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Mass Relationships in
Chemical Reactions
Chapter 3
4 - 5 Lectures
Dr. Ali Bumajdad
Chapter 3 Topics
Stoichiometry
• Average Atomic Masses
•The Mole
•Molar Mass (M.m.) and Molecular mass (M.w.)
•Percent Composition
•Calculation Empirical & Molecular Formula
•Balancing Chemical Equation
•Stoichiometric Calculations & Limiting Reactant
•Theoretical yield, actual yield and percentage yield
Dr. Ali Bumajdad
Atomic mass & Average Atomic Masses
Macro World
grams
Molar mass
Micro World
atoms & molecules
Amu
Atomic mass
•Atomic mass is the mass of an atom in
atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
(0)
Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + …….
100
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
Dr. Ali Bumajdad
Average atomic mass (6.941)
Sa Ex 3.1: Cu vaporized in a mass spectrometer and
It was found to have two isotopes one of mass 62.93 amu
and abundance 69.09% and the other of mass 64.93 amu
and abundance 30.91%
Find its average mass.
Av. m = 63.55 amu
•The Mole
SI unit of amount of substance
number
number
•mole : amount of a substance that contains
6.0221367 x 1023 objects.
•mole : number of carbon atoms in 12 g of 12C
1 mol = NA = 6.0221367 x 1023
number
Avogadro’s number (NA)
Mass of 1 mole
Number of atoms in 1 mole
12.00 grams of 12C contains 6.022 x 1023
1.008 grams of H contains 6.022 x 1023
Mass of 1/2 mole
Number of atoms in 1/2 mole
6.00 grams of 12C contains 3.011 x 1023
Dr. Ali Bumajdad
Molar Mass (M.m.) and Molecular mass (M.w.)
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass in amu = molar mass in grams/mol
•From the periodic table
C
12.01 amu
12.01 g
mass of 1 atom of C
mass of 1mol of C
mass of 6.022 x 1023 atoms of C
One Mole of atoms
S
C
12.01 g
32 g
200.59 g
Hg
55.85 g
63.55 g
Cu
Fe
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
For any compound
Molecular mass in amu = molar mass in grams/mol
M.w. of SO2 = 64.07 amu
M.w. of C8H10N4O2 = 194.20 amu
One Mole of molecules
Q) 1 mol of apple contains 6.022 x 1023 apple
Q) 1 mol of CH3OH contains 6.022 x 1023 CH3OH
Q) 1 mol of CH3OH contains 6.022 x 1023 O atoms
Q) 1 mol of CH3OH contains 4 x 6.022 x 1023 H atoms
Q) 1/2 mol of CH3OH contains (6.022 x 1023)/2 CH3OH
Q) 1/2 mol of CH3OH contains (4 x 6.022 x 1023)/2 H atoms
 N = n × NA
 N = no. of dozen × 12
Dr. Ali Bumajdad
n=
m
(1)
M.m.
n
(2)
n=
= number of mole
M.m. = molar mass in g/mol
m = mass
NA = Avogadro’s number
N = number of objects
N
NA
AaBb
×a
No. of A atoms
No. of AaBb molecules
a
No. of AaBb molecules
× a  6.022×1023
6.022×1023 × a
No. of A moles

Q) 1.0 ×103 CH4 molecules contain:
1) How many H atoms.
2) How many moles of H atoms.
Q) How many atoms are in 0.551 g of potassium (K) ?
(2)
I need n and NA
n=
Find n using:
n=
N
NA
m
(1)
M.m.
n = 0.551 g / 39.10 g mol-1
n = 0.0141 mol
N = (0.0141) (6.022 x 1023) = 8.49 x 1021 K atoms
Expected
Dr. Ali Bumajdad
Q) How many H atoms are in 72.5 g of C3H8O ?
Answer = 5.82 x 1024 atoms H
Dr. Ali Bumajdad
Because
m of 1 apple in g =
m of 1 dozen in g
12

m of 1 atom in g =
m of 1 mol in g
(3)
NA
m of 1 atom in g =
M.m.
NA
(3)
Sa Ex. 3.2: Calculate the mass in gram of 6 atoms of
Americium (Am)
Method 1: Using Eq. 3
Method 2:
243 g
6.022 x 1023 atoms
X
6 atoms
Sa Ex. 3.3: Calculate the number of moles and the number
of atoms in 10.0 g sample of Al.
