Download Topic # 3

Percent Composition, Empirical and
Molecular Formulas
Courtesy www.lab-initio.com
Calculating Percent Composition
of a Compound
 Like all percent problems:
part
x 100 % = percent
whole
1) Find the mass of each of the
components (the elements),
2) Divide by the total mass of the
compound; then x 100
Calculating Percent Composition
Calculate the percent of oxygen in
magnesium carbonate, MgCO3.
U: % O
K: MgCO3
part
P:
whole x 100 % = percent
S: Part oxygen =
3 x (16.00 g) = 48.0 g O
whole mass = 24.3 g
+
12 g
+ 3 x (16.00 g) =
84.3 g MgCO3
Calculating Percent composition
Calculate the percent of oxygen in
magnesium carbonate, MgCO3.
part
x 100 % = percent
whole
S:
48.0g
84.3 g
x 100 %
=
56.9 % O
Examples
 Calculate the percent
composition of C2H4?
85.7% C, 14.3 % H
 U: % C; % H
 K: C2H4, assume 1 mole
 P: : (part/whole) * 100
whole = 2(12.0g) + 4(1.0g) = 28.0 g C2H4
24.0g C
X 100 = 85.7 % C
28.0 g C2H4
4.0 g H
X 100 = 14.3 % H
28.0 g C2H4
Total = 100 %
Percent Composition
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG
monosodium glutamate), a compound used
to flavor foods and tenderize meats?
a) 8.22 %C
b) 24.3 %C
c) 41.1 %C
Percent Composition
What is the percent carbon in C5H8NO4 (the
glutamic acid used to make MSG
monosodium glutamate), a compound used
to flavor foods and tenderize meats?
a) 8.22 %C
b) 24.3 %C
c) 41.1 %C
• How could you
determine the percent of
sugar in a piece of
bubble gum?
Formulas
Empirical formula: the lowest whole
number ratio of atoms in a compound.
Molecular formula: the true number
of atoms of each element in the
formula of a compound.
 molecular formula = (empirical formula)n
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas (continued)
Formulas for ionic compounds are
ALWAYS empirical (lowest whole
number ratio).
Examples:
NaCl
Al2(SO4)3
MgCl2
K2CO3
Formulas (continued)
Formulas for molecular compounds
MIGHT be empirical (lowest whole
number ratio).
Molecular:
H2 O
C6H12O6
C12H22O11
Empirical:
H2 O
CH2O
C12H22O11
C6H12O6
C3 H 6 O3
To obtain an Empirical Formula
1. Determine the mass in grams of each
element present, if necessary.
2. Calculate the number of moles of each
element.
3. Divide each by the smallest number of
moles to obtain the simplest whole
number ratio.
4. If whole numbers are not obtained* in
step 3), multiply through by the smallest
number that will give all whole numbers
* Be
careful! Do not round off numbers prematurely
A sample of a brown gas, a major air
pollutant, is found to contain 2.34 g N and
5.34g O.
Determine a formula for this
substance.
Formula:
2.34 g N
1 mole N
14.0 g N
5.34 g O
1 mole O
16.0 g O
= 0.167 mole N
0.167 mole
=1
= 0.334 mol O
0.167 mole
=2
NO2
Empirical Formula Determination
Adipic acid contains 49.3% C, 6.9% H, and
43.8% O by mass. What is the empirical
formula of adipic acid?
1.Assume there is 100 grams so you can
treat % as mass
2. Convert grams to moles
49.3 g carbon 1 mol carbon
 4.11mol carbon
12.0 g carbon
6.9 g hydrogen 1 mol hydrogen
 6.9 mol hydrogen
1.0 g hydrogen
43.8 g oxygen 1 mol oxygen
 2.74 mol oxygen
16.0 g oxygen
Empirical Formula Determination
3. Divide each value of moles by the
smallest of the values.
Carbon:
4.11 mol carbon
1.50
2.74 mol
Hydrogen: 6.9 mol hydrogen
2.74 mol
Oxygen:
 2.51
2.74 mol oxygen
1.00
2.74 mol
Empirical Formula Determination
4. Multiply each number by an integer to
obtain all whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Empirical Formula Determination
A compound contains 75% C and 25% H.
What is the empirical formula of this
compound? (#1 on worksheet 5)
1.Assume there is 100 grams so you can
treat % as mass, and convert grams to
moles
75 g carbon 1 mol carbon
 6.3 mol carbon
12.0 g carbon
25 g hydrogen 1 mol hydrogen
 25 mol hydrogen
1.0 g hydrogen
Empirical Formula Determination
2. Divide each value of moles by the
smallest of the values.
6
.
3
mol
carbon
Carbon:
1.00
6.3 mol
25
mol
hydrogen
Hydrogen:
4
6.25 mol
Since the mole ratios are whole number there is
no need to multiply to find whole numbers.
Empirical Formula:
CH4
Finding the Molecular Formula
The empirical formula for adipic acid was just
calculated as C3H5O2. The molecular mass of
adipic acid is 146 g/mol. What is the
molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
Finding the Molecular Formula
The empirical formula for adipic acid was
just calculated as C3H5O2. The molecular
mass of adipic acid is 146 g/mol. What is the
molecular formula of adipic acid?
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
2. Divide the molecular mass by the mass given
by the empirical formula.
146
2
73.0
Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2.
The molecular mass of adipic acid is 146 g/mol.
What is the molecular formula of adipic acid?
146
 2 (C3H5O2) x 2 =
73.0
C6H10O4
3. Multiply the empirical formula by this
number to get the molecular formula.
Molecular Formula
• The empirical formula for vitamin C is
C3H4O3. However, experimental data
indicates that the molecular mass of
vitamin C is about 180 g/mol.
What is the molecular formula of vitamin C?
Calculation of the Molecular Formula
2
C6H8O6
Calculation of the Molecular Formula
Interestingly, the colorless liquid,
used in rocket engines has the same
empirical formula as the brown air
pollutant calculated previously, NO2.
However, this compound has a molar
mass of 92.0 g/mole. What is the
molecular formula of this substance?
Calculation of the Molecular Formula
2
N2O4
Empirical Formula from % Composition
A substance has the following composition by
mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the
substance?
• Consider a sample size of 100 grams
•This will contain 60.8g Na, 28.60 g B and
10.60 g H
•Determine the number of moles of each
•Determine the simplest whole number ratio
A substance has the following composition
by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
A substance has the following composition
by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
60.80 g Na
1 mole Na = 2.64 mole Na
23.0 g Na
28.60 g B
1 mole B
10.81 g B
10.6 g H
1 mole H
1.0 g H
2.64 mole
= 2.64 mol B
2.64mole
= 10.6 mol H
2.64mole
NaBH4
=1
=1
=4