Chemistry
10.3
“Percent Composition and
Chemical Formulas”
I. Percent Composition
A. Def – the percent by mass of each
element in a compound.
1. Must = 100%
2. Formula…
Grams of element X
% mass of element X =
Grams of compound
X 100%
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 = 13.60 g
% Mg = 8.20 g
X 100%
13.60 g
= 60.3%
% O = 5.40 g
13.60 g
= 39.7%
X 100%
*Does this make sense? Yes
39.7% + 60.3% = 100%
4. Ex: 9.03 g of Mg combine completely with
3.48 g of N to form a compound. What is the %
composition of this compound?
72.2 % = Mg
27.8 % = N
5. 29.0 g Ag completely react with 4.30 g of S
to form a compound. What is the %
composition of this compound? 87.1 % = Ag
12.9 % = S
B. Percent Composition from the Chemical Formula
1. Formula
%Mass = grams of element in 1 mol of compound
X 100%
molar mass of compound
2. Must = 100%
3. Add molar mass + # of moles
4. Ex: Calculate the % composition of propane (C3H8)
molar mass =
% C = 36.0
44.0
% H = 8.0
44.0
44.1 g /mol
X 100%
X 100%
% C = 81.8%
% H = 18 %
5. Calculate the % composition of the
following compounds:
%C = 79.9
%H = 20.1
a. C2H6 =
b. NaHSO4 = %Na = 19.1 %H = .8 %S = 26.7 %O = 53.3
c. NH4Cl = %N = 26.2 %H = 7.6 %Cl = 66.3
d. CO(NH2)2 = %C = 20.0 %O = 26.6 %N = 46.6
%H = 6.7
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
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)
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
14.0 g N
74.1 g O x 1 mol O
16.0 g O
= 1.85 mol N
= 4.63 mol O
(Step 2)
1.85 mol N
1.85
4.63 mol O
1.85
= 1 mol N
= 2.5 mol O
(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
Check answer by finding % composition for N
and O.
Calculate the empirical formulas for the
following…
OH
1. 94.1% O, 5.9% H
2. 79.8% C, 20.2% H CH3
3. 67.6% Hg, 10.8% S, 21.6 O HgSO4
4. 27.59% C, 1.15% H, 16.09% N, 55.17% O
C2HNO3
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
Examples:
1. Calculate the molecular formula of
the compound whose molar mass is
60.0 g and empirical formula is CH4N.