mass defect

1. Mass Relationships in
Chemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. A Look Ahead…
• Atomic mass and atomic mass unit
• Avogadro’s number and the molar mass of an element
• Molecular mass and formula mass
• Percent composition of compounds
• Experimental determination of empirical formulas and
molecular formulas
• Chemical reactions and chemical equations
• Amounts of reactants and products
• Limiting reagents, excess reagents
2

3. By definition:
1 atom 12C “weighs” 12 amu
On this scale:
1H = 1.008 amu; 16O = 16.00 amu
Atomic mass (atomic weight) is the mass of an atom in atomic
mass units (amu).
One atomic mass unit: a mass exactly equal to one-twelfth the
mass of one carbon-12 atom.
a H atom is 8.400% as massive as 12C atom;
atomic mass of H is 0.084 x 12 amu = 1.008 amu
Atomic Mass & Atomic Mass Unit (amu)
3
Micro World:
atoms & molecules
Macro World:
grams

4. The average atomic mass is the weighted average
of all of the naturally occurring isotopes of the
element.
Average Atomic Mass
4

5. Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016
100
= 6.941 amu
Average atomic mass of lithium:
Average Atomic Mass
5

6. Average atomic mass (6.941)
6

7. Class Work – 3.5
The atomic masses of 35Cl (75.53%) and
37Cl (24.47%) are 34.968 amu and 36.956 amu,
respectively. Calculate the average atomic mass of
chlorine.
Ans:
34.968 x 75.53 + 36.956 x 24.47
100
= 35.45 amu
7

8. The mole (mol) is the amount of a substance that contains
as many elementary entities as there are atoms in exactly
12.00 grams of 12C.
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
Dozen = 12
Pair = 2
Mole (mol): A unit to count numbers of particles
8

9. How big is the Avogedro’s number
• It would take over 19 million years to spend
Avogadro's number of dollars if the money were
spent at the rate of one billion dollars per second.
• Counting at a rate of one atom per second, for 48
hours per week, it would take the entire population
of the world 10 million years in order to reach
Avogadro's number.
• In order to obtain Avogadro's number of grains of
sand, it would be necessary to dig the entire surface
of the Sahara desert (whose area of 8 x 106 km2 is
slightly less than that of the United States) to a
depth of 2 metres.
9

10. Molar mass is the mass of 1 mole of in grams.
eggs
shoes
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
For any element
atomic mass (amu) = molar mass (grams)
Molar Mass
1 atom O = 16.00 amu
1 mole O = 16.00 g O
10

11. One Mole of:
C (12 g) S (32 g)
Cu (64 g) Fe (56 g)
Hg (201 g)
11

12. 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
1 12C atom
12.00 amu
x
12.00 g
6.022 x 1023 12C atoms
=
1.66 x 10-24 g
1 amu
M = molar mass in g/mol
NA = Avogadro’s number
Relationship between amu & gram
12

13. Class Work – 3.7
What is the mass in grams of 13.2 amu?
Ans:
13

14. Class Work – 3.8
How many amu are there in 8.4 g?
Ans:
14

15. x
6.022 x 1023 atoms K
1 mol K
=
How many atoms are in 0.551 g of potassium (K)?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
0.551 g K
1 mol K
39.10 g K
x
8.49 x 1021 atoms K
15

16. Class Work – 3.20
How many atoms are present in 3.14 g of copper
(Cu)?
Ans:
1 mol Cu = 63.55 g Cu
1 mol = 6.022 × 1023 particles (atoms)
grams of Cu → moles of Cu → number of Cu atoms
16

17. Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
SO2
Molecular Mass
17

18. How many H atoms are in 72.5 g of C3H8O?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023 atoms H
5.82 x 1024 atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O
1 mol C3H8O
60 g C3H8O
x
8 mol H atoms
1 mol C3H8O
x
6.022 x 1023 H atoms
1 mol H atoms
x =
18

19. Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na 22.99 amu
1Cl + 35.45 amu
NaCl 58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
NaCl
Formula Mass
19

20. What is the formula mass of Ca3(PO4)2?
1 formula unit of Ca3(PO4)2
3 Ca 3 x 40.08
2 P 2 x 30.97
8 O + 8 x 16.00
310.18 amu
20

21. Class Work – 3.26
How many molecules of ethane (C2H6) are present in
0.334 g of C2H6?
Ans:
molar mass of C2H6 = 2(12.01 g) + 6(1.008 g) = 30.068 g
1 mol = 6.022 × 1023 particles (molecules)
grams of ethane → moles of ethane → number of ethane
molecules
21

22. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of
the compound
C2H6O
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
Percent Composition of Compounds
2x12.01 g + 6x1.008 g + 16.00 g = 46.07 g
22

