1. Mass Relationships in Chemical Reactions
Chapter 3
4 - 5 Lectures
Dr. Ali Bumajdad

2. 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

3. Atomic mass is the mass of an atom in atomic mass units (amu)
Atomic mass & Average Atomic Masses
Micro World
atoms & molecules
Amu
Atomic mass
Macro World
grams
Molar 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 = amu
16O = amu

4. Average atomic mass of lithium:
Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + …….
100
(0)
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x x 7.016
100
= amu
Dr. Ali Bumajdad

5. Average atomic mass (6.941)

6. Sa Ex 3.1: Cu vaporized in a mass spectrometer and
It was found to have two isotopes one of mass amu
and abundance 69.09% and the other of mass amu
and abundance 30.91%
Find its average mass.
Av. m = amu

7. Avogadro’s number (NA)
SI unit of amount of substance
The Mole
Pair = 2
Dozen = 12
number
number
mole : amount of a substance that contains x 1023 objects.
mole : number of carbon atoms in 12 g of 12C
1 mol = NA = x 1023
number
Avogadro’s number (NA)

8. Number of atoms in 1/2 mole
Mass of 1 mole
Number of atoms in 1 mole
12.00 grams of 12C contains x 1023
1.008 grams of H contains x 1023
Mass of 1/2 mole
Number of atoms in 1/2 mole
6.00 grams of 12C contains x 1023
Dr. Ali Bumajdad

9. Molar mass is the mass of 1 mole of in grams shoes marbles atoms
Molar Mass (M.m.) and Molecular mass (M.w.)
eggs
Molar mass is the mass of 1 mole of in grams
shoes
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole lithium atoms = g of Li
For any element
atomic mass in amu = molar mass in grams/mol

10. From the periodic table
C amu mass of 1 atom of C
12.01 g mass of 1mol of C
mass of x 1023 atoms of C

11. One Mole of atoms S C
12.01 g
32 g g Hg
55.85 g
63.55 g Cu Fe

12. Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
1 molecule SO2 = amu
1 mole SO2 = g SO2

13. Molecular mass in amu = molar mass in grams/mol
For any compound
Molecular mass in amu = molar mass in grams/mol
M.w. of SO2 = amu
M.w. of C8H10N4O2 = amu

14. One Mole of molecules

15. Q) 1 mol of apple contains 6.022 x 1023 apple
Q) 1 mol of CH3OH contains x CH3OH
Q) 1 mol of CH3OH contains x O atoms
Q) 1 mol of CH3OH contains 4 x x H atoms
Q) 1/2 mol of CH3OH contains (6.022 x 1023)/2 CH3OH
Q) 1/2 mol of CH3OH contains (4 x x 1023)/2 H atoms
 N = n × NA
 N = no. of dozen × 12
Dr. Ali Bumajdad

16. M.m. NA n = number of mole M.m. = molar mass in g/mol m = mass
(1) NA
n = N
(2)
n = number of mole
M.m.
= molar mass in g/mol
m = mass
NA = Avogadro’s number
N = number of objects

17. Q) 1.0 ×103 CH4 molecules contain: How many H atoms.
AaBb
× a
No. of A atoms
No. of AaBb molecules
a 
× a  6.022×1023
No. of AaBb molecules
No. of A moles
6.022×1023 × a 
Q) 1.0 ×103 CH4 molecules contain:
How many H atoms.
2) How many moles of H atoms.

18. Q) How many atoms are in 0.551 g of potassium (K) ?
I need n and NA NA
n = N
(2)
M.m.
n = m
(1)
Find n using:
n = g / g mol-1
n = mol
N = (0.0141) (6.022 x 1023) = 8.49 x 1021 K atoms
Expected
Dr. Ali Bumajdad

19. Q) How many H atoms are in 72.5 g of C3H8O ?
Answer = 5.82 x 1024 atoms H
Dr. Ali Bumajdad

20.  12 NA NA Because m of 1 dozen in g m of 1 apple in g = (3)
m of 1 atom in g =
m of 1 mol in g
(3)  NA
m of 1 atom in g =
M.m.
(3)

21. Sa Ex. 3.2: Calculate the mass in gram of 6 atoms of
Americium (Am)
Method 1: Using Eq. 3
Method 2:
243 g x 1023 atoms
X 6 atoms

22. 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 = mol Al
N = 2.23 × Al atoms

23. Sa Ex. 3.4: How many Si atoms in 5.68 mg of silicon
computer chip?
N = 1.22 × 1020 Si atoms

24. 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

25. (1) M.m.= 174.1g (2) No. of moles = 8.96 × 10-5 mol
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

26. 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. = g/mol
mass of the CO32- = 292 g

27. (1) no. of molecules= 5×1015 molecules
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

