1. Chapter 10
Chemical Quantities

2. What are practical ways to measure a container full of sand??
Mass
Volume
How many (Count)

3. Matter is generally measured in one of three ways:
By count:
1 dozen apples = 12 apples
By mass:
1 dozen apples = 2.0 kg of apples
By volume:
1 dozen apples = .20 bushels of apples

4. Converting units
If you know how count, mass, and volume relate to each other, you can do conversions between them.
EX: How many bushels of apples are in 32.7 kg of apples?

5. Numbers as Symbols
It is not practical to individually count out small objects like sand grains, so instead we use a value that represents a larger quantity
Ex: 120 eggs = 10 dozen eggs

6. Chemical Numbers
Atoms, molecules, and ions are extremely small, so chemists use a unit similar to a “dozen” to measure a specific number of particles.

7. The MOLE
A MOLE is 6.02 x 1023 representative particles of whatever is being talked about
Avogadro's number = 6.02x1023
Representative particles refer to the type of substance that you are talking about
Abbreviated mol
You talking about me?

8. Examples of Representative Particles
Substance
Representative Particle
A balloon full of Neon
Neon Atoms
A cup full of Water
Water Molecules
A spoon full of NaCl
NaCl Formula Units
Number of Iron ions in Water
Fe Ions
How much oxygen is in the air
O2 molecules

9. Measuring by Count
If you know the number of moles of something or how many particles of something there are, you can convert to the other.
Ex: How many moles of magnesium are there in 1.23 x 1023 atoms of magnesium?

10. Measuring by Count
Ex: How many atoms are in 1.22 mol of Iron?

11. *Measuring by Count
Ex: How many atoms of carbon are in 3.21 mol of C8H18?

12. Homework
Pg. 296
# 10, 13, 14,
Pg. 315
# 49, 52,

13. Measuring by Mass
The mass of a mole of atoms of a given element will always be the same.
It equals the atomic mass expressed in grams
Molar Mass: the mass of one mole of a given particle (atom, compound, etc.)

14. Finding Molar Mass (Important notes below…write this down)
Find the element;
Find the molar mass;
Multiply by the number of atoms of that element;
Write this number down;
Repeat for each element;
Take the sum.

15. Finding Molar Mass C: O2: H2O: C: 1 x 12.00g = 12.00 g/mol C
O: 2 x g = g/mol O2
H2O:
H: 2 x 1.01 g = g H
O: 1 x g = g O
18.02 g/mol H2O

16. Finding Molar Mass C12H22O11: C: 12 x 12.01g = 144.12 g C
H: 22 x 1.01g = g H
O: 11 x 16.00g = g O
g/mol C12H22O11

17. Measuring by Mass
If you know how many moles of a substance you have, you can find its total mass using the molar mass of that substance.
Ex: What is the mass of 2.42 mol of H2O?
= 43.6 g H2O

18. Your turn!

19. Measuring by Mass
If you are given the mass, you can also find the number of moles
Ex: How many moles are there in g of Na2CO3?
106.0 g/mol Na2CO3

20. Measuring by Volume
The volume of 1 mole of a solid and liquid are very different depending on the substance
However, the volume of one mole of a gas is equal for every gas at the same physical conditions

21. Avogadro’s Hypothesis
Equal volumes of gases at the same temperature and pressure contain equal number of particles.
Even though particles are different sizes, they still occupy the same amount of space because there is a great deal of space between gas particles

22. Standard Temperature and Pressure
In order to compare the volume of a gas to one mole of the gas, we must define the temperature and pressure
As Temp goes up, Volume goes up
As Volume goes down, Pressure goes up

23. Standard Temperature and Pressure
Standard Temperature and Pressure (STP): temperature of 0 ˚C and a pressure of 1 atm
STP is used as a universal standard at which we compare things

24. Measuring by Volume At STP one mole of a gas occupies 22.4 L of space
Ex: How many moles of oxygen (O2) would be present in a 63.2 L container at STP?

25. Practice
How many moles are present in a sample of 4.02 x 1024molecules of NH3?
What is the mass of 2.14mol of C2H4O2?
How many molecules can be found in 3.42mol of BaSO4?
6.68mol
129g
2.06x10^24

26. Practice makes perfect
Under STP, what volume does 4.95mol of CH4 take up?
If a gas takes up 34.71L of space under STP, how many moles of that gas is present?
How many moles are present in a 134g sample of HBr?
111L
1.55mol
1.66mol

27. Mo’ Practice = Mo’ Perfect
How many cookies are in 24.1moles of Oreo’s?
How many/much _________ can be found in __________ while under STP?
1.45 x 10^25 cookies

28. Quiz Practice
HCl
CO2

29. Quiz Practice
How many moles are in 4.03 𝑥 molecules of silver (I) oxide?
4.03 𝑥 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 2 𝑂 1𝑚𝑜𝑙 6.02 𝑥 =
= 𝑚𝑜𝑙 𝐴𝑔 2 𝑂

30. Quiz Practice How much mass does 3.79 moles of 𝐵 𝐹 3 have?
B: 10.8 x 1=10.8g
F: 19.0 x 3= 57.0g
Total: 67.8g
3.79 𝑚𝑜𝑙 67.8𝑔 1𝑚𝑜𝑙 =
=257.0g 𝐵𝐹 3

31. Quiz Practice

32. Quiz Practice
How much volume in liters does moles of NaCl fill at STP?
6.307𝑚𝑜𝑙 22.4𝐿 1𝑚𝑜𝑙 =
141𝐿 NaCl

33. Quiz Practice
How much mass does moles of Lithium Bromide (LiBr) have?
Li: 6.9 x 1= 6.9g
Br: 79.9 x 1= 79.9g
Total= 86.8g
1.805𝑚𝑜𝑙 86.8𝑔 1𝑚𝑜𝑙 =
= 156.7𝑔 1 𝑚𝑜𝑙 𝐿𝑖𝐵𝑟