23. Percent Composition and Empirical Formulas
Determine the empirical formula of a
compound that has the following
percent composition by mass:
K 24.75, Mn 34.77, O 40.51 percent.
nK = 24.75 g K x = 0.6330 mol K
1 mol K
39.10 g K
nO = 40.51 g O x = 2.532 mol O
1 mol O
16.00 g O
nMn = 34.77 g Mn x = 0.6329 mol Mn
1 mol Mn
54.94 g Mn
23

24. Percent Composition and Empirical Formulas
K : ~
~ 1.0
0.6330
0.6329
Mn :
0.6329
0.6329
= 1.0
O : ~
~ 4.0
2.532
0.6329
nK = 0.6330, nMn = 0.6329, nO = 2.532
KMnO4
24

25. Class Work – 3.40
Calculate the percent composition by mass of the
compound chloroform (CHCl3).
Ans:
The molar mass of CHCl3 = 12.01 g/mol + 1.008 g/mol + 3(35.45
g/mol) = 119.4 g/mol.
25

26. Class Work – 3.49
What is the empirical formula of the compound with
the following compostions? 2.1 % H, 65.3 % O, and
32.6 % S.
Ans: Assume 100 g of compound.
This gives the formula H2.08S1.017O4.081. Dividing by 1.017 gives
the empirical formula, H2SO4.
26

27. g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O = g of sample – (g of C + g of H)
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C = 0.5 mol C
1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Moles of each element C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Experimental Determination of Empirical Formulas
27

28. Determination of Molecular Formulas
A sample of a compound contains 1.52 g of nitrogen and 3.47
g of oxygen. The molar mass of this compound is between 90
g and 95 g. Determine the molecular formula and the
accurate molar mass of the compound.
nN = 1.52 g N x = 0.108 mol N
1 mol N
14.01 g N
nO = 3.47 g O x = 0.217 mol O
1 mol O
16.00 g O
N0.108O0.217 ; smallest subscript 0.108
Empirical formula NO2
28
To calculate molecular formula, we must know the approximate
molar mass of the compound in addition to its empirical formula.

29. Determination of Molecular Formulas (contd.)
Empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
Ratio between molar mass and empirical molar mass,
Molar mass
Empirical molar mass
90 g
46.01 g
=  2
Molar mass is twice the empirical molar mass.
There are two NO2 units in each molecule of the compound.
Molecular formula is (NO2)2 or N2O4.
29

30. Class Work – 3.54
Monosodium glutamate (MSG) has the following composition
by mass: 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60%
Na. What is its molecular formula if its molar mass is about
169g?
Ans:
30

31. Class Work – 3.54
31

32. reactants products
Writing Chemical Equations
2H2 + O2 2H2O
Reactants are the starting materials in a chemical
reaction, e.g., H2 & O2; written on the left of the arrow.
Chemical equation is the chemist’s shorthand description
of a reaction.
Products are the substances formed as a result of a
chemical reaction, e.g., H2O; written on the right of the
arrow.
reactants products
⇌
32

33. Writing Chemical Equations
Physical states are represented by g(gas), l(liquid) & s(solid):
2CO(g) + O2(g) 2CO2(g)
2HgO(s) 2Hg(l) + O2(g)
NaCl(s) NaCl(aq)
H2O
KBr(aq) + AgNO3(aq) KNO3(aq) + AgBr(s)
[No reaction in solid phase]
33

34. How to “Read” Chemical Equations
2Mg(s) + O2(g) 2MgO(s)
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 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
34

35. Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the
left side and the correct formula(s) for the product(s)
on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of each
element the same on both sides of the equation. Do
not change the subscripts.
2C2H6 NOT C4H12
35

36. Balancing Chemical Equations
3. Start by balancing those elements that appear in only
one reactant and one product.
C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbon
on left
1 carbon
on right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
36

37. Balancing Chemical Equations
4. Balance those elements that appear in two or more
reactants or products.
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen
(3x1)
multiply O2 by
7
2
= 7 oxygen
on right
C2H6 + O2 2CO2 + 3H2O
7
2
remove fraction
multiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O
37

38. Balancing Chemical Equations
5. Check to make sure that you have the same number of
each type of atom on both sides of the equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2) 4 C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)
38

39. Class Work – 3.59
Balance the following equations:
(a) C + O2 → CO
(b) CO+ O2 → CO2
(c) H2 + Br2 → HBr
(d) K + H2O → KOH + H2
(e) Mg + O2 → MgO
(f) O3 → O2
(g) H2O2 → H2O + O2
(a) 2C + O2 → 2CO
(b) 2CO + O2 → 2CO2
(c) H2 + Br2 → 2HBr
(d) 2K + 2H2O → 2KOH + H2
(e) 2Mg + O2 → 2MgO
(f) 2O3 → 3O2
(g) 2H2O2 → 2H2O + O2
39

