% Composition or %Mass
Percent Composition: the mass of each
element, in percent, that makes up a
compound
 % composition = part/whole x 100

% Composition or % Mass


% Composition: part/whole x 100
% mass = g desired element X 100
total g of compound
 Make sure the % add up to 100 %
Don’t confuse % Composition
with Molar Mass

% mass = g desired element X 100 = %
total g of compound
Molar Mass = atomic mass from periodic table = g/mol
Determine the percentage composition of the following
compound
HNO3
Determine the percentage composition of the following
compound
HNO3 has 1 mol H =
1 mol N =
3 mol O =
Determine the percentage composition of the following
compound
HNO3 has 1 mol H = 1.01 g
1 mol N = 14.01 g
3 mol O = 48.00 g
63.02 g HNO3
Next we take each individual element and divide by the
mass of compound
Part/whole x 100 = % composition
Determine the percentage composition of the following
compound HNO3 has 1 mol H = 1.01 g
1 mol N = 14.01 g
3 mol O = 48.00 g
63.02 g HNO3
Next we take each individual element and divide by the mass of
compound
Part/whole x 100 = % composition
1.01 g H
x 100 = 1.60%
63.02 g HNO3
Determine the percentage composition of the following
compound HNO3 has 1 mol H = 1.01 g
1 mol N = 14.01 g
3 mol O = 48.00 g
63.02 g HNO3
Part/whole x 100 = % composition
1.01 g H
x 100 = 1.60%
63.02 g HNO3
14.01 g N
x 100 = 22.23%
63.02 g HNO3
Determine the percentage composition of the following
compound HNO3 has 1 mol H = 1.01 g
1 mol N = 14.01 g
3 mol O = 48.00 g
63.02 g HNO3
Part/whole x 100 = % composition
1.01 g H
x 100 = 1.60%
63.02 g HNO3
14.01 g N
x 100 = 22.23%
63.02 g HNO3
48.00 g O
x 100 = 76.17%
63.02 g HNO3
% Composition or % Mass

% mass = g desired element X 100
total g of compound
 Make sure the % add up to 100 %
If 8.20 g Mg combines with 5.40g O, what is
the % of each compound?
%mass Mg=
% Composition or % Mass

% mass = g desired element X 100
total g of compound
 Make sure the % add up to 100.
If 8.20 g Mg combines with 5.40g O, what is
the % of each compound?
%mass Mg= 8.2 g Mg = 8.2 g Mg x 100=60.3%
8.2 + 5.4 13.6 g MgO
% Composition

% mass = g desired element X 100
total g of compound
 Make sure the % add up to 100.
If 8.20 g Mg combines with 5.40g O, what is
the % of each compound?
%mass Mg= 8.2 g Mg = 8.2 g Mg x 100=60.3%
8.2 + 5.4
13.6 g
%mass O = 5.4 g O = 5.4 g O x 100 = 39.7%
8.2 + 5.4
13.6 g
% Composition

% mass = g desired element X 100
total g of compound
 Make sure the % add up to 100.
If 8.20 g Mg combines with 5.40g O, what is
the % of each compound?
%mass Mg= 8.2 g Mg = 8.2 g Mg x 100=60.3%
8.2 + 5.4
13.6 g
%mass O = 5.4 g O = 5.4 g O x 100 = 39.7%
8.2 + 5.4
13.6 g
100 - 60.3= 39.7
% Composition % Mass

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
% Composition % Mass

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
C = 3(12.01) = 36.03 g C
% Composition

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
% Composition

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
% mass C = 36.03 g C = 36.03 g C x 100 = 81.7%
36.03 +8.08
44.11 g
% Composition

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
% mass C = 36.03 g C = 36.03 g C x 100 = 81.7%
36.03 +8.08
44.11 g
% mass H = 8.08 g C = 8.08 g C x 100 = 18.3%
36.03 +8.08
44.11 g
% Composition

% mass = g desired element X 100
total g of compound
Calculate the % composition of C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
% mass C = 36.03 g C = 36.03 g C x 100 = 81.7%
36.03 +8.08
44.11 g
% mass H = 8.08 g C = 8.08 g C x 100 = 18.3%
36.03 +8.08
44.11 g
total 100%

Calculate mass C in 82.0g C3H8
C = 3(12.01) = 36.01 g C
H = 8(1.01) = 8.08 g H
???????

