These high school chemistry worksheets are full of pictures, diagrams, and deeper questions covering all aspects of stoichiometry! This unit is meant to cover the basics of stoichiometry, the mole concept, empirical and molecular formulas, percent composition, limiting reactant problems, and percent yield problems.
This unit is designed to help students practice these skills that are important for the rest of the year in chemistry.
This unit is part of my Differentiated Chemistry Whole Year Homework Bundle.
Frequently Asked Questions:
1. Why should you purchase this bundle?
- This unique homework sheet unit (as part of my whole year bundle) will help you teach a cohesive and consistent program all year.
- These homework pages were carefully designed with a wealth of images! I have created many of the images myself and other images are licensed from professional designers. These homework pages are not plain worksheets with text questions that can be easily copied from a textbook. Each page is meant to help students to learn chemistry in a very visual way. Students circle, color, and analyze pictures and diagrams in ways that are far superior to plain text textbook questions.
2. What does this unit contain?
This unit contains 16 pages (I am working on posting them to be sold separately as well):
1. Introduction to the Mole
2. Calculating Molar Mass from a Chemical Formula
3. Mass to Moles
4. More Mole Calculations
5. Writing Empirical Formulas from Chemical Formulas
6. Percent Composition from Chemical Formula
7. Decipher the Empirical Formula: Two Ways
8. Molecular Formula from the Empirical Formula and Molar Mass
9. Percent Composition, Empirical Formula, and Molecular Formula
10. Stoichiometry: Moles to Moles
11. Stoichiometry Word Problems: Given Moles of Reactants
12. Stoichiometry Word Problems: Given Mass
13. Stoichiometry Word Problems: Given Volumes and Densities
14. Limiting Reactant Problems: Step by Step
15. Limiting Reactant: A Big Word Problem
16. Percent Yield from Actual and Expected Yields
3. How many pages does this unit contain?
This unit contains 16 student pages plus an answer key for every page.
4. What will the format of each page be?
Each page will be unique. Each is designed to roughly cover the material that I would teach in an hour long class period. These are terrific for daily homework assignments because they don’t take too long to complete.
These pages have been carefully designed in Illustrator. I have created a unique set of questions to help students to review material taught in class and think deeper about the material. Many of the pages ask students to highlight or color something, to identify items in a diagram, to match related concepts, or interact with a topic in a new way. Many of the pages ask students to connect more than one concept; they are intended to help students see the bigger picture in each unit. A few pages ask students to use the internet to do a little research.
If you own any of my other resources, don’t worry about repeat pages. These homework pages are truly unique and separate from my activities. These homework pages will truly complement any activities or resources you already have or use in your class.
5. How do I handle homework?
First of all, I don’t grade it. I learned in my early teaching years that when I grade homework, I am rewarding students who copied off of their one studious friend the period before my class, and I am penalizing students who have limited educational time outside of school. I often give time at the end of the period to work on “homework” pages. Often, I start off the next day’s class with the answer key projected onto some sort of screen (ELMO or projector) so that students can check their answers as they walk in. My students know that they will do better in my class if they do the homework and I care about effort more than being correct.
6. What if you want to grade homework? Are answer keys included? Are they easy to grade?
Answer keys are included (for almost all of the pages, where it makes sense to have an answer key). I designed these pages to be pretty simple to grade, if you want to do that.
7. Why is each homework page only one page?
In my time as a teacher, I have noticed that for some reason, homework assignments that have more than one side of a page are just neglected by students. If I hand out a one sided homework page and tell them, here’s your homework, they say, yay, it’s just 1 page! They will often at least start it if not finish it before the end of the day. I really think there is a psychological barrier to starting an assignment with two sides. Call me crazy, but test it out! Try giving my homework assignments and watch your class actually do their homework! If you want to give two pages, you can print these out and then copy 2 back to back for your students.
A way to save paper would be to print all of the homework assignments and copy them as a packet. This is great to give students all at once in the beginning of the unit, so they have every page in advance, which works great if they’re absent!
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Now, we're going to go back to our diagram showing us the relationship between
the unknown and the known substances.
Only now, instead of stopping and
starting with moles of our substances, we're going to go one step further out and
look at the masses of our known and our unknown.
Note, that we're still going to go to moles of our known.
Use the coefficients of our balanced chemical equation.
Find the moles of B, and then convert to the mass of B.
This step here in the middle,
converting from moles of A to moles of B using our coefficients for
balanced chemical equation, is going to be done in every single problem.
So let's look at an example of how we would do this.
First, we have an equation.
We do want to check to make sure it's balanced.
And what I see, is that it's not balanced.
Because immediately, I notice that there are two nitrogens on the right side of
the equation but only one on the left side.
So the first thing I need to do, is go through and balance the equation.
I'm going to make a short list to tally up what I
have on either side of the reaction.
So I have nitrogen, oxygen, and hydrogen.
And I have 1 nitrogen on the left, I have 3 oxygens on the left and 2 hydrogens.
On the right side of my equation, I have 2 nitrogens,
I have 4 oxygens and I have just 1 hydrogen.
So, because I see oxygen in multiple substances,
I'm actually going to balance that last.
I'm going to start with my nitrogen, and I'm going to put a 2 in front of the NO2,
so that changes my nitrogens to two, and it also changes my oxygens to now
a total of five oxygens, because I have two times two is four, plus one is five.
And I see that it doesn't have any effect on the hydrogen.
If I put a 2 in front of HNO3 to balance my hydrogens,
I get the 2 hydrogens on the right side.
I also see that it changes my nitrogens to 3 and it changes my oxygens to 7.
So I have, 2 hydrogens on the right, two nitrogens plus one nitrogen for
three nitrogens, six plus one, seven oxygens.