Use Eq. 1 and 2
n = 0.371 mol Al
N = 2.23 × 1023 Al atoms
Sa Ex. 3.4: How many Si atoms in 5.68 mg of silicon
computer chip?
N = 1.22 × 1020 Si atoms
Sa Ex. 3.5: Calculate the number of moles and the mass in
gram of a sample of Co containing 5.00 ×1020 atoms
Method 1:
m of 5.00 ×1020 atoms = 4.89 ×10-2g
n = 8.30 ×10-4 mol
Method 2:
n = 8.30 ×10-4 mol
m of 5.00 ×1020 atoms = 4.89 ×10-2g
Sa Ex. 3.6: (1) Calculate the molar mass of C10H6O3.
(2) A sample of 1.56 ×10-2g of C10H6O3,
how many moles does this sample represent
(1) M.m.= 174.1g
(2) No. of moles = 8.96 × 10-5 mol
Sa Ex. 3.7: (1) Calculate the molar mass of CaCO3.
(2) A sample of 4.86 moles of CaCO3,
what is the mass of the CO32- ions present?
M.m. = 100.09 g/mol
mass of the CO32- = 292 g
Sa Ex. 3.8: Bees release Isopentyl acetate (C7H14O2) when
sting. The amount release is about 1 g
(1) How many molecules of C7H14O2 are released
(2) How many atoms of Carbon are present
(3) How many atoms are present
(1) no. of molecules= 5×1015 molecules
(2) no. of C atoms = 4× 1016 C atoms
(3) no. of atoms = 1× 1017 atoms
We should use Eq.1 and Eq. 2 and …
n=
m
M.m.
(1)
(2)
n=
N
NA
Number of H atoms = 1.03 ×1024 H atoms
We should use Eq.1
n=
m
(1)
M.m.
Mass of Zn = 23.3 gram
We should use Eq.1 and Eq. 2
n=
m
M.m.
(1)
(2)
n=
N
NA
Number of S atoms = 3.06 × 1023 S atoms
Heavy
Light
Heavy
Light
•Percent Composition
A) If I know M.F.
Suppose I have AaBb molecule
(M.m.A) × a
%A=
M.m. AaBb
x 100%
(4)
(12.01 g) x 2
x 100% = 52.14%
46.07 g
(1.008 g) x 6
%H =
x 100% = 13.13%
46.07 g
(16.00 g) x 1
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
Dr. Ali Bumajdad
B) If I do not know M.F. but I know the masses
Suppose I have AaBb molecule
%A=
x 100%
%C =
24 g
32 g
x 100% = 75 %
%H =
8g
32 g
x 100% = 25 %
CyHx
m of C =24 g
m of H = 8g
(5)
mA
mA +mB
Dr. Ali Bumajdad
%H=
%P=
%O=
= 3.086%
= 31.61%
= 65.31%
•Calculation Empirical & Molecular Formula
A) Empirical formula from combustion of known amount
1) Find the moles of H and C from mass of H2O and CO2
2) Find mass of O (or any other element) using:
Mass of O = mtotal – (mH+mC)
3) Find the moles of O from mass of O
4) Divide by the smallest mole value
Combust 11.5 g
ethanol
22.0 g CO2
13.5 g H2O
g CO2
mol CO2
mol C
0.5 mol C
g H2O
mol H2O
mol H
1.5 mol H
gO
mol O
0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Dr. Ali Bumajdad
Q) 0.1156g f compound contains C, H and N only. The H2O
Absorber increase by 0.1676g and the CO2 absorber increase
by 0.1638g what is the empirical formal of the compound?
The E.F. is CH5N
Dr. Ali Bumajdad
B) Empirical Formula from element masses or %mass
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value
C) Molecular Formula from element masses or %mass and
M.m.M.F.
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value. Now you know E.F.
3) Use:
M.F. = E.F. ×
M.m. M.F.
M.m. E.F.
(6)
Dr. Ali Bumajdad
M.F. = N2O4
M.m. = 90.02 g/mol
Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%C
And 4.07% H and its molar mass = 98.96 g/mol
a) Determine its E.F.
b) Determine its M.F.
(Assume that we have 100 grams of compound)
a) E.F. = CH2Cl
b) M.F. = C2H4Cl2
Dr. Ali Bumajdad
Sa. Ex 3.12: White powder contain 43.64% P and 56.36% O.
The compound molar mass = 283.88 g/mol.
a) What is E.F.
b) What is M.F.
