Mole Unit
Moles
Molarity
Percent Composition
Empirical Formulas
In Your Textbook
Moles: pp. 171 – 186
Molarity: pp. 509 – 513
Percent Composition: pp. 188 – 191
Empirical Formulas: pp. 192 - 194
Assignment 1: Defining a Mole
Type
Of
Particles
Formula
Molar
Mass
# Particles
in a
mole
Copper
Atoms
Cu
63.5g
6.02 x 1023
Carbon
tetrachloride
Molecules
CCl4
153.8g
6.02 x 1023
Hydrochloric Acid
Molecules
HCl
36.5g
6.02 x 1023
Neon
Atoms
Ne
20.2g
6.02 x 1023
Oxygen Gas
Molecules
O2
32.0g
6.02 x 1023
Sodium Hydroxide
Formula
Units
NaOH
40.0g
6.02 x 1023
20 grams of sodium hydroxide would be:
0.5
moles
3 x 1023
rep. particles
What are the particles (atoms, molecules, or formula units)? Why?
The particles are formula units because NaOH is an ionic compound.
40.4 grams of neon would be:
2.00
moles
12.0 x 1023
rep. particles
44.8
liters
What are the particles? Why?
The particles are atoms because Ne is an element.
3.01 x 1023 atoms of neon would be:
0.500
moles
10.1
grams
11.2
liters
Why atoms?
Neon is an element.
4.5 x 1023 atoms of copper would be:
0.75
moles
48
grams
trick question liters
Copper is not a gas!!
Why atoms?
Copper is an element.
6.02 x 1023 atoms of oxygen in a sample of oxygen gas would be:
0.500 moles of oxygen gas
16.0
grams of oxygen gas
11.2
liters of oxygen gas
Explain:
Oxygen gas is diatomic, so 1 mole of oxygen gas is really 2 moles of
oxygen atoms. Therefore, 1 mole of atoms is only 0.500 moles of O2.
Assignment 2: Solving mole problems using factor label.
1.
What is the mass of one mole of the following?
Sodium sulfate
Na2SO4 = 142 grams
Carbon monoxide
CO = 28 grams
2.
What is the mass of 0.30 moles of aluminum?
0.30 moles Al x
3.
How many atoms of iron are in 0.250 grams of iron?
0.250 g Fe x
4.
6.03 x 1023 atoms Fe
1 mole
1 mole Fe x
55.85 g Fe
= 2.69 x 1021 atoms
How many moles is 6.04 grams of manganese(II) sulfate?
6.04 g MnSO4 x
5.
27.0 g = 8.1 g
1 mole
1 mole MnSO4
151.0 g MnSO4
= 0.0400 moles MnSO4
What is the volume of a 25.0 gram sample of sulfur dioxide at STP?
25.0 g SO2 x
1 mole SO2
64.06 g SO2
x
22.4 L SO2
1 mole SO2
= 8.74 L SO2
Assignment 3: More solving mole problems using factor label.
1.
How many moles of lead is 2.04 x 1023 atoms of lead?
2.04 x 1023 atoms Pb x
2.
1 mole NaCl
58.44 g NaCl
x 6.02 x 1023 formula units = 2.06 x 1023 f. u.
1 mole NaCl
What is the mass of 0.62 moles of barium chloride?
0.62 moles BaCl2 x 208.23 g BaCl2
1 mole BaCl2
4.
= 0.339 moles Pb
How many formula units of sodium chloride are in a 20.0 gram sample?
20.0 g NaCl x
3.
1 mole Pb
6.02 x 1023 atoms Pb
= 130 g BaCl2 (2 significant digits)
How many atoms of hydrogen are in a 13.0 gram sample of carbon tetrahydride?
13.0 g CH4 x 1 mole CH4 x 6.02 x 1023 molecules x 4 H atoms = 1.95 x 1024 atoms
16.04 g CH4
1 mole CH4
1 molecule
5.
What is the mass of one molecule of sucrose?
1 molecule C12H22O11 x
6.
1 mole
x 342.3 g = 6 x 10-22 g
6.02 x 1023 molecules
1 mole
How many grams does a sample of sulfur trioxide gas that occupies a volume of
350.0mL at STP weigh? What is the density of the gas at STP?