Calculate mass C in 82.0g C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
total = 44.11 g C3H8
36.03 g x 100 = 81.7% C
44.11 g

Calculate mass C in 82.0g C3H8
C = 3(12.01) = 36.03 g C
H = 8(1.01) = 8.08 g H
36.03 g x 100 = 81.7% C
44.11 g
82.0 g C3H8 x 81.7% = 66.994 = 67.0 g C
Find the percent composition of a compound
that contains 20.4 g Nitrogen and 27.3 g
Oxygen in a 47.7 g sample
Find the percent composition of a compound
that contains 20.4 g Nitrogen and 27.3 g
Oxygen in a 47.7 g sample
20.4 g N
x
47.7 g Sample
100 =
Find the percent composition of a compound
that contains 20.4 g Nitrogen and 27.3 g
Oxygen in a 47.7 g sample
20.4 g N
x
47.7 g Sample
100 = 42.8% Nitrogen
Find the percent composition of a compound
that contains 20.4 g Nitrogen and 27.3 g
Oxygen in a 47.7 g sample
27.3 g O
x
47.7 g Sample
100 = 57.2 % Oxygen
Find the percent composition of a compound
that contains 20.4 g Nitrogen and 27.3 g
Oxygen in a 47.7 g sample
20.4 g N
x
47.7 g Sample
100 = 42.8% Nitrogen
27.3 g O
x
47.7 g Sample
100 = 57.2 % Oxygen
100%
Find the percent composition of MgNO3
Mg =
N=
3O=
Find the percent composition of MgNO3
Mg = 24.31 g
N = 14.01 g
3 O = 3 x 16.00 = 48.00 g O
Find the percent composition of MgNO3
Mg =
24.31 g
N=
14.01 g
3 O = 3 x 16.00 = 48.00 g O
86.32 g MgNO3
Find the percent composition of MgNO3
Mg =
24.31 g
+
N=
14.01 g
+
3 O = 3 x 16.00 = 48.00 g O
86.32 g MgNO3
Find the percent composition of MgNO3
Mg =
24.31 g Mg
x 100= 28.16%
86.32 g MgNO3
N=
14.01 g N
x 100= 16.23%
86.32 g MgNO3
O=
48.00 g O
x 100= 55.61%
86.32 g MgNO3
Find the percent composition of MgNO3
Mg =
24.31 g Mg
x 100= 28.16%
86.32 g MgNO3
N=
14.01 g N
x 100= 16.23%
86.32 g MgNO3
O=
48.00 g O
x 100= 55.61%
86.32 g MgNO3
100%


Empirical Formula- smallest whole number
ratio of elements in a compound
HO is empirical formula for H2O2
Empirical Formula
 Smallest
whole number ratio of atoms
in a compound
C2H6
reduce subscripts
CH3

Empirical Formula- smallest whole number
ratio of elements in a compound
Molecular formula- Actual # of atoms in a
molecule as it appears in nature
 May or may not be the same as empirical
For H2O2
 HO is empirical formula
 H2O2 is molecular formula




Empirical Formula- smallest whole number
ratio of elements in a compound
Molecular formula- Actual # of atoms in a
molecule as it appears in nature
 May or may not be the same as empirical
For CO2 empirical & molecular are the same
Molecular Formula
 “True
Formula” - the actual number of atoms
in a compound
empirical
formula
CH3
molecular
formula
C2H6

Empirical Formula- smallest whole number
ratio of elements in a compound
Molecular formula- # of atoms in a
molecule as it appears in nature
 May or may not be the same as empirical
Empirical? Molecular? Or Both?
 H2O ?
 CH4 ?
 C3H6 ?







Empirical Formula- smallest whole number ratio of
elements in a compound
Molecular formula- Actual # of atoms in a
molecule as it appears in nature
 May or may not be the same as empirical
H2O ? Empirical and Molecular
CH4 ? Empirical and Molecular
C3H6 = Molecular
CH2 = Empirical


Empirical Formula- smallest whole number ratio of
elements in a compound
Molecular formula- Actual # of atoms in a
compound as it appears in nature
 May or may not be the same as empirical
The mole ratio of the elements in a compound’s
Molecular formula is a multiple of the mole ratio of
the empirical formula
Ex.
C6H6 and C2H2 are both multiples of the empirical
formula CH

Determine the empirical formula of a
compound that contains35.98%Al &
64.02%S.
Determine the empirical formula of a
compound that contains35.98%Al &
64.02%S.
Imagine you have 100 g of formula then
35.98 g Al and 64.02 g S

Determine the empirical formula of a
compound that contains35.98%Al &
64.02%S.
Imagine you have 100 g of sample formula then
35.98 g Al and 64.02 g S
35.98 g Al x 1 mole Al = 1.33 moles Al
26.98 g Al

Determine the empirical formula of a
compound that contains35.98%Al &
64.02%S.
Imagine you have 100 g of formula then
35.98 g Al and 64.02 g S
35.98 g Al x 1 mole Al = 1.33 moles Al
26.98 g Al
64.02 g S x 1 mole S = 2.00 moles S
32.07 g S

