How to Do Solution Stoichiometry Using Molarity as a Conversion Factor | How to Pass Chemistry
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Molarity is just another word for
concentration. The formula for molarity is equal to the moles of your solute,
divided by the liters of your solution. Now molarity can be used as a conversion
factor because it has two different units. Allowing you to go from one unit
to the next. Alright, let’s head over to the blackboard for our example. It takes
12.5 milliliters of a 0.300 molar HCl solution to
neutralize 285 milliliters of NaOH solution. What is the concentration of
the NaOH solution? And we’re given our balanced equation. Let’s start with
identifying our given and what we’re finding. So we’re given 12.5 milliliters
of HCl solution and then we’re also given our molarity of our HCl solution
as well. Now whenever you see that molarity, that’s going to be used as your
conversion factor. I want you to split that up automatically to moles over
liters. So the 0.300 will go with the moles on top and this
will be per every one liter of HCl. We’re also given 285 milliliters of NaOH
solution. Alright, now we’re finding the concentration of NaOH solution or really
molarity. And once again molarity is mole’s over liters, so we first have to
find the moles of NaOH and then we’ll divide that by the leaders of NaOH. A lot
of times students don’t know where to start so my trick for you is, whatever
you are not looking for, that’s what you’re going to start with. So since
we’re looking for molarity of NaOH. We’re gonna start with HCl. So we’re gonna
start with the milliliters of HCl and then make our way to moles of NaOH. So
our first step is to find the moles NaOH. And let’s plan this out, we said
we’re gonna always start with the compound or really, we’re starting with
whatever we’re not looking for. So since we’re finding NaOH we’re gonna start
with the milliliters of HCl. Our plan will be, starting with the milliliters of
HCl then convert that to liters of HCl using the metric system. From there we’re
going to go from liters of HCl to moles of HCl using the molarity of HCl that
was initially given and then we’ll get to moles of HCl then we’ll use a
mole to mole ratio from our balanced equation and get to moles of NaOH. Okay
let’s set this up. So starting with our 12.5 milliliters of HCl let’s align
the milliliters of HCl so they can cancel and our first step is to change
this to litres of HCl using the metric system. So I’m going to put a 1
milliliter on the bottom and 10 to the negative third liters on top of HCl. From
here our milliliters of HCl would then cancel and we’re at liters of HCl. Next
step is to change this to moles of HCl using the molarity that was provided. So
I’ll put the 0.300 moles on top of HCl and on the bottom
would be one liter of HCl. The liters of HCl would then cancel and we’re now at
moles of HCl. Now we want to find moles of NaOH so we’re going to use a mole to mole
ratio and we’ll take that from our balanced equation. So note that going
back to our balanced equation, everything is a one-to-one relationship meaning,
there is no coefficient in front there’s no number in front of that compound, so
it’s just a one mole relationship to everything. So I’m going to put one mole
of NaOH on top and one mole of HCl on the bottom. Then once we do that our
moles of HCl would then cancel and we’re finally left with moles of NaOH. Make
sure to multiply straight across and divide. Really in this case it’s just
divided by one. And our answer would be 0.00375
moles of NaOH. Our next step is to convert the milliliters to liters of the
NaOH that was provided, so in our given we were given the two hundred and eighty
five milliliters of NaOH. So since we’re finding molarity, this is going to be the
second part where we have moles that we just found in step one and now we’re
finding the liters of NaOH and at the end we’ll divide. So I’m going to take
the milliliters of NaOH and align milliliters and milliliters so they can
cancel and then on top I’ll place 10 to the negative third liters of NaOH.
So our milliliters and milliliters cancel and now we’re at liters of NaOH. There’s
another way to do this. So if you prefer putting 1,000 milliliters on the bottom
and a 1 liter on top that is completely fine. Both of these will give you the
same exact answer. So now we have 0.285 liters of NaOH. Our last step is to
divide moles of NaOH that we found in step one by the liters of NaOH that
we found in step two. Because we’re finding molarity, which was the moles of
NaOH divided by the liters of NaOH. This is why we first have to start with
finding our moles of NaOH, then finding our liters of NaOH, dividing and that’s
our final answer. So I’m going to put the moles of NaOH on top that we found in
step one. And divide that by the liters of NaOH that we found in step two. Once
we do this, this would give us 0.0132 molar solution of
NaOH. So by molar solution I really just mean molarity. That capital M means
molarity, you could also put moles over liters as your final answer but you’ll
commonly see a capital M. Why don’t you try one out on your own, I placed a
practice problem at the very top of the description box and at the bottom you’ll
find the step-by-step solution. Let me know how you do in the
comments and guys my personal mission is to help millions of students with
chemistry so if I’ve helped you in any way let me know by subscribing and
liking this video. Alright I wish you the absolute best, I’m gonna go and record
the voiceover now. So I’ve been working on these molarity notes for a week, I had
to make sure to put how to find molarity, different examples of solution
stoichiometry, I even covered dilutions. These notes are
so much easier to understand than the average textbook. You know I wish I had
something like these notes because when I was in class there wasn’t enough time
and I was always rushing to write everything down. So check the link in the
description, download the notes and master molarity. Alright guys I’ll see
you in the next video

52 thoughts on “How to Do Solution Stoichiometry Using Molarity as a Conversion Factor | How to Pass Chemistry

  1. I don't know who you are but I just saw you on Video Creators and GIRL!!! You found your calling. I seriously want to know more about chem too!

  2. You are amazing. After hours of watching youtube videos yours is the only one that has helped me grasp the Molarity concept. Thank you !!!

  3. Between you and Kahn academy, I jut might be able to make it to the second year of my Biotechnology program lol… Thank you!

  4. can you help me with the following problem an 8%(w/v) solution of HCL has what molarity(M)? this is the type of questions my teacher uses .

  5. Woooow! I really appreciate your help, thank you for your hard work and dedication, you’re a really good teacher, I’m late studying for my final exam in the AM and I feel way more confident! 🙏🏾

  6. When 7 minutes of teaching from YouTube is better than a month in high school…..holy fucking shit thank you 😯😤😫

  7. I have a dilution lab exam tomorrow and this is saving my ass . Since it’s “advanced” professor did 1 example lmfao thanks ! Keep up the chemistry . You think you can do a bit of analytical chem like PH related

  8. You are AMAZING!!!! My prayers have been heard you made the exact video I needed THANK YOU!!!!! You are so sweet and calm which is great for learning :))))))

  9. Thank you so much! I love how you broke down exactly what the equations is asking and why. This helped tremendously. Now when I look at these type of equations I know what they are asking to be solved and why! Again, thanks!!!!!

  10. I absolutely loved the way you explained the lesson and the practice problem in your description box was brilliant. It helped me a lot and I thought it was worth mentioning that this video helped me so much. Thanks!!!

  11. I am 54 years old finishing my Bachelors in nursing (after this class). After not having chemistry for 35 years I had a lot of catching up to do. My class is 7 weeks long, and that's a lot of information! Thanks so much for making it easy. You are amazing!

  12. You are an amazing teacher. I sit in an hour and a half lecture walk away knowing very little then I watch your video for six minutes and completely understand the concept

  13. thank you sm for teaching me something in 8 minutes that my chem teacher couldn’t in 2 weeks 😂💛

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