Name: _____________________________________ Date: __________________ Period: ____ #: ____
STOICHIOMETRY
Example #1:
How many grams of sodium hydroxide are needed to completely react with 50.0 grams of sulfuric acid
to form sodium sulfate and water?
Balanced equation:
2NaOH + H2SO4  Na2SO4 + H2O
The number from
the question with
unit and substance.
50.0 g H2SO4
1
The mole ratio…H2SO4 goes on the bottom to cancel the substance,
thus NaOH (the one asked for) goes on top. The numbers in front
come from the COEFFICIENTS in the balanced equation.
1 mol H2SO4
98.076 g H2SO4
2 mol NaOH
1 mol H2SO4
The H2SO4 molar mass…grams goes on the
bottom to cancel and 1 mol always goes on top.
39.998 g NaOH
= 40.8 g NaOH
1 mol NaOH
The NaOH molar mass…1 mole goes on the bottom
to cancel and the grams always goes on top.
Limiting Reactants
Identify? Given the mass of two reactants and asked for the mass of a product.
1. Each number in the problem starts its own calculation.
2. Do the full mass-to-mass conversion for each…convert to grams of the product asked to find.
3. Now you’ll have two numbers for grams of the product…the smaller number is how much
product is really made. (theoretical yield)
The reactant used in that stoichiometry that had the smaller number is the limiting reactant or reagent.
The other reactant would be the excess reactant or reagent.
If you are asked to find how much of the excess reactant remains, first you must determine how much
was actually used (remember, there was extra).
1. Start another calculation with the mass of the limiting reactant given in the problem.
2. Convert it into grams of the excess reactant. That will tell you how much was used.
3. To determine how much remains, you must subtract the amount used from the amount given.
Example #2:
Determine the mass of tetraphosphorus decoxide formed if 50.0 g of phosphorus (P4) and 50.0 g of
oxygen are combined. Identify the limiting reactant and determine how many grams of the excess
reactant will remain after the reaction is complete.
Balanced equation: P4 + 5O2  P4O10
50.0 g P4
mol P4
1 mol P4 O10 283.88 g P4 O10
×
×
×
= 115 g P4 O10
1
123.88 g P4
1 mol P4
mol P4 O10
50.0 g O2
mol O2
1 mol P4 O10 283.88 g P4 O10
×
×
×
= 88.7 g P4 O10
1
32.00 g O2
5 mol O2
mol P4 O10
50.0 g O2
mol O2
1 mol P4
123.88 g P4
×
×
×
= 38.71 g P4
1
32.00 g O2 5 mol O2
mol P4
50.0 g P4 initial −38.71 g P4 used = 11.29 g P4
Mass of P4O10 = 88.7 g
Limiting Reactant = O2
Excess remaining = 11.3 g P4
Name: _____________________________________ Date: __________________ Period: ____ #: ____
Combustion Analysis
This technique requires that you burn a sample of the unknown substance in a large excess of oxygen
gas. The combustion products will be trapped separately from each other and the weight of each
combustion product will be determined. From this, you will be able to calculate the empirical formula of
the substance. This technique has been most often applied to organic compounds.
Some points to make about combustion analysis:
1. The elements making up the unknown substance almost always include carbon and hydrogen.
Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but
problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N.
2. We must know the mass of the unknown substance before burning it.
3. All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O.
4. If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the
oxides. The mass of oxygen in the sample will almost always be determined by subtraction.
5. Often the N is determined via a second experiment and this introduces a bit of complexity to the
problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen
product is N2, sometimes NH3.
6. Sometimes the problem asks you for the empirical formula and sometimes for the molecular
formula (or both). Two points: (a) you have to know the molar mass to get to the molecular
formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for
it.
1) Convert the grams of carbon dioxide into grams of carbon.
2) Convert the grams of water into grams of hydrogen.
3) Subtract the grams of carbon and grams of hydrogen from the grams of the organic compound to
determine the grams of oxygen in the organic compound.
Now you will have grams of carbon, hydrogen, and oxygen. Follow the steps for determine empirical
formulas.
Example #3:
A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete
combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this
compound?
0.3664 g CO2
12.01 g C
×
= 0.1000 g C
1
44.01 g CO2
0.1000 g C
mol C
×
= 0.008326 mol C
1
12.01 g C
0.1500 g H2 O
2.016 g H
×
= 0.01678 g H
1
18.018 g H2 O
0.01678 g H
mol H
×
= 0.01664 mol H
1
1.008 g H
mol O
0.2500 g [CHO] - 0.1000 g C - 0.01678 g H = 0.1332 g O 0.1332 g O
×
= 0.008325 mol O
1
16.00 g O
0.008326 mol C
=1 mol C
0.008325
Empirical Formula = CH2O
0.01664 mol H
=2 mol H
0.008325
0.008325 mol O
=1 mol O
0.008325