Percent Composition

- The % by mass of the elements in a compound is called the percent composition

How to find the percent composition: 1. Calculate molar mass

                                                            2. Calculate each element percentage of that total

                                                                Give percentage to 1decimal place

-Percent composition tells what part of a compound each component makes up

               %composition= (Mass of element)/(Mass of compound)×100%

 Ex1. What is the % composition of CO2?

1 mole C= 1×12.0= 12.0g

2mole O= 2 ×16.0= 32.0g

                                  44.0g
                        %C= (mass of carbon)/(mass of CO2)=(12.0×100%)/44.0g=27.3%
                        %O= (mass of oxygen)/(mass of CO2)= (32.0×100%)/44.0g=72.7%

                        Check to see if the result is close to 100%:27.3%+72.7%= 100%

Because of rounding answer may also be shown as 99.9 and 100.1 which are    both OK

Ex2. What is the % composition of the entire element in glucose: C6H12O6

C = 72.0g        H=12.0g          O= 96.0g         Total Mass=180.0g

          %C= (72.0g×100%)/180.0g= 40.0%
          %H= (12.0g×100%)/180.0g= 6.7%
          %O= (96.0g×100%)/180.0g= 53.3%
          40.0+6.7+53.3=100%

To find the percent composition of a group of atoms (Al2(SO4)3), the total of the mass of the group atom is used for the percentage

Ex3. Find the % SO4 in Al2(SO4)4

2Al× 27.0g= 54g

3S × 32.1g= 96.3g                              96.3(Sulphur)+192.0(Oxygen)=288.3g

12O× 16.0g= 192.0g

342.3g

          %SO4= 288.3g/342.3×100%= 84.2%SO4
From the percent composition, ratio of moles can be determined and this will result in an empirical formula

            Step 1: Assume 100.0g of material

            Step 2: Convert all %s to grams            Step 3: Convert grams to moles

Step 4: divide all moles by the smallest; this will give you a ratio to determine the empirical formula

Ex4. Sample of compound: 45.8% Sulphur and 54.2% Flouride. Give the empirical formula

Step 1: Assume the 100.0g of material

Step 2: Therefore: 45.8g S and 54.2g F

Step 3: 45.8/32.1g= 1.4268molS     54.2/19.0g= 2.8526molF 

Step 4: 2.8526/1.4268= 1.99993/(1moleS)= SF2
Therefore the empirical formula for 45.8% Sulphur and 54.2% Flouride is SF2
How to calculate percent composition1: V=xbEeyT8nK84
How to calculate percent composition2: V=_H009sTvYE0