Determine the empirical formula of a compound that
contains35.98%Al & 64.02%S.
Imagine you have 100 g of formula then
35.98 g Al and 64.02 g S
35.98 g Al x 1 mole Al = 1.33 moles Al
26.98 g Al
64.02 g S x 1 mole S = 2.00 moles S
32.07 g S

1.33 = 1 Al
1.33
2.00 = 1.5 S
1.33
for every 1 Al need 1.5 S
Determine the empirical formula of a compound that
contains35.98%Al & 64.02%S.
Imagine you have 100 g of formula then
35.98 g Al and 64.02 g S
35.98 g Al x 1 mole Al = 1.33 moles Al
26.98 g Al
64.02 g S x 1 mole S = 2.00 moles S
32.06 g S

1.33 = 1 Al
2.00 = 1.5 S
for every 1 Al need 1.5 S
1.33
1.33
But can’t be decimal has to be whole number therefore
2 Al and 3 S  Al2S3

Determine the empirical formula of a
compound that contains 10.89%Mg,
31.77%Cl & 57.34%O?
Determine the empirical formula of a
compound that contains 10.89%Mg,
31.77%Cl & 57.34%O?
If 100 g then
10.89 g Mg x 1 mole Mg = .4480 mole Mg
24.31 g Mg

Determine the empirical formula of a
compound that contains 10.89%Mg,
31.77%Cl & 57.34%O?
If 100 g then
10.89 g Mg x 1 mole Mg = .4480 mole Mg
24.31 g Mg
31.77 g Cl x 1 mole Cl = .8962 mole Cl
35.45 g Cl

Determine the empirical formula of a
compound that contains 10.89%Mg,
31.77%Cl & 57.34%O?
If 100 g then
10.89 g Mg x 1 mole Mg = .4480 mole Mg
24.31 g Mg
31.77 g Cl x 1 mole Cl = .8962 mole Cl
35.45 g Cl
57.34 g O x 1 mole O = 3.584 mole O
16.00 g O

Determine the empirical formula of a compound that
contains 10.89%Mg, 31.77%Cl & 57.34%O?
If 100 g then
10.89 g Mg x 1 mole Mg = .4480 mole Mg
24.31 g Mg
31.77 g Cl x 1 mole Cl = .8962 mole Cl
35.45 g Cl
57.34 g O x 1 mole O = 3.584 mole O
16.00 g O
.4480 = 1.00 Mg .8962 = 2.00 Cl
3.584 = 8.00 O
.4480
.4480
.4480

Determine the empirical formula of a compound that
contains 10.89%Mg, 31.77%Cl & 57.34%O?
If 100 g then
10.89 g Mg x 1 mole Mg = .4480 mole Mg
24.31 g Mg
31.77 g Cl x 1 mole Cl = .8962 mole Cl
35.45 g Cl
57.34 g O x 1 mole O = 3.584 mole O
16.00 g O
.4480 = 1.00 Mg .8962 = 2.00 Cl
3.584 = 8.00 O
.4480
.4480
.4480

MgCl2O8
Molecular Formula
1. Determine empirical formula mass.
2. Divide molecular mass by empirical mass.
3. Multiply each subscript by the answer from
step 2
Step 2
MF mass
n
EF mass
Step 3
EF n

Calculate molecular formula of compound
whose molar mass is 60.0g & the empirical
formula is CH4N.
Calculate molecular formula of compound
whose molar mass is 60.0g & the empirical
formula is CH4N.
Molar mass of Molecular Formula is 60.0 g
Molar Mass of Empirical Formula is
CH4N = 12.01 + 4(1.01) + 14.01g = 30.06 g

Calculate molecular formula of compound
whose molar mass is 60.0g & the empirical
formula is CH4N.
Molar mass of Molecular Formula is 60.0 g
Molar Mass of Empirical Formula is
CH4N = 12.01 + 4(1.01) + 14.01g = 30.06 g

MF mass = n
EF mass
(EF)n
Calculate molecular formula of compound
whose molar mass is 60.0g & the empirical
formula is CH4N.
Molar mass of Molecular Formula is 60.0 g
Molar Mass of Empirical Formula is
CH4N = 12.01 + 4(1.01) + 14.01g = 30.06 g

