Welcome to Stoichiometry With Mass brought to you by Ketzbook
In this video, we are going to learn how to solve stoichiometry problems that involve
mass.
When the amounts of reactants and products are quoted in grams or some other mass units,
the stoichiometry problem becomes more complex because you will need to first convert the
mass to moles.
Then, you can convert from moles of one chemical to moles of another chemical.
Last, you may need to convert from moles back to mass.
These three steps may seem like a bit much, but they are necessary because you must be
in MOLES in order to convert from one chemical to another.
Let’s see how these three steps work for a problem involving food.
Suppose that you are a chef at a grill restaurant, and you are also in charge of ordering all
the ingredients you need.
Unfortunately, you are completely out of hot dog buns, and your supplier sells everything
in pounds.
If you have 11 pounds of extra hot dog meat, how many pounds of buns do you need to purchase?
Start the problem by writing down the balanced reaction, that is, your recipe.
One bun and one piece of hot dog meat are needed to make one hot dog.
Now, in order to figure out how many buns you want to buy, you need to first calculate
the number of hot dogs that you have, and in order to do that, you need to know how
much one hot dog weighs.
It’s best to write down the individual mass of every reactant and product.
One bun weighs 0.1 pounds, one dog weighs 0.125 pounds, and a hot dog with bun weighs
0.225 pounds.
Next, we need to calculate the number of hot dogs we have.
This is a unit conversion problem, so we first write down the quantity that we know, 11 pounds
of hot dogs.
Multiply this by a conversion factor fraction.
The weight of one hot dog is our conversion factor.
Because we are starting with pounds, we put the 0.125 pounds on the bottom of the fraction,
and because we are solving for the number of hot dogs, we put 1 hot dog on top.
Remember that our conversion factor fraction is always equal to one.
Because 1 hot dog weighs 0.125 pounds, the fraction 1 hot dog over 0.125 pounds is equal
to one.
Now, let’s do the calculation.
The number of hot dogs is eleven times one over 0.125.
Because the one is on top, we type in our calculator 11 divided by 0.125, and the answer
is 88.
As for the units, pounds on top cancels with pounds on the bottom, and we are left with
88 hot dogs.
Next, we need to calculate the number of hot dog buns we need.
This is also a unit conversion problem, so we start by writing down the number we know,
88 hot dogs.
Multiply this by a conversion factor fraction.
This time, our conversion factor is the balanced reaction or recipe.
From the recipe, we know that one hot dog needs one bun.
Hot dogs on the top and bottom cancel, and our answer is 88 buns.
We need 88 buns, but our supplier wants to know the weight of buns we need, so we are
going to convert this back into pounds.
The number we know is 88 buns, and we multiply this by a conversion factor fraction.
The weight of a bun is our conversion factor.
Because we are starting with number of buns, we put 1 bun on the bottom of the fraction
and 0.1 pounds on the top.
Buns on the top and bottom cancel out to give us pounds as our final units.
For the calculation, the one in the fraction is on the bottom, so in your calculator type
88 times 0.1, which gives us our final answer of 8.8 pounds of buns.
This is the typical three step calculation for most stoichiometry problems.
Let’s try this now with a chemical reaction.
How many grams of oxygen do you make from the electrolysis of 100 grams of water?
The electrolysis of water is a decomposition reaction in which water breaks down into it’s
elements, oxygen and hydrogen.
Because the reaction needs a continuous supply of electricity, it is an endothermic reaction.
The first thing that we need to do to solve any stoichiometry problem is to write down
the balanced reaction.
H2O breaks down to form H2 and O2.
Both hydrogen and oxygen are diatomic elements.
Pause the video and check this reaction to see if it is balanced or not.
The oxygen is not balanced, with one on the left and two on the right, so we put a 2 in
front of H2O.
That now makes 4 hydrogens on the left, so we need a two in front of H2.
The reaction is now balanced.
Step 2, calculate the molar masses of each chemical in the reaction.
To calculate the molar mass of water, we add together the molar masses of the all the atoms
in water.
Since there are two hydrogens in every water molecule, we multiply 2 times 1.008 to get
2.016.
There is only one oxygen, so that is simply one times 16, which is 16.
Next, add 2.016 and 16 to get a total mass of 18.02.
In other words, one mole of water has a mass of 18.02 grams.
Notice that the big 2 in front of H2O in the balanced equation did NOT appear in our calculation
of the molar mass of water.
The coefficients in the balanced reaction do NOT affect the molar masses of any chemical
because molar mass is the mass of only one mole of that substance.
Go ahead and calculate the molar mass of every chemical in the reaction and write them below
each chemical.
Hydrogen’s molar mass is 2.016 g/mol, and oxygen’s molar mass is 32 g/mol.
Step 3, convert the mass of your known quantity to moles.
Re-read the question.
What is the known quantity?
That’s right, 100 grams of water.
Write this down first.
To convert this to moles, multiply it by a fraction.
The molar mass of water is the conversion factor.
Because we started with grams, we write 18.02 grams on the bottom and 1 mol on the top.
Grams on the top and bottom cancel out.
Because the one is on the top of the fraction, in your calculator type 100 divided by 18.02.
The answer works out to be 5.55 moles of water.
Step 4, convert the moles of water to moles of whatever you are solving for.
Write down the answer from the previous step, 5.55 moles of water, and multiply this by
a fraction.
This time, the balanced equation is our conversion factor.
The coefficients in the balanced equation tell us the ratio of the chemicals in the
reaction.
Because we started with moles of H2O, write 2 H2O on the bottom.
Because we want to convert this to oxygen, write 1 O2 on the top.
Remember that if there is no number in front of the chemical, the coefficient is really
one.
H2O on the top and bottom cancel out.
Because the one is on the top of the fraction, in your calculator type 5.55 divided by 2,
which works out to be 2.775 moles of oxygen.
The last step is to convert the moles of oxygen to grams.
Write down the answer from the previous step, 2.775 moles of O2, and multiply this by a
fraction.
The molar mass of O2 is our conversion factor.
Because we started with moles, 1 mole goes on the bottom, and 32 grams goes on the top.
Moles on the top and bottom cancel out.
Because the one is on the bottom of the fraction, in your calculator type 2.775 times 32, which
works out to be 88.8 grams of oxygen.
This is our final answer to the problem, but we should check to see if it makes sense.
Because water is decomposing into two products, we expect our answer to be less than 100,
which it is.
Because oxygen is much heavier than hydrogen, most of the mass should go into O2 rather
than H2, so our answer should be greater than 50, which it also is.
Therefore, 88.8 grams is quite reasonable.
Remember to follow these steps when solving a stoichiometry problem that involves mass,
and the only way to get good at it is to practice, practice, practice.
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