1) Find the moles of elements (for %mass assume you have 100 g)
2) Divide by the smallest mole value
E.F. = P2O5
M.F. = P4O10
Dr. Ali Bumajdad
Balancing Chemical Equation
Reactant
Product
2 Mg + O2
2 MgO
Coefficent
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
1 moles Mg + 1/2 mole O2 makes 1 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Balancing Chemical Equations
C2H6 + O2
CO2 + H2O
1.Start by balancing those elements that appear in
only one reactant and one product.
start with C or H but not O
C2H6 + O2
2CO2 + 3H2O
2. Balance those elements that appear in two or
more reactants or products.
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
Dr. Ali Bumajdad
Q) Balance:
1) 1C2H5OH (L) + 3O2 (g)
2) Na2CO3 (s) + 2HCl (aq)
3) 1C6H6 (L) + 7.5 O2 (g)
2
15
2CO2 (g) +
3H2O (g)
2NaCl (s) + H2O (L) + CO2 (g)
6 CO2 (g) + 3H2O (g)
12
6
Sa Ex. 3.14: balance
(NH4)2Cr2O7 (s)
Cr2O3 (s) +
N2(g) + 4 H2O (g)
Sa Ex. 3.15: balance
2NH3 (g) + 5/2O2 (g)
4
5
2NO (g) +
4
3H2O (g)
6
Dr. Ali Bumajdad
4Al + 3O2
2Al2O3
Stoichiometric Calculations & Limiting Reactant
Calculation involving masses
( Coefficient provide moles ratios not exact amount)
1. Balanced the equation
2. Find moles of one reactant or product
3. Use coefficient ratio to calculate the moles of the
required substance
4. Convert moles of required substance into masses
Dr. Ali Bumajdad
Q) How many moles of C atoms are needed to combined
with 4.87 mol Cl to form C2Cl6
1) Balanced the equation
2C
+
3Cl2
C2Cl6
2) Find moles of one reactant or product
2C
?
+
3Cl2
C2Cl6
2.435 mol
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of C
X
3 mol Cl2
2.435 mol of Cl2
= 1.62 mol C
Dr. Ali Bumajdad
Q) 209 g of methanol burns in air according to the
equation:
CH3OH + O2
CO2 + H2O
Find mass of water formed.
1) Balanced the equation
2CH3OH + 3O2
2CO2 + 4H2O
2) Find moles of one reactant or product
2CH3OH + 3O2
2CO2 + 4H2O
209 g
= 6.52 mol
n=
32.04 g/mol
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of CH3OH
4 mol H2O
6.52 mol of CH3OH
X = 13.04 mol H2O
4) Convert moles of required substance into masses
m
(1) m H O =n ×M.m.=13.04×18.016 = 235 g H O
n=
2
2
M.m.
Dr. Ali Bumajdad
Q) How many grams of Ca must react with 83.0 g of Cl2
to form CaCl2
1) Balanced the equation
Ca
+
Cl2
CaCl2
2) Find moles of one reactant or product
Ca
+
Cl2
CaCl2
83.0 g
?
83.0 g
= 1.17 mol
n=
70.9 g/mol
3) Use coefficient ratio to calculate the moles of the required substance
1 mol of Ca
X
1 mol Cl2
1.17 mol of Cl2
= 1.17 mol Ca
4) Convert moles of required substance into masses
m
(1)
n=
m Ca =n ×M.m.= 1.17×40.08 = 46.9 g Ca
M.m.
Sa. Ex. 3.16:
LiOH(s) + CO2 (g)
Li2CO3(s) + H2O(L)
What is the mass of CO2 require to react with 1.00 Kg of LiOH?
1) Balanced the equation
2LiOH(s) + CO2 (g)
Li2CO3(s) + H2O(L)
2) Find moles of one reactant or product
2LiOH(s) + CO2 (g)
Li2CO3(s) + H2O(L)
?
1.00 × 103 g
1.00 × 103 g
= 41.75 mol (expected)
n=
23.95 g/mol
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of LiOH
1 mol CO2
41.75 mol of LiOH
X = 20.88 mol CO2
4) Convert moles of required substance into masses
m
(1) m CO =n ×M.m.= 20.88×44.0 = 920 g CO
n=
2
2
M.m.
Dr. Ali Bumajdad
Limiting Reactant (the reactant the completely consumed
in a chemical reaction)
6 green used up
6 red left over
Dr. Ali Bumajdad
e.g.
