on this side.

And now gold
is not balanced anymore.

So I have to go back,
and I have to put a two here

to balance the gold,

and that gives me two gold.

And so now
the whole thing is balanced.

I go back
and just double check to make sure.

But these--
this is the end result.

This is my balanced reaction.

Okay, let's try this one.

Go ahead with the same procedure

that I did before

and see if you can balance it
that way.

Okay, I am going to rewrite
this again

to give myself some space
to write

because there isn't a lot
between the atoms

for those coefficients here.

Okay, so on each side,

I have carbon, I have hydrogen,

I have nitrogen,
and I have oxygen.

Carbon, hydrogen, nitrogen,
and oxygen.

Okay, let's count
how many we have of each.

Over here I have 6 carbon,

and I have 12 plus 2,

14 hydrogen,

4 nitrogen,

and 1 oxygen, okay?

On this side, I have one carbon,

five hydrogen,

one nitrogen,

and one oxygen.

I know that I don't want to start

by balancing hydrogen.

Hydrogen is in everything.

So I'm going to avoid that one

and wait for the hydrogen
until last.

Let's first look at the carbon
and see what we can do.

We have six on the left.

Let's get six on the right.

Okay, so that will change the number
of carbons that I have.

That will change the number
of hydrogens that I have.

And it will change the number
of oxygens that I have.

So I now have

six carbons.

I have 12 hydrogens plus 3,

that is 15.

And I have six oxygen.

Okay.

So those are balanced.

Let's go then--
we're skipping the hydrogen,

let's go to the nitrogen.

We have four on this side,

and we have one
on the other side.

Well, let's go

and put a four here,

and see what happens.

So now that changes nitrogen,

that also changes hydrogen.

So,

we now have four nitrogens.

But we have 12 hydrogens

plus 12 hydrogens,

we have 24 hydrogens.

Okay, so now we have oxygen left
to balance

before we get to that hydrogen

that we want to save for last.

We have six on the right
and only one on the left.

So we want to put a six here.

And let's--
that will change oxygen,

and it will change the amount
of hydrogen that we have.

So we would then have 6 hydrogen,

and we would have 12--

or 6 oxygen, I'm sorry.

And we would have 12 hydrogen

plus another 12 hydrogen

to give us 24 hydrogens.

And so then I just go back
and double check

that all of those are correct.

And it looks like every one
is balanced.

That is
our balanced chemical reaction.

There is a one
in front of the first one,

and then a six, six, four.

Okay, let's try--

let's describe this one in words.

How would you do that?

I would say you can use,
remember,

you can use individual units.

So I could say two

gold atoms...

react...

with three...

chlorine...

molecules...

to form...

two...

gold--

and we do need
a Roman numeral here,

oops,

gold (III) chloride...

molecules.

So, okay.

That's one way we could put it.

Remember, we could--
you could have also answered

that you wanted to describe it
in terms of moles,

and that's much more common,

that you would have two moles...

...gold.

And I'm not going to write
the words out this time,

just to make it a little quicker,

plus three moles...

of chlorine...

to form...

two moles...

gold (III)...

chloride.

And so that would be another way

that you could write
that reaction,

either in individual units
or in terms of moles.

Okay,
here is another example to try,

our last example.

Let's go ahead
and balance this one.

And I'm again
going to rewrite this,

just to give myself some space.

Okay, so I have carbon, hydrogen,

and oxygen on each side,

Carbon, hydrogen, and oxygen,
let's count them.

I have 4 carbons here, 10 hydrogens,
and 2 oxygens.

And on the product side,
I have one carbon,

two hydrogens, and three oxygens.

So I know I want to wait
to balance the oxygens

until the last bit
because there--

it is in--
because oxygen is in both products.

So I have four carbons
on the reactant side.

I want to have four
on the product side.

I'm going to put a four there
in front of the CO2.

That changes carbon to four.

That will also change
the number of oxygen.

So I have eight plus one is nine.

And then I have 10 hydrogen.

So, they come in pairs,

so I only actually want five
of those pairs.

So, I will have 10 hydrogen
over here.

And that gives me, let's see,

4 is 8,

plus 5 is 13.

This is where
it gets a little tricky.

So, with these kind of--

this is a combustion reaction.

And anytime you have a molecule

reacting with oxygen

to give the oxides
of that molecule--

so the oxide
of carbon is carbon dioxide,

and the oxide
of hydrogen is water, H2O.

So anytime you have

a carbon-hydrogen compound

reacting with 02
to give CO2 and water,

that's a combustion reaction.

And oftentimes you will get
this odd number of oxygens,

but the oxygens
only come in pairs,

which is really frustrating.

Because what happens?

What do you do?

And sometimes you will see
that this is the only time

that you will ever be able

to use fractions in--

in reaction coefficients.

You will see
that some books allow you

to do 13/2 here,

so that you can get 13 atoms,

even when
they come only in pairs.

And that is one way
to answer this problem.

The other way to do this,
when you come across this,

and it will always be
this oxygen right here

that this happens to.

The other way to answer this

is to multiply everything--
to write the 13/2,

and then
multiply everything by 2.

So if I multiplied everything
by two,

I would end up

with 2C4H10,

plus 13O2,

gives me 8CO2,

plus 10H2O.

And if you add up the number
of atoms on each side,

you'll still see
that they are balanced,

but this is the smallest
whole number answer

that you can get
in this particular case.