Do you like chemistry worksheets?  Did you not notice that they’re all listed on the right sidebar of this website?  If you answered “yes” to both of these questions, then you’re in the right place to do some practice chemistry worksheets!  Here they are:

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Science and math myths and facts

It’s often said that “math is the language of science”, which means that in order to do sciencey stuff, you have to know a lot of math.  On this page, we’re going to figure out exactly what the relationship is between math and science.  Let’s get started!

 1.  Studying math without studying science is a waste of time.

Imagine a math problem – any math problem.  How about 2 + 2 = 4?  It’s simple enough to understand, isn’t it?

Sure it is.  It’s also irrelevant.  If you say that 2 + 2 = 4 by itself, you might as well have memorized the equation that “b + q = 34.” I imagine that you’re probably bright enough that you could memorize any number of relationships like this without any trouble – even if they were meaningless.

And that’s the point:  By itself, math doesn’t actually do anything.  In order to make sense of 2 + 2 = 4, you need to understand that if you add two of something to two of something else, you’ll have four of those things.  This simple relationship (and some of the less simple relationships you’ve doubtlessly seen in meth class) are studied not because they’re fun to look at, but because they allow us to figure out how things work in the real world.  Not all of these things are as easy to understand as “two things plus two things equals four things”, but they can be manipulated so that they have meaning.

Without science, learning math is like learning Klingon.  Fun for weirdos, but very handy.

2.  Science without math won’t get you very far

Let’s say that I want to launch a bunch of people into space.  I know that if I add energy to a mixture of hydrogen and oxygen gases that they’ll give me a huge quantity of energy.  Let’s build a rocket and fire it up!

There’s only one problem:  We don’t want to kill the people inside the rocket.  Let’s say that we dump a bunch of hydrogen and oxygen into a rocket and light it.  Maybe we’ll have too much energy and the thing will blow up.  Maybe we’ll have slightly less energy and just send the people on a one-way trip to Pluto.  Maybe the rocket will go 10 km up and then run out of fuel.  Or maybe the rocket will make a quiet “pop” and sit on the pad.  There’s a reason we have rocket scientists:  It’s hard to make a rocket actually work.

If we don’t have math, we’ll never be able to predict how much of the fuel to combine to make the rocket do the most unlikely of possible things:  Work normally.  We’ll forever be blowing up rockets and astronauts because we’re just using trial-and-error to make the rocket work.

On the other hand, if we use math, we can figure out the relationship of hydrogen and oxygen that will be needed to make the reaction work properly.  Once this is done, we’ll be able to figure out how much of each will be needed to make it happen.  And once the astronauts get into space, we’ll be able to predict how many Twinkies will be needed to feed them.

This isn’t to say that you can’t do science without math.  However, if you want to move past the “what happens if I hit my neighbor over the head with a brick” phase, you may may want to get studying.

3.  Science is messy and math is not.

Let’s say that you’ve got a bunch of 1 cm pieces of wood and you want to simultaneously fit as many as possible into an opening in another piece of wood.  How many pieces of wood will this take?

If you’re a mathematician, you’ll say that if you have 9 pieces of 1 cm wide wood, then they’ll fit into a 9 cm wide opening.  For those of you that aren’t good with the math, 9 x 1 = 9.

Which is the wrong answer.  We don’t live in a perfect world where things fit perfectly together.  No matter how awesome I am, I won’t be able to make these pieces of wood 9 cm thick nor will I be able to make an opening exactly 9 cm across.  In the real world, I’ll get nine pieces of wood, and then use either glue to fit them if they’re too small or shave one of them a little thinner to make it fit.

This is a perfect example of how science and mathematics are related.  If I were a mathematician who had no interest in science, I’d be confused why the wood didn’t fit correctly.  If I were a scientist who had no interest in math, there’s an excellent chance I’d pick the wrong number of pieces of wood to start with.

This is why we say that math is the language of science.  It’s not just an analogy – it’s literal truth.  When my son was first born, I joked with my wife that it might be funny to teach him Pig Latin as his first language.  My wife didn’t find this as amusing as I did.

If we’d done this, it would have been the same thing as teaching him math without ever exposing him to science.  Sure, he’d have interesting and creative ideas, but without a useful language he’d have no way of making anybody else understand what he was talking about.

Likewise, let’s say that my son had an enormous capacity for scientific thought.  Unfortunately, let’s imagine that we never actually spoke around him so he had no first language at all.  How far do you think he’d get in his studies, and if he succeeded in doing something, how would he ever tell anybody.