0.3500 L x
1 mole SO3 x 80.07 g SO3
22.4 L SO3
1 mole SO3
Density = 1.251 g SO3
350.0 mL
= 1.251 g SO3
= 0.003574 g/mL
Assignment 4: Molarity problems
1.
A salt solution has a volume of 250 mL and contains 0.70 mol of sodium chloride.
What is the molarity of the solution?
0.70 moles NaCl = 2.8 M
0.25 L
2.
A solution of glucose has a volume of 2.0 L and contains 36.0 g of solute. What
is the molarity of the solution?
36.0 g C6H12O6 x
3.
1 mole
180.16 g
= 0.20 moles
M = 0.20 moles = 0.10 M
2.0 L solution
How many moles of solute are in 250 mL of 2.0 M calcium chloride solution?
How many grams would that be?
2.0 M =
x moles
0.25 L
x = 0.50 moles Ca Cl2
0.50 moles CaCl2 x
4.
110.98 g CaCl2 = 55 g CaCl2
1 mole CaCl2
Describe how you would make a 0.25 M solution of calcium hydroxide. (Use
sketch to illustrate.)
Put 0.25 moles (19 grams) of Ca(OH)2 in a volumetric flask and add water to the
1 L mark.
5.
What is the molarity of the sugar solution in a can of coke? (Assume sugar to be
sucrose)
39 g C12H22O11 x
1 mole
= 0.11 mole
M = 0.11 mole =
0.31M
342 g
.355 L
Assignment 5: More molarity problems
1.
What is the molarity of the following solutions:
82.0g of calcium nitrate is dissolved in 500.0 mL of solution
82.0 g Ca(NO3)2 x 1 mole Ca(NO3)2 = 0.500 moles
164.1 g Ca(NO3)2
M =
0.500 moles
= 1.00 M
0.500 L solution
250.0mL that contains 50.0g of copper(II) sulfate
50.0 g CuSO4 x 1 mole CuSO4
= 0.3133 moles CuSO4
159.6 g CuSO4
M =
2.
0.3133 moles
= 1.253 M
0.2500 L solution
Calculate the mass of solute in the following solution:
250.0mL of sodium sulfate that is 2.00M
2.00M = x moles
0.2500 L
3.
4.
x = 0.500 moles x
142.04 g
1 mole
= 71.0 g Na2SO4
How many milliliters of solution can be made from the following?
0.100M solution using 117g of sodium chloride
117 g NaCl x 1 mole = 2.00 moles
0.100M = 2.00 moles
58.44 g
x liters
x = 20.0 L
How would you prepare 1.500L of nitric acid that is 4.00M by diluting a 15.8M
stock solution?
4.00M = x moles HNO3
15.8M = 6.00 moles HNO3
1.500 L soln
x liters soln
x = 6.00 moles HNO3
(so you need 6.00 moles)
5.
x = 0.380 L
**So, take 380.mL of the 15.8Msolutions
and add water until you have 1.500L
How many grams of lead(II) acetate must be used to make 500.0 mL of a solution
that is to contain 10.0 mg/mL of lead ions? What is the molarity of the solution?
10.0mg x 1 g = 0.010g Pb+2 x 500.0mL = 5.00 g Pb+2 needed
1000mg
5.00 g Pb+2 x 325 g Pb(C2H3O2)2
207 g Pb+2
7.85 g x
1 mole = 0.0242 mole
325 g
= 7.85 g Pb(C2H3O2)2
M = 0.0242 moles = 0.484M
0.5000 L
Assignment 6: Calculate the percent composition of the
following substances.
1.
carbon dioxide (CO2)
12g C x 100 = 27%
44g
2.
aluminum sulfate
Al2(SO4)3
54g Al x 100 = 16%
342g
3.
96g S x 100 = 28%
342g
192g O x 100 = 56%
342g
A compound is found to have 20.0 grams of calcium and 35.5 grams of chlorine.
20.0g Ca x 100 = 36.0%
55.5g
4.
32g O x 100 = 73%
44g
35.5g Cl x 100 = 64.0%
55.5g
A compound is found to contain 74.0 grams of mercury and 6.0 grams of oxygen.
74.0g Hg x 100 = 92.5%
80.0g
6.0g O x 100 = 7.5%
80.0g
How many grams of oxygen are in 24.5 grams of this compound?
