http://www.chem.tamu.edu/class/majors/tutorialnotefiles/percentcomp.htm
- The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.
To calculate the percent composition of a component in a compound:
1.
Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator .
2.
Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms.
3.
Divide the mass due to the component by the total molar mass of the compound and multiply by 100.
Calculate the percent composition of carbon in each of the following:
2
Molar mass of compound:
Mass due to carbon: 12.01 g/mol
Percent composition of carbon:
6
12
6
Molar mass of compound:
Mass due to carbon:
Percent composition of carbon:
- A formula that gives the simplest whole-number ratio of atoms in a compound.
1.
Start with the number of grams of each element, given in the problem. o If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given .

2.
Convert the mass of each element to moles using the molar mass from the periodic table .
3.
Divide each mole value by the smallest number of moles calculated.
4.
Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula. o If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple.

 e.g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3. e.g. If one solution is 1.25, then multiply each solution in the problem by 4 to get
5.
Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known. Simply calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to find the ratio between the molecular formula and the empirical formula. Multiply all the atoms (subscripts) by this ratio to find the molecular formula. (See Example #2)
A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?
Start with the number of grams of each element, given in the problem.
Convert the mass of each element to moles using the molar mass from the periodic table .

Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number.
This is the mole ratio of the elements and is represented by subscripts in the empirical formula.
NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of
NutraSweet and find the molecular formula. (The molar mass of NutraSweet is 294.30 g/mol)
Start with the number of grams of each element, given in the problem.

If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given .

Convert the mass of each element to moles using the molar mass from the periodic table .
Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number.

This is the mole ratio of the elements and is represented by subscripts in the empirical formula.

If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple.
Now, we can find the molecular formula by finding the mass of the empirical formula and setting up a ratio:

- The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.
- The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react.
No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made.
Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.
A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?
First, we need to create a balanced equation for the reaction:
3(g)
2(g)
(g)
2
(g)
Next we can use stoichiometry to calculate how much product is produced by each reactant. NOTE: It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.

The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the
"limiting reactant."
Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant
(ammonia) actually did react with the limiting reactant (oxygen).
We're not finished yet though. 1.70 g is the amount of ammonia that reacted , not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.
1. A sample contains 27.1 g of calcium oxide. How many moles of calcium oxide are in the sample?
NOTE: Use the Periodic Table to find the molecular mass (grams per mole)
2. How many atoms are in 0.652 mol of iron?
NOTE: A mole is by definition 6.0220 x 10 23 particles which can generally be rounded to 6.02 x 10
23
.
3. How many liters does 3.8 moles of O
2
occupy at STP (standard temperature and pressure)?
NOTE: At STP, 1 mole of any gas = 22.4L. STP is 273K (0 C) and 1 atm.
4. A solution of NaCl has a molarity of 0.549 M. How many moles are in 350. mL of this solution?
NOTE: Molarity is MOLES per LITER . The volume in milliliters must be converted to Liters.

5. How many grams of sodium hydroxide are needed to make 250. mL of a 0.200 M solution?
6. How many grams of carbon dioxide are there in a container with a volume of 4.50L at
STP?
7. How many moles of nitrogen are there in a 7.45 mol sample of ammonium phosphate?
8. If 120. g of propane, C
3
H
8
, is burned in excess oxygen, how many grams of water are formed?
9. 90.0 g of FeCl
3 reacts with 52.0 g of H
2
S. What is the limiting reactant? What is the mass of HCl produced? What mass of excess reactant remains after the reaction?
NOTE: The limiting reactant is the reactant that limits the amount of product that can be formed and is completely consumed during the reaction. The excess reactant is the reactant that is left over once the reaction has stopped due to the limiting reactant. See notes limiting reactant .