1. Chapters 10
Chemical Quantities

2. Conversion Factors
Conversion factor – A fraction equal to 1 that is used to change one unit into another.
(When the numerator = denominator, a fraction equals 1.)

3. Dimensional Analysis
Dimensional Analysis – A problem solving method where conversion factors are used to cancel unwanted units.

4. Conversion Examples a) Convert $25 to nickels. $25 4 quarters $1
25 * 4 * 5 / 1 / 1 =
500 nickels

5. Conversion Example 2 Convert 180 days to seconds. 180 days 24 hours
60 minutes
1 hour
60 seconds
1 minute
180 * 24 * 60 * 60 / 1 / 1 / 1 =
15,552,000 seconds
or 1.6 x 107 seconds

6. Common Conversions to Know
1 meter = 100 centimeters
1 meter = 1000 millimeters
1 kilometer = 1000 meters
Convert 125 cm to km.
1 m
100 cm
1 km
1000 m
125 cm
125 / 100 / 1000 =
km or 1.25 x 10-3 km

7. Conversion Example 4 Convert 15 m/s to km/hr. 15 m 1 s 1 km 1000 m
60 seconds
1 minute
60 minutes
1 hour
15 * 60 * 60 / 1000 =
54 km/hr

8. Measuring Matter How do we describe how much of something we have?
By count, by mass, by volume.
We use words like “dozen” to talk about an amount.
In chemistry, we use the MOLE.

9. Mole Mole – SI unit for measuring an amount of a substance.
Avogadro’s Number = 6.02 x 1023
Representative particles = smallest unit that still has all the characteristics of that substance.
1 mole = 6.02 x 1023 representative particles

10. Representative Particles
What is the representative particle of :
Element (ex. Cu): ___________
Exception: The representative particle of the 7 diatomic elements is a molecule. (ex. H2)
Covalent compound (ex. H2O): _________
Ionic Compound (ex. NaCl): ___________
atom
molecule
formula unit

11. Conversions 4 moles Ca = atoms Ca. 4 moles Ca 6.02 x 1023 atoms Ca
1 mole Ca
= 2.41 x 1024 atoms Ca

12. Conversions 5 x 1018 atoms Cu = moles Cu. 5 x 1018 atoms Cu 1 mole Cu
= 8.3 x 10-6 moles Cu

13. Conversions 9.2 moles F2 = molecules F2?
9.2 moles x 1023 molecules F2
1 mole
= 5.5 x 1024 molecules F2

14. Conversions 9.2 moles F2 = atoms F?
9.2 moles F x 1023 mlcls F atoms F
1 mole molecule F2
= 1.1 x 1025 atoms F

15. Conversions 3.4 moles C2H4 = total atoms?
3.4 moles C2H x 1023 mlcls C2H atoms
1 mole C2H mlcl C2H4
= 1.22 x 1025 atoms

16. Molar Mass
Molar Mass – The mass of one mole of an element or compound.
Molar mass of a compound = the sum of the masses of the atoms in the formula.
Use the atomic masses in grams/mol on the periodic table.

17. Molar Mass Find the molar mass of each: 87.6 g/molSr
MgBr2
Ba3(PO4)2
87.6 g/mol
(24.3) + (2 x 79.9) =
184.1 g/mol
(3 x ) + (2 x 30.97) + (8 x 16) =
602.1 g/mol

18. Mole–Gram Conversions
1 mole = molar mass (in grams)
5.3 moles LiOH = ___________ grams LiOH
(Molar mass LiOH : = 24 g/mol)
5.3 moles LiOH g LiOH
1 mole LiOH
= grams LiOH

19. Gram-Mole Conversions
68 grams F2 = moles F2?
68 grams F mole F2
38 grams
68 / 38 = 1.8 moles F2

20. STP STP = Standard Temperature & Pressure Standard Temp  0oC
Standard Press  1 atm
(See Reference Tables)

21. Molar Volume of a Gas
Avogadro’s Hypothesis: equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
At STP, 1 mole of any gas occupies a volume of 22.4 L.
1 mole = 22.4 L (of a gas at STP)

22. Mole-Volume Conversions
5.4 moles He = L He at STP?
5.4 moles He L He
1 mole He
5.4 x 22.4 = L He

23. Mole-Volume Conversions
5.4 moles CH4 = L CH4 gas at STP?
5.4 moles CH L CH4
1 mole CH4
5.4 x 22.4 = L CH4

24. Volume-Mole Conversion
560 L SO3 = mol SO3 at STP
560 L SO mole SO3
22.4 L SO3
560 / 22.4 = 25 mole SO3

25. Molar Mass-Density Conversions
grams
liters
grams
mole
Density = Molar Mass =
A gaseous compound composed of sulfur and oxygen has a density of 3.58 g/L at STP. What is the molar mass of this gas?
3.58 g L
L mole
3.58 x 22.4 = 80.2 g/mole

26. Molar Mass-Density Conversion
What is the density of Krypton gas at STP?
83.8 grams Kr mole
mole Liters
83.8 / 22.4 = 3.74 g/L Kr

27. 22.4 Liters at STP (gases only)
Molar Mass
6.02 x 1023 particles
22.4 Liters at STP (gases only)
1 mole
1 mole
Grams
(use Per.Tble)
6.02 x 1023
particles

