Transcript 1 Mole

The
Mole
Calculating Formula/Molar Mass
Calculate the molar mass of carbon dioxide,
CO2.
12.01 g + 2(16.00 g) = 44.01 g
 One mole of CO2 (6.02 x 1023 molecules)
has a mass of 44.01 grams
Mole Relationships
Atoms
or
molecule
s
1 Mole
6.02 X 1023 Atoms
1 Mole
Molar Mass (g)
Mole
1 Mole
6.02 X 1023 Atoms
Mass
Molar Mass (g)
1 Mole
How would you convert,
Atoms  Moles
atoms
mole
1
Moles  Atoms
= moles
moles
6.02 x 1023 atoms
1
Moles  Mass (g)
moles
g
= g
g
g
6.02 x 1023 atoms
= atoms
1 mole
mole
mole
1
= moles
g
mole
Atoms  Mass (g)
Mass (g)  Atoms
1 mole
= atoms
Mass (g)  Moles
g
1
6.02 x 1023 atoms
atoms
1 mole
g
= g
6.02 x 1023 atoms 1 mole
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles
of lithium?
3.50 mol Li
7 g Li
1 mol Li
= 24.5
g Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams
of lithium?
18.2 g Li
1 mol Li
7 g Li
= 2.60 mol Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 3.50
moles of lithium?
3.50 mol
6.02 x 1023 atoms
1 mol
= 2.11 x 1024 atoms
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
7 g Li
6.02 x 1023 atoms Li
1 mol Li
(18.2)(6.02 x 1023)/6.94 = 1.57 x 1024 atoms Li
• How many moles is 5.69 g of NaOH?
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/ mol
40.00 g/mol
5.69 g NaOH
1 mol NaOH
40.00 g NaOH
= 0.142 mol NaOH
• How many grams are in 9.45 mol of
dinitrogen trioxide?
N2O3 = 2(14.01) + 3(16.00) = 76.02 g/mol
9.45 mol N2O3 76.02 g N2O3
= 718.2 =718 g N2O3
1 mol N2O3
Find the number of sodium ions, Na+, in
3.00 mol of Na2CO3
3.00 mol
Na2CO3
2 mol Na+
1mol Na2CO3
6.022 x 1023 ions
1 mol Na+
= 3.61 x 10 24 Na+ ions
Find the number of sodium ions, Na+, in
3.00 mol of Na4P2O7
3.00 mol
Na4P2O7
4 mol Na+
1mol Na4P2O7
6.022 x 1023 ions
1 mol Na+
= 7.23 x 1024 Na+ ions
Percent Composition, Empirical and
Molecular Formulas
Courtesy www.lab-initio.com
7.3: Calculating Percent
Composition of a Compound
• Like all percent problems:
Part
whole
x 100
• Find the mass of each component,
• then divide by the total mass (assume
one mole).
mass of element
% mass of element 
 100
mass of cmpd
Percent Composition
• What is the % composition of water.
– Find MM: 2(1.01) + 16.00 = 18.02 g/mol
%H =
%O =
2.02 g H
18.02 g H2O
16.00 g O
18.02 g H2O
 100 =
11.2 % H
 100 = 88.8 % O
• Double check: %’s should add up to 100.
Practice Problem
What is the percent carbon in C5H8NO4
(MSG monosodium glutamate), a
compound used to flavor foods
and
tenderize meats?
1. Find mass of C
1. 5 x 12.0 g = 60.0 g C
2. Find mass of MSG
1. 5(12.0) + 8(1.0) +14.0 +4(16.0)=146.0 g
3.
Mass of element/ mass of cmpd x 100
1.
60.0 g C/146.0 g x100 = 41.1% C
Percent Composition
• Find the percentage composition of a
sample that is 28 g Fe and 8.0 g O.
– Determine total mass: 28 g + 8.0 g = 36 g
%Fe =
%O =
28 g
36 g
8.0 g
36 g
 100 = 78% Fe
 100 = 22% O
Calculating Percentage Composition
Calculate the percentage composition
of magnesium carbonate, MgCO3.