So, now I see that again my nitrogens are out of balance.
I change that to a three for the nitrogen.
Now I have six, seven high oxygens, I still have
2 hydrogens on the left and now it appears that my equation is balanced, but
I'm always going to go back through and double check to make sure it's correct.
So, we have three nitrogens on the left.
We have 3 nitrogens on the right.
We have six plus one, seven oxygens on the left.
We have six plus one equals seven oxygens on the right,
we have two hydrogens on the left, and we have two hydrogens on the right.
So, now that I know I have a balanced chemical equation,
I can proceed with the stoichiometry calculation.
In this case it's asking how many grams of nitric acid
are produced from 100 grams of nitrogen dioxide with excess water.
So the first thing I want to do,
is pull out the information that seems to be useful in this problem.
I see that I have HNO3 grams equals question mark.
That's just a reminder that that's the answer I'm trying to find.
It tells me I'm starting with 100 grams of NO2 is my starting material.
Plus I have excess water.
And the only reason we have to worry about it being excess water,
is to know that we're not going to run out of the water.
The NO2 will be able to react completely to form the maximum amount of HNO3.
Now, I see that I'm going to have to convert between grams and
moles of substances, so I want to find the molar mass.
And I can do this with the information on the periodic table.
Looking at the molar mass of nitrogen and oxygen, considering that I
have two oxygens, I find the molar mass of NO2 is 46.01 grams per mole.
And that, for HNO3, the molar mass is 63.01 grams per mole.
Note, that I find the molar mass for
the substance as written, excluding any coefficients.
I'll take the coefficients into account when I do my calculation.
But the molar mass of a substance is going to be the same regardless of
what its coefficient is or isn't in a chemical equation.
Now, I look and see which of these but,
substances will allow me to find the moles of that substance.
If I look at HNO3, I don't know the mass, I only know the molar mass.
So, I'm not going to be able to determine the moles of HNO3.
So, that's not going to be my starting point.
For water, I know we have excess water, but
I don't know a specific amount of water.
So, that's not going to help me get to moles of some known quantity.
The only thing that remains is my NO2.
I have 100 grams of NO2.
Now, given the molar mass of NO2,
I can actually go from grams of NO2 to moles of NO2.
And so I use the molar mass,
putting the 46.01 grams on the bottom, so that my units will cancel out.
So now, I can cancel out grams of NO2 with grams of NO2.
If I stop my calculation at this point, I would have mols of NO2.
Which tells me how much I started with, but doesn't tell me anything about
the mass of HNO3, which is what is being asked for in the question.
So now, I need to use my mol to mol ratio.
Just as we did in the previous calculation,
where we were going from moles of one substance to moles of another.
I still have used by mass to get to moles.
Now I can put in my mole ratio.
Now I'm going to use the coefficients from my balanced equation, so I have 3 moles
of NO2, from the coefficient of 3 in my equation, on the bottom.
And I put that on the bottom, so
that moles of NO2 will cancel with moles of NO2.
On the top, I'm going to have 2 moles of HNO3,
because that's what I'm trying to find.
So now, moles of NO2 cancels with moles of NO2.
And if I stopped here, I'd have moles of HNO3 as my answer.
But I don't want moles of HNO3, I want grams of nitric acid or grams of HNO3.
So, I need to do one more step to get from moles of HNO3 to grams of HNO3.
And again, I'm going to use my molar mass, this time for the HNO3.
And I know 63.01 grams per mole of HNO3.
Now, my moles of HNO3 cancels with my moles of HNO3.
I check back over my units and
I see that the only units I have left remaining are grams of HNO3.
Now, I can do the calculation by taking
100 times two times 63.01 divided
by 46.01 divided by three, and
what I find is that I get 91.30 grams of HNO3.
I notice that my answers are on the same order of magnitude,
100 grams of NO2 to 91.3 grams of HNO3, so that seems to be a reasonable answer.
Now let's look at an example problem involving the decomposition of
sodium azide, which is used in airbags, and
it produces as large volume of gas in a very short time frame.
Let's look at our plan to figure out how we
can get the mass of N2 produced from a given mass of sodium azide, or NAN3.
Remember, that the only relationship we know between amounts of
two different substances in an equation are the molar amounts.
So, if I want to look at a relationship between two substances,
I first have to convert to moles.
Because if I know the moles of one substance, I can use my mole ratio for
my balanced chemical equation to find the moles of the other substance, and
then I can convert to grams.
Remember, that nitrogen exists as a diatomic or N2.
Therefore, when I'm looking for my molar mass of N2,
I'm going to want to put it in there as 28.02 grams per mole.
If we were referring to atomic nitrogen, we would need to specify that.
Remember, for diatomic substances such as nitrogen or hydrogen, oxygen, when we
say the name of the element we assume that it is in the form of the diatomic element.
We find that we can produce 6.47 grams of nitrogen.
Remember, that we're given our 10 grams of sodium azide, and
we're asked to find the grams of nitrogen.
Here we can set up our 10.0 grams of NaN3.
We need the molar mass of sodium azide, so
we use the values on the periodic table, and
that gets us 65.01 grams per mole of NaN3.
So now, our grams cancels with grams.
Now we have moles of sodium azide.
And now we need to use our mole to mole ratio from our balanced equation.
Notice, that our coefficient tells us, we have 2 moles of sodium azide and
we have 3 moles of Nitrogen.
So, our moles of sodium azide will cancel.
Now we have moles of nitrogen, and we're going to use our molar mass of 28.02 grams
per mole of nitrogen to find the grams of nitrogen.
And so what we find, is that we get 6.47 grams.
Note, that I had to convert to moles first, before I could use my mole ratio.
These numbers come from the balanced chemical equation.
In the next module, we're going to look at limiting and excess reagents.