## Introduction: Stoichiometry

I was looking around Instructables and saw many chemistry related Instructables, so I thought one on stoichiometry would help. Basically stoichiometry (my definition) is the study of amounts in relation to a chemical reaction. Stoichiometry is the base for all modern chemistry. The method I'm going to use is like a flow chart opposed to the "railroad tracks" that is taught by most teachers and books.

This Instructable is good for people who are just learning stoichiometry and those who want an easier way. This is also great for all those people who like to use chemical reactions to blow stuff up. You can find the right amount to optimize the reaction and save reactants so you don't waste money.

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## Step 1: The Mole

At the center of stoichiometry is the mole. The mole allows a chemist to find what masses of substances to use in a reaction.

One mole is an amount of a substance that contains 6.022×10^^{23} atoms. To help you understand how astronomically big this number is if I gave everyone on Earth (estimated 7 billion) $3 million dollars a day; I could keep handing out money for 78564 years. Yes that is right. Almost 80 thousand years.

There are many different ways to find moles depending on what you are working with. For solids or liquids that aren't solutions its sample mass/molecular weight=moles. Use a periodic table to find the molecular weight. For example to find one mole of lets say carbon-14 the equation is x/14=1

x=14grams

Another example: Find 5 moles of H2O.

x/18=5

x=90grams

Find how many moles are in 20g. of NaCl.

20/58=x

x=.345

I think you can get it from that.

## Step 2: Balancing Chemical Equations

Now that you know the mole the second thing you need to do stoichiometry is balanced chemical equations. This is just the basic way. There are more complex equations that can't be balanced this way.

First lets take a simple and common reaction. Combustion. The definition of combustion is a fuel when burned with oxygen produces only H2O and CO2.

Here we have the combustion reaction of sugar (which happens in your cells and powers your body).

C6H12O6 + O2 --> H2O + CO2

To balance this all you need to do is get the same number of atoms on both sides. First off just pick a molecule(any of them work but the biggest is usually the best) and assign it a number (again any number works but to keep it easy use one).

So far we have:

C6H12O6 + O2 --> H2O + CO2

(1)

So on the left side we have 6 carbon atoms 12 hydrogen atoms and an unknown amounts of oxygen atoms.

To balance we need the same number on each side. Since we know the left sides number of carbon and hydrogen atoms we know the right sides number. We have to balance the molecules to have the same number of atoms.

6 H2O- Since we have 12 hydrogen atoms and each molecule has 2 atoms we 6 molecules of water.

6 CO2- Since we have 6 carbon atoms and each molecule has one atom in it we need 6

molecules of CO2

So now the equation looks like that and we are almost done:

C6H12O6 + O2 --> H2O + CO2

(1) x (6) (6)

Now that we know the number of molecules on the right we know the number of oxygen atoms in the equation. We have six H2O molecules each with one oxygen atom, so we have 6 oxygen atoms in those molecules. We also have six CO2 molecules each with two oxygen atoms in it, so we have 12 oxygen atoms in those molecules. So we have a total of 18 oxygen atoms on the left side.

6 H2O= 6 oxygen atoms

6 CO2= 12 oxygen atoms

6+12=18 atoms

If we know the number on the right side we know the number on the left side. Since we have one molecule of sugar that needs six atoms of oxygen; we subtract six from 18.

18-6=12

We have an unknown amount of oxygen molecules each with 2 oxygen atoms each and we have 12 oxygen atoms left. To finish just divide the number of oxygen atoms left by the number you need in each molecule.

12/2=6 oxygen molecules

That is the last number we need and the equation is balanced

C6H12O6 + O2 --> H2O + CO2

(1) (6) (6) (6)

That one was really easy but sometimes you will get an equation that doesn't work out so well. Sometimes one of the number wont come out as a whole number and that just wont work, because you can't have half of a molecule. Don't despair though all your hard work wasn't in vain just multiple by whatever makes that number a whole number

Example:

C3H7NO2 + O2 --> CO2 + H2O + NO2

(1)

3 carbon, 7 hydrogen, 1 nitrogen dioxide, x oxygen

C3H7NO2 + O2 --> CO2 + H2O + NO2

(1) x (3) (3.5) (1)

3 CO2- 6 oxygen atoms 3.5 H2O- 3.5 oxygen atoms

6+3.5=9.5 oxygen atoms

9.5/2=4.75 oxygen molecules

C3H7NO2 + O2 --> CO2 + H2O + NO2

(1) (4.75) (3) (3.5) (1)

As you can see not all the numbers came out as whole numbers, but it can easily be fixed by multiplying it by the least common multiple. In this case its 4.

