From WolfWikis

Stoichiometry problems

Stoichiometry problems center on

1. the use of the molar mass of the elements
2. the use of balanced equations.
3. the use of the number of Avogadro

Moles of things

A mole is essentially a group of like items, like a football team. If you have 26 football teams how many players do you have? Well, you would have to multiply by the team size. You know that number better than I do, we play soccer mostly, there it would be eleven, so that the answer would be 26x11=286 players.

Unfortunately, a chemical football team is rather large. It has N_av=602,200,000,000,000,000,000,000 players (or so). Because the team size (the Number of Avogadro) is ridiculously large in this sport, chemists often think in whole teams (moles) rather than in individual players (items).

Moreover, what these 'items' are exactly, varies, sometimes they are atoms, sometimes molecules and sometimes formula units (that is: whatever collection of atoms I just wrote down as formula).

E.g. a mole of oxygen gas consists of N_av molecules O₂, therefore it has twice as many (2N_av) atoms O.

Just think of how many pairs of football shoes the team has as opposed to the total number of shoes. In our analogy, the pairs correspond to molecules, the atoms to individual shoes. Notice that because the team size is so humongous chemists tend to apply the thinking-in-whole-teams concept to everything: players, shoes, what have you.

Let's apply that to soccer: 26 teams-of-soccer-players would be wearing 26 teams-of-shoe-pairs, which is 26*2=52 teams-of-shoes which makes 52x11= 572 shoes.

Notice that it is only the last step that would make the number gigantic in chemistry, because instead of 11 we need to write 602,200,000,000,000,000,000,000. This is exactly why chemists try to think in terms of teams (moles) as much as they can.

Sometimes therefore we do need to make explicit what we mean when we talk about moles. Usually when we say a mole of oxygen we mean a mole of oxygen molecules (O₂) (as I did in the previous statement), because that is the most common form in which this element is available to us. It is what we are breathing. Obviously this mole of pairs would correspond to two moles of O atoms.

When dealing with this issue it is often useful to write out a (possibly 'fake') chemical equation in which we assume that a compound (or formula unit) can be decomposed into its elemental atoms:

Ca₃(PO₄)₂ ==> 3Ca + 2P + 8O

We may not be able to actually do this reaction, but writing it helps us keeping track of things. Provided we balance this correctly -notice e.g. that we need to multiply the subscripts 4 and 2 to arrive at 8O- we can immediately see that one mole of calcium phosphate contains 8 moles of oxygen atoms. To synthesize it from the elements P₄ (white phosphorous) and O₂ (as contained in air), we would need 0.5 mole of white phosphorous and four moles of atmospheric oxygen.

Molar masses

For a stoichiometric compound you can find the molar mass from its chemical formula and the atomic masses of the elements

Atomic masses

Atomic masses can be found on the inside covers of most text books and in most periodic tables. They are not quite integer, because elements are composed of mixtures of isotopes but in many cases we can do a rough calculation where they are taken as integer (+ one digit). E.g.:

The dimension is always [g/mol]

Notice that in chemistry the teams are gigantic by number of players, but the players are tiny. A whole team of C-players only weighs 12 grams! Once again, it is easier to think in terms of the whole team (a mole) than in terms of the mass of a single player (an atom). A C-atom only weighs 2x10^-23 grams (0.000,000,000,000,000,000,000,02 grams). C-players are lightweights!

Chemical formulas

For stoichiometry calculations we need to know the chemical formula of a compound in the form A_xB_yC_z, e.g. benzene is C₆H₆

Exercises

Exercise 1

Balance the following equations

C₆H₆ + O₂ => CO₂ + H₂O

C₆H₁₂ + O₂ => CO₂ + H₂O

CS₂ + O₂ => CO₂ + SO₂

Al(OH)₃ + SO₂ => Al₂(SO₃)₃ + H₂O

Exercise 2

Compute the molar mass for

Al₂O₃

AuCl₃

Rb₂SeO₄

Ho(OH)₃

C₁₂(OH)₃N₂O₃F₃

Exercise 3

I have one gram of the following compounds; how many moles do I have?

Ca(OH)₂

AuCl

Rb₂SO₄

HoBr₃

C₁₂(OH)₃N₂O₃F₃

Exercise 4

I add one mole of oxygen in the following reactions and it is exactly enough to burn up the other reactant. How many grams reactant do I need?

Warning: the equations are not balanced!

C₆H₁₄ + O₂ => CO₂ + H₂O

C₅H₁₂ + O₂ => CO₂ + H₂O

CSH₄ + O₂ => CO₂ + SO₂ + H₂O