Stoichiometry | Chemical reactions and stoichiometry | Chemistry | Khan Academy
- Articles, Blog

Stoichiometry | Chemical reactions and stoichiometry | Chemistry | Khan Academy

We know what a chemical equation
is and we’ve learned how to balance it. Now, we’re ready to learn
about stoichiometry. And this is an ultra fancy word
that often makes people think it’s difficult. But it’s really just the study
or the calculation of the relationships between
the different molecules in a reaction. This is the actual definition
that Wikipedia gives, stoichiometry is the calculation
of quantitative, or measurable, relationships
of the reactants and the products. And you’re going to see in
chemistry, sometimes people use the word reagents. For most of our purposes you can
use the word reagents and reactants interchangeably. They’re both the reactants
in a reaction. The reagents are sometimes for
special types of reactions where you want to throw a
reagent in and see if something happens. And see if your belief about
that substance is true or things like that. But for our purposes
a reagent and reactant is the same thing. So it’s a relationship between
the reactants and the products in a balanced chemical
equation. So if we’re given an unbalanced
one, we know how to get to the balanced point. A balanced chemical equation. So let’s do some
stoichiometry. Just so we get practice
balancing equations, I’m always going to start with
unbalanced equations. Let’s say we have iron
three oxide. Two iron atoms with three
oxygen atoms. Plus aluminum, Al. And it yields Al2
O3 plus iron. So remember when we’re doing
stoichiometry first of all, we want to deal with balanced
equations. A lot of stoichiometry problems
will give you a balanced equation. But I think it’s good practice
to actually balance the equations ourselves. So let’s try to balance
this one. We have two iron atoms here
in this iron three oxide. How many iron atoms do we have
on the right hand side? We only have one. So let’s multiply this
by 2 right here. All right, oxygen, we have
three on this side. We have three oxygens
on that side. That looks good. Aluminum, on the left hand
side we only have one aluminum atom. On the right hand side we have
two aluminum atoms. So we have to put a 2 here. And we have balanced
this equation. So now we’re ready to do
some stoichiometry. There’s not just one type of
stoichiometry problem, but they’re all along the lines of,
if I give you x grams of this how many grams of aluminum
do I need to make this reaction happen? Or if I give you y grams of this
molecule and z grams of this molecule which one’s
going to run out first? That’s all stoichiometry. And we’ll actually do those
exact two types of problems in this video. So let’s say that we were given
85 grams of the iron three oxide. So my question to you
is how many grams of aluminum do we need? Well you look at the equation,
you immediately see the mole ratio. So for every mole of this, so
for every one atom we use of iron three oxide we need
two aluminums. So what we need to do is figure
out how many moles of this molecule there are in 85
grams. And then we need to have twice as many moles
of aluminum. Because for every mole of the
iron three oxide, we have two moles of aluminum. And we’re just looking at the
coefficients, we’re just looking at the numbers. One molecule of iron three
oxide combines with two molecule of aluminum to make
this reaction happen. So lets first figure out how
many moles 85 grams are. So what’s the atomic mass
or the mass number of this entire molecule? Let me do it down here. So we have two irons
and three oxygens. So let me go down and figure out
the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it’s fair enough
to round to 56. Let’s say we’re dealing with
the version of iron, the isotope of iron, that
has 30 neutrons. So it has an atomic
mass number of 56. So iron has 56 atomic
mass number. And then oxygen, we already
know, is 16. Iron was 56. This mass is going to be 2
times 56 plus 3 times 16. We can do that in our heads. But this isn’t a math
video, so I’ll get the calculator out. 2 times 56 plus 3 times
16 is equal to 160. Is that right? That’s 48 plus 112,
right, 160. So one molecule of iron three
oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10
to the 23 molecules of iron oxide is going to have
a mass of 160 grams. So in our reaction we said we’re
starting off with 85 grams of iron oxide. How many moles is that? Well 85 grams of iron
three oxide is equal to 85 over 160 moles. So that’s equal to, 85 divided
by 160 equals 0.53125. Equals 0.53 moles. So everything we’ve done so far
in this green and light blue, we figured out how
many moles 85 grams of iron three oxide is. And we figured out
it’s 0.53 moles. Because a full mole would
have been 160 grams. But we only have 85. So it’s point 0.53 moles. And we know from this balanced
equation, that for every mole of iron three oxide we
have, we need to have two moles of aluminum. So if we have 0.53 moles of the
iron molecule, iron three oxide, then we’re going to need
twice as many aluminum. So we’re going to need 1.06
moles of aluminum. I just took 0.53 times 2. Because the ratio is 1:2. For every molecule of this, we
need two molecules of that. So for every mole of this, we
need two moles of this. If we have 0.53 moles, you
multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to
figure out how many grams is a mole of aluminum and then
multiply that times 1.06 and we’re done. So aluminum, or aluminium as
some of our friends across the pond might say. Aluminium, actually
I enjoy that more. Aluminium has the atomic
weight or the weighted average is 26.98. But let’s just say that the
aluminium that we’re dealing with has a mass of 27
atomic mass units. So one aluminum is 27
atomic mass units. So one mole of aluminium is
going to be 27 grams. Or 6.02 times 10 to 23 aluminium atoms
is going to be 27 grams. So if we need 1.06 moles, how many
is that going to be? So 1.06 moles of aluminium is
equal to 1.06 times 27 grams. And what is that? 1.06 times 27. Equals 28.62. So we need 28.62 grams of
aluminium, I won’t write the whole thing there, in order to
essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62
grams of aluminium, then they’ll be left over after
this reaction happens. Assuming we keep mixing it
nicely and the whole reaction happens all the way. And we’ll talk more about
that in the future. And in that situation where we
have more than 28.63 grams of aluminium, then this molecule
will be the limiting reagent. Because we had more than enough
of this, so this is what’s going to limit
the amount of this process from happening. If we have less than 28.63 grams
of, I’ll start saying aluminum, then the aluminum will
be the limiting reagent, because then we wouldn’t be able
to use all the 85 grams of our iron molecule, or our
iron three oxide molecule. Anyway, I don’t want to confuse
you in the end with that limiting reagents. In the next video, we’ll do
a whole problem devoted to limiting reagents.

