Tuesday 20 December 2011

how to balance chemical equations

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Being able to balance chemical equations is a very important skill for students of chemistry, making this one of our most popular chemistry tutorials. All chemical calculations require you to work with a balanced equation. Here we will show you a simple, easy way to balance all chemical equations you will meet at introductory level chemistry courses.
(A completely different method is used if you have to work with Half Equations. If you don't know what a Half Equation is, then don't worry, as that means they are not in your course and you will not have to worry about balancing them.) The key to balancing chemical equations is to apply the rules below.
If you already know the rules, then try these worked examples for revision practice.
What is a balanced equation?

A chemical equation is balanced when the number of atoms of each type on each side of the equation is the same. Which means if you have 12 hydrogens on the left hand side of the equation, you must have 12 hydrogens on the right hand side, if there are 4 oxygens on the left, there must 4 oxygens on the right, and so on. This is because of the law of conservation of mass - you can't make or destroy atoms during a chemical reaction. But you can't just add atoms at random to each side, you have to work with the molecules of the reactants. Also, you will find it very tricky to try to balance a word equation, it is very much easier to use a chemical equation with chemical symbols, as then you will be able to see how many atoms of each type are in each chemical.

Example 1
Unbalanced Equation:- C3H8 + O2 ---> H2O + CO2

There are three carbons on the left, but only one on the right.
There are eight hydrogens on the left but only two on the right.
There are two oxygens on the left but three on the right.

Balanced Equation:- C3H8 + 5O2 ---> 4H2O + 3CO2
How do we balance the equation?

Balancing chemical equations isn't difficult, once you know the way to do it. Start by finding out how many atoms of each type are on each side of the equation. Some teachers recommend making a little table listing the numbers of each atom for the left hand side and for the right hand side.

Next, look for an element which is in only one chemical on the left and in only one on the right of the equation. (But it is usually a good idea to leave hydrogen and oxygen until you've done the others first.)

To balance that element, multiply the chemical species on the side which doesn't have enough atoms of that type by the number required to bring it up to the same as the other side. The number is called the coefficient.
BUT
If you have to multiply by, say, 2 1/2, do so, THEN multiply EVERYTHING on each side of the equation by two to get rid of the half.
We don't like having halves in equations, as you can't get half a molecule.

Now look for the next element or species that is not balanced and do the same thing.

Repeat until you are forced to balance the hydrogen and oxygens.

If there is a complex ion, sometimes called a polyatomic ion, on each side of the equation that has remained intact, then that can often be balanced first, as it is acts as a single species. The ions NO3- and CO32- are examples of a complex ion.

A VERY useful rule is to leave balancing oxygen and hydrogen to the last steps as these elements are often in more than one chemical on each side , and it is not always easy to know where to start. Some people also say you should leave any atom or species with a valancy of one one until the end, and also generally leave anything present as an element to the end.

In Example 1 above, you would balance the carbons first, by putting a 3 in front of the CO2, then balance the hydrogens by putting a 4 in front of H2O and finally the oxygens (which are in more than one compound on the right, so we leave them until last) by putting a 5 in front of the O2.

Example 2
Unbalanced equation:- H2SO4 + Fe ---> Fe2(SO4)3 + H2

Balance the SO4 first (as it is a complex ion and it is in one chemcial species on each side)

3H2SO4 + Fe ---> Fe2(SO4)3 + H2

Now balance the Fe (which is also in one chemical on each side)

3H2SO4 + 2Fe ---> Fe2(SO4)3 + H2

Finally, balance the hydrogen (although it is in one chemical species on each side, it is usually a good idea to leave it until last)

Balanced Equation:- 3H2SO4 + 2Fe ---> Fe2(SO4)3 + 3H2

We alter the coeficients in the equation.
Do NOT touch the subscripts for the atoms in a chemical species, or you will change it into an different chemical. That would be a bit like saying I want six chicken legs for a meal, so I'll go get a six-legged chicken. As chickens have two legs, you will need three normal, two-legged, chickens, not a six-legged mutant monster, probably from outer space.

If you start by trying to balance something which is in more than one species on one side, you can't easily tell which species you should have more of, and so can end up going round in circles, continually altering things. If this happens, just start again, but balancing atoms or complex ions that are in one species on each side. (This is important or it will not work.)
What about these halves I mentioned?

Example 3
Unbalanced Equation:- Al + O2 ---> Al2O3

We can start with either the Al or the O, but we will start with Al, as we normally leave O to the end. Even though it wouldn't matter where we started in this case.