Formula mass of magnesium carbonate:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 24.31g 
  100  28.83%
Mg  
 84.32 g 
 12.01g 
  100 14.24%
C  
 84.32 g 
 48.00 g 
  100  56.93%
O  
 84.32 g 
100.00
Using Percent as a Conversion
Factor
• How many grams of copper are in a
38.0-gram sample of Cu2S?
– Multiply mass of cmpd with % element
Cu2S is 79.8% Cu (Sample 7-10 p
227)
(38.0 g Cu2S)x(0.79852) = 30.3 g Cu
B. Empirical Formula
• Smallest whole number ratio of atoms
in a compound
C2H6
reduce subscripts
CH3
EF vs MF
• Empirical Formula
(EF)
• Def’n: lowest whole #
ratio
• H2O
Water
peroxide
• HO
• CH
benzene
• CH2O
sugar
• Molecular Formula
(MF)
• Def’n: actual ratio
•
•
•
•
H2O
H2O2
C6H6
C6H12O6
Calculating Empirical Formula
• It is not just the ratio of atoms, it is also
the ratio of moles of atoms.
• In 1 mole of CO2 there is 1 mole of
carbon and 2 moles of oxygen.
Steps for Calculating the Empirical
Formula
We get a ratio from the percent composition.
Assume you have a 100 g sample.
1. % = g
2. Convert grams to moles (÷ atomic mass)
3. Find lowest whole number ratio.
(÷ by smallest # of moles)
(if this step gives a decimal, multiply by 2, 3,
or 4 to get whole #’s)
4. Use ratio to write the EF
• Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and
45.11 %N.
Part 1 Calculate the number of moles
Assume 100 g
• 38.67 g C x 1mol C
= 3.220 mole C
12.01 g C
• 16.22 g H x 1mol H
= 16.09 mole H
1.01 g H
• 45.11 g N x 1mol N = 3.219 mole N
14.01 g N
Part 2 calculate the formula!
Divide by the smallest
number of moles
3.220 mol C
3.219
= 1 mol C
16.09 mol H
3.219
=
3.219 mol N
3.219
= 1 mol N
5 mol H
CH5N
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
=1N
1.85 mol
14.01 g
74.1 g 1 mol
= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Try this!
Adipic acid contains 49.32% C, 43.84% O,
and 6.85% H by mass. What is the
empirical formula of adipic acid?
1. Convert to grams, calculate # of moles
49.32 g carbon 1 mol carbon
 4.107 mol carbon
12.01 g carbon
6.85 g hydrogen 1 mol hydrogen
 6.78 mol hydrogen
1.01 g hydrogen
43.84 g oxygen 1 mol oxygen
 2.74 mol oxygen
16.00 g oxygen
2. Divide each value of moles by the
smallest of the values.
Carbon:
4.107 mol carbon
1.50
2.74 mol
Hydrogen: 6.78 mol hydrogen
2.74 mol
Oxygen:
 2.47
2.74 mol oxygen
1.50
2.74 mol
3. Multiply each number by an integer to
obtain all whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Empirical formula: C3H5O2
Oxygen: 1.00
x 2
2
Molecular Formula
• “True Formula” - the actual number
of atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Steps to calculating 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?
1. Find the empirical formula and its mass
empirical mass = 14.03 g/mol
2. Divide the molecular mass by the empirical
mass.
28.1 g/mol
14.03 g/mol
= 2.00
3. Multiply each subscript by the answer from
step 3.
(CH2)2  C2H4
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?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
2. Divide the molecular mass by the empirical
mass.
146
2
73
3. Multiply each subscript by the answer from
step 3.
(C3H5O2) x 2 = C6H10O4
Example
• A compound is known to be
composed of 71.65 % Cl, 24.27% C
and 4.07% H. Its molar mass is
known (from gas density) to be 98.96
g. What is its molecular formula?
–EF: CClH2
–MF: C2Cl2H4