MF mass = n
EF mass
60.0 g = 2
30.06 mass
(EF)n
(CH4N)2 = C2H8N2
A compound is composed of 40.68% C, 5.08% H
& 54.24%O. It has a molecular mass of
118.1g/mol. Find the empirical & molecular
formula.
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C
12.01 g C
5.08 g H x 1 mole H = 5.03 mole H
1.01 g H
54.24 g O x 1 mole O = 3.39 mole O
16.00 g O
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C = 1 mole C
12.01 g C
3.387
5.08 g H x 1 mole H = 5.03 mole H = 1.49 mole H
1.01 g H
3.387
54.24 g O x 1 mole O = 3.390 mole O = 1 mole O
16.00 g O 3.387
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C = 1 mole C
12.01 g C
3.387
5.08 g H x 1 mole H = 5.03 mole H = 1.49 mole H
1.01 g H
3.387
54.24 g O x 1 mole O = 3.390 mole O = 1 mole O
16.00 g O 3.387
CH1.5O x 2 to eliminate decimal= C2H3O2 = EF
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C = 1 mole C
12.01 g C
3.387
5.08 g H x 1 mole H = 5.03 mole H = 1.49 mole H
1.01 g H
3.387
54.24 g O x 1 mole O = 3.390 mole O = 1 mole O
16.00 g O 3.387
CH1.5O x 2 to eliminate decimal= C2H3O2 = EF
EF mass of C2H3O = 59.05 g
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C = 1 mole C
12.01 g C
3.387
5.08 g H x 1 mole H = 5.03 mole H = 1.49 mole H
1.01 g H
3.387
54.24 g O x 1 mole O = 3.390 mole O = 1 mole O
16.00 g O 3.387
CH1.5O x 2 to eliminate decimal= C2H3O2 = EF
EF mass of C2H3O = 59.05 g
118.1 g = 2
59.05 g
A compound is composed of 40.68% C, 5.08% H & 54.24%O. It
has a molecular mass of 118.1g/mol. Find the empirical &
molecular formula.
40.68 g C x 1 mole C = 3.387 mole C = 1 mole C
12.01 g C
3.387
5.08 g H x 1 mole H = 5.03 mole H = 1.49 mole H
1.01 g H
3.387
54.24 g O x 1 mole O = 3.390 mole O = 1 mole O
16.00 g O 3.387
CH1.5O x 2 to eliminate decimal= C2H3O2 = EF
EF mass of C2H3O2 = 59.05 g
118.1 g = 2 = 2
59.05 g
(C2H3O2)2 = C4H6O4





Lab:
Title: % Composition of Gum
Purpose: To find the % of sugar in gum
Materials: Watch, gum, balance
Procedure:
 1. Get balance from cabinet
 2. Mass gum IN WRAPPER
 3. Take gum out of wrapper & chew for 15
minutes
 4. While chewing gum, mass wrapper
 5. After 15 min, put gum back in wrapper & get
mass






Observations:
 Mass of gum in wrapper _______
 Mass of wrapper __________
 Mass of gum after chewing _______
Calculations:
1. Find true mass of gum by subtracting the
mass of the wrapper from the chewed &
unchewed gum ___________
2. Subtract unchewed mass from chewed
mass (this is the mass of the sugar)
Find the % sugar in the gum
Conclusions: Write a sentence summing up
your data
Percentage Composition
 the
percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass
Percentage Composition
 Find
%Cu =
%S =
the % composition of Cu2S.
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 =
79.852% Cu
 100 =
20.15% S
Percentage Composition
 Find
the percentage composition of a
sample that is 28 g Fe and 8.0 g O.
%Fe =
%O =
28 g
36 g
8.0 g
36 g
 100 = 78% Fe
 100 = 22% O
Percentage Composition
 How
many grams of copper are in a
38.0-gram sample of Cu2S?
Cu2S is 79.852% Cu
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
Percentage Composition
 Find
the mass percentage of water in
calcium chloride dihydrate,
CaCl2•2H2O?
%H2O =
36.04 g
147.02 g
 100 = 24.51%
H2O
Empirical Formula
 Smallest
whole number ratio of atoms
in a compound
C2H6
reduce subscripts
CH3
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find
subscripts.
4. When necessary, multiply subscripts by
2, 3, or 4 to get whole #’s.
Empirical Formula
 Find
the empirical formula for a sample of
25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
1.85 mol
14.01 g
=1N
74.1 g 1 mol
= 4.63 mol O
16.00 g
1.85 mol
= 2.5 O
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers
 multiply by 2
N2O5
Molecular Formula
 “True
Formula” - the actual number of atoms
in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the empirical
mass.
4. Multiply each subscript by the answer from
step 3.
MF mass
n
EF mass
EF n
Molecular Formula
 The
empirical formula for ethylene is CH2.
Find the molecular formula if the molecular
mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4

To find empirical formula:
 1. Change % to g
 2. Change g to mol
 3. Find ratio of moles
 4. Ratios become subscripts in formula
 Tricks- if molar ratio is:
– .5 X by 2
– .3 or .6 X 3
– .25 or .75 X 4
– .2 or lower, round down
– .8 or above round up