Put a two in front of the Al on the left
2Al + O2 ---> Al2O3

Put a 1 1/2 in front of the O2 on the left
2Al + 1 1/2O2 ---> Al2O3

We don't like halves, so multiply everything on BOTH sides by two
4Al + 3O2 ---> 2Al2O3

It's now a balanced equation. Easy, eh?

Example 4
Unbalanced equation:- Na2CO3 + HCl ---> NaCl + CO2 + H2O

Well, the sodium (Na) is not balanced yet and is in one chemical species on each side, so we need two NaCl on the right, agreed? Remember, we tend to leave H and O until the end.

Na2CO3 + HCl ---> 2NaCl + CO2 + H2O

There is only one Cl on the left, but two on the right, so let's sort that one next, by making it 2HCl on the left

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O

And now if you look carefully, you'll see that the number of carbon, hydrogen and oxygen atoms on each side are the same, so it's balanced.

Example 5
Unbalanced equation: - Mg3N2 + H2O ---> MgO + NH3

We can see that there are 3 Mg atoms on the left but only one on the right, so let's sort that one first - Mg3N2 + H2O ---> 3MgO + NH3

We could continue by balancing the O or the N next as we can see they don't add up. But let's stick with our useful rule of leaving O and H to the end if possible, and go for the nitrogen next. There are two on the left side and only one on the right, so put a two in front of the NH3

Mg3N2 + H2O ---> 3MgO + 2NH3

At this point we can go for either the oxygen or the hydrogens, so we'll do the oxygens, just because I want to. We need to add a three in front of the H2O to get three oxygens on each side

Mg3N2 + 3H2O ---> 3MgO + 2NH3

And a quick count shows that we have balanced everything now! For practice, try this one by balancing the O before the N, it will give you the same answer, of course.

Once you have had a bit of pracise at balancing chemical equations, lots of other questions asked in chemistry become much easier to solve.

There are also a set of worked examples of balancing chemical equations, complete with answers for you to practice with. The answers are hidden so you can try them yourself before clicking to reveal each one in turn.
 

A chemical equation describes what happens in a chemical reaction. The equation identifies the reactants (starting materials) and products (resulting substance), the formulas of the participants, the phases of the participants (solid, liquid, gas), and the amount of each substance. Balancing a chemical equation refers to establishing the mathematical relationship between the quantity of reactants and products. The quantities are expressed as grams or moles.

It takes practice to be able to write balanced equations. There are essentially three steps to the process:

    Write the unbalanced equation.

        Chemical formulas of reactants are listed on the lefthand side of the equation.
        Products are listed on the righthand side of the equation.
        Reactants and products are separated by putting an arrow between them to show the direction of the reaction. Reactions at equilibrium will have arrows facing both directions.

    Balance the equation.

        Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation. Tip: Start by balancing an element that appears in only one reactant and product.
        Once one element is balanced, proceed to balance another, and another, until all elements are balanced.
        Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because this will change the formulas.

    Indicate the states of matter of the reactants and products.

        Use (g) for gaseous substances.
        Use (s) for solids.
        Use (l) for liquids.
        Use (aq) for species in solution in water.
        Write the state of matter immediately following the formula of the substance it describes.

Worked Example Problem

Tin oxide is heated with hydrogen gas to form tin metal and water vapor. Write the balanced equation that describes this reaction.

    Write the unbalanced equation.

    SnO2 + H2 → Sn + H2O

    Refer to Table of Common Polyatomic Ions and Formulas of Ionic Compounds if you have trouble writing the chemical formulas of the products and reactants.

    Balance the equation.

    Look at the equation and see which elements are not balanced. In this case, there are two oxygen atoms on the lefthand side of the equation and only one on the righthand side. Correct this by putting a coefficient of 2 in front of water:

    SnO2 + H2 → Sn + 2 H2O

    This puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four hydrogen atoms on the right. To get four hydrogen atoms on the right, add a coefficient of 2 for the hydrogen gas. Remember, coefficients are multipliers, so if we write 2 H2O it denotes 2x2=4 hydrogen atoms and 2x1=2 oxygen atoms.

    SnO2 + 2 H2 → Sn + 2 H2O

    The equation is now balanced. Be sure to double-check your math! Each side of the equation has 1 atom of Sn, 2 atoms of O, and 4 atoms of H.

    Indicate the physical states of the reactants and products.