We scientists like to make fun of mathematicians as being lame number crunchers who never do anything useful, but the truth is that without their work we’d never be able to figure anything out.  At this point in history, the fastest we can travel is light speed (and then only if we’re made of light).  If there’s a way to build a faster-than-light drive, do you really think that we’ll figure it out without learning a lot of new math?  Not likely.

Likewise, mathematicians think that we scientists needless get our hands dirty with a lot of phenomena that make everything more complicated than they should be.  And they’re right – the world is a messy place and nothing is ever perfect.  Mathematics describes how things should be, while science takes into account how they really are.

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Reaction quotients

If you let a chemical reaction run for long enough, it will reach equilibrium.  That’s just a fact of life.  Reaction + time = equilibrium.

However, how can you tell if a reaction has reached equilibrium yet?  The answer: Calculate the reaction quotient, Q.


Or you could ask this guy.

Let’s learn how.

What is Q?

Let’s consider the equation we’ve been looking at for the last few tutorials and take a look at the equilibrium expression:


Now, let’s take a look at the expression for the reaction quotient:


See anything familiar?  If not, go take a look again.  Yep, you got it:  K and Q are the same thing.  This leads to an obvious question:  Why do we have the same  expression for two different values?

Not exactly.  Check it out:

  • K describes a system at equilibrium.  For the reaction above, K describes the concentrations of every chemical in your beaker when everything has reached equilibrium.
  • Q describes a system which may or may not be at equilibrium.  The very instant you throw your reagents into a flask, you can start calculating Q.  Not surprisingly, Q will change over the course of the reaction as the concentrations of the reagents and products shift.  However, as the equation suggests, once the reaction finally reaches equilibrium, K and Q are the same.

Essentially, Q is just a way of monitoring the progress of a chemical reaction that hasn’t yet achieved equilibrium.

If I look at Q, what does it tell me?

It tells you whether a chemical reaction has reached equilibrium.

  • If Q > K, this means that there are more products in your particular mixture than there will be at equilibrium.  As a result, as time progresses, you can expect to see some of the products revert back to reagents.
  • If Q < K, this means that there are more reagents in your mixture than there will be at equilibrium.  As time passes, you’ll see more products being formed.
  • If Q = K, you’ve reached equilibrium.  You’ve reached your final concentrations of products and reagents.

In a practical sense, Q allows you to figure out if your reaction is done yet.  When you throw a bunch of chemicals into a beaker, you can monitor the concentrations of each chemical to figure out what Q is.  When the Q you calculate is equal to the K value you know for the reaction, then you’re done!

Photo credits:

John de Lancie, the man who played “Q” on several Star Trek TV series:  By “Pinguino” (“Pinguino’s” flickr account) [CC BY 2.0 (, via Wikimedia Commons.

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What do equilibrium constants tell us?

You may be wondering why the tutorial about what equilibrium constants are good for is after the tutorial about how to calculate them.  The answer for this is simple:  Because if you don’t know what they are, you can’t really understand what they mean. Fortunately, your knowledge is now to the point where you know what they are, so we can get going with all of this.


The company I use for clip art thinks that this is what a knowledgeable person looks like.  I’ll wait while you go put on a suit.

So, without further ado, let’s get going!

What Equilibrium Constants Mean

When you see an equilibrium constant, you probably think to yourself, “Hey, there’s a number!”  And if you see another one, you probably think to yourself, “Hey, there’s another number!”  This, of course, assumes that  you’ve got a lot of free time to look at numbers.

As it turns out, equilibrium constants tell us quite a bit about what’s going on with an equilibrium.  Check it out, using our example from the last tutorial:


  • If K is a lot bigger than 1, the reaction contains mostly products.  In our example, we can see that a huge value for K would correspond to a large concentration of C and small concentrations of A and B.  Our reaction, then, has mostly converted the A and B into the product, C.
  • If K is a lot smaller than 1, the reaction contains mostly reagents.  In our example, K can only be small if either C is very small or A and B are huge.  Either way, this means that the reaction hasn’t done much since it got started.
  • If K is 1, then there’s roughly the same amount of product and reagent.  This may, depending on the equation, be a little bigger or smaller for one or the other, but it’s about right.

And that’s it!

Image credits:

Suit guy with computer:  Image courtesy of Ambro at

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Equilibrium Constants

Now that you understand what equilibria are (and if you don’t, check this out), it’s time to start doing some math.  For those of you who aren’t big fans of math, don’t worry – it’s not as bad as it sounds.


Get a grip, dude.

Before we get started, it’s time to go back to our friend from the kinetics tutorials, the rate constant.  As you’ll recall, the rate constant is a value called “k” that allows us to figure out the relationship between concentration and reaction rate.  For a reaction with the general format A + B → C + D, the rate of the reaction can be expressed as:

rate = k[A]ˣ[B]ʸ

Where “x” is the order of the reaction in A and “y” is the order of the reaction in B.  If you don’t know what all this means, think back to the kinetics stuff that you learned a while back.

