1. “Percent Composition and Chemical Formulas”
Chemistry 10.3
“Percent Composition and Chemical Formulas”

2. I. Percent Composition
A. Def – the percent by mass of each element in a compound.
1. Must = 100%
2. Formula…
Grams of element X
Grams of compound
% mass of element X =
X 100%

3. 3. Ex: An 8. 20 g piece of magnesium combines with 5
3. Ex: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the % composition of this compound?
-mass of compound =
% Mg = 8.20 g
13.60 g
= 60.3%
13.60 g
X 100%

4. % O = 5.40 g 13.60 g = 39.7% *Does this make sense?
39.7% % = 100%
X 100%
Yes

5. 4. Ex: 9. 03 g of Mg combine completely with 3
4. Ex: g of Mg combine completely with 3.48 g of N to form a compound. What is the % composition of this compound?
g Ag completely react with 4.30 g of S to form a compound. What is the % composition of this compound?
72.2 % = Mg
27.8 % = N
87.1 % = Ag
12.9 % = S

6. B. Percent Composition from the Chemical Formula 1. Formula
%Mass = grams of element in 1 mol of compound
molar mass of compound
2. Must = 100%
3. Add molar mass + # of moles
4. Ex: Calculate the % composition of propane (C3H8)
X 100%

7. molar mass = % C = 36.0 44.0 % H = 8.0 44.1 g /mol X 100% % C = 81.8%

8. 5. Calculate the % composition of the following compounds: a. C2H6 =
b. NaHSO4 =
c. NH4Cl =
d. CO(NH2)2 =
%C = %H = 20.1
%Na = %H = .8 %S = 26.7 %O = 53.3
%N = %H = %Cl = 66.3
%C = %O = %N = %H = 6.7

9. II. Empirical Formalas
A. Def – gives the lowest whole # ratio of atoms in a compound.
1. can be used for atoms or moles
2. Ex: HO  H2O2 (hydrogen peroxide)
3. Empirical formula can be the same as molecular formula.
-Ex: CO2 (1:2 ratio)
4. What are the empirical formulas for the following?
a. C6H12O6
b. C4H8
c. N4O10
CH2O
CH2
N2O5

10. 5. Ethyne (C2H2) also called acetylene, is a gas used in welder’s torches. Styrene (C8H8) is used in making polystyrene.
-What is similar about these 2 compounds?
both have the same empirical formula (CH)

11. B. Calculating Empirical Formulas
1. Ex: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?
a. Step 1 – determine the # of moles (assume %’s are in grams)
b. Step 2 – divide by smallest mole #
c. Step 3 – Check for decimals (2.5, 3.75, 1.25)
2. Solve (step 1)
25.9 g N x 1 mol N
74.1 g O x 1 mol O
= 1.85 mol N
14.0 g N
= 4.63 mol O
16.0 g O

12. (Step 2)
1.85 mol N
1.85
4.63 mol O
= 1 mol N
= 2.5 mol O

13. (Step 3)
*What can we multiply by to get a whole # ratio between N and O?
= 2
1 mol N x 2 = N2
2.5 mol O x 2 = O5
Empirical Formula = N2O5

14. Check answer by finding % composition for N and O.
Calculate the empirical formulas for the following…
% O, 5.9% H
% C, 20.2% H
% Hg, 10.8% S, 21.6 O
% C, 1.15% H, 16.09% N, 55.17% O OH
CH3
HgSO4
C2HNO3

15. III. Calculating Molecular Formulas (includes ionic compounds)
A. Determine empirical formula
B. Determine molar mass of compound
C. Divide… Molar Mass
Emp. Formula Mass
D. Take each element in the empirical formula times “step C”
E. Write out new formula

16. Examples:
1. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N.