24.5g x
5.
7.5g = 1.8g Oxygen
100.0g
A 39.2 gram sample of a compound is found to contain 29.4 grams of carbon.
The only other element found to be present is hydrogen. Is this compound CH4?
Explain.
29.4g C x 100 = 75% C
CH4 would be: 12g C x 100 = 75%
39.2g
16g
Yes. The compound is CH4 It has the same % composition.
Assignment 7: Determine the empirical formula of the
following. If possible, name the compound.
1.
2.
3.
A compound is found to be 38.67 % potassium, 13.85 % nitrogen, and 47.48 %
oxygen.
38.67g K x 1 mole =
0.9893 mole / 0.9886 = 1
39.09g
13.85g N x 1 mole =
0.9886 mole / 0.9886 = 1
KNO3
14.01g
47.48g O x 1 mole =
2.968 mole / 0.9886 = 3
potassium nitrate
16.00g
A compound contains 67.6% mercury, 10.8 % sulfur, and 21.6 % oxygen. What
is the empirical formula?
67.6g Hg x 1 mole = 0.337 mole / 0.337 = 1
200.6g
10.8g S x 1 mole = 0.337 mole / 0.337 = 1
HgSO4
32.07g
21.6g O x 1 mole = 1.35 mole / 0.337 = 4
mercury(II) sulfate
16.00g
A compound contains 75.0 grams of carbon and 25.0 grams of hydrogen. What is
the empirical formula?
75.0g C x
1 mole
12.01g
= 6.24 mole / 6.24 = 1
25.0g H x
1 mole
1.008g
= 24.8 mole / 6.24 = 4
CH4
4.
100 grams of a compound containing aluminum and oxygen is found to have
52.94 grams of aluminum.
52.94g Al x 1 mole = 1.962 mole / 1.962 = 1 x 2 = 2
26.981g
Al2O3
47.06g O x 1 mole = 2.941 mole / 1.962 = 1.5 x 2 = 3
15.999g
aluminum oxide
5.
50.0 grams of a compound containing only oxygen and hydrogen is found to have
47.05 grams of oxygen. What is the empirical formula?
47.05g O x 1 mole = 2.941 mole / 2.93 = 1
15.999g
HO
2.95g H x 1 mole = 2.93 mole / 2.93 = 1
1.008g
Assignment 8: Determine the empirical formula of the
following. If possible, name the compound.
1.
A compound is determined to contain 1.00 grams of hydrogen, 6.0 grams of
carbon, and 8.0 grams of oxygen. What is the % composition of the compound?
1.00g H x 100 = 6.67 % hydrogen
15.0 g
6.0g C x 100 = 40.% carbon
15.0g
8.0g O x 100 = 53 % oxygen
15.0g
What is the empirical formula of the compound?
1.00g H x 1 mole = 0.992 mole / 0.50 = 2
1.008g
6.0g C x 1 mole = 0.50 mole / 0.50 = 1
CH2O
12.01g
8.0g O x 1 mole = 0.50 mole / 0.50 = 1
16.00g
If the molar mass of the compound is 60.0 grams, what is the molecular formula?
The molar mass of the
30g = 2 so C2H4O2
empirical formula is 30g
60g
2.
Which of the following are an empirical formula?
Ribose: C5H10O5 no
Chlorophyll: C55H72MgN4O5 yes
Ethyl butyrate: C6H12O2 no
DEET: C12H17ON yes
If the molar mass of ribose is approximately 150 grams, what is the molecular
formula?
Because the mass of the formula above is 150, that is the molecular
formula. The empirical formula is C1H2O1.
3.
15.5 grams of a hydrate is heated to drive off the water. The anhydrous
compound remaining is found to weigh 11.7 grams. The anhydrous compound is
then decomposed revealing 4.2 grams of calcium and 7.5 grams of chlorine.
What is the empirical formula of the compound? 15.5g – 11.7g = 3.8g of H2O
3.8g H2O x
1mole = 0.21 mole / 0.10 = 2
18.0g
4.2g Ca x 1mole = 0.10 mole / 0.10 = 1
40.1g
7.5g Cl x 1 mole = 0.21 mole / 0.10 = 2
35.5g
CaCl2 * 2 H2O is the formula