C3H7NO2 + O2 --> CO2 + H2O + NO2

(1) (4.75) (3) (3.5) (1) * 4

=

C3H7NO2 + O2 --> CO2 + H2O + NO2

(4) (19) (12) (14) (4)

Always remember to check your work. Once you get this down its on to the actual stoichiometry. Oh and one last thing once you have balanced the equation write it like this:

4 C3H7NO2 + 19 O2 --> 12 CO2 + 14 H2O + 4 NO2

## Step 3: Stoichiometry of Solids or Liquid Non-solutions

First the labels of each part of a chemical equation.

4 C3H7NO2 + 19 O2 --> 12 CO2 + 14 H2O + 4 NO2

^ ^ ^ ^

coefficient reacts produces chemical

with symbol

The basic outline:

A,B,C,D- coefficients

CHEM- chemical substance

Sm- sample mass

Mw- molecular weight

n-moles

A CHEM + B CHEM --> C CHEM + D CHEM

Sm Sm

l l

Sm/Mw Mw*n

l l

n-----------------n/A=n/D----------------n

Now to describe what this means. If you start with the mass of A CHEM then you divide the mass of that substance by its molecular weight(found using the periodic table) to find the moles. Now multiple the number of moles by the coefficient of substance you find the moles of and divide it by the coefficient of the substance you are translating it into. This gives you the number of moles. You can now find find the mass of the product produced using the Sm/Mw=n equation. If you rework the equation its Mw*n=Sm. That may look difficult but its extremely easy. Some examples using the sugar combustion equation from the last step.

You have 26.3g. of sugar how many grams of water will be produced.

C6H12O6 + 6 O2 --> 6 H2O + 6 CO2

26.3g 38.59g

l l

26.3/180 44*.877

l l

.146 moles-- .146/1 = .146*6 = .877 -- .877 moles

38.59g of CO2 will be produced

This also works if you know the moles of one of the substances. Here is another using the electrolysis of water.

Balance the equation and then find the moles and grams of oxygen produced if you have 3 moles of water.

H2O ---> H2 + O2

1 H2O ---> 1 H2 + .5 O2 *2

2 H2O ---> 2 H2 + 1 O2

x 48g

l l

x/18 32*1.5

l l

3 ----- (3/2)1 -----1.5

1.5moles 48g of oxygen produced

Again remember to check your work. You can use the law of conservation of mass to do this. Mass on both sides of the equation must be the same. If you add up the mass for one side it should equal the sum of the mass on the other side. If not you messed up somewhere and need to review your work.

You may have noticed I said this is for solids and liquids, but I have been using gases. Theoretically you can use this for any type of stoichiometry if you know the chemical equation and the mass of one of the substances, but in practice gases are difficult to work with, so finding the mass becomes problematic. You need a different mole equation to work with gases.

## Step 4: Stoichiometry of Gases

Stoichiometry with gases everything stays the same but the equation you use to find the moles. It only takes basic knowledge of the gas laws to know why this is. Gas changes volume depending on temperature, and pressure. The relationship between volume and temperature is Charles's law which is at constant pressure a given mass of gas's volume increases as a factor of its temperature. Basically as it gets hotter its volume increases. The equation is V/T=k.

V- volume

T- temperature

k- constant

Charles's law must be done in Kelvin because it is an direct variation.

The other law is Boyle's law which shows the relationship between pressure and volume. The law states that at a constant temperature a give mass of gas's volume decreases as pressure goes up. The equation is PV=k

P- pressure

V- volume

k- constant

Gay-Lussac's law states the relationship between pressure and temperature. Again it must be done in Kelvin, since its a direct variation. In simple words it states as pressure increases so does temperature. the equation is P/T=k.

