1. Atoms, Molecules, and Stoichiometry (Lecture 3)
Conversion between the mole and related quantities; Determining empirical / molecular formula

2. Why are they important??? The mole Avogadro constant Molar mass8
Why are they important???
The mole
Avogadro constant
Molar mass
Molar volume

3. number of particles (N)
Why are they important???
mass (m)
number of particles (N)
number of moles (n)
Related by Avogadro constant
volume (V)

4. Relation between n and N
(because when N increase, n also increase)
 N = kn
When n = 1, N = 6.02 x 1023 = L
 k = 6.02 x 1023 = L
 N = Ln
Rearranging,

5. number of particles (N)
Why are they important???
mass (m)
number of particles (N)
number of moles (n)
Related by Avogadro constant
volume (V)

6. number of particles (N)
Why are they important???
mass (m)
Related by molar mass
number of particles (N)
number of moles (n)
Related by molar volume
volume (V)

7. Relation between n and V / m
V  n
(because when V increase, n also increase)
 V = Vmn
m  n
 m = Mn

8. number of particles (N)
Why are they important???
mass (m)
number of particles (N)
number of moles (n)
volume (V)

9. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?

10. Conversion between various quantities
mass (m)
number of particles (N)
number of moles (n)
volume (V)

11. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?

12. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?
Since

13. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?

14. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?

15. Conversion between various quantities
Example 1: In 4.0 g of oxygen gas, how many oxygen atoms are there?
There are 1.51 x 1023 oxygen atoms in 4.0 g of oxygen

16. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

17. Conversion between various quantities
mass (m)
number of particles (N)
number of moles (n)
volume (V)

18. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?
Answer: mass of SO2 is g

19. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

20. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

21. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

22. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

23. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

24. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?

25. Conversion between various quantities
Example 2: What is the mass of 300 cm3 of sulphur dioxide measured at r.t.p.?
The mass of sulphur dioxide is g

26. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.
Answer: 6.00 dm3 9

27. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

28. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

29. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

30. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

31. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

32. Conversion between various quantities
Example 3:
Calculate the volume of 7 g of nitrogen gas at r.t.p.

33. Conversion between various quantities
Example 3:
Calculate the number of nitrogen molecules in this sample
Answer: 1.53 x 1023

34. Conversion between various quantities
Example 3:
Calculate the number of nitrogen molecules in this sample

35. Conversion between various quantities
Example 3:
Calculate the number of nitrogen molecules in this sample

36. Conversion between various quantities
Example 3:
Calculate the number of nitrogen molecules in this sample

37. Conversion between various quantities
Example 3:
Calculate the number of nitrogen molecules in this sample

38. Conversion between various quantities
Example 3:
Calculate the average mass of an atom of nitrogen
Answer 2.33 x g

39. Conversion between various quantities
Example 3:
Calculate the average mass of an atom of nitrogen
Mass of nitrogen gas
Average mass of an atom =
Number of nitrogen atoms

40. Conversion between various quantities
Example 3:
Calculate the average mass of an atom of nitrogen
Mass of nitrogen gas
Average mass of an atom =
Number of nitrogen atoms

41. Conversion between various quantities
Example 3:
Calculate the average mass of an atom of nitrogen
Mass of nitrogen gas
Average mass of an atom =
Number of nitrogen atoms

42. Empirical and Molecular Formula
Definitions:
Empirical formula shows the simplest whole number ratio for the atoms of different elements present in the compound
Molecular formula shows the actual number of atoms of different elements in one molecule of the compound

43. Empirical and Molecular Formula
Example: 4 3 2 1
Empirical formula
Molecular formula
Compound No
Ethene
C2H4
CH2
Propene
C3H6
CH2
Cyclohexane
C6H12
CH2
methane
CH4
CH4

44. Determining Empirical Formula
Empirical formula shows the simplest whole number ratio for the atoms of different elements present in the compound
If you are given the mass ratio, 3 steps:
Write down the mass ratio of the elements involved
Convert the mass ratio to mole ratio
Determine the smallest whole number values for the mole ratio obtained 10

45. Determining Empirical Formula
Example 1: Given that a compound contains 65.31% oxygen, 32.65% sulphur and 2.04% hydrogen, what is its empirical formula? O S H
Mass ratio
65.31
32.65
2.04
Mole ratio
Simplest whole number ratio

46. Determining Empirical Formula
Example 1: Given that a compound contains 65.31% oxygen, 32.65% sulphur and 2.04% hydrogen, what is its empirical formula?
Hence the empirical formula is H2SO4

47. Determining Empirical Formula
Example 2: A sugar was found to contain 40.0% by mass of carbon, 6.7% by mass of hydrogen, and 53.3% by mass of oxygen. What is its empirical formula?
Answer: The empirical formula of the sugar is CH2O

48. Determining Empirical Formula
Example 2: A sugar was found to contain 40.0% by mass of carbon, 6.7% by mass of hydrogen, and 53.3% by mass of oxygen. What is its empirical formula? C H O
Mass ratio
40.0
6.7
53.3
Mole ratio
Simplest whole number ratio

49. Determining Molecular Formula
2 steps are required:
M(molecular formula) = M(empirical formula)n, determine n
Write down the molecular formula 11

50. Determining Molecular Formula
Example 1: Given that an organic compound having an empirical formula of CH2 has a molar mass of 56.0 gmol-1, determine the molecular formula
Since M(molecular formula) = M(empirical formula)n,
= M(CH2)n
56.0 = n ( )
56.0 = 14 n
n = 4
Hence molecular formula is (CH2)4 = C4H8.

51. Determining Molecular Formula
Example 2: Vitamin C (ascorbic acid) contains 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen by mass. The experimentally determined molar mass is gmol-1. What is the empirical formula and molecular formula for ascorbic acid?
Answer: Empirical formula: C3H4O Molecular formula: C6H8O6

52. Determining Molecular Formula
Example 2: Vitamin C (ascorbic acid) contains 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen by mass. The experimentally determined molar mass is gmol-1. What is the empirical formula and molecular formula for ascorbic acid? C H O
Mass ratio
40.92
4.58
54.50
Mole ratio
Simplest whole number ratio

53. Determining Molecular Formula
Example 2: Vitamin C (ascorbic acid) contains 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen by mass. The experimentally determined molar mass is gmol-1. What is the empirical formula and molecular formula for ascorbic acid? C H O
Mass ratio
40.92
4.58
54.50
Mole ratio
Simplest whole number ratio
1 x 3
1.343 x 3
1 x 3

54. Determining Molecular Formula
Example 2: Vitamin C (ascorbic acid) contains 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen by mass. The experimentally determined molar mass is gmol-1. What is the empirical formula and molecular formula for ascorbic acid? C H O
Mass ratio
40.92
4.58
54.50
Mole ratio
Simplest whole number ratio 3 4 3

55. Determining Molecular Formula
Example 2:
Hence, empirical formula is C3H4O3
Since M(molecular formula) = M(empirical formula)n,
= n x M(C3H4O3)
176 = n {(12.0 x 3) + (4.0) + (16.0 x 3)
n = 2
Hence the molecular formula of ascorbic acid is (C3H4O3)2 = C6H8O6.

56. What have I learnt?
Converting between amount, number of particles, molar volume, and molar mass
Determining empirical formula and molecular formula of a compound

57. “If you can’t win, make the person infront break the record” Anonymous
End of Lecture 3
“If you can’t win, make the person infront break the record” Anonymous