    To do this, you need to be familiar with the properties of various compounds or you need to be told what the phases are for the chemicals in the reaction. Oxides are solids, hydrogen forms a diatomic gas, tin is a solid, and the term 'water vapor' indicates that water is in the gas phase:

    SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g)

    This is the balanced equation for the reaction. Would you like to see another example of balancing equations?
Why do we need to balance equations, anyway?

Remember that at the beginning of the year, we did a lab where we added baking soda to vinegar and collected the whole mess in a rubber glove? Well, most of you were able to show that the mass of the stuff that we made was the same as the mass of the stuff we started with. (If you weren't one of those lucky people, then let me be the first to tell you this: The mass of the stuff that you make in a chemical reaction is the same as the mass of the stuff that you start with).  This is called the Law of Conservation of Mass.

Now, this shouldn't really be all that surprising, considering that this is true for most everything else in life. For example, when I make my world-famous chili, the weight of the chili that I make is the same as the weight of all the ingredients added together. As it is with chili, so it is with chemical reactions.

Now, when we write chemical equations, we need to have the formulas for the reagents on the left side (the stuff that's going to do the chemical reaction) and the formulas for the products (the stuff you make) on the right. If we were to simply put the formulas of the chemicals on the left and right without saying how much of it was going to react, then we would run the risk of saying that the mass of what we end up with is different than the mass of what we started with. This would be the same thing as writing a recipe where we didn't specify how much of each ingredient is needed to make the chili.

The bottom line: You need to balance the equations by sticking numbers in front of the chemicals on the left and right sides of the equation, like it or not. How can you do this? Check out the next section, titled...

... Balancing Chemical Equations

OK. You know why you need to balance chemical equations, but you don't yet know how to do it. It turns out that I'm star who knows how to explain things in a way that even the dumbest people know how to follow. And, hey, if the dumbest people can figure it out, so can you!

Listen: There are four easy steps that you need to follow to make this work. Here they are:

1. Get yourself an unbalanced equation. I might give this to you, or I might make you figure it out. Either way, if you don't have an equation with all the chemical formulas and the arrow and all that other stuff, then you're out of luck.

2. Draw boxes around all the chemical formulas. Never, ever, change anything inside the boxes. Ever. Really. If you do, you're guaranteed to get the answer wrong.

3. Make an element inventory. How are you going to know if the equation is balanced if you don't actually make a list of how many of each atom you have? You won't. You have to make an inventory of how many atoms of each element you have, and then you have to keep it current throughout the whole problem.

4. Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Whenever you change a number, make sure to update the inventory - otherwise, you run the risk of balancing it incorrectly. When all the numbers in the inventory balance, then the equation can balance, and you can relax and enjoy a delicious bowl of Mr. Guch's chili.

Read on for an example...

An example of equation balancing:

Let's say I ask you the following thing on a test: "Balance the equation that takes place when sodium hydroxide reacts with sulfuric acid to form sodium sulfate and water." How do we solve this using the steps above?

1. Get yourself an unbalanced equation. Here's where you use your knowledge of formulas to help you out. If you know what the formula of sodium hydroxide, sulfuric acid, sodium sulfate, and water are, you'd be able to write the following unbalanced equation:

These are simply the formulas for the chemicals named in the problem



2. Draw boxes around all the chemical formulas. This is the step that people frequently don't do because they feel that it's a stupid thing to do. Those people are morons. Ignore them. You're drawing those boxes so that you'll be sure not to mess around with the formulas to balance the equation. While they all suffer in the pits of academic hell, you'll be laughing from the honor roll. Here's what the equation looks like:

All I did was put boxes around the formulas.



3. Make an element inventory. In this inventory, your job is to figure out how many atoms of each element you have on the left and right sides of the equation. Now, if you look at the equation, you should be able to see that on the left side of the equation there is one sodium atom, five oxygen atoms (one from the sodium hydroxide, four from the sulfuric acid), three hydrogen atoms (one from the sodium hydroxide, two from the sulfuric acid), and one sulfur atom. On the right side of the equation, there are two atoms of sodium, one atom of sulfur, five atoms of oxygen (four from the sodium sulfate and one from the water), and two atoms of hydrogen. Thus, your element inventory should look like this:



4. Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. So, looking at the inventory, what should we do?