28. We should use Eq.1 and Eq. 2 and …
M.m.
n = m
(1) NA
n = N
(2)
Number of H atoms = 1.03 ×1024 H atoms

29. We should use Eq.1
M.m.
n = m
(1)
Mass of Zn = 23.3 gram

30. We should use Eq.1 and Eq. 2 Number of S atoms = 3.06 × 1023 S atoms
M.m.
n = m
(1) NA
n = N
(2)
Number of S atoms = 3.06 × 1023 S atoms

31. Percent Composition
Heavy
Light

32. Suppose I have AaBb molecule
A) If I know M.F.
Suppose I have AaBb molecule
(4)
% A =
(M.m.A) × a
M.m. AaBb
x 100%
%C =
(12.01 g) x 2
46.07 g
x 100% = 52.14%
%H =
(1.008 g) x 6
46.07 g
x 100% = 13.13%
C2H6O
%O =
(16.00 g) x 1
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%
Dr. Ali Bumajdad

33. Suppose I have AaBb molecule
B) If I do not know M.F. but I know the masses
Suppose I have AaBb molecule
(5)
% A =
m A
mA +mB
x 100%
%C =
24 g
32 g
x 100% = 75 %
CyHx
%H =
8 g
32 g
x 100% = 25 %
m of C =24 g
m of H = 8g
Dr. Ali Bumajdad

34. % H = = 3.086%
% P = = 31.61%
% O = = 65.31%

35. Calculation Empirical & Molecular Formula
A) Empirical formula from combustion of known amount
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)
Find the moles of O from mass of O
4) Divide by the smallest mole value

36. Divide by smallest subscript (0.25)
Combust 11.5 g
ethanol
22.0 g CO2
13.5 g H2O
g CO2
mol CO2
mol C
g H2O
mol H2O
mol H
g O mol O
0.5 mol C
1.5 mol H
0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Dr. Ali Bumajdad

37. Q) 0.1156g f compound contains C, H and N only. The H2O
Absorber increase by g and the CO2 absorber increase
by g what is the empirical formal of the compound?
The E.F. is CH5N
Dr. Ali Bumajdad

38. 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. E.F.
M.m. M.F.
(6)
Dr. Ali Bumajdad

39. M.F. = N2O4
M.m. = g/mol

40. Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%C
And 4.07% H and its molar mass = g/mol
Determine its E.F.
Determine its M.F.
(Assume that we have 100 grams of compound)
E.F. = CH2Cl
M.F. = C2H4Cl2
Dr. Ali Bumajdad

41. Sa. Ex 3. 12: White powder contain 43. 64% P and 56. 36% O
Sa. Ex 3.12: White powder contain 43.64% P and 56.36% O. The compound molar mass = g/mol.
What is E.F.
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

42. 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
Balancing Chemical Equation
Reactant
Product
2 Mg + O 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 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

43. 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.
C2H O2
2CO2 + 3H2O 7 2
2C2H6 + 7O2
4CO2 + 6H2O
Dr. Ali Bumajdad

44. 1) 1C2H5OH (L) + 3O2 (g) 2CO2 (g) + 3H2O (g)
Q) Balance:
1) 1C2H5OH (L) + 3O2 (g) CO2 (g) H2O (g)
2) Na2CO3 (s) + 2HCl (aq) NaCl (s) + H2O (L) + CO2 (g)
3) 1C6H6 (L) O2 (g) CO2 (g) + 3H2O (g)
Sa Ex. 3.14: balance
(NH4)2Cr2O7 (s) Cr2O3 (s) N2(g) + 4 H2O (g)
Sa Ex. 3.15: balance
2NH3 (g) /2O2 (g) NO (g) H2O (g)
Dr. Ali Bumajdad

45. 4Al + 3O Al2O3

46. Calculation involving masses
Stoichiometric Calculations & Limiting Reactant
Calculation involving masses
( Coefficient provide moles ratios not exact amount)
Balanced the equation
Find moles of one reactant or product
Use coefficient ratio to calculate the moles of the required substance
Convert moles of required substance into masses
Dr. Ali Bumajdad

47. Q) How many moles of C atoms are needed to combined
with 4.87 mol Cl to form C2Cl6
1) Balanced the equation
2C Cl C2Cl6
2) Find moles of one reactant or product
2C Cl C2Cl6
2.435 mol ?
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of C mol Cl2
X mol of Cl2
= 1.62 mol C
Dr. Ali Bumajdad

48. M.m. Q) 209 g of methanol burns in air according to the equation:
Find mass of water formed.
CH3OH + O CO2 + H2O
1) Balanced the equation
2CH3OH + 3O CO2 + 4H2O
2) Find moles of one reactant or product
2CH3OH + 3O CO2 + 4H2O
209 g
32.04 g/mol
n =
= 6.52 mol
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of CH3OH mol H2O
6.52 mol of CH3OH X
= mol H2O
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m H2O =n ×M.m.=13.04× = 235 g H2O
Dr. Ali Bumajdad

