Chemistry 121
Worksheet 8 Notes
1.
Given the following reaction: C3H8
Oregon State University
Dr. Richard L Nafshun
+ 5 O2  3 CO2 + 4 H2O
Assume that 10.00 g of propane react with excess oxygen. Calculate:
(A) the number of moles of propane that react
 1mol 
 = 0.2267 mol propane
10.00 g C3H8 
 44.11g 
(B) the number of molecules of propane that react;
 6.022 x10 23 molecules 
 = 1.365 x 1023 propane molecules
0.2267 mol propane 
1mol


(C) the number of moles of carbon dioxide and water that are formed;
 3 mol CO 2 
 = 0.6801 mol CO2
0.2267 mol propane 
 1 mol C 3 H 8 
 4 mol H 2 O 
 = 0.9068 mol H2O
0.2267 mol propane 
 1 mol C 3 H 8 
(D) the number of grams of carbon dioxide and water that are formed;
 44.01g 
 = 29.93 g CO2
0.6801 mol CO2 
1
mol
CO
2 

 18.02 g 
 = 16.34 g H2O
0.9068 mol H2O 
1
mol
H
O
2


(E) the number of molecules of carbon dioxide and water that are formed;
 6.022 x10 23 molecules 
 = 4.096 x 1023 CO2 molecules
0.6801 mol CO2 
1
mol


 4 mol H 2 O 
 = 0.9068 mol H2O
0.2267 mol propane 
 1 mol C 3 H 8 
 6.022 x10 23 molecules 
 = 5.461 x 1023 H2O molecules
0.9068 mol H2O 
1mol


(F) the number of moles, grams and molecules of oxygen that are used up.
 5 mol O 2 
 = 1.134 mol O2
0.2267 mol propane 
 1 mol C 3 H 8 
 6.022 x10 23 molecules 
 = 6.826 x 1023 O2 molecules
1.134 mol O2 
1mol


 32.00 g 
 = 36.29 g O2
1.134 mol O2 
 1 mol O2 
(G) a student collects 23.50 grams of CO2. What is the percent yield?
 actual mass 
 23.50 g 
100% = 
 (100%) = 78.52%
Percent Yield = 
29
.
93
g
theoretica
l
mass




2.
Calculate the molarity of the following aqueous solutions.
(A)
6.00 mol HCl in 2.50 L of solution.
 mol   6.00 mol 
 = 2.40 M
M= 
 =
 L   2.50 L 
(B)
45.00 grams NaOH in 10.00 L of solution.

 1 mol  
 45.00 g 

 40.01 g  
 mol  


M= 
 =
 = 0.1125 M
10.00 L
 L  





3.
How many grams of NaOH are required to prepare 50.0 mL of 0.125 molar NaOH?

 1 mol  
 X g

 40.01 g  
 mol  

 = 0.125 M
M= 
 =

L
0
.
0500
L

 





X = 0.251 g NaOH
4.
(A)
What is the mass percent composition of Na, S and O in the compound Na2SO4?
The molar mass of Na2SO4 is 142.06 g/mol
The mass of sodium in Na2SO4 is 2 x 23.00 g/mol = 46.00 g/mol
The mass percent of sodium in Na2SO4 is the part "over the whole."
 46.00 g/mol
Mass percent of sodium in Na2SO4 = 
 142.06 g/mol

 (100%) = 32.38%

The mass of sulfur in Na2SO4 is 1 x 32.06 g/mol = 46.00 g/mol
The mass percent of sulfur in Na2SO4 is the part "over the whole."
 32.06 g/mol
Mass percent of sulfur in Na2SO4 = 
 142.06 g/mol

 (100%) = 22.57%

The mass of oxygen in Na2SO4 is 4 x 16.00 g/mol = 64.00 g/mol
The mass percent of oxygen in Na2SO4 is the part "over the whole."
 64.00 g/mol
Mass percent of oxygen in Na2SO4 = 
 142.06 g/mol
(B)

 (100%) = 45.05%

What is the mass percent composition of Ca, N and O in the compound
Ca(NO3)2?
The molar mass of Ca(NO3)2 is 164.10 g/mol
The mass of calcium in Ca(NO3)2 is 1 x 40.08 g/mol = 40.08 g/mol
The mass percent of calcium in Ca(NO3)2 is the part "over the whole."
 40.08 g/mol
Mass percent of calcium in Ca(NO3)2 = 
 164.10 g/mol

