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Chapter 7.1a Writing and Balancing Chemical Equations

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This content explains how to write and balance chemical equations, which represent chemical reactions by showing the identities and relative quantities of reactants and products.

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hi everyone my name is Dr Michelle

banogan and I'm going to be doing the

videos for this section we are talking

about chemical reactions and it's quite

exciting because so far in this semester

we've learned about atoms and molecules

and now we're ready to put them all

together and have some reactions so in

this particular video we are learning

about how to write and balance chemical

equations our learning goals are to

derive chemical equations from narrative

descriptions of chemical reactions and

then also to learn to write and balance

chemical equations in the molecular format

so when we write chemical equations

we're going to find that they need to be

balanced and we'll talk about what that

means later but these reactions will

represent the identities and the

relative quantities of the substances

involved right the substances undergoing

the chemical or physical change so we

have a chemical reaction written here we

can see that the symbols for our

molecules and our atoms are written so

we've got

methane CH4 reacting with oxygen O2

and these two are our reactants they

occur on the left hand side of the equation

equation

on the right hand side we have our

products those substances that are

formed in our reaction so these include

carbon dioxide CO2 and also water H2O as

our products

we can also see that there are what we

call coefficients and coefficients are

the numbers in front of the different

chemical species so we see on the left

hand side for our reactant O2 we have a

2 in front of it which means that we

need to have two oxygen molecules in

order to react with one methane and so

we can see this shown by these space

filling models for the molecules right

here the black sphere is carbon and the

four white spheres are hydrogen so this

is a methane molecule and it would need

to react with two oxygen molecules right

so here the red spheres represent oxygen

atoms so together we have 202 molecules

and we see that we also have a

coefficient on our product side the

product water will be generated to give

us two water molecules for every one

molecule of carbon dioxide that's formed

so we learned a few new important

vocabulary words here for our chemical

reactions we have substances that are on

the left hand side of our chemical

equation and we call these our reactants

these are the substances undergoing a reaction

reaction

on the right hand side of our chemical

equation we discussed that these are our

products these are the substances that

are generated by the reaction

and on both sides the substances are

separated by plus signs so we see this

on the reactant and the product side and

then between them we have an arrow right

the arrow separates the reactant side

from the product side

and then we also discussed that there

are coefficients we see that there are

twos right in front of the oxygen and

Waters there's no coefficient in front

of methane or CO2 and that's because

it's implied that there would be a

coefficient of one so typically we omit

the one and we do not write it out and

we're going to talk more about balancing

chemical equations and that will involve

trying to figure out the correct

coefficients in our chemical equations

but we need to know how to interpret

what these coefficients mean so a few

things about coefficients we need to

make sure that we use the smallest

possible whole number coefficients in

our chemical equation

and that simply means that if all of our

coefficients are say a multiple of two

then I could divide them all by two and

that would give me the smallest possible

coefficients if all of my coefficients

are factors of 10 right then they're

going to be very large coefficients so

I'll need to divide all of them by 10 so

that I have the lowest the lowest

possible coefficients also the

coefficients are there to represent the

relative numbers of reactants and

products and so we can interpret them

really like ratios

so there's different ways for us to

interpret this particular chemical

reaction right we could think about this

reaction happening on the molecular

scale so we could have one molecule of

methane reacting with two molecules

of oxygen right and this would give us

one molecule of CO2 and two molecules

of water as product we could also think

of a larger scale right we could think

about moles so this would represent one

mole of methane reacting with two moles

of oxygen to generate one mole

of carbon dioxide and two moles

of water in each case whether we're

talking about molecules or moles right

it's really about having the correct

ratio of reactants and products

we could also think about this being in

dozens of molecules right so it's really

just about making sure we have the right ratio

to drive this home here we have just a

snapshot of our reactant and product

side mixtures at a given point in our

chemical reaction

and if these ratios need to be true for

our reaction then what we should see in

our reaction mixture should also

represent that ratio so here we said

that the black and white molecules are

methane so you can see that we have

three methane

molecules in our reactant mixture and

the red molecules are the oxygen so one

so even though we don't have the exact

numbers that are represented in our

chemical reaction

the three methane molecules to six

oxygen molecules is still a one to two

ratio which is the ratio that we see in

our chemical reaction and if we look at

our product mixture we see the same

thing that the ratio of molecules is the

same as what we see in our balanced

chemical equation

so in our reactant mixture the carbon

dioxide is represented by the black

sphere in the Middle with the two

oxygens on the side the two red spheres

so we have one two three carbon dioxide molecules

molecules

and the water molecules are the red

sphere with the two white right so this

is oxygen and then two hydrogens so one

two three four five six water molecules

and again you can see that three to six

is the same as a one to two ratio which

is the ratio that we see in our balanced

so how now do we balance chemical equations

equations um

when we have a balanced chemical

equation we have equal numbers of atoms

for each element on both the reactant

and the product sides

to achieve this we need to make sure

that the numbers of atoms for a given

element is multiplied by the coefficient

of any formula containing that element

by the element subscript in the formula

so here we can see if we look at this

chemical reaction the oxygen has a

coefficient of 2 and it has a subscript

of 2 in the molecular formula so in

order to calculate the number of oxygens

that are present as reactant here we

would need to multiply two by two

and then finally if an element appears

in more than one formula on a given side

of the equation the number of atoms

represented in each must be computed and

then added together

so here we have our formula that we've

been working with right the reaction of

methane with oxygen so let's make sure

that this is a balanced chemical

equation first we're going to look at Carbon

Carbon

we'll consider the number of carbon

atoms on the reactant side and then the

product side and if they are equal then

we will know that the carbons are

balanced so on the reactant side we have

the carbon in methane there's only one

carbon atom in each methane and we only

have one methane molecule so not

multiplying the coefficient by the

number of methane atoms is one times one

meaning we only have one carbon atom as

a reactant but if we look on the product

side we have carbon present in the

carbon dioxide the coefficient is one

and there's only one carbon in each

carbon dioxide so this is also one times

one and we can see that yes the carbons

are balanced when we consider the

hydrogens on the reactant side we have

four hydrogens in each one carbon each

one methane molecule the coefficient is

one the subscript for hydrogen is four

so we're Computing one times four we

have four hydrogens as reactants and on

the right hand side our product side we

have hydrogen present in water so the

coefficient is two we multiply this by

the subscript for hydrogen in the

molecule which is also two

so 2 times 2 gives me 4 hydrogen atoms

since both give us four we can see that

this is balanced as well

and then finally for the oxygen we see

that oxygen is present as O2 at the

reactant side the coefficient is two we

multiply that by its subscript which is

also 2 to give us four oxygen atoms

present on the reactant side on the

product side we see that oxygen is

present both in the CO2 and also in the

water so we're going to figure out how

much oxygen is in each and then we will

add them together

so in the carbon dioxide the coefficient

is 1

the subscript for the oxygen in this

molecular formula is 2 which gives us

two atoms of oxygen on the product side

in the carbon dioxide and then for the

water molecule the coefficient is 2 and

the subscript for oxygen in this

molecular formula is one

so this is also two when I add these two

together this is going to give me 4.