Well, we can see that on the left side of the inventory, there is one atom of sodium and on the right there are two. The solution: Stick a "2" in front of the sodium hydroxide on the left side of the equation so that the numbers of sodium atoms are the same on both sides of the equation. When we do this, the new atom inventory should look like this: (I'll let you figure out how this is done)



Now what? Well, looking at the new inventory, we can see that we now have two sodium atoms on both the left and the right sides, but the others still don't match up. What to do?

You can see from the inventory that on the right side of the equation, there are two hydrogen atoms and on the left there are four. Using your amazing powers of mathematics (and hopefully not needing to use a calculator), you can see that two multiplied by the number two becomes four. That's what you need to do. How? Put a "2" in front of the water on the right side of the equation to make the hydrogens balance out. Now that this is done, you should make a new inventory that looks something like this:



Since both sides of the inventory match, the equation is now balanced!  All other equations will balance in exactly the same way, though it might take a few more steps in some cases.

Problems you might encounter:

Is it all as easy as I made it look above? Well, yes and no. Yes, it should work all the time. No, sometimes you need to do some tricks to find the right numbers to add into the equation.

For example, what happens when you do the inventory, and you find that there are two atoms of element X on the left side of the equation and three on the right. How can you make those match?

When you run into this problem, find the lowest common denominator of those two numbers, and then put the numbers in front of those two boxes which allow the inventory on both sides to match. In the element X example, the lowest common denominator of two and three is six, so you'd put a "3" in front of the molecule on the left, and a "2" in front of the one on the right. Element X will then match up, and you can use a new inventory to see what else needs to be done.

Another common problem: What happens when the only way you can get a problem to work out is to make one of the numbers a decimal or fraction?

When this happens, find the largest molecule in the equation and stick a "2" in front of it. Then start the problem over. Will this work all the time? Well, no. But it will work sometimes, and give you a new strategy for hard problems.

Most importantly: Always remember to keep the inventory of the elements current! If you try to keep it in your head, you'll screw it up. Nobody can keep a bunch of changing numbers in their head for very long. I certainly can't, and you can't either.

Some practice problems:

Here are some practice problems. The solutions are in the section below this one.

1. __NaCl + __BeF2 --> __NaF + __BeCl2

2. __FeCl3 + __Be3(PO4)2 --> __BeCl2 + __FePO4

3. __AgNO3 + __LiOH --> __AgOH + __LiNO3

4. __CH4 + __O2 --> __CO2 + __H2O

5. __Mg + __Mn2O3 --> __MgO + __Mn

Solutions for the practice problems:

1. 2 NaCl + 1 BeF2 --> 2 NaF + 1 BeCl2

2. 2 FeCl3 + 1 Be3(PO4)2 --> 3 BeCl2 + 2 FePO4

3. 1 AgNO3 + 1 LiOH --> 1 AgOH + 1 LiNO3

4. 1 CH4+ 2 O2 --> 1 CO2 + 2 H2O

5. 3 Mg + 1 Mn2O3 --> 3 MgO + 2 Mn
Instructions on balancing chemical equations:
  • Enter an equation of a chemical reaction and press the 'Balance!' button. The answer will appear below
  • Always use the upper case for the first character in the element name and the lower case for the second character.
     Examples: Fe, Au, Co, Br, C, O, N, F.     Compare: Co - cobalt and CO - carbon monoxide
  • To enter an electron into a chemical equation use {-} or e
  • To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
    Example: Fe{3+} + I{-} = Fe{2+} + I2
  • Substitute immutable groups in chemical compounds to avoid ambiguity.
    For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
    but XC2H5 + O2 = XOH + CO2 + H2O will
  • Compound states [like (s) (aq) or (g)] are not required.
  • If you do not know what products are enter reagents only and click 'Balance!'. In many cases a complete equation will be suggested.
 Examples of complete chemical equations to balance:
  • Fe + Cl2 = FeCl3
  • KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2
  • K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO
  • C6H5COOH + O2 = CO2 + H2O
  • K4Fe(CN)6 + KMnO4 + H2SO4 = KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O
  • Cr2O7{-2} + H{+} + {-} = Cr{+3} + H2O
  • S{-2} + I2 = I{-} + S
  • PhCH3 + KMnO4 + H2SO4 = PhCOOH + K2SO4 + MnSO4 + H2O
  • CuSO4*5H2O = CuSO4 + H2O
 Examples of the chemical equations reagents (a complete equation will be suggested):
  • H2SO4 + K4Fe(CN)6 + KMnO4
  • Ca(OH)2 + H3PO4
  • Na2S2O3 + I2
  • C8H18 + O2