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Molar Mass
Molar mass =
Mass in grams of one mole of any element,numerically equal to its atomic weight
Molar mass of molecules can be determined from thechemical formula and molar masses of elements
Each H
2
O molecule contains 2 H atoms and 1 O 2
atom
Each
mole
of H
2
O molecules contains 2 moles of H
and 1 mole of O
One
mole
of O atoms corresponds to 15.9994 g
Two
moles
of H atoms corresponds to 2 x 1.0079 g
Sum =
molar mass
= 18.0152 g H
2
O per mole
The relative number of moles of each element ina substance can be used as a conversion factorcalled the molar ratio.
Molar ratio =
moles element A
Molar Ratios
Chapter 3
Molar
ratio =
moles element Amole of substance
Molar ratio =
moles element Amoles element B
or
H
2
O:
- Molar Ratio = 2 moles of H
mole H
2
O
Molar Ratios
Chapter 3
mole H
2
O
mole O
C
6
H
12
O
6
6 moles of C mole C
6
H
12
O
6
M l
R ti
l
H
Molar Ratios
M
olar Ratio =
12 moles Hmole C
6
H
12
O
6
6 moles Omole C
6
H
12
O
6
6 moles C =
1 mol C
12 moles H
2 moles H
Molar ratios can be used to determine thenumber of moles of a particular element in agiven substance.
How many moles of oxygen are contained in
Molar Conversions
Chapter 3
y
yg
70.0 g of C
6
H
12
O
6
Moles
O
Moles
C
6
H
12
O
6
MolarRatio
Molar Conversions
Chapter 3
Given: 70.0 g of C
6
H
12
O
6
Find: Moles of O
Molar mass
Molar ratio
Strategy:
grams of glucose
moles of glucose
moles of oxygen
Moles of O = 70.0 g of C
6
H
12
O
6
x 1mol C
6
H
12
O
6
x 6 mol O
mol C
6
H
12
O
6
180.1 g
= 2.33 mol O
Molar Conversions
glucose = C
6
H
12
O
6
(Molar mass = 180.1 g)
How many moles of glucose are contained in 70.0 gof C
6
H
12
O
6
Grams x
= moles
1 mol grams
How many moles of oxygen are contained in 70.0 gof C
6
H
12
O
6
How many grams of H are contained in 70.0 g ofC
6
H
12
O
6
Molar
Conversions
How many moles of nitrogen are contained in 70.0 g ofC
6
H
5
NO
3
How many grams of oxygen are contained in 1.5 molesof C H NO?
Chapter 3
of
C
6
H
5
NO
3
How many atoms of C are contained in 70.0 g ofC
6
H
5
NO
3
Empirical formula: smallest whole-number ratio of atomspresent in a compound
Molecular formula: actual number of each type of atompresent in a given compound
Many molecular compounds have different empirical and
Empirical and Molecular Formulas
Chapter 3
Many
molecular compounds have different empirical and
molecular formulas
Molecular Formula
Empirical Formula
C
6
H
6
CH
C
6
H
12
O
6
CH
2
O
N
2
H
4
NH
2
N
2
O
4
NO
2
Percent Composition
Empirical formulas are generally obtained by determiningthe percent composition of a compound:
Percent composition:
the percentage of the mass contributed by eachelement in a substance element
in a substance
% Element X = (# atoms of X)(AW) x 100%
FW of compound
Formula Weight (FW): the sum of the atomic weightsof all of the atoms in a chemical formula
Note: Molar mass
is a
mass
in grams that is numerically the
same as the
formula weight
Percent Composition
Calculate the % composition of H
2
O (i.e find %H and %O).
First, find the FW of H
2
O:
FW = 2(1.0079 amu) + 1(15.9994 amu)
Chapter 3
= 18.0152 amu
% H = 2(1.0079 amu) x 100% = 11.2% H
18.0152 amu
% O = 1(15.9994 amu) x 100% = 88.8 % O
18.0152 amu
6.7 Mass Percentage
What is the mass percentage of each element in urea(CH
4
N
2
O)?