since I had four oxygens on the reactant

side and four in the product side for

so let's get a little bit more practice

balancing chemical equations

we have an unbalanced chemical equation

involving water being broken up into its

elements hydrogen H2 and oxygen O2 we're

told that this is unbalanced so we're

going to first check and see how many of

each species we have on the reactants

and product side and then we'll inspect

the reaction to figure out how we can

change it to make it balanced

so starting with the hydrogens on the

reactant side we see that we've got one

molecule of water and there are two

hydrogens in each molecule giving me a

total of two on the reactant side the

hydrogen is present as H2 the

coefficient is one the subscript is two

which means that I have two hydrogen

atoms on the product side so the

hydrogens are balanced when we look at

the oxygens we see that oxygen is

present on the reactant side as water

our coefficient is one for the oxygen

the subscript would be one

so I only have one oxygen atom on the

product side I have oxygen presidents O2

my coefficient is 1 and I'm multiplying

that by my subscript which is 2 to give

me two oxygen atoms on the right hand side

side

so we can see that no the oxygens are

not balanced

in order to balance this we can look at

our reaction and think how can we change

it so that the oxygens would become

balanced we need more oxygen on the

reactant side right because we're

comparing one to two so we need

so it says to achieve balance the

coefficients of the equation must be

changed as needed we know that the

subscripts cannot be changed right that

would change the actual identity of the

species involved right it would change

the molecules so what we can change is

are the coefficients of these species

so we can try to increase the amount of

oxygen on the reactant Side by adding a

2 in front of the water so we're going

two water molecules

coming apart to give us hydrogen gas and

oxygen gas now let's inspect this to see

if it has solved our problems and

balanced the oxygen so I'm going to use

a little bit of a shortcut to write how

many reactant and product atoms we have

for each so I'm going to first inspect

for hydrogen and I'm going to draw two

dashes on the left hand side I'll write

how many hydrogens I have as reactant

and then on the right I'll write how

many I have as product

so again I'm doing the 2 write the

coefficients for the water molecule

multiplied by the subscript for the

hydrogen so I can see that I have four

hydrogens on the reactant side

for the product side I can see I have

one as my coefficient for the hydrogen

gas my subscript is 2. so 1 times 2

gives me only two hydrogen atoms

doing the same for oxygen we see our

coefficient of 2 multiplied by 1 gives

me only two oxygen atoms and on the

right hand side I have a coefficient of

1 for the oxygen O2 gas and a subscript

of two multiplying those together gives

me two oxygen atoms so we've solved our

problem for the oxygen and balanced it

but our hydrogens are not balanced

so looking at our reaction we can see

that we need more hydrogen gas on the

product side the hydrogen on the product

side is in the form H2 so if we add a

coefficient in front of that then that

will allow us to hopefully balance our

equation so I'm going to write in a 2 here

and then we can see how this changed the

number of hydrogen atoms right now we'll

have 2 times our subscript 2 to give us

four hydrogens nothing else should have

changed so now both my hydrogens and my

let's try another example

so here we have nitrogen and oxygen

reacting to give me dinitrogen pentoxide

n2o5 we're going to do the same thing so

first I'll look at my nitrogens to see

if they're balanced and my oxygens on

the left hand side I have two nitrogens

as reactants on the right hand side for

products I also have two

for the oxygens as reactant I have two

and on the right hand side I have 5. so

we're balanced with respect to nitrogen

but not with respect to oxygen but I can

see that I can balance these if I'm able

to make the number of oxygens equal on

both the left and right sides

what I can do is notice that if I

multiply the number of oxygens on the

left hand side by 5 and I multiply the

oxygens on the right hand side by 2

then I should have 10 on both sides

right I'm using the lowest common

multiple of the numbers we had initially

so I can do this by changing my reaction

so then I would have

added a 5 in front of the O2 and A2 in

front of the n2o5

so now in terms of the nitrogen I have 2

on the left hand side and I have 2 times

2 or 4 nitrogens on the left on the

right hand side

for the oxygens on the left hand side I

have 10 right 5 times 2 and on the right

hand side I have 2 times 5 also 10. so

now we've succeeded in balancing our

oxygens but the nitrogens are not balanced

