1. Chapter 6 Chemical Quantities
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2. Section 6.1 The Mole: A Measurement of Matter
OBJECTIVES:
Describe how Avogadro’s number is related to a mole of any substance.

3. Section 6.1 The Mole: A Measurement of Matter
OBJECTIVES:
Calculate the mass of a mole of any substance.

4. What is a Mole? 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.

5. Moles (abbreviated: mol)
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 Avogadro’s number.

6. Representative particles
The smallest pieces of a substance.
For a molecular compound: it is the molecule.
For an ionic compound: it is the formula unit (ions).
For an element: it is the atom.
Remember the 7 diatomic elements (made of molecules)

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

8. Types of questions
How many molecules of CO2 are there in 4.56 moles of CO2 ?
4.56 mol CO2
6.02x1023molecules CO2
2.75x10 24molecules CO2
1 mol CO2

9. How many moles of water is 5.87 x 1022 molecules?
5.87x1022 molecules H2O
1 mol H2O
mol H2O
6.02x1023 molecules H2O

10. How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
1.23 mol C6H12O6
6.02x1023 molecules C6H12O6
6 atoms C
1 mol C6H12O6
1molecule C6H12O6
= 4.44 x1024 atoms C

11. How many moles is 7.78 x 1024 formula units of MgCl2
7.78x1024 fun MgCl2
1 mol MgCl2
12.9 mol MgCl2
6.02x1023 fun MgCl2

12. 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.

13. Gram Atomic Mass (gam)
Equals the mass of 1 mole of an element in grams
12.01 grams of C has the same number of pieces as grams of H and grams of iron.
We can write this as g C = 1 mole C
We can count things by weighing them.

14. Examples How much would 2.34 moles of carbon weigh? 2.34 mol C
12.0 g C
28.1 g C
1 mol C

15. How many moles of magnesium is 24.31 g of Mg ?
24.31 g Mg
1 mol Mg
1.000 mol Mg
24.3 g Mg

16. How many atoms of lithium is 1.00 g of Li?
1.00 g Li
1 mol Li
6.02x10 23 atoms Li
8.72x10 22 atoms Li
6.9 g Li
1 mol Li

17. How much would 3.45 x 1022 atoms of U weigh?
3.34x10 22 atoms U
1 mol U
238.0 g U
13.2 g U
6.02x10 23 atoms U
1 mol U

18. 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

19. What about compounds? What is the mass of one mole of CH4?
1 mole of C = g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = = 16.05g
The Gram Molecular Mass (gmm) of CH4 is 16.05g
this is the mass of one mole of a molecular compound.

20. Gram Formula Mass (gfm)
The mass of one mole of an ionic compound.
Calculated the same way as gmm.
What is the GFM of Fe2O3?
2 moles of Fe x g = g
3 moles of O x g = g
The GFM = g g = g

21. Molar Mass Mass of 1 mole of substance
The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu).

22. Section 6.2 Mole-Mass and Mole-Volume Relationships
OBJECTIVES:
Use the molar mass to convert between mass and moles of a substance.

23. Section 6.2 Mole-Mass and Mole-Volume Relationships
OBJECTIVES:
Use the mole to convert among measurements of mass, volume, and number of particles.

24. Molar Mass
Molar mass is the generic term for the mass of one mole of any substance (in grams)
The same as: 1) gram molecular mass, 2) gram formula mass, and 3) gram atomic mass- just a much broader term.

25. Examples
Calculate the molar mass of the following and tell what type it is:
Na2S
N2O4 C
Ca(NO3)2
C6H12O6
(NH4)3PO4
Na 2 atoms x 22.9 = 45.8
S 1 atom x = 32.1
77.9 g/ mol

26. 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.

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

28. For example How many moles is 5.69 g of NaOH? 1 mole
= mole of NaOH
39.9 g
Na =
O =
H =
22.9
16.0
1.0
X 1 atom
= 22.9
= 16.0
= 1.0
=39.9 g

29. Examples How many moles is 4.56 g of CO2? 4.56 g CO2 1 mol CO2
C 1 atom x = 12.0
O 2 atoms x 16.0 = 32 .0
= 44.0 g

30. How many grams is 9.87 moles of H2O?
9.87 mol H2O
18.0 g H2O
178 g H2O
1 mol H2O
H 2 atom x = 2.0
O 1 atom x 16.0 = 16.0
= 18.0 g

31. How many molecules is 6.8 g of CH4?
6.8 g CH4
1 mol CH4
6.02x1023 molecules CH4
16.0 g CH4
1 mol CH4
C 1 atom x 12.0 = 12.0
H 4 atoms x 1.0 =
16.0 g
= 2.6 x1023 molecules CH4

32. 49 molecules of C6H12O6 weighs how much?
49 molecules C6H12O6
1 mol C6H12O6
180 g C6H12O6
6.02x1023 molecules C6H12O6
1 mol C6H12O6
C 6 atoms x 12.0 = 72.0
H 12 atoms x 1.0 = 12.0
O 6 atoms x 16.0 = 96.0
180 g
= 1.5 x 10 – 20 g C6H12O6

33. 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
We need to compare them at the same temperature and pressure.

