% Composition,
Empirical/Molecular Formula!!
Percent Composition of a Compound
• The relative amounts of the elements in a
compound are expressed as the percent
composition or the percent by mass of each
element in the compound.
• The percent composition of a compound
consists of a percent value for each different
element in the compound.
Calculating the Percent Composition
• The percent by mass of an element in a
compound is the number of grams of the
element divided by the mass in grams of the
compound, multiplied by 100%
% mass of element = grams of element x 100
grams of compound
EXAMPLE #1
• When a 13.60 g sample of a compound
containing only magnesium and oxygen is
decomposed, 5.40 g of oxygen is obtained.
What is the percent composition of oxygen in
this compound?
% mass of element = grams of element x 100
grams of compound
5.40 g/ 13.60 g x 100% = 39.7%
EXAMPLE #2
• A compound is formed when 9.03 g Mg
combines completely with 3.48 g N. What is
the percent composition of each element in
this compound?
% mass of element = grams of element x 100
grams of compound
Solution #2
% mass of element = grams of element x 100
grams of compound
Mass of the compound: 9.03 g + 3.48 g = 12.51 g
9.03 g Mg/ 12.51 g x 100% = 72.2% Mg
3.48 g N / 12.51 g x 100% = 27.8% N
Percent Composition from the
Chemical Formula
• You can also calculate the percent composition of a
compound if you know only its chemical formula.
% mass = (# atoms of element)(mass of 1 mole of element) x 100
(mass of 1 mole of compound)
EXAMPLE #3
• Propane (C3H8), the fuel commonly used in gas
grills, is one of the compounds obtained from
petroleum. Calculate the percent composition
of each element of propane.
Solution #3
Propane (C3H8),
• %C = (# atoms in cmpd)(mass of C)/ mass of
propane x 100%
= (3)(12.0 g) / 44.0g x 100%
• %H = (# atoms in cmpd)(mass of H) / mass of
propane x 100%
= (8)(1.0 g) / 44.0g x 100%
Try #11 on worksheet
H2O
% mass = (# atoms of element)(mass of 1 mole of element) x 100
(mass of 1 mole of compound)
%H = ?
%O = ?
Empirical Formulas
• The basic ratio, called the empirical formula,
gives the lowest whole-number ratio of the
atoms of the elements in a compound.
• IN other words, if you know the %
composition of a compound, you can figure
out the formula of the compound. This is
VERY useful when you are developing new
chemicals!!
Calculating Empirical Formulas
Assume you have 100g sample of the compound
Step 1: % composition of each element = grams
of each element
Step 2: Change grams of each element into
moles
Step 3: Calculate the mole ratio
Step 4: Ta-da! Empirical formula!!
Example #4
A compound is analyzed and found to contain
58.5 % Ra, 41.5 % Br . What is the empirical
formula of the compound?
Step 1: % composition of each element = grams
of each element
58.5 grams Ra
41.5 grams Br
Example #4 Solution-Step 2
Step 2: Change grams of each element into
moles
58.5 g Ra x (1 mol Ra/ 226 g Ra) = 0.257 mol Ra
41.5 g Br x (1 mol Br/ 79.9 g Br) = 0.519 mol Br
Example #4 Solution-Step 3
Step 3: Calculate the mole ratio
The formulas must be whole-number ratios.
Divide both mol quantities by the smallest
mol quantity.
Ra: 0.257 mol / 0.257 mol = 1.00
Br: 0.519mol / 0.257 mol = 2.02
Example #4 Solution-Step 4
Step 4: Ta-da! Empirical formula!!
Ratio of Ra= 1.00
Ratio of Br= 2.02
The ratios are pretty close to whole-number
ratios.
– Soooooooo, RaBr2
Example #5
• A compound is analyzed and found to contain
26.8 % Sn, 16.0 % Cl, 57.2 % I. What is the
empirical formula of the compound?
Step 1: % composition of each element = grams
of each element
Step 2: Change grams of each element into moles
Step 3: Calculate the mole ratio
Step 4: Ta-da! Empirical formula!!
Solution to Example #5
Step 1: Assume that 100 g contains 26.8 g Sn, 16.0 g Cl, and
57.2 g O
Step 2: Change grams into moles
26.8 g Sn x (1 mol Sn/ 118.71 g Sn) = 0.226 mol Sn
16.0 g Cl x (1 mol Cl/ 35.45 g Cl) = 0.451 mol Cl
57.2 g I x (1 mol I/126.90 g I) = 0.451 mol I
Step 3: The formulas must be whole-number ratios. Divide
both mol quantities by the smallest mol quantity
Sn: 0.226 mol / 0.226 mol = 1.00
Cl: 0.451 mol / 0.226 mol = 2.00
I: 0.451 mol/0.226 mol = 2.00
Step 4: The ratios are pretty close to whole-number ratios.
Soooooooo, SnCl2I2
EXAMPLE #6
• A compound is analyzed and found to contain
25.9% N and 74.1% O. What is the empirical
formula of the compound?
Step 1: % composition of each element = grams
of each element
Step 2: Change grams of each element into moles
Step 3: Calculate the mole ratio
Step 4: Ta-da! Empirical formula!!
Solution #6
Step 2: Change grams into moles
Step 1: Assume
contains
25.9
g N and
25.9 g Nthat
x (1100
molgN/
14.0 g N)
= 1.85
mol74.1 g O
74.1 g O x (1 mol O/ 16.0 g O) = 4.63 mol O
Step 3: The formulas must be whole-number ratios. Divide
both mol quantities by the smallest mol quantity
N: 1.85 mol / 1.85 mol = 1
O: 4.63 mol / 1.85 mol = 2.5
Step 4: They are still not in whole-number ratios. Multiply by
the smallest number possible to make both numbers whole.
In this case 2.
1x2=2N
2.5 x 2 = 5 O
N2O5
Molecular Formula
• The molecular formula of a compound is
either the same as its experimentally
determined empirical formula, or it is a simple
whole-number multiple of its empirical
formula.
Empirical Formulas
• An empirical formula may or may not be the
same as the molecular formula.
– Example: the lowest ratio of hydrogen to oxygen
in hydrogen peroxide is 1:1. The empirical
formula would be HO. However the actual
molecular formula of hydrogen peroxide is H2O2.
Notice that the ratio is still the same.
Empirical vs. Molecular Formulas
• The empirical formula of a compound shows
the smallest whole-number ratio of the atoms
in the compound.
• The molecular formula tells the actual
number of each kind of atom present in a
molecule of the compound.