Video Transcript
In this video, we will learn how to
apply what we know about balanced chemical equations to calculate the masses of
reactants and products using ratios, molar quantities, and molar masses. Now, before we dive into this
topic, let’s take a moment to review what we know about chemical equations.
A chemical equation is kind of like
a recipe that you might follow to make some baked goods. When we’re following both a recipe
and a chemical equation, we combine specific amounts of our starting materials to
create something new. When we’re following a recipe, we
combine specific amounts of our ingredients, like flour, sugar, and eggs. But when we’re following a balanced
chemical equation, we’re combining specific amounts of chemical compounds, which we
call our reactants. And then the reactants chemically
combine to form the products.
Now, when we follow a chemical
equation, it specifically needs to be a balanced chemical equation. This means that the chemical
equation follows the law of conservation of mass, which is sometimes historically
referred to as the law of indestructibility of matter. The law of conservation of mass
states that, in a closed system, the mass is constant, a closed system being
something that matter cannot escape from, like a sealed container.
The law of conservation of mass was
famously demonstrated by the French scientist Antoine Lavoisier in the late
1700s. Lavoisier was investigating a type
of reaction where a metal, such as mercury, is heated to form a new substance,
called a metal calx. Heating mercury, for instance,
would form a red powder, which is known as mercury calx. All of these reactions have the
curious property that the calx that was formed was heavier than the original mass of
the metal that was started off with.
But when Lavoisier performed the
reaction in a sealed jar, making the system a closed system, the mass of the system
remained constant. It turns out that the metals were
reacting with the oxygen in the air to form metal oxides. And when you consider the mass of
the metal plus the mass of the oxygen it was reacting with, the mass was in fact
conserved.
As we can see from looking at the
balanced chemical equation for this reaction, the law of conservation of mass
reflects that we have the same amount and type of atoms on both the reactants and
products side of our balanced chemical equation. In chemical reactions, the atoms
that make up the reactants just recombine to form new substances. This means that the numbers that
are in front of the chemical species in our chemical equations, which are called the
stoichiometric coefficients, reflect the amount of each substance that participates
in the reaction, given in moles.
So knowing that these
stoichiometric coefficients reflect the amount of the substance in moles that
participates in the reaction, we can calculate the total mass of the products and
the total mass of the reactants for this reaction and see for ourselves that the law
of conservation of mass would in fact hold for this reaction.
Plus a quick aside, it’s worth
mentioning that the conservation of mass is not actually held for all processes. In nuclear reactions, for instance,
mass can be converted into energy and vice versa, a fact that is reflected by
Einstein’s famous equation 𝐸 equals 𝑚𝑐 squared. However, for these kinds of
processes, if we consider the energy and the mass together, it is held constant. But we don’t need to worry about
this too much as the conservation of mass will hold true for the vast majority of
processes we encounter in chemistry.
So now that we’ve reviewed balanced
chemical equations, how can we use them to solve problems? Oftentimes in chemistry, we want to
know how much of a reactant we’ll need to produce a certain amount of our
products. Or we will want to know how much of
our product will be form given the amount of reactants that we start off with. Or we’ll want to know how much of a
reactant we’ll need given how much of a different reactant we’re using. In other words, we need to know how
to convert between amounts of chemical substances that are participating in the same
chemical reaction.
Let’s go back to our muffin recipe
real quick. Sometimes when we follow a recipe,
we don’t quite have the right amount of ingredients to follow the recipe as
written. Or you might want to scale up the
recipe to make more. For instance, we might wanna make
18 muffins instead of the 12 muffins that this recipe makes, though we might have
done something like this intuitively in our heads in the past.
If we write out our work
rigorously, we would want to use ratios to scale the recipe to figure out how much
of all our ingredients we’ll need to make these 18 muffins. For instance, from the original
recipe, we can see that we need one cup of milk to make 12 muffins. So the ratio of cups of milk to
number of muffins is one to 12. Then, to find the amount of milk
that we need to make 18 muffins, we just multiply this ratio by the number of
muffins that we want to make, which would give us one and a half cups of milk.
