Welcome to Stoichiometry Grams to Grams Tricks brought to you by Ketzbook.
This is my third tutorial in the series, and in this video,
we will solve a grams to grams stoichiometry problem,
learn how to combine three steps into one step, and discuss tricks for making the solution easier.
Let’s start with a problem.
How many grams of carbon dioxide are produced when 2800 grams of octane are burned in excess air?
2800 grams is about 1 gallon of gasoline, so this problem is essentially asking us how
much carbon dioxide does your car gives off for every gallon of gas it consumes.
This is a grams to grams stoichiometry problem.
Notice that the problems says, “excess air.”
This means that there is more than enough oxygen for the combustion reaction,
so we are dealing with the complete combustion of octane.
The first thing that we’ll need to do is write out the balanced chemical reaction.
Octane is C8H18, and because this is a combustion reaction done in air, the other reactant is O2.
Remember that oxygen is diatomic.
For complete combustion reactions, all of the carbon is turned into carbon dioxide,
and all of the hydrogen is turned into water.
That’s it for the reactants and products.
Now, let’s balance it.
We are going to balance oxygen last because it is in three different chemicals.
Let’s start with carbon.
There are 8 carbons on the left side, so we put an 8 in front of CO2.
There are 18 hydrogens on the left side, so we put a 9 in front of H2O.
That leaves us with 25 oxygens on the right side.
The only way we can get 25 oxygens on the left side is to put 12.5 in front of O2.
However, we want the coefficients to all be whole numbers,
so we can fix this by multiplying the entire reaction by two.
We put a 2 in front of octane, the 12.5 turns into 25, the 8 become 16, and the 9 becomes 18.
And it’s all balanced.
Next, write down the molar mass of each chemical in the problem.
The problem asks us to convert between octane and carbon dioxide,
so we need the molar mass of both of them.
8 carbons and 18 hydrogens works out to be 114.2 grams per mole.
Remember that the molar mass has nothing to do with the coefficients in the balanced reaction.
Carbon dioxide has a molar mass of 44.01 grams per mole.
The third step in our problem is to convert from grams to moles for the amount we know.
The problems states that we are starting with 2800 grams of octane, so write that down first.
We multiply this by a conversion factor fraction, using the molar mass of octane as our conversion factor.
Since we started with grams, write 114.2 grams on the bottom;
and because we are solving for moles, write 1 mole on the top.
Grams on the top and bottom cancel each other out,
and because the 114.2 is on the bottom, we divide 2800 by 114.2,
which works out to be 24.5 moles of octane.
In the next step, we will convert from moles of octane to moles of carbon dioxide.
Go ahead and write down our answer from the previous step, 24.5 moles of octane,
and multiply it by a conversion factor fraction.
This time, the balanced reaction is our conversion factor.
Since we are starting with octane, write 2 C8H18 on the bottom;
and because we are solving for carbon dioxide, write 16 CO2 on the top.
Notice that all I’m doing is writing down the chemical with its coefficient from the balanced reaction.
Now, C8H18 on the top and bottom cancel out.
To do the calculation, remember that numbers on the top of fractions are multiplied and
numbers on the bottom of fractions are divided.
In your calculator type 24.5 times 16 divided by 2, which gives us 196 moles of carbon dioxide.
In the last step, we will convert moles of carbon dioxide to grams.
Go ahead and write down our answer from the previous step, 196 moles of carbon dioxide,
and multiply it by a conversion factor fraction.
The molar mass of carbon dioxide is our conversion factor.
Since we are starting with moles, write 1 mole on the bottom;
and because we are solving for grams write 44.01 grams on the top.
Moles on the top and bottom cancel out,
and we multiply 196 times 44.01 to get our final answer of 8600 grams of carbon dioxide.
I rounded the number to two significant figures.
Now that is a lot of carbon dioxide.
In terms of mass, it is more than three times the amount of gasoline that we started with.
Where did all that extra mass come from?
The oxygen, of course.
Okay, back to the problem.
There were a lot of steps in this problem.
Is there any way that we could simplify it?
Yes!
Instead of having three separate calculations,
we can combine those three conversions into one big calculation.
Start off, once again, by writing down 2800 grams of C8H18.
Only this time, multiply it by three conversion factor fractions.
What goes inside those fractions is exactly the same as before,
and there are a few tricks that can help us remember what goes where.
No matter what, one mole always goes in the top left and the bottom right.
That reminds us that the other half of those fractions are molar masses.
The number on the bottom left, 114.2 grams, is the molar mass of what we are starting with.
The number on the top right, 44.01 grams, is the molar mass of what we are solving for.
In the middle fraction we always get the numbers from the balanced reaction.
The chemical we are solving for goes on the top, so we put 16 CO2 on the top,
and the chemical we are starting with goes on the bottom, that is 2 C8H18.
In order to calculate this, we don’t need a fancy calculator.
First type the number at the beginning, 2800, and then divide by numbers on the bottom of
fractions and multiply by numbers on the top of fractions.
That is, 2800 divided by 114.2 times 16 divided by two times 44.01.
The answer, of course, is still 8600 grams of carbon dioxide.
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