Now that we’ve spent a couple episodes talking about the basics of the mole,
we’re hopefully feeling comfortable with unit conversions using dimensional analysis,
Avogadro’s number, and molar mass.
So far, we’ve converted moles to grams or atoms and vice versa.
Now that we know how to do those kinds of calculations, we can use the same skills during
problems that are a bit more complicated.
Our process for getting our answers won’t change, we’ll just be doing these steps
multiple times instead of once.
In chemistry, it’s common to use multiple conversion factors like stepping stones to
get to the units we’re after.
And that’s the river we’re crossing in today’s episode.
Hi, I’m Will Comar, and welcome back to Study Hall Chemistry, presented by Arizona
State University and Crash Course.
Let’s dive in!
One situation we might run into that requires multiple conversion factor stepping stones
is converting atoms to grams, or vice versa.
When we use a conversion factor, we’re only moving from one unit to another, like from
atoms to moles, or moles to grams.
Since we don’t have a conversion factor that gets us directly from atoms to grams,
we need to use two factors to get us there, with moles in the middle.
It’s like if we’re taking a flight with a layover in one city on our way to our final
destination, because we can’t get a direct flight.
Let’s see how this would play out in an example.
If we wanted to know how many grams are in 2.30 x 10^24 atoms of gallium, we’d list
our givens as normal.
We’d also identify what we need to find: the amount in grams.
Here’s where things are a little different.
When we pick out our conversion factors, we’ll need a first one to convert from atoms to
moles, then a second one to convert from moles to grams.
For the first one, we can use Avogadro’s number, 6.022 x 10^23 atoms per mole.
For the second, we can use the molar mass of gallium, which is about 70.0 grams per mole if we round.
That way, we have conversion factors that include all the units we have and the units
we’re trying to get to.
Let’s do our conversions one stepping stone at a time with some help from TestTube.
When we write out all our terms as stepping stones, we start with the units we have,
which would be atoms.
To get to the next one, we’d multiply by our first conversion factor, Avogadro’s number.
Like with any conversion, we want the units we need on top, and the units we have on the bottom.
Since Avogadro’s number is written as atoms over moles, this means we need to flip it
around so the atoms are on the bottom, with moles on top.
It might help to think of this as dividing our given by Avogadro’s number, not multiplying this time.
Then, onto the next stone, where we’ll do the same thing again with our second conversion factor.
This time, the units we need are grams, and we’re trying to convert from moles, so we
can leave the molar mass of gallium, 70.0 grams per mole, right-side up.
Alright, now our stones are in place, and we can step from one conversion factor to
the next as the units cancel out.
So atoms would cancel, and moles would cancel, and we’ll be left with grams of gallium.
We can plug our numbers into a calculator and multiply to get our answer, 267.353 grams of gallium.
All we need to do next is to express our answer with the proper rounding to appropriate
significant figures, or sig figs.
We’d round to three sig figs: 267.35 grams of gallium are present in 2.30 x 10^24 atoms.
Now our multiple unit conversions are complete, and we’ve made it to the other side of the river.
Success!
The key to working with these more involved conversions is keeping our units straight.
So instead of writing out the full equation all together, another method could be to write
out the units first and fill the numbers in after.
If we were to try to find the number of atoms in 12.47 grams of calcium, we could use this
technique to make our process a little easier.
Calcium’s a great element to practice with, since its molar mass is a nice round number,
and since it’s one of the most abundant elements, it shows up in a lot of places--including exams!
Our given here is in grams, and we need to find the amount in atoms.
Like our last example, we need two conversion factors: the molar mass of calcium to convert
from grams to moles, and Avogadro’s number to convert from moles to atoms.
As we write out our expression, we’ll focus on the units only to start.
We want our given units, grams, to cancel, and we can make that happen with a conversion
factor that’s in grams per mole.
In order to cancel, though, grams should be on the bottom of our conversion factor.
So we know we need to write it in terms of moles over grams.
For the second conversion, if we write the units as atoms over moles, our moles will
cancel, leaving us with our answer in atoms.
Now that we have our units written out, we can fill in our numbers:
12.47 for the grams of calcium, 1 over 40.0 for the molar mass, and
6.022 x 10^23 over 1 for Avogadro’s number.
Then we’ll multiply the expressions, and end up with our answer,
which gets rounded to three sig figs.
There’s 1.88 x 10^23 atoms of calcium in 12.47 grams.
Sometimes we'll be solving mole problems that deal with substances in their gaseous form.
To do that, we need to understand another important equation: the Ideal Gas Law.