49. M.m. Q) How many grams of Ca must react with 83.0 g of Cl2
to form CaCl2
1) Balanced the equation
Ca Cl CaCl2
2) Find moles of one reactant or product
83.0 g ?
Ca Cl CaCl2
70.9 g/mol
n =
= 1.17 mol
3) Use coefficient ratio to calculate the moles of the required substance
= 1.17 mol Ca
1 mol of Ca mol Cl2
X mol of Cl2
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m Ca =n ×M.m.= 1.17×40.08 = 46.9 g Ca

50. 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
1.00 × 103 g ?
23.95 g/mol
n =
= mol (expected)
2LiOH(s) CO2 (g) Li2CO3(s) H2O(L)
3) Use coefficient ratio to calculate the moles of the required substance
2 mol of LiOH mol CO2
41.75 mol of LiOH X
= mol CO2
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m CO2 =n ×M.m.= 20.88×44.0 = 920 g CO2
Dr. Ali Bumajdad

53. (the reactant the completely consumed in a chemical reaction)
Limiting Reactant
6 green used up
6 red left over
Dr. Ali Bumajdad

54.  Limiting reactant determine the amount of product
2H O H2O
e.g.
2 mol 1mol mol
1 mol 1/2mol mol
2.1 mol 1mol mol
limiting
1.9 mol 1mol mol
limiting
 Limiting reactant determine the amount of product
Dr. Ali Bumajdad

55. 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

56. Calculation involving a limiting reactant
Balanced the equation
Find moles of the two reactants
Identify the limiting reactant (How?)
Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance
Convert moles of required substance into masses

57. Q) In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
1) Balanced the equation
2Al + Fe2O Al2O3 + 2Fe
2) Find moles of the two reactants
2Al + Fe2O Al2O3 + 2Fe
124g ?
124 g
26.98 g/mol
n =
= 4.60 mol
601g
601 g
g/mol
= 3.76 mol
limiting
3) Identify the limiting reactant (How?)
4.60 2
= 2.3
3.76 1
= 3.76
limiting
Dr. Ali Bumajdad

58. M.m. and the required substance to find out the number of mole
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 mol AL2O3
4.60 mol of Al X
= 2.30 mol Al2O3
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m Al2O3 =n ×M.m.= 2.30× = 235 g Al2O3
Dr. Ali Bumajdad

59. Sa.Ex. 3.18: 18.1 g of NH3 and 90.4g of CuO reacted
NH CuO N2 + Cu + H2O
Which is the limiting reactant?
How many grams of N2 will be formed?
1) Balanced the equation
2NH CuO N Cu + 3H2O
2) Find moles of the two reactants
18.1g ?
18.1 g
17.03 g/mol
n =
= 1.06 mol
90.4g
90.4 g
79.55 g/mol
= 1.14 mol
2NH CuO N Cu + 3H2O
limiting
3) Identify the limiting reactant (How?)
1.06 2
= 0.53
1.14 3
= 0.38
limiting
Dr. Ali Bumajdad

60. M.m. and the required substance to find out the number of mole
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 mol N2
1.14 mol of Al X
= mol N2
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m N2 =n ×M.m.= 0.380×28.0 = 10.6 g N2
Dr. Ali Bumajdad

62. Theoretical yield, actual yield and percentage yield
Mass of product
Maximum yield
Calculated using
stiochiometry
Mass of product
Known by experiment
Never more that
theoretical yield
% Yield =
Actual Yield
Theoretical Yield
x 100
(7)
Dr. Ali Bumajdad

63. Sa.Ex. 3.19: 8.60 kg of H2 and 68.5kg of CO reacted H2 + CO CH3OH
Theoretical yield?
% yield if the actual yield = 3.57 × 104 g CH3OH
1) Balanced the equation
2H CO CH3OH
2) Find moles of the two reactants
8.60 ×103g ?
2.016 g/mol
n =
= 4.27×103 mol
68.5 × 103g
28.02 g/mol
= 2.44×103 mol
2H CO CH3OH
limiting
3) Identify the limiting reactant (How?)
4.27×103 2
= 2135
2.44×103 1
= 2440
limiting
Dr. Ali Bumajdad

64. M.m. and the required substance to find out the number of mole
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 H mol CH3OH
4.27×103 mol of H X
= 2135 mol CH3OH
4) Convert moles of required substance into masses
M.m.
n = m
(1)
m CH3OH =n ×M.m.= 2135×32.04 = 6.86×104 g CH3OH
% yield =
6.86×104 g
3.57 × 104 g
= 52.0%
× 100
Dr. Ali Bumajdad