2H2 + O2
2 mol 1mol
2H2O
2mol
1 mol
1/2mol
1mol
2.1 mol
1mol
limiting
2mol
1.9 mol
limiting
1mol
1.9mol
 Limiting reactant determine the amount of product
Dr. Ali Bumajdad
• Why limiting reactant is important?
Because in the stoichiometric calculation I should only
use the coefficient of the limiting reactant
• How do I know that a reaction contains a limiting reactant ?
If I've been given the masses or number of moles of two
reactant then I might have limiting reactant
Dr. Ali Bumajdad
Calculation involving a limiting reactant
1. Balanced the equation
2. Find moles of the two reactants
3. Identify the limiting reactant (How?)
4. Use the coefficient ratio between the limiting
reactant and the required substance to find out
the number of mole of the required substance
5. Convert moles of required substance into masses
Q) In one process, 124 g of Al are reacted with 601 g of
Fe2O3
2Al + Fe O
Al O + 2Fe
2
3
2
3
Calculate the mass of Al2O3 formed.
1) Balanced the equation
2Al + Fe2O3
Al2O3 + 2Fe
2) Find moles of the two reactants
2Al + Fe2O3
Al2O3 + 2Fe
124g 601g
?
limiting
n=
124 g
= 4.60 mol
26.98 g/mol
n=
601 g
= 3.76 mol
159.69 g/mol
3) Identify the limiting reactant (How?)
4.60
= 2.3
2
limiting
3.76
1
= 3.76
Dr. Ali Bumajdad
3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
2 mol of Al
4.60 mol of Al
1 mol AL2O3
X = 2.30 mol Al2O3
4) Convert moles of required substance into masses
m
(1)
n=
m Al2O3 =n ×M.m.= 2.30×101.96 = 235 g Al2O3
M.m.
Dr. Ali Bumajdad
Sa.Ex. 3.18: 18.1 g of NH3 and 90.4g of CuO reacted
NH3 + CuO
N2 + Cu + H2O
1) Which is the limiting reactant?
2) How many grams of N2 will be formed?
1) Balanced the equation
2NH3 + 3CuO
N2 + 3Cu + 3H2O
2) Find moles of the two reactants
2NH3 + 3CuO
N2 + 3Cu + 3H2O
18.1g
90.4g
?
limiting
n=
18.1 g
= 1.06 mol
17.03 g/mol
n=
90.4 g
= 1.14 mol
79.55 g/mol
3) Identify the limiting reactant (How?)
1.06
2
= 0.53
1.14
3
= 0.38
limiting
Dr. Ali Bumajdad
3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
3 mol of CuO
1.14 mol of Al
1 mol N2
X = 0.380 mol N2
4) Convert moles of required substance into masses
m
(1) m N =n ×M.m.= 0.380×28.0 = 10.6 g N
n=
2
2
M.m.
Dr. Ali Bumajdad
•Theoretical yield, actual yield and percentage yield
•Mass of product
•Maximum yield
•Calculated using
stiochiometry
•Mass of product
(7)
•Known by experiment
Actual Yield
x 100
% Yield =
•Never more that
Theoretical Yield
theoretical yield
Dr. Ali Bumajdad
Sa.Ex. 3.19: 8.60 kg of H2 and 68.5kg of CO reacted
H2 + CO
CH3OH
1) Theoretical yield?
2) % yield if the actual yield = 3.57 × 104 g CH3OH
1) Balanced the equation
2H2 + CO
CH3OH
2) Find moles of the two reactants
2H2 + CO
CH3OH
8.60 ×103g 68.5 × 103g
?
limiting
8.60 ×103g
68.5 × 103g
3
= 4.27×10 mol n =
= 2.44×103 mol
n=
2.016 g/mol
28.02 g/mol
3) Identify the limiting reactant (How?)
4.27×103
= 2135
2
limiting
2.44×103
= 2440
1
Dr. Ali Bumajdad
3) Use the coefficient ratio between the limiting reactant
and the required substance to find out the number of mole
of the required substance
2 mol of H2
4.27×103 mol of H2
1 mol CH3OH
X = 2135 mol CH3OH
4) Convert moles of required substance into masses
m
(1)
n=
M.m.
m CH3OH =n ×M.m.= 2135×32.04 = 6.86×104 g CH3OH
% yield =
3.57 × 104 g
× 100 = 52.0%
6.86×104 g
Dr. Ali Bumajdad