28. Multi-step Problem: Example 1
If you had 5.0 L of CO2 how many grams would that be?
Step 1: L  moles
Step 2: moles  grams
5.0 L CO2 1 mole CO g CO2 =
22.4 L CO mole CO2
9.8 g CO2

29. Multi-step Problem: Example 2
How many molecules are in 60.0 grams of water?
Step 1: grams  moles
Step 2: moles  molecules
60.0 g H2O 1 mole H2O x 1023 mlcls =
18.0 g H2O 1 mole H2O
= 2.0 x 1024 molecules of H2O

30. Percent Composition
Percent Composition - % by mass of each element in a compound
Percent =
Part
Whole
x 100

31. Percent Composition Percent Comp =
Example: Find the mass percent composition of Al2(SO4)3
Mass of 1 element
Mass of compound
x 100 54
342
x 100
= 15.8%
% Al:
Al: x 27 = 54
S: x 32 = 96
O: x 16 = 192
Total Comp. = 342 96
342
% S:
x 100 = 28.1%
192
342
%O:
x 100 = 56.1%

32. Percent Example Find the percent composition of NiSO3.
Ni: g %Ni: 58.7
S: g
O: (3 x 16) = 48 g
Total Comp
x 100 = 42.3 % 32
138.7
%S:
x 100 = 23.1 % 48
138.7
%O:
x 100 = 34.6%

33. More Percents
Which of the following shows a compound that is 92.3%C and 7.7%H?
a) C2H4 b) C3H6
c) CH4 d) C6H6

34. Empirical Formulas Empirical Formula – The simplest formula.
Shows the smallest whole number ratio of elements in a compound.
Covalent formulas will not always be empirical.
Example: CH
Molecular Formula – The actual formula.
For ionic compounds – it will be the simplest ratio.
For molecular compounds – it will NOT always be the simplest ratio.
Example: C6H6

35. To Calculate Empirical Formula
Calculate the empirical formula of a 2.5 gram compound containing 0.90g Ca and 1.60g Cl.
Step 1: Convert GRAMS to MOLES.
Ca: 0.90g 1 mole = mole Ca
40.1 g
Cl: 1.60g mole = mole Cl
35.5 g

36. Calculating Empirical Formula
Step 2: DIVIDE the # of moles of each substance by the smallest number to get the simplest mole ratio.
Ca: = Cl: = 2.01 ~ 2
CaCl2

37. Calculating Empirical Formulas
Step 3: If the numbers are whole numbers, use these as the subscripts for the formula. If the numbers are not whole numbers, multiply each by a factor that will make them whole numbers.
Look for these fractions:
0.5  x 2
0.33  x 3
0.25  x 4

38. Empirical Formula Example
What is the empirical formula of a compound that is 66% Ca and 34% P?
(Assume you have 100 grams of a compound and replace % with grams.)

39. Empirical Formula Example
Step 1: grams  moles
Ca: 66g 1 mole = mole Ca
40.1 g
P g 1 mole = mole P
31.0 g
Step 2: Divide by the smallest.
Ca: = P: = 1

40. Empirical Formula Example
Step 3: Multiply until you get whole numbers.
(If you multiply one factor by a number, you have to multiply ALL the factors by that number!)
Ca: 1.5 x 2 = P: 1 x 2 = 2
Ca3P2

41. Determining Molecular Formulas
A compound has an empirical formula of CH2O. Its molecular mass is 180g/mol. What is its molecular formula?
Step 1: Find the mass of the empirical formula.
C: 1 x 12 = 12
H: 2 x 1 = 2
O: 1 x 16 = 16
Total:

42. Determining Molecular Formula
Step 2: Divide the molecular mass by the mass of the empirical formula to get the “multiplying factor”.
Step 3: Multiply each of the subscripts in the empirical formula by this factor to get the molecular formula.
6 (CH2O)  C6H12O6
180 30
= 6

43. Determining Molecular Formula
Find the molecular formula of ethylene glycol (CH3O) if its molar mass is 62 g/mol.
Step 2: 62 / 31 = 2
Step 3: 2 (CH3O)  C2H6O2
Step 1: (3 x 1) + 16 = 31 g/mol

44. Empirical/Molecular Example
The percent composition of methyl butanoate is 58.8% C, 9.8% H, and 31.4 % O and its molar mass is 204 g/mol.
What is its empirical formula?
What is its molecular formula?

45. Empirical/Molecular Example
58.8 g C mol C
12.0 g C
4.9 = 2.5
1.96
9.8 = 5
= 1
x 2 = 5
x 2 = 10
x 2 = 2
= 4.9 mol C
9.8 g H mol H
1.0 g H
= 9.8 mol H
31.4 g O mol O
16.0 g O
= 1.96 mol O
Empirical Formula = C5H10O2

46. Empirical/Molecular Example
Empirical Formula  C5H10O2 Mass = 5(12) + 10(1) + 2(16) = 102 g/mole Molecular mass  204 g/mol = 2 Empirical mass  102 g/mol So molecular formula is 2 x emp. form: 2(C5H10O2) = C10H20O4