Chapter 3
FW = 12.011 amu + 4(1.0079 amu) + 2 (14.0067 amu) + 1(15.9994 amu)
= 60.0554 amu
% C = 1(12 011 amu) x 100% = 20 00% C% C = 1(12.011 amu) x 100% = 20.00% C
60.0554 amu
% H = 4(1.0079 amu) x 100% = 6.71 % H
60.0554 amu
% N = 2(14.0067amu) x 100% = 46.65% C
60.0554 amu
% O = 1(15.9994 amu) x 100% = 26.64 % H
60.0554 amu
Calculating Empirical Formulas
Step 3: Divide by smallest
(this gives the molar ratio of the elements)
C: 6.164 moles C = 4.
1.234 moles N
H: 8.631 mol H = 6.
1.234 mol N
N: 1.234 mol N = 1.
1.234 mol N
Calculating Empirical Formulas
Step 4: Multiply ‘til Whole
(this step is necessary only when step 3 does not give whole number molar ratios)
x 1
= 5.000 mol C
x 1 = 5.000 mol C
C: 6.164 moles C = 4.
Chapter 3
x 1 = 7.000 mol Hx 1 = 7.000 mol Hx 1 = 1.000 mol Nx 1 = 1.000 mol N
x 1
5.000 mol C
x 1
5.000 mol C
C: 6.164 moles C
1.234 moles C
H: 8.631 mol H = 6.
1.234 mol C
N: 1.234 mol N = 1.
1.234 mol C
Calculating Empirical Formulas
5.000 mol C5.000 mol C
¾
Use these numbers to write the empiricalformula:
Chapter 3
7.000 mol H7.000 mol H1.000 mol N1.000 mol N
CC
55
HH
77
NN
¾
An iron compound contains 69.943 % Fe
and
30.057 % O. Calculate its empirical formula. ¾
Step 1: % to mass
Calculating Empirical Formulas
¾
Step
1: % to mass
Æ
69.943 g Fe
Æ
30.057 g O
¾
Step 2: Mass to Moles
mol Fe = 69.943 g Fe x
mol Fe
= 1.2524 mol Fe
55 845 g Fe
Calculating Empirical Formulas
Chapter 3
.845 g Fe
mol O = 30.057 g O x
mol O
= 1.8786 mol O
15.9994 g O
¾
Step 3: Divide by smallest
O:
1.8786 mol O
= 1.
Calculating Empirical Formulas
Chapter 3
1.2524 mol Fe
Fe: 1.2524 mol Fe = 1.
1.2524 mol Fe
Step 1:Step 1:
Empirical formula of nicotine: CEmpirical formula of nicotine: C
55
HH
77
NN
Step 2:
Formula Weight = 81.
Using Empirical Formulas to Find
Molecular Formulas
Molar mass = Molecular weight = 162.23Step 3:
Ratio= MW = 162.23 = 2.
FW
Step 4:
Molecular formula: C
(5x2)
H
(7x2)
N
(1x2)
= C
10
H
14
N
2
Using Mass Percent Data
Ethylene glycol analyzes as 38.70 %C, and 9.74 %Hwith the remainder being oxygen.
Determine the empirical formula of ethylene glycol
Chapter 3
Combustion Analysis
Chapter 4
A 1.125 g sample of a hydrocarbon was burned to produce 3.447 g of CO
2
and 1.647 g of H
2
O. Determine the empirical
formula.
Empirical and Molecular Formulas
Molecular weight : Mass spectrometer% C, H, O, N : Combustion analysis
Empirical and Molecular Formulas
of an unknown compound
Moles and Chemical Reactions
We have used the mole concept to calculate massrelationships in chemical formulas
Molar mass of ethanol (C
2
H
5
OH)?
Chapter 4
Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.
= 46.069 g/mol
Mass percentage of carbon in ethanol?
g
% C = 2 x 12.011 x 100 %
