Here are the steps to solve this practice problem: A. C2H4 + 3O2 → 2CO2 + 2H2O B. C2H4 = 28 g/mol, O2 = 32 g/mol, CO2 = 44 g/mol, H2O = 18 g/mol C. If we want 2.25 mol CO2 and the ratio is 2:1, then we need 2.25 * 2 = 4.5 mol C2H4 4.5 mol C2H4 * 28 g/mol = 126 g of C2H4 D. The ratio of H2O to O2 is 2:3. If we have 12 g

1. Conservation of Mass
Chemical Reactions

2. Today
• Post-laboratory
• Conservation of Mass
• Balancing Equations
– Empirical formula
• Stoichiometry
– Nomenclatures
– Mole-mole
– Mole-mass

5. How many s’mores can you make…
if you vary the number of biscuits?
if you vary the number of marshmallows?
if you vary the number of chocolate coating?

6. Conservation of Mass
Law of Conservation of Mass dates
from Antoine Lavoisier's 1789 discovery
that mass is neither created nor destroyed
in chemical reactions.

7. Antoine Lavoisier
1743 - 1794
A French nobleman who also was
an amateur chemist, carried out a
series of experiments designed to
allow him to measure not just the
mass of the metal and the calx (now
known as oxides) but also the mass
of the air surrounding the reaction.
His results showed that the mass
gained by the metal in forming the
calx was equal to the mass lost by
the surrounding air.
© dersimiz.com

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9. ©

10. Number of atoms which take part in a
chemical reaction is EQUAL to the number
of atoms in the products.

11. Law of Conservation of Mass
The total mass of the reactants
is always EQUAL
to the total mass of the products.
reactants products

12. Law of Conservation of Mass
Hydrogen and Oxygen forming Water
H
H
O
O
+ 
Are the number of atoms on both sides of the reaction the same?
O
H
H
O+
H2 + O2  H20

13. Law of Conservation of Mass
H
H
O
O
+ 
A "balanced" equation, following the law of conservation of mass.
O
H
H
+
2 H2 + O2  2 H20
H
H
O
H
H

14. During a chemical reaction, matter is
neither created nor destroyed.

15. Balancing Equations
Example:
Nitrogen gas reacts with a Dihydrogen

16. Write down your given equation.
N2 + H2 → NH3
STEP
1

17. Write down the number of atoms per each element
that you have on each side of the equation.
N = 2
H = 2
N = 1
H = 3
Look at the subscripts next to each atom to find the number of atoms in the equation.
STEP
2

18. Option 1: Leave the gases (hydrogen and/or
oxygen) for last.
N = 2
H = 2
N = 1
H = 3
STEP
3

19. Time to work on the Nitrogen!
N2 + H2 → NH3
If you have more than one element left to balance: select
the element that appears in only a single molecule of
reactants and in only a single molecule of products.
STEP
4

20. "No one path is required to get to the finish line."
N2 + H2 → NH3
Or you can have Hydrogen first as your "target."
STEP
4

21. For nitrogen, get the least common multiple (LCM) of the two
number on both sides of the reaction
N = 2 N = 1
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
1 x 4 = 42
STEP
5

22. To balance both sides, each atom must have the LCM as their
total number. Therefore,
N2  2N
This remains as 2
because it's the
LCM already
This should be
multiplied by 2 to
get to the LCM.
Simply put "2" as
its coefficient
STEP
6

23. N2 + H2 → 2NH3
So the equation becomes (temporarily) like this:
NH3 is a compound therefore the coefficient 2 is also a
multiplier for H3

24. Hence the number of atoms will change
accordingly
N = 2
H = 2
N = 2
H = 6
This means that Step 5 will be done to Hydrogen as well.

25. Get the LCM of Hydrogen for both sides.
H = 2 H = 6
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
6 x 1 = 6
6 x 2 = 12
6 x 3 = 18
6 x 4 = 246
STEP
5
Part
2

26. Hydrogen atoms therefore will have:
3H2  2H3
"3" must be placed
as a coefficient to
suffice the 6 LCM
shown prior to this
slide
This remains
like from the
previous
solution
STEP
6
Part
2

27. N2 + 3H2 → 2NH3
Thus the equation finally should look like this:
N2 has a coefficient 1 but we normally do not write it in any
equations and/or chemical reactions.

28. Activity Worksheet:
Balancing Act

29. Example
Nitrogen gas reacts with a Dihydrogen
N2 + 3H2 → 2NH3
"balanced"
N2 + H2 → NH3
"unbalanced"

30. Empirical Formula
A formula giving the proportions of the
elements present in a compound but not the
actual numbers or arrangement of atoms.

31. Empirical Formula
N2 + 3H2 → 2NH3

1 : 3 : 2

2N2 + 6H2 → 4NH3
4N2 + 12H2 → 8NH3

32. Sample Equations
A. Sodium Chloride reacts with Fluorine
NaCl + F2 → NaF + Cl2
B. Octane is burned in the presence of Oxygen
C8H18 + O2 → CO2 + H2O
C. Lead(II) Hydroxide reacts with Hydrochloric acid
Pb(OH)2 + HCl → H2O + PbCl2

33. Sample Equations
A. Sodium Chloride reacts with Fluorine
2 NaCl + 1 F2 → 2 NaF + 1 Cl2
B. Octane is burned in the presence of Oxygen
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
C. Lead(II) Hydroxide reacts with Hydrochloric acid
1 Pb(OH)2 + 2 HCl → 2 H2O + 1 PbCl2

34. STOICHIOMETRY

35. Stoichiometry
A subject matter in chemistry that tells the quantity
of one substance that reacts with some quantity of
anything else in a chemical reaction
Calculations using balanced equations are called stoichiometric
calculations.