The Ideal Gas Law is an equation that represents the relationship between pressure, volume,
and temperature for a gaseous substance.
Obviously, real world chemistry is hardly ever ideal, but this law represents gases
without taking their pesky complicating factors into account.
Ideal gases, as we refer to them when we do this, are easier to work with since we’ve
simplified the math involved.
The law is written as PV = nRT.
In other words, pressure times volume equals the number of moles of a substance multiplied
by the temperature and R, the universal gas constant.
The universal gas constant is equal to 8.3145 joules per mole Kelvin
or 0.0821 liter atmospheres per mole Kelvin.
If that sounds like every unit of measurement there is, that’s because, well, there are
a lot of units in the universal gas constant.
We don’t need to worry too much about where they come from, especially since most will
cancel out during Ideal Gas Law calculations anyway, but they’ll be helpful to have around later.
One common way to use this equation is to figure out the pressure, volume, or temperature
of one mole of gas in certain circumstances.
So we can use our dimensional analysis and mole problem solving skills to work with this equation!
And, if we presume the gas is at standard temperature and pressure, or STP, we get two
bonus given numbers to plug into our equation.
At standard temperature, T would equal 0 degrees Celsius, or 275.15 Kelvin.
At standard pressure, P would equal 1.00 atmosphere, 760 millimeters of mercury, or 101.3 kilopascals.
Those are some wild numbers and units, but don’t worry about having to memorize any of these.
Most of the time, this information is given to us, or we can always look it up.
At standard temperature and pressure, we can plug both of these constants in for T and
P in our equation and solve for the volume of 1.00 mole of any gas.
The reason this is important is because we can use the answer we’ll get as a conversion
factor in all kinds of future reactions.
Let’s start by rearranging the Ideal Gas Law equation so that we’re solving for volume.
Next we'll need a conversion factor to get an answer in units for volume that we actually recognize.
Something in, for instance, liters, would be awesome.
Luckily, we can use an R with liters in it!
Like I mentioned, R is the ideal gas constant, and can be equal
to 0.0821 liter atmospheres per mole Kelvin.
Since this R has atmospheres and Kelvins in it, we’ll want to express P in atmospheres
and T in Kelvins to get all of our units to match up. So, our initial written out equation would then be
volume = 1 mol * (0.0821 L*atm/(mol*K))(273.15 K) / 1.00 atm.
And, since we picked our units strategically, our moles, Kelvins, and atmospheres all cancel out,
leaving us with liters like we wanted.
Told you, having all those units around would come in handy!
Now all that’s left is to multiply and divide our numbers, and round down our answer to
the appropriate number of sig figs.
In this case, our answer will have 3 of them.
We can now say that for one mole of any gas at STP, the volume of that gas is 22.4 liters.
We’ve just proven the standard gas volume at one mole.
Now we can apply STP and the standard gas volume to other kinds of multi-step mole calculations
dealing with gaseous substances.
For example, we can use it to find out the mass of a gas.
Let’s say we work for an airline that uses diatomic nitrogen to fill airplane tires,
and we want to find the mass in grams of 900 liters of diatomic nitrogen at STP.
We can use the standard gas volume as our conversion factor.
Since it’s expressed as liters per mole, it will get us from units of volume to units of moles.
To get from moles to grams, we can also count the molar mass of diatomic nitrogen, 28.0
g/mol, as another conversion factor.
So, we have 900 liters of N2 multiplied by our first conversion factor, one mole of N2 over 22.4 liters,
which we’ll multiply by the second conversion factor, our molar mass.
Our liters and moles cancel, leaving us with our desired units in grams.
So there are 1,125 grams in 900 liters of diatomic nitrogen gas.
We’d then round to three sig figs, so we’d have 1.13 x 10^3 grams as our final answer
in scientific notation.
The standard gas volume got us here in half the time!
Using dimensional analysis to solve mole problems like these gives us a lot of power.
Sure, there’s a lot of math, but we can do really cool things with that math.
We can unlock the amount of moles in a substance, convert it to grams or to atoms, and, if it’s an ideal gas,
determine its temperature, volume, and pressure.
These are powerful tools used every day by modern chemists and scientists to manufacture
aerosols, safely compress gases, or send up weather balloons!
Next time, we’ll venture into one of the most important things we’ll
cover in this series: stoichiometry,
which uses mole ratios to determine the amount of a reactant or product in relation to the
other substances in a chemical reaction.
We’ll need all the things we’ve learned up to this point to solve those kinds of problems,
so it’ll be a good opportunity to refresh what we’ve covered so far.
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and the Crash Course team at Complexly.
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See you all next time!