34. Mixed Mole Conversions
We now know:
1 mole = 6.02 x 1023 particles = molar mass = 22.4 L of gas at STP
Because of this, we can convert from count, mass or volume to count, mass or volume.
BUT, you must first convert to moles

35. Mixed Mole Conversions

36. Practice!
CO2
44.01g

37. Practice, part dos!

38. Ugh, More? 
180.16

39. Practice??
60.05

40. Even more practice??? (mind=blown)
32.00

41. And I thought my mind was already blown…

42. Practice Extraordinaire!
What is the mass of 3.72 x 1024 molecules of C3H8?
=273g C3H8
3.72 x 1024 molecules
1mol
44.11g
6.02 x 1023 molecules

43. To practice or not to prac… well let’s just practice
What is the volume of 57.9g of methane CH4 gas at STP?
= 80.8L CH4
57.9g
1mol
22.4L
16.05g

44. Oh the practice! What is the mass of 1.5 x 1025 molecules of CO2?
= 1097g CO2 = 1.10 x 103g CO2
1.5 x 1025 molecules
1mol
44.01g
6.02 x 1023 molecules

45. Percentages Matter! Fertilizers A man walks into a restaurant….
Higher nitrogen content  greener grass
Higher potassium content  stronger roots
A man walks into a restaurant….

46. Percent Composition The percent by mass of each element in a compound
Usually used to determine the chemical experimentally
If the formula is known, the percent can be easily determined from each atom’s molar mass
A good video which explains percent composition:

47. Percent Composition Ex: What is the percent composition of H2O
H: 2 x 1.01 g/mol = g/mol
O: 1 x g/mol = g/mol
18.02 g/mol H2O
% H = (2.02 g/mol) / (18.02 g/mol) x 100 = 11.2 % H
% O = (16.00 g/mol) / (18.02 g/mol) x 100 = 88.79% O

48. Percent Composition Ex: CH2Cl2 C: 1 x 12.01 g/mol = 12.01 g/mol
H: 2 x 1.01 g/mol = g/mol
Cl: 2 x g/mol = g/mol
84.93 g/mol CH2Cl2
% C = (12.01g/mol) / (84.93g/mol) x 100 = % C
% H = (2.02g/mol) / (84.93g/mol) x 100 = % H
% Cl = (70.90g/mol) / (84.93g/mol) x 100 = % Cl

49. More Practice?!?! What is the percent composition of ZnBr2?
Zn: 65.39g/mol x 1 = 65.39g/mol
Br: 79.90g/mol x 2 = 159.8g/mol
Total: 225.2g/mol
Zn: (65.39g/mol) / (225.2g/mol) x 100 = 29.04%
Br: (159.8g/mol) / (225.2g/mol) x 100 = 70.96%

50. “You’re killing me smalls”
If a nickel ore sample that had a mass of g was found to contain 48.97g of nickel, what is the percent composition of nickel in the ore sample?
(48.97g Ni) / (159.62g ore) x 100 = 30.68% Ni

51. Ah-Yea-ah!! What is the percent composition of 𝐶𝑎 ( 𝑁𝑂 3 ) 2 ?
Ca: 40.08g/mol x 1 = 40.08g/mol
N: 14.01g/mol x 2 = 28.02g/mol
O: 16.00g/mol x 6 = 96.00g/mol
Total: g
Ca: (40.08g/mol) / (164.10g/mol) x 100 = 24.42%
N: (28.02g/mol) / (164.10g/mol) x 100 = 17.07%
O: (96.00g/mol) / (164.10g/mol) x 100 = 58.50%

52. Yes-sir-ee Bob! What is the percent composition of NaOH?
Na: 22.99g/mol x 1 = 22.99g/mol
O: 16.00g/mol x 1 = 16.00g/mol
H: 1x 1 = 1.01g/mol
Total: 40.00g/mol
Na: (22.99g/mol) / (40.00g/mol) x 100 = 57.48%
O: (16.00g/mol) / (40.00g/mol) x 100 = 40.00%
H: (1.01g/mol) / (40.00g/mol) x 100 = 2.53%

53. Bubblegum and Magic Cards!

54. Empirical Formula The ratio of atoms in a compound
NOT the chemical formula!

55. Calculating the Empirical Formula from Percent Composition
Assume you have 100 g of the compound.
Change “%” to “g”
Convert grams to moles for each element
Divide each mole amount by the smallest mole amount, these numbers are the coefficients for the E.F.
If the numbers from step 4 are not all whole numbers, multiply the step 4 numbers by a whole number so all step 4 numbers are whole numbers
.5 moles  x moles  x 4
.333 moles  x moles  x 5

56. Empirical Formulas
Ex: A compound is determined to be 26.56% K, 35.41% Cr, and % O. What is the empirical formula of the compound?

57. Molecular Formula of a Compound
Molecules can have formulas that are not the simplest ratio. They can be whole number multiples of the Empirical Formula

58. Calculating Molecular Formula
Start with the E.F.
Determine molar mass of E.F.
Divide the molar mass of the unknown molecule by the molar mass of the E.F.
This tells how many E.F.’s are in the M.F
Multiply the subscripts of the E.F. by the number found in step 3

59. Molecular Formulas
Ex: What is the molecular formula of a compound that has an empirical formula of P2O5 and a molar mass of g?
P: 2 x 31.0g = 62.0 g
O: 5 x 16.0g = 80.0 g
142.0g P2O5
283.89g =1.999 ≈ 2 therefore, MF=P4O10
142.0g