## Step 3: Stoichiometry of solids or liquid non-solutions

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.