We’ll follow a very similar process
when we’re doing calculations involving chemical reactions. So let’s take a look at this
balanced chemical equation to get some practice. In this reaction, we have nitrogen
and hydrogen reacting to form ammonia. Instead of using the amounts of
ingredients in a recipe to convert between different amounts of substances, we’re
going to create molar ratios using the stoichiometric coefficients from our balanced
chemical equation. So let’s practice this by creating
molar ratios for each pair of substances in our equation.
Let’s start off with the molar
ratio between nitrogen and hydrogen. As we can see from our balanced
chemical equation, every one mole of nitrogen needs three moles of hydrogen to
react. So the molar ratio between nitrogen
and hydrogen is one to three. Now, let’s look at nitrogen and
ammonia. From our balanced chemical
equation, we can tell that every one mole of nitrogen produces two moles of
ammonia. So the molar ratio here is one to
two.
Finally, let’s look at hydrogen and
ammonia. Every three moles of hydrogen
produces two moles of ammonia, like we can see from our balanced equation. So the molar ratio of hydrogen to
ammonia is three to two. It didn’t come up in this example,
but it’s worth pointing out the molar ratio should always be expressed as the
simplest whole number ratio. So, for instance, if we ended up
with a molar ratio of two to two from looking at our equation, we would actually
want to express the smaller ratio as one to one since two to two can be reduced
further by dividing it by two. So now let’s put these molar ratios
to work to convert between different amounts of substances.
So what amount of H2, in moles, is
needed to produce four moles of NH3? So in a problem like this, we’re
given an amount of one chemical species that’s participating in the reaction, in
this case ammonia. And we want to convert it into an
amount of a different chemical species in the reaction, in this case hydrogen. We’ll tackle this kind of problem
just like we saw with the muffin example earlier. We’ll multiply the amount of NH3
that we’re given by the molar ratio of H2 to NH3, which is three to two, which gives
us six moles of H2. Notice that we want to arrange our
molar ratio so that the moles of ammonia will cancel. We want to make sure that we aren’t
arranging our molar ratio the other way around.
So now we know how to make a molar
ratio from a balanced chemical equation and use it to find the amount of products
and reactants we’ll consume and produce in a reaction. But just to reiterate, to perform
these calculations, the chemical equation needs to be balanced. Otherwise, our work just won’t make
sense.
Now, molar ratios are incredibly
useful to find the amounts of all our chemical species that we need. But when we perform chemical
reactions in the lab, we usually don’t know the amount of moles of a substance that
we have directly. Rather, we’ll usually weigh out the
mass of our reactants on a balance. So we’re going to need to know how
to convert between mass and moles and vice versa. We can do this by using the molar
mass of a substance, because the amount of a substance in moles is equal to its mass
divided by its molar mass. So if we have a mass of a substance
and we want to know the amount of it in moles, we can convert between the two by
dividing by the molar mass. And we can go the other way around
and convert from moles to mass by multiplying the amount of the substance in moles
by the molar mass.
So now that we’ve learned all these
different ways to convert between different quantities, let’s sum things up before
we move on to some example problems. If we wanna convert between
different amounts of substances that are participating in the same chemical
equation, we can use molar ratios. If we wanna convert between the
amount of a substance and its mass, we can use the molar mass of that substance. Though it didn’t come up in this
video, sometimes problems we will ask us to solve for the number of entities,
meaning the number of atoms or molecules, that are in a sample of a substance, which
we can do by using Avogadro’s constant, which tells us the number of entities that
are in a mole.
If we ever get stuck solving a
problem, we can use this chart by starting with what we’re given and following it
around until we have the quantity that we’re interested in solving for. So now let’s apply everything that
we’ve learned to some examples.
Methane burns in oxygen according
to the following equation: CH4 plus 2O2 reacts to form CO2 plus 2H2O. What is the molar ratio of methane
to oxygen?
