1. % Composition, Empirical and Molecular Formulas
Chemical Quantities
The Mole,
% Composition, Empirical and Molecular Formulas

2. How you measure how much?
You can measure mass, or volume, or you can count pieces.
We measure mass in grams.
We measure volume in liters.
We count pieces in MOLES.

3. Moles Defined as the number of carbon atoms in exactly
12 grams of carbon-12.
1 mole is 6.02 x particles.
Treat it like a very large dozen
6.02 x is called Avagadro’s number.

4. Representative particles
The smallest pieces of a substance.
For a molecular compound it is a molecule.
For an ionic compound it is a formula unit.
For an element it is an atom.

5. Types of questions 3 12 3 2 5 How many oxygen atoms in the following?
CaCO3
Al2(SO4)3
How many ions in the following?
CaCl2
NaOH 3 12 3 2 5

6. Types of questions using the equality; 1 mole = 6.02 x 1023
How many molecules of CO2 are the in 4.56 moles of CO2 ?
4.56 mole x x1023 mc =
mole
How many moles of water is 5.87 x molecules?
5.87 x mc x mole =
x1023 mc
2.75x1024mc
mole

7. Types of questions using the equality; 1 mole = 6.02 x 1023
How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
1.23 moles x x1023 mc x 6 atoms =
mole mc
How many moles is 7.78 x 1024 formula units of MgCl2?
7.78x1024 FU x 1 mole = mole
x1024 FU
4.44x1024 atoms

8. Measuring Moles Remember relative atomic mass?
The amu was one twelfth the mass of a carbon 12 atom.
Since the mole is the number of atoms in 12 grams of carbon-12,
the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

9. Gram Atomic Mass The mass of 1 mole of an element in grams.
12.01 grams of carbon has the same number of pieces as grams of hydrogen and grams of iron.
We can right this as g C = 1 mole
We can count things by weighing them.

10. Examples How much would 2.34 moles of carbon weigh?
2.34 moles C x 12 g C
mole
How many moles of magnesium in g of Mg?
24.31 g Mg x mole =
g Mg
= g
1.013 mole

11. How many atoms of lithium in 1.00 g of Li?
1.00 g Li x 1 mole x 6.02x1023 atoms
g Li mole
How much would 3.45 x 1022 atoms of U weigh?
3.45x1022 atoms U x 1 mole x g U
x1023atoms 1 mole
8.60x1022 atoms
13.6 g

12. What about compounds?
in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms
To find the mass of one mole of a compound
determine the moles of the elements they have
Find out how much they would weigh
add them up

13. What about compounds? What is the mass of one mole of CH4?
1 mole of C = 12 g
4 mole of H x 1 g = 4g
1 mole CH4 = = 16g
The Gram Molecular mass of CH4 is 16.05g
The mass of one mole of a molecular compound.

14. Gram Formula Mass The mass of one mole of an ionic compound.
Calculated the same way.
What is the GFM of Fe2O3?
2 moles of Fe x 56 g = 112 g
3 moles of O x 16 g = 48 g
The GFM = 112 g + 48 g = 160g

15. Molar Mass The generic term for the mass of one mole.
The same as gram molecular mass, gram formula mass, and gram atomic mass.

16. Examples
Calculate the molar mass of the following and indicate what type it is.
Na2S
2 (23) + 32 =
N2O4
2(14) + 4(16) = C
= 78g
1mole Na2S = 78g
Gram Formula Mass
= 92g
1 mole N2O4 = 92g
Gram Molecular Mass
12g
1 mole C = 12 g
Gram Atomic Mass

17. Molar Mass Cont. Ca(NO3)2 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g
1 mole Ca(NO3)2 = 164g
C6H12O6
6(12) + 12(1) + 6(16) = = 180g
1 mole C6H12)6 = 180g Gram Molecular Mass
(NH4)3PO4
3(14) + 12(1) (16) = = 149g
1 mole (NH4)3PO4 = 149g Gram Formula Mass
Gram Formula Mass

18. Finding moles of compounds Counting pieces by weighing
Using Molar Mass
Finding moles of compounds
Counting pieces by weighing

19. Molar Mass The number of grams of 1 mole of atoms, ions, or molecules.
We can make conversion factors from these. To change grams of a compound to moles of a compound.

20. For example
How many moles is 5.69 g of NaOH?

21. For example
How many moles is 5.69 g of NaOH?

22. For example How many moles is 5.69 g of NaOH?
need to change grams to moles

23. For example How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH

24. For example How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH
1mole Na = 23g 1 mol O = g 1 mole of H = 1 g

25. For example How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH
1mole Na = 23g 1 mol O = 16 g mole of H = 1 g
1 mole NaOH = 40 g

26. For example How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH
1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g
1 mole NaOH = 40 g

27. For example How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH
1mole Na = 23g 1 mol O = 16 g mole of H = 1 g
1 mole NaOH = 40 g

28. Examples How many moles is 4.56 g of CO2? 4.56g CO2 x 1 mole 1 44gCO2
How many grams is 9.87 moles of H2O?
9.87 moles x 18g H2O
mole
= moles
= 178g

29. Examples How many molecules in 6.8 g of CH4?
6.8g CH4 x 1 mole x 6.02x1023 mc
g CH mole
49 molecules of C6H12O6 weighs how much?
49 mc x 1 mole x g C6H12O6 =
x1023 mc mole
8820 g____= 1.5x10-20 g
6.02x1023
= 2.56x1023 mc

30. Gases and the Mole

31. Gases Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure
Scientists compare gases at Standard Temperature and Pressure

32. Standard Temperature and Pressure
0ºC and 1 atm pressure
abbreviated STP
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