36. Atomic Mass
The mass of an atom of a chemical element
expressed in atomic mass units or amu.
It is approximately equivalent to the number of
protons and neutrons in the atom (the mass
number).

37. Atomic Mass
Element Symbol Atomic Mass (amu)
Carbon C 12
Helium He 4
Hydrogen H 1
Nitrogen N 14
Oxygen O 16
Potassium K 39
Sodium Na 23
Sulfur S 32
Zinc Zn 65

38. Atomic Mass
Element Symbol Atomic Mass (amu)
Aluminum Al 27
Chlorine Cl 36
Copper Cu 64
Fluorine F 19
Iodine I 127
Iron Fe 56
Lead Pb 207
Phosphorus P 31
Scandium Sc 45

39. Formula Mass
The sum of the atomic weights (mass) of the atoms
in the empirical formula
Example:
C3H6
3 (12 amu) + 6 (1 amu) = 42 amu
CH2
12 amu + 2 (1 amu) = 14 amu

40. Formula Mass
N2 + 3H2 → 2NH3
M: {2 x (14)} + {3 x [2 x (1)]} → {2 x [(14)+ [3 x (1)]}
M: {28} + {3 x 2} → {2 x [14 + 3}
M: 28 + 6 → 2 x 17
M: 34 amu = 34 amu

41. Molecular (Molar) Mass
Mass of a given substance (chemical element or
chemical compound) divided by the amount of
substance.
Sum of the atomic weights of the atoms in the
molecular formula
SI unit for molar mass is g/mol.

42. Molecular Mass
3N2 + 9H2 → 6NH3
M: {3 x [2 x (14)]} + {9 x [2 x (1)]} → {6 x [(14)+ [3 x (1)]}
M: {3 x 28} + {9 x 2} → {6 x [14 + 3]}
M: 84 + 18 → 6 x 17
M: 102 amu = 102 amu

43. We shall use Molar Mass more often.

44. Sample Equations (Molar Mass)
A. Sodium Chloride reacts with Fluorine
2 NaCl + 1 F2 → 2 NaF + 1 Cl2
B. Octane is burned in the presence of Oxygen
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
C. Lead(II) Hydroxide reacts with Hydrochloric acid
Pb(OH)2 + 2 HCl → 2 H2O + 1 PbCl2

45. Mole to Mass Conversion

46. 2 NaCl + 1 F2 → 2 NaF + 1 Cl2
2 mol x
(23 g + 36 g)
1 mol x
(2 x 19 g)
2 mol x
(23 g + 19 g)
1 mol x
(2 x 36 g)
118 g/mol 38 g/mol 84 g/mol 72 g/mol
Mole to Mass Conversion

47. 2 NaCl + 1 F2 → 2 NaF + 1 Cl2
118 g/mol 72 g/mol
Sample Equation
If you have 118 g of NaCl, you can form 72 g of Cl2

48. 2 NaCl + 1 F2 → 2 NaF + 1 Cl2
118 g/mol
Sample Equation
If you have 118 g of NaCl, you can form 102 g of NaF
102 g/mol

49. 2 NaCl + 1 F2 → 2 NaF + 1 Cl2
38 g/mol
Sample Equation
If you have 38 g of F2, you can form 72 g of Cl2
72 g/mol

50. 2 NaCl + 1 F2 → 2 NaF + 1 Cl2
Sample Equation
If you have 38 g of NaCl, you can form 102 g of NaF
102 g/mol38 g/mol

51. How many grams of Cl2 are produced when 2.50 moles
of sodium chloride are used?

52. Note the ratio of the moles from the given equation.
2 : 1
Ratio from the question, 2.5 : x
x  unknown (number of moles of Cl2)
So,
2
1
=
2.5
𝑥

53. To get the molar mass of Cl2,
2
1
=
2.5
𝑥
→
𝑥
1
=
2.5
2
→ 𝑥 = 1.25
1.25 moles of Cl2 is produced
Where Cl2 is 72 g/mol (36 g/mol x 2)
Therefore,
1.25 moles x 72 g/mol = 90 g

54. To get the molar mass of Cl2,
2.5 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 ×
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑙2
2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙
×
72 𝑔 𝑜𝑓 𝐶𝑙2
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑙2
= 𝟗𝟎 𝒈
Mole ratio
Mole to mass
conversion

55. How many grams of Cl2 are produced when 2.50 moles of
sodium chloride are used?
90 g of Cl2 are produced when 2.5 moles of NaCl are used

56. If 3 moles of NaF are produced, how many grams of
fluorine must be consumed?
57 g of F2

57. How many moles of sodium chloride must be used,
given the data in the previous question?
3 moles of NaCl

58. What about its molar mass?
408 g of NaCl

59. Practice at Home
1. Ethylene reacts with Oxygen (gas)
C2H4 + O2 → CO2 + H2O
A. Balance the equation
B. Get the molar masses of all the compounds
C. If we want to produce 2.25 mol of carbon dioxide, how much in
grams of ethylene should we use?
D. How many moles of oxygen gas was present when 12 g of water was
observed after the reaction?