P- pressure

T- temperature

k- constant

You can combine all three of those laws to get the combine gas law. Its equation is PV/T=k. To use this with stoichiometry you need to combine it with one last law, which is Avogadro's law. It states that at constant temperature and pressure equal volumes of gas contain the same number of moles. The equation is V/n=k

V- volume

n- moles

k- constant

When all four of these laws are combined together they make the ideal gas law. The equation for that (this is the important part since we will be using it for stoichiometry) is PV=nRT.

R is the constant. The number changes depending on what units you measure it in. The basic outline of the units is PV/nT. I will be using the units kPa*L/n*K.

kPa- kilo Pascals

L- liters

n- moles

K- Kelvin

The n*K part will never change since the temperature must be measured in Kelvin and there are no other units for moles. The constant I am using is 8.314 L*kPa/n*K. To make it easier to solve stoichiometry problems arrange the equation

so it looks like this PV/8.314*K.

Now on to the actually stoichiometry. Nothing is really different just a different equation.

Here is an example using the reaction between oxygen and iron to produce ferric oxide.

You have 5L. of oxygen in a lab that is 300 Kelvin and 22kPa. How many grams of iron(III) oxide can you produce.

3 O2 + 4 Fe --> 2 Fe2O3

l 4.64g

22*5/8.314*300= l

.044 .029*160

l l

.044----(0.044/3)2=.029---.029

4.64g of ferric oxide can be produced

See exactly the same just using PV/RT.

Note that this will work for all gases, but if the gas is near is condensation point it will not work for it.

## Step 5: Stoichiometry of Solutions

Stoichiometry with solutions is the same as before just with a different equation for moles (starting to see a pattern here?). When doing doing stoichiometry with solutions you need to know the concentration of reactants in your solvent. Specifically you need to know the moles per unit of solvent. There are many different ways of doing this, but I'm going to use molarity. Molarity is simply moles per liter.

To find molarity of a solution we use n/L=M (M stands for molarity). To use it for stoichiometry arrange it so it looks like M*L=n.

I'm going to describe how to make a 50mL. .75 molarity solution of NaCl in water. First start by doing the math.

.75*.05 = .0375 moles

So I need .0375 moles of salt. Do the math to find what that is in grams.

.0375*28 = 1.05g of salt

Next is to make the actual solution. Start by filling a graduated cylinder with around 40mL of water. Now add the salt and let it completely dissolve. Finally add water up to 50mL.

Always when making a solution using molarity use less solvent then what you want the final solution to be dissolve the solute and then add the solvent to the final amount. This is because the solute takes up space too; if you add the solute to the measured amount of solvent it would come out as taking up more volume then you want, and you will have the wrong molarity.

Using solutions is great for chemical reactions, since it speeds them up. This is because the solvent breaks apart the ionic bonds of reactants. Besides that is also allows more atoms to get close enough to react. Since the solvent breaks apart the ionic compounds the chemical equation must be wrote different to reflect this.

Example:

AgNO3 + NaCl --> NaNO3 + AgCl

When dissolved in a solvent it would look like this:

Ag^{+}(aq) + NO3^{-} (aq) + Na^{+} (aq) + Cl^{-}(aq) --> AgCl (s) + NO_{3}^{-} (aq) + Na^{+} (aq)

Since the NO_{3}^{-} + Na^{+} don't participate in the reaction they can be removed. The equation would look like this.

Ag^{+}(aq) + Cl^{-}(aq) --> AgCl(s)

The letters in parenthesis represent the state the substance is in

(s)- solid

(l)- liquid

(g)- gas

(aq)- aqueous or in solution

Here is an example of solution stoichiometry using the reaction above.

You have 4L of 10 molarity solution of AgNO3. How many grams AgCl can be produced.

AgNO3 + NaCl --> NaNO3 + AgCl

10mole/L 5720g.

l l

10*4=40 40*143

l l

40------------(40/1)1------------40

5720 grams would be produced.

You may have noticed I didn't use the ionic equation I showed above. That is because it doesn't have to be used, but it should be. I didn't because I thought this more clearly demonstrated the stoichiometry. That and I am lazy.