balanced

so we can look at our equation and see

that we need more nitrogen on the left

hand side nitrogen is present as N2 on

the left hand side so if I'm able to add

a 2 in front of it then that should balance

balance

all right so going back and looking at

my number of nitrogen atoms it should be

2 times 2 or 4 on my left hand side and

the right hand side should now not have

been changed so we now are balanced with

respect to both the nitrogen and the oxygen

oxygen

let's do one more example we're going to

see in this example that is sometimes

convenient to use fractions while we're

balancing but then in the end we want

our coefficients to be the lowest whole

numbers and so we'll need to multiply by

something in order to get rid of our

fractions and I'll show you what I mean

so let's look at our reaction here we

have a reaction involving carbon

hydrogen and oxygen so we're reacting

c2h6 with oxygen O2 to generate water

and carbon dioxide

initially we can go through and see how

many of each species we have

we've got carbon hydrogen and oxygen

in terms of carbon on the left hand side

I see I have two atoms of carbon on the

right hand side I have only one in the

carbon dioxide

for hydrogen I have six hydrogens on the

left hand side

on the right hand side I have two hydrogens

hydrogens

and for the oxygen on the left hand side

I have two and on the right hand side I

have oxygen present in both products so

I have one oxygen present in water and I

also have two

oxygens in the carbon dioxide to give me

a total of three

so none of our atoms are none of our

elements are balanced so we need to set

about how to do this

um for reactions evolving carbon

hydrogen and oxygen I would recommend

that you start by balancing the carbons

the carbons

so let's look at the carbons we said we

have two on the left and one on the

right so it would seem that we could add

a coefficient of 2 for the CO2 and that

indeed now I've got two carbons on the

right hand side but because there is

oxygen also in the carbon dioxide this

has changed the number of oxygens I have

for my product side as well

so now I have one oxygen in the water

and a coefficient of 2 multiplied by the

subscript of 2 for the oxygen gives me

four so now I have a total of five

after balancing our carbons I want to

now look at the hydrogens

so next

I see that I've got six hydrogens on the

left and two on the right

if I add a 3 in front of my water

molecule then that should indicate that

right so we didn't change anything on

the Loft so we still have six hydrogens

on the left and on the right we have

three times two

to give us a total of six hydrogens for

our last step

is to balance

see that we've got two oxygens on the

left hand side

and since I changed

the number of water molecules and that

also contains oxygen I need to

reevaluate how many oxygens I have on

the product side

so in the water I have three

right three times a subscript of one for

the oxygen

and in the carbon dioxide I have 2 times

a subscript of two so that would be four

so I have a total of seven oxygens on

the right hand side

now in order to balance this I need to

add more oxygens on the left right on

the reactant side

I can see that if I multiply the 2 by 7

halves then I would achieve a number of

7 on the left and that's the number that

I have on the right so if I can achieve

this then the oxygens will also be

balanced so I can use 7 halves as my

coefficient for the oxygens

and when I do this I have seven halves

times two or seven oxygens on the left

which equals

the number of oxygens I said I have on

the right so now we are balanced with

respect to all of our elements

but I have this fractional coefficient

which is not preferred

so when a balance is achieved with

fractions you can then multiply all of

your coefficients by some whole number

in order to convert the fraction to an integer

integer

um and multiplying through all of your

coefficients by the same number will

keep the same ratios right which

achieved the balanced chemical equation

but will just help us to have them in

integer format

so to get rid of my seven halves I can

see I just need to multiply by the

number that I have in my denominator so

because 2 is in my denominator If I

multiply all of my coefficients by 2

then this should get rid of my fraction

so I'm going to multiply

by 2

so I'm going to write my balanced

chemical equation down here now so if I

multiply each coefficient by 2 I will

have 2

for the C2 H6

If I multiply 7 halves by 2 I'm going to

get 7

as my coefficient for the O2

If I multiply 2 by 3

for my coefficient for the waters I will

get 6 and then finally 2 multiplied by 2

for the carbon dioxide will give me four

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YouTube Transcript:Chapter 7.1a Writing and Balancing Chemical Equations