This question is asking us to
create molar ratios, which are ratios that help us relate amounts of different
chemical species that are participating in the same chemical reaction. Molar ratios are useful to us
because they help us convert between amounts of one substance and amounts of
another. To create molar ratios, we’ll use
the stoichiometric coefficients from a balanced chemical equation, which refer to
the amount of each chemical species that participates in the reaction in moles.
So before we create our molar
ratios, let’s quickly ensure that our chemical equation is balanced. There is one carbon on each side of
the equation, four hydrogens, and four oxygens. So our equation is balanced and
good to go. This question is asking us to
create a molar ratio between methane and oxygen. And as we can see from looking at
our balanced chemical equation, the stoichiometric coefficient in front of methane
is one and the stoichiometric coefficient in front of oxygen is two. So the molar ratio of methane to
oxygen is one to two, which means that for every one mole of methane, two moles of
oxygen is needed to react with it.
What is the molar ratio of methane
to carbon dioxide?
Now, we want to do the same thing,
but for a ratio of methane to carbon dioxide. As we can see from our balanced
chemical equation, we need one mole of methane to produce one mole of carbon
dioxide, which gives us a molar ratio of one to one.
What is the molar ratio of oxygen
to water?
Now, we need the ratio of O2 to
H2O. The stoichiometric coefficients in
front of both oxygen and water are two. So we get a ratio of two to
two. But you’ll notice that we can
simplify this ratio by dividing it by two, which would give us a ratio of one to
one, which is the correct molar ratio, as we should always express these molar
ratios in the simplest whole numbers that we can. So the molar ratio of oxygen to
water is one to one.
So now that we’ve worked this
problem where we find the molar ratios from a chemical equation, let’s try a harder
problem where we apply them.
Sulfur burns in oxygen to form
sulfur dioxide. What is the mass of sulfur needed
to produce four grams of sulfur dioxide?
First, let’s create a balanced
chemical equation for the reaction described in this problem. We have sulfur, which has the
symbol S, burning in oxygen, which would be O2, which is reacting to form sulfur
dioxide, or SO2. If we take a quick look at the
equation that we’ve created, we have one sulfur on each side of the equation and two
oxygens. So our equation is balanced.
This question gives us a mass of
sulfur dioxide, and it’s asking us to convert that into a mass of sulfur. But we can’t go directly from a
mass of one chemical species to a mass of another. So instead, we’re going to have to
go from the mass of sulfur dioxide to moles of sulfur dioxide, then convert into
moles of sulfur, and then from there solve our problem by finding the mass of
sulfur. We can convert from mass to moles
or moles to mass by using the molar mass of a substance, because the amount of a
substance in moles is equal to the substance’s mass divided by its molar mass.
We can calculate the molar mass for
sulfur dioxide using the periodic table. Sulfur has a molar mass of 32.06
grams per mole, and oxygen has a molar mass of 15.999 grams per mole, which gives us
64.058 grams per mole for the molar mass of sulfur. Now, we can convert from grams of
SO2 to moles of SO2 by dividing by the molar mass, which we can see from our formula
and we can also see from the fact that the units of grams of SO2 will cancel.
Now, we can either perform the
calculation here, giving us 0.0624 moles of SO2, or we can continue the calculation
on in one line. Either way, our next step will be
to convert from moles of SO2 into moles of S. We can do this conversion by using a
molar ratio that we create from the balanced chemical equation. As we can see, every one mole of
sulfur produces one mole of sulfur dioxide.
Now, our final step is to convert
from moles of sulfur into the mass of sulfur so we can solve this problem, which we
can do by multiplying by the molar mass of sulfur, 32.06 grams per mole. Multiplying everything out, this
gives us 2.002 grams of sulfur. But the mass of sulfur dioxide that
was given in the problem was given to one significant figure. So we should state our answer to
one significant figure as well, which gives us two grams of sulfur needed to produce
four grams of sulfur dioxide.
Now, let’s summarize what we
learned in this video with the key points. We can create molar ratios by using
the stoichiometric coefficients found in the balanced chemical equation. These molar ratios are then used to
convert between amounts of reactants and products. Finally, we can use the molar mass
to convert between mass and moles and vice versa.