34. 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
1 mole = 22.4 L of any gas at STP

35. Examples What is the volume of 4.59 mole of CO2 gas at STP?
4.59 mol CO2
22.4 L CO2
103 L CO2
1 mol CO2

36. How many moles is 5.67 L of O2 at STP?
1 mol O2
0.253 mol O2
22.4 L O2

37. What is the volume of 8.8 g of CH4 gas at STP?
8.8 g CH4
1 mol CH4
22.4 L CH4
12 L CH4
16.0 g CH4
1 mol CH4
C 1 atom x 12.0 = 12.0
H 4 atoms x 1.0 =
16.0 g

38. Density of a gas D = m / V for a gas the units will be g / L
We can determine the density of any gas at STP if we know its formula.
To find the density we need the mass and the volume.
If you assume you have 1 mole, then the mass is the molar mass (from PT)
At STP the volume is 22.4 L.

39. Examples Find the density of CO2 at STP. 40.0 g CO2 1 mol CO2
= 1.79 g / L CO2
1 mol CO2
22.4 L CO2
C 1 atom x = 12.0
O 2 atoms x 16.0 = 32.0
= g

40. Find the density of CH4 at STP.
16.0 g CH4
1 mol CH4
0.714 g CH4
= g / L CH4
1 mol CH4
22.4 L CH4
L CH4
C 1 atom x 12.0 = 12.0
H 4 atoms x 1.0 =
16.0 g

41. The other way
Given the density, we can find the molar mass of the gas.
Again, pretend you have 1 mole at STP, so V = 22.4 L.
m = D x V
m is the mass of 1 mole, since you have 22.4 L of the stuff.
What is the molar mass of a gas with a density of g/L?
2.86 g/L?

42. Summary These four items are all equal: a) 1 mole
b) molar mass (in grams)
c) 6.02 x 1023 representative particles
d) 22.4 L at STP
Thus, we can make conversion factors from them.

43. Section 6.3 Percent Composition and Chemical Formulas
OBJECTIVES:
Calculate the percent composition of a substance from its chemical formula or experimental data.

44. Section 6.3 Percent Composition and Chemical Formulas
OBJECTIVES:
Derive the empirical formula and the molecular formula of a compound from experimental data.

45. Calculating Percent Composition of a Compound
Like all percent problems:
Part whole
Find the mass of each component,
then divide by the total mass.
x 100 %

46. Example:
What is the percent composition of H2O
H 2atoms x = g/mol
O 1atom x = g/mol
18.02 g/mol
% H = (2.02/18.02) x 100 = 11.2%
% O = (16.00/18.02) x 100 = 88.8 %

47. Example
Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.
29.0 g Ag g S = 33.3 g Total
% S = (4.30g / 33.3g) x 100 = 12.9%
% Ag = (29.0g / 33.3g) x 100 = 87.1%

48. Getting it from the formula
If we know the formula, assume you have 1 mole.
NaCl = I mole
Then you know the mass of the pieces and the whole.
Na 1 x = g/mol
Cl 1 x = g/mol
58.44 g/mol

49. Examples Calculate the percent composition of C2H4?
C 2 atoms x 12.0 = 24.0
H 4 atoms x = 4.0
28.0 g
%C = 24.0/28.0 x 100 = 85.7 % C
%H = 4.0 / 28.0 x 100 = 14.3 % H

50. How about Aluminum carbonate? Sample Problem 6-13, p.132
Al and CO is written Al2(CO3)3
Al 2 atoms x 27.0 =
C 3 atoms x 12.0 =
O 9 atoms x = 144.0
= g
% Al = 54.0 / x 100 = % Al
% C = 36.0 / x 100 = % C
% O = / x 100 = 61.5 % O

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

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
We can get a ratio from the percent composition.
Assume you have a 100 g.
The percentages become grams.
Convert grams to moles.
Find lowest whole number ratio by dividing by the smallest.

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 gC
16.22 g H x 1mol H = mole H gH
45.11 g N x 1mol N = mole N gN

55. Example The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N
The ratio is mol H = 5 mol H molN mol N
= C1H5N1

56. A compound is 43.64 % P and 56.36 % O. What is the empirical formula?
1 mol
1.41 mol
43.64g P 1
1.41
30.96g
3.5 mol
56.36g O
1mol
2.5
16.00g
1.41
P2O5
Empirical formula
And molar mass
203.88

57. 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 empirical formula mass.

58. Example
Laurium has a molar mass of g. Laurium is known to be composed of % Li, 24.27% C and 4.07% O.. What is Laurium’s molecular formula?
1mol
71.65g Li
10.32 mol
41.25
X4 =165
0.25 mol
6.94g
2.02 mol
24.27g C 8
X4 =32
1mol
12.01g
0.25 mol 1
4.07g O
X4 =4
1mol
0.25 mol
16.00g
0.25 mol
Empirical formula and molar mass
of Laurium is
Li165C32O4 ( g)

59. Molar mass of the molecular formula for Laurium is 4760.46 g
The empirical formula of Laurium is Li165C32O4
Molar mass of the Laurium empirical formula is g
Molar mass of molecular formula(given) g =
= 3
Molar mass of empirical formula(calculated) g
Empirical formula
Molecular Formula
Li165C32O4
Li495C96O12 x3