40. Class Work – 3.59
Balance the following equations:
(h) N2 + H2 → NH3
(i) Zn+ AgCl → ZnCl2 + Ag
(j) S8 + O2 → SO2
(k) NaOH + H2SO4 → Na2SO4 + H2O
(l) Cl2 + NaI → NaCl + I2
(m) KOH + H3PO4 → K3PO4 + H2O
(n) CH4 + Br2 → CBr4 + HBr
(h) N2 + 3H2 → 2NH3
(i) Zn+ 2AgCl → ZnCl2 + 2Ag
(j) S8 + 8O2 → 8SO2
(k) 2NaOH + H2SO4 → Na2SO4 + 2H2O
(l) Cl2 + 2NaI → 2NaCl + I2
(m) 3KOH + H3PO4 → K3PO4 + 3H2O
(n) CH4 + 4Br2 → CBr4 + 4HBr
40

41. How much reactant? or How much product?
Stoichiometry: the quantitative study of reactants and products
in a chemical reaction.
We use moles to calculate the amount of products formed.
Mole method: the stoichiometric coefficients in a chemical
equation can be interpreted as the number of moles of each
substance.
2CO(g) + O2(g) → 2CO2(g)
2 molecule 1 molecule 2 molecule
2(6.022x1023 molecules) 6.022x1023 molecules 2(6.022x1023 molecules)
2 mol 1 mol 2 mol
Grams of CO → moles of CO → moles of CO2 → grams of CO2
Amounts of Reactants and Products: Mole method
41

42. 1) Write balanced chemical equation.
2) Convert quantities of known substances into moles.
3) Use coefficients in balanced equation to calculate the
number of moles of the sought quantity.
4) Convert moles of sought quantity into desired units.
Amounts of Reactants and Products: Mole method
42
2A → 3B

43. Methanol burns in air according to the equation
2CH3OH + 3O2 → 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
209 g CH3OH
1 mol CH3OH
32.0 g CH3OH
x
4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x =
235 g H2O
43

44. Limiting Reagent, Excess Reagent, Analogy
44
• The main goal with
chemical reactions,
especially in manu-
facturing plants, is to
produce maximum
quantity of useful
products from the
original materials at
the minimum cost.
• To achieve this, an excess of one of the original materials is
usually supplied to ensure that the more expensive material is
completely used up.

45. Limiting Reagent
2NO + O2 → 2NO2
NO is the limiting reagent.
• Stops the reaction
• Limits the amount of
product that can be form
O2 is the excess reagent.
Reactant used up first in the
reaction.
Reactants present in quantities
greater than necessary to react
with the quantities of the limiting
reagent.
45
Equation
NO:O2
2:1
Taken
1:1 (7:7)
O2 excess
NO Used
up first
Limiting
O2 Excess

46. In one process, 124 g of Al are reacted with 601 g
of Fe2O3:
2Al + Fe2O3 → Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al mol Al mol Fe2O3 needed g Fe2O3 needed
OR
g Fe2O3 mol Fe2O3 mol Al needed g Al needed
124 g Al
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
x
160. g Fe2O3
1 mol Fe2O3
x = 367 g Fe2O3
Start with 124 g Al need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent.
46

47. Use limiting reagent (Al) to calculate amount of product that
can be formed.
124 g Al
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
x
102. g Al2O3
1 mol Al2O3
x = 234 g Al2O3
2Al + Fe2O3 → Al2O3 + 2Fe
At this point, all the Al is consumed and
Fe2O3 remains in excess.
47
g Al → mol Al → mol Al2O3 → g Al2O3

48. Class Work – 3.83
Nitric oxide (NO) reacts with oxygen gas to form
nitrogen dioxide (NO2), a dark brown gas:
2NO(g) + O2(g) → 2NO2(g)
In one experiment 0.886 mole of NO is mixed with
0.503 mole of O2. Calculate which of the two
reactants is the limiting reagent. Calculate also the
number of moles of NO2 produced.
48

49. Class Work – 3.83
Ans: Let's calculate the moles of NO2 produced assuming
complete reaction for each reactant.
2NO + O2 → 2NO2
NO is the limiting reagent; it limits the amount of product
produced. The amount of product produced is 0.886 mole NO2.
49

50. Class Work – 3.83
Consider the reaction:
MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
If 0.86 mole of MnO2 and 48.2 g of HCl react, which
reagent will be used up first? How many grams of Cl2 will
be produced?
Ans:
50

51. Class Work – 3.83
Ans:
51