## Step 6: Thermochemical Stoichiometry

Yes stoichiometry can even be used with thermodynamics. It shows the movement of energy through out a reaction. There is two types of energy that can used. There is enthalpy (heat), or free energy. Entropy (randomness) can also be used. Free energy is just the combination of entropy and enthalpy.

There are two ways you find these. There are tables that list all three. These can be in many different units, so make note of it. The energy differs for every substance so you have to look them all up. The other way to find it is using the equation G=H-S*K.

G- free energy

H- enthalpy

S- entropy

K- temperature in Kelvin

Using this method you still need two of the three, so I think you should use a table for all three unless you want to practice your algebra.

The important thing to remember here (and relates most to stoichiometry) is that its measured in units of energy per mole. That lets use do stoichiometry with it. The units I'm using are joules/mole or J/mole.

For example I'll use a reaction that very relevant to you. The combustion of sugar using enthalpy.

You have 30g. of sugar. How much energy can you get from it.

C6H12O6 + 6 O2 --> 6 H2O + 9 CO2

C6H12O6 + 6 O2 - 1250 kJ/mol --> 6 H2O + 6 CO2 + 209 kJ

30g. l l

l l l

30/180 l l

l l l

.167 moles---- .167*1250-----------------------------------209 kJ

You get 209 kJ of energy.

The first thing you may have noticed is that I changed the sign from a - to a +. That is because the number in the equation is the standard enthalpy of formation or the energy needed to create it. Since we are destroying it we get that energy back. All you have to really know is switch the signs. The second thing you make have noticed is that I didn't include the energy from the oxygen. That is because oxygen is an element and, all elements have a standard enthalpy of formation of 0.

This part may be wrong so correct me if I am. Its been awhile since I have used it, but I'm pretty sure this work.

## Step 7: Limiting Reactant

One of the best things about stoichiometry is using it to find the limiting reactant. You can find what reactant is used up first and stops the reactions. It is very simple you just find the moles of both reactants, and convert them both to moles of one of the products, but remember it has to be the same product. Whichever is the lowest is the limiting reactant.

Here is an example using the reaction of Cl2 and Na to make table salt.

You have 20g. of Na and 35g. of Cl which is the limiting reactant.

2 Na + Cl2 --> 2 NaCl

20 35

l l

20/11 35/17

l l

l 2.06- (2.06/1)2 - 4.12

1.82--------- (1.82/2)1 ------------- .91

Sodium is the limiting reactant.

## Step 8: Now You Know Stoichiometry

If there is anything I need to make clearer or something I did wrong please tell me. This is my first instructable so go easy on me. :)

I hope this helps someone with stoichiometry. Now go use it to do some awesome chemical reactions. I suggest blowing something up or something involving fire. Maybe that is just me?

## 8 Discussions

11 months ago on Step 7

Under Step 7, the Limiting Reactant. You used the atomic number again instead of the molecular weight. Therefore, for Na, that should be 20 g/23 mw = .87 mols of Na, and for Cl should be 35 g/35.45 mw = .99 mols of Cl. Then, using the coefficient ratio of D/A, for Na its 2/2 =1, so it remains at .87 mols. For Cl, it's 2/1, so it's 1.97 mols of Cl. The end result is the same: Na (sodium) is the limiting reactant.

5 years ago on Step 2

This site has balancing chemical equation worksheets for extra practice balancing equations: http://stemsheets.com/science/balancing-equations-worksheet

9 years ago on Step 1

"6.022×1023" 6.022×10

23^Reply 9 years ago on Step 1

Yeah didn't notice that. When I wrote it up I used an exponent, but it did show up I guess. Thank you :)

9 years ago on Step 1

In the NaCl example, how did you determine the molecular weight of NaCl to be 28?

Reply 9 years ago on Step 1

Uh oops thank you :) I was going through it fast and used the number instead of weight.

Reply 9 years ago on Step 1

Ok, so 58?

Reply 9 years ago on Step 1

Yep, that one should be right unless my arithmetic is way off. If you are wondering about any molecules just look on a periodic table for the weights of each element and add them up.