In any case, equilibrium processes have forward and reverse rates that are the same.  Or, in math terms:

rate(forward) = rate(reverse)

Given that there are two reactions, one in which A + B → C + D and the other where C + D → A + B, we end up with two rate expressions.  Namely:


Where “z” is the order of the reaction in C and “n” is the order of the reaction in D.  “kf” and “kr” refer to the rate constants of the forward and reverse reactions, respectively, while “ratef” and “rater” equal to the rate of the reaction in the forward and reverse directions.

At equilibrium, we run into a cool situation where the forward and reverse rate of the reaction are the same.  This allows us to come up with a single expression to relate the equilibrium to the concentrations of every chemical involved:


Using the power of algebra, we can just go ahead and, for what seems like no reason at all, separate some of the terms from each other.  Not to worry, there’s a reason for doing this:


And, given that both “k” terms are constants (they are, after all, called “rate constants”), we can just combine them together in a term we’ll call the equilibrium constant:


This, my friends, is the equation that we use to do math using equilibria.

An aside:  Figuring out z, n, x, and y

You may be wondering, given the complexity of rate laws, how you can figure out the z, n, x, and y superscripts in this equation.  For complicated reactions, this can be very difficult, but for chemistry that you’re likely to run into, just use the coefficients in front of each compound in the equation to figure it out.  For example, in the case of the reaction:


The equilibrium expression would be:


Where the superscript over nitrogen is 1 because there’s no number in front of it in the equation, the superscript in front of water is 2 because there’s a 2 in front of it in the equation, and so forth.

How can we use this expression?

Well, let’s say that the following chemical reaction takes place with the following rate law:


Now, let’s say that the equilibrium constant for this reaction is 3.6 x 10⁻¹¹ for this process.  Given all this cool info, here’s a practice problem:

Example:  I have a solution that contains 0.50 M of compound A and 0.20 M of compound B.  If I start with no C at all and let the reaction run for a while, what will the equilibrium concentrations of each compound be?

Answer:  Plug it all into the equilibrium expression.

Here’s how:

  • Initially, you’ve got 0.50 M of compound A.  At equilibrium, some of it will have reacted to form C.  How much?  We don’t know, so let’s just say that the concentration has decreased by “x”.  This means that at equilibrium, the concentration of A will be “0.50 – x” M.
  • Initially, you’ve got 0.20 M of compound B.  At equilibrium, some of it will have reacted to form C.  Again, the concentration has decreased by “x”, which means that the equilibrium concentration will be “0.20 – x” M.
  • Initially, there’s no C at all.  However, when A and B react, some unknown quantity of C will form.  We don’t know how much this is, but we do know from the chemical equation that the amount that forms will be twice as much as the amount of A and B that reacted.  Our equilibrium concentration, then, will be 2x M.

From here, it’s a simple matter of plugging all of these values into the expression for the equilibrium constant to figure out what this mystery x value is:


And, using our friend the quadratic equation, we find that x = 0.0043, which means that:

[A] = 0.50 – 0.0043 = ~0.50 M

[B] = 0.20 – 0.0043 = ~0.20 M

[C] = 2(0.0043)² = 3.7 x 10⁻⁵ M

Wasn’t that fun?

No.  It was not fun.

Most of us don’t like using the quadratic equation much.  It’s kind of annoying to memorize and anybody with any sense just goes online to find something that will calculate it for you (here’s where I did it).  Not fun.

However, let’s consider this:  In many equilibria, the amount of product that’s formed isn’t very much.  As a result, we can assume that “x” is going to be much smaller than the concentrations of the reagents we started with.  What this means is that the term [0.50 – x] can just be written as [0.50], since subtracting a very tiny number from 0.50 still gives you basically 0.50.  Using this simplification, we can simplify the equation above to give us:


Which, when solved, gives us an x value of 0.0042 M.  This isn’t identical to what we found above, but it’s a pretty small difference.  Most of the time, that’s A-OK with us.

Image credits: 

Screaming guy:  Image courtesy of imagerymajestic at

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Acids and bases

Acids and bases:

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Colligative properties and Ksp

Colligative properties and Ksp values:

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Solutions and concentration

Quick introductory worksheet for solutions:

  • Solutions practice sheet:  Suspensions, colloids, solutions.  And solvents and solutes.  And the magic of drinking seawater.

How solutions are born – Polarity and IM forces in solutions:

Units of concentration:

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Types of chemical reaction

Types of reaction:

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Balancing equations worksheets:

Writing complete equations:

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