About James Carlton

Read All Posts By James Carlton

100 thoughts on “Stoichiometry | Chemical reactions and stoichiometry | Chemistry | Khan Academy

  1. Great video, although I did wince when he stoichiometry like that…it's said "Stoke-e-om-etry" in my experience.

  2. I am so going to stop watching this if I hear aluminium one more time
    …. ugh it drives me nuts. Stop it.

  3. As Greek everytime I hear the pronunciation of the stoichiometry word I have this urge to rip out my ears. The oi sound is one ( vowel) οι which is the same as η, ι, οι, υ, ει. In this particular words sounds like the English e as in (E)aster and it comes from the Greek word στοιχείο which is element. Note that both οι and ει are double vowels and sound exactly the same. As E

  4. This man explained Stoichiometry better in under 10 minutes than my college chemistry teacher could in an hour and a half

  5. Can someone help me with this excersice, It says: During the complete combustion of a paraffinic hydrocarbon at 20 degrees Was got a volumetric contraction equal to 1/2. Find the weight of two mols of the hydrocarbon.

  6. Thank you so much you just saved my life man! I would make myself a promise to never fall asleep in class again, however we all know I would never be able to do that

  7. Can anyone tell me why we only found out the amount needed for one mole of Aluminium? Shouldn't the final value be multiplied by two or shouldn't the AMU be multiplied by two?

  8. The issue is that our chem teacher forces us to write it in a specific way that I don’t understand at all, if I could do it like this I’d be fine and I’d understand the questions 😂

  9. Sorry i canot understand why you times 56 with 2and 3 with 16 .I understand how to find the atomic mass. But atomic mass is what we times with numbers or what?

  10. "Often makes people think it's difficult" On God chemistry teachers go out of their way to make things as complicated and difficult as humanly possible.

  11. Just thought i would mention before i watch that my uni lecturer sent me here haha, must be doing some thing right.

  12. I'm confused with the last part… Why did we use 1 aluminum x 1mol not the fixed equation where it has 2 aluminum.. So at the end why wasn't it 2al = 54g/mol x 1.06g

  13. I'm confused i thought you need 2 al but at the end you only use 1 al to get your mole ratio can you please explain further ?

  14. Well this helped a lot. My teacher was doing it with dimensional analysis and it seemed complicated but this cleared it up and makes it look easier

  15. Random guy showing Stoichiometry definition using Wikipedia


  16. i'm currently failing chem with an F and i've got a test on this tomorrow and i couldn't pay enough attention to understand the lessons throughout the month that it took my teacher to teach it and after this video i STILL can't grasp the concept so uh rip my grade guess i'm definitely taking that final

  17. I missed out on 850 energy points because I watched this video here instead of the KA website >:(

  18. Sal, You made it real Hard tbh for a revisor
    As one can do Cross Multiplication by Taking Mr and given Grams
    Mr Of Fe203 X Mr Of Al2
    85 grams x

  19. it’s so funny how here are comments from 3 years ago where people are like oh no i’ve got an exam tomorrow and etc. and they probably don’t even remember those tests

Leave a Reply

Your email address will not be published. Required fields are marked *