33. Examples What is the volume of 4.59 mole of CO2 gas at STP?
4.59mole x 22.4L
mole
How many moles is L of O2 at STP?
5.67L x 1 mole L
What is the volume of 8.80g of CH4 gas at STP?
8.80g CH4 x 1mole x 22.4L
g CH mole
= = 103L
= .523moles
= = 12.3L

34. We have learned how to change moles to grams moles to atoms
moles to formula units
moles to molecules
moles to liters
molecules to atoms
formula units to atoms
formula units to ions

35. Mass
Periodic Table
Moles

36. Mass
Volume
Periodic Table
Moles

37. Mass
Volume
Periodic Table
22.4 L
Moles

38. Mass
Volume
Periodic Table
22.4 L
Moles
Representative
Particles

39. Mass Volume Moles 6.02 x 1023 Representative Particles 22.4 L
Periodic Table
22.4 L
Moles
6.02 x 1023
Representative
Particles

40. Mass Volume Moles 6.02 x 1023 Representative Particles Atoms 22.4 L
Periodic Table
Moles
6.02 x 1023
Representative
Particles
Atoms

41. Mass Volume Moles 6.02 x 1023 Representative Particles Ions Atoms
Periodic Table
Moles
6.02 x 1023
Representative
Particles
Ions
Atoms

42. Percent Composition Like all percents Part x 100 % whole
Find the mass of each component,
divide by the total mass.

43. Example
Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S.
Ag g
S g
33.3g
/33.3
= x 100 = 87.09%
/33.3
= x 100 = 12.91%

44. Getting % from the formula
If we know the formula, assume you have 1 mole.
Then you know the pieces and the whole.

45. Examples Calculate the percent composition of C2H4? C 2(12g)=24
H 4(1g) = +4
28g
/28
= x 100 = 85.71%
/28
= x 100 = 14.29%

46. Example Calculate the percent composition of Aluminum carbonate.
Al2(CO3)3
Al 2(27g)= 54
C 3(12g)= 36
O 9(16)= 144
234g
/234
= x 100 = 23.08%
/234
= x 100 = 15.38%
/234
= x 100 = 61.54%

47. • Step 2: Multiply the elements %, by the mass of the compound given.
You can also calculate the mass of an element in a given amount of a compound using % composition.
• Step 1: calculate the % comp. only of the element you want to find the mass of.
• Step 2: Multiply the elements %, by the mass of the compound given.
Example: Calculate the mass of sulfur in 3.54g of H2S.
MM of H2S = H 2 (1) =
S 1(32) = +32
34g H2S
% S = 32/34 x 100 = 94.1% S
94.1% x 3.54g = 3.33g S

48. Calculate the mass of nitrogen in 25g of (NH4)2CO3.
H 8(1g) =
C 1(12g) =
O 8(16g) = +128
176g (NH4)2CO3
%N = 28/176 x 100 = 15.91%
15.91% x 25g = 4.0g N

49. Calculate the mass of magnesium in 97.4g of Mg(OH)2.
Mg 1 (24g) = 24
O 2(16g) = 32
H 2 ( 1g) = +2
58g Mg(OH)2
%Mg = 24/58 x 100 = 41.38%
41.38% x 97.4g = 40.3g Mg

50. From percentage to formula
Empirical Formula
From percentage to formula

51. The Empirical Formula
The lowest whole number ratio of elements in a compound.
The molecular formula the actual ration of elements in a compound.
The two can be the same.
CH2 empirical formula
C2H4 molecular formula
C3H6 molecular formula
H2O both

52. Calculating Empirical
Just find the lowest whole number ratio
C6H12O6
CH4N
It is not just the ratio of atoms, it is also the ratio of moles of atoms.
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

53. Calculating Empirical
Means we can get ratio from percent composition.
Assume you have a 100 g.
The percentages become grams.
Can turn grams to moles.
Find lowest whole number ratio by dividing by the smallest moles.

54. Example
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g so
38.67 g C x 1mol C = mole C g C
16.22 g H x 1mol H = mole H g H
45.11 g N x 1mol N = mole N g N

55. Example The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N
The ratio is mol H = 5 mol H mol N mol N
C1H5N1

56. A compound is 43.64 % P and 56.36 % O. What is the empirical formula?
43.64 g P x 1mol P = mole P g P
56.36 g O x 1mol O = mole O g O
The ratio is mol O = 2.5 mol O mol P mol P
Can not have 2.5 atoms! Double
P2O5

57. Caffeine is 49. 48% C, 5. 15% H, 28. 87% N and 16. 49% O
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
49.48 g C x 1mol C = mole C g C
5.15 g H x 1mol H = 5.15 mole H g H
28.87 g N x 1mol N = mole N g N
16.49 g O x 1mol O = mole O g O

58. The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O
The ratio is 5.15 mol H = 5 mol H mol O mol O
The ratio is mol N = 2 mol N mol O mol O
C4H5N2O1

59. Empirical to molecular
Since the empirical formula is the lowest ratio the actual molecule would weigh more.
By a whole number multiple.
Divide the actual molar mass by the the mass of one mole of the empirical formula.
Caffeine has a molar mass of 194 g. what is its molecular formula?
C4H5N2O1 = 97g
194/97 = 2
Molecular Formula = C8H10N4O2

60. Example 71.65 g Cl x 1mol Cl = 2.047 mole Cl 35 g Cl
A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be g. What is its molecular formula?
71.65 g Cl x 1mol Cl = mole Cl g Cl
24.27 g C x 1mol C = mole C g C
4.07 g H x 1mol H = 4.07 mole H g H

61. The ratio is 2.047 mol Cl = 1 mol Cl 2.023 mol C 1 mol C
The ratio is 4.07 mol H = 4 mol H mol C mol C
Empirical Formula = CClH4
= 51g
98.96/51 = 2
Molecular Formula = C2Cl2H8