 (100%) = 24.42%

The mass of nitrogen in Ca(NO3)2 is 2 x 14.01 g/mol = 28.02 g/mol
The mass percent of nitrogen in Ca(NO3)2 is the part "over the whole."
 28.02 g/mol
Mass percent of nitrogen in Ca(NO3)2 = 
 164.10 g/mol

 (100%) = 17.07%

The mass of oxygen in Ca(NO3)2 is 6 x 16.00 g/mol = 96.00 g/mol
The mass percent of oxygen in Ca(NO3)2 is the part "over the whole."
 96.00 g/mol
Mass percent of oxygen in Ca(NO3)2 = 
 164.10 g/mol
5.
Given the following reaction: C3H8

 (100%) = 58.50%

+ 5 O2  3 CO2 + 4 H2O
Assume that 88.22 g of propane react with excess oxygen. Calculate:
(A) the number of moles of propane that react
 1mol 
 = 2.000 mol propane
88.22 g C3H8 
 44.11g 
(B) the number of molecules of propane that react;
 6.022 x10 23 molecules 
 = 1.204 x 1024 propane molecules
2.000 mol propane 
1
mol


(C) the number of moles of carbon dioxide and water that are formed;
 3 mol CO2 
 = 6.000 mol CO2
2.000 mol propane 
 1 mol CH 4 
(D) the number of grams of carbon dioxide and water that are formed;
 44.01g 
 = 264.1 g CO2
6.000 mol CO2 
 1 mol CO2 
(E) the number of molecules of carbon dioxide and water that are formed;
 6.022 x10 23 molecules 
 = 3.612 x 1024 CO2 molecules
6.000 mol CO2 
1mol


 4 mol H 2 O 
 = 8.000 mol H2O
2.000 mol propane 
1
mol
CH
4 

 6.022 x10 23 molecules 
 = 4.816 x 1024 H2O molecules
8.000 mol H2O 
1
mol


(F) the number of moles, grams and molecules of oxygen that are used up.
 5 mol O2 
 = 10.00 mol O2
2.000 mol propane 
 1 mol CH 4 
 6.022 x10 23 molecules 
 = 6.02 x 1024 O2 molecules
10.00 mol O2 
1
mol


 32.00 g 
 = 320.0 g O2
10.00 mol O2 
 1 mol O2 
(G) a student collects 55.90 grams of CO2. What is the percent yield?
 actual mass 
 55.90 g 
100% = 
 (100%) = 21.17 %
Percent Yield = 
 264.1g 
 theoretical mass 
6.
Balance:
1 C11H24
C11H24
+
17 O2
+
O2
 CO2
 11 CO2
+
+
H2O
12 H2O
How many moles of water are produced if four moles of C11H24 are consumed?
48 (twelve moles of water are produced for every mole of C11H24 consumed.)
 12 mol H 2 O 
 = 48 mol H2O
4 moles C11H24 
 1 mol C11 H 24 
7.
Balance:
Li + O2  Li2O
4 Li + O2  2 Li2O
How many moles of Li2O are produced if eight moles of Li are consumed?
4 (2 moles of Li2O are produced for every 4 moles of Li consumed.)
 2 mol Li2 O 
 = 4 mol Li2O
8 mol Li 
 4 mol Li 
8.
Determine the mass percent composition of lithium sulfate.
Li2SO4 = 109.94 g/mol
% Li = (2 x 6.941 g/mol)/109.94 g/mol = 12.63 %
% S = (1 x 32.06 g/mol)/109.94 g/mol = 29.16 %
% O = (4 x 16.00 g/mol)/109.94 g/mol = 58.21 %
_______
100.00 %
9.
20.5 g NaCl are dissolved in 500.0 mL of total solution. Calc M.
NaCl = 58.45 g/mol
M = grams/L = (20.5 g/58.45 g/mol) / 0.5000 L = 0.702 M or mol/L
10.
2.50 L of 2.00 M NaF solution is diluted to 4.00 L what is the molarity of the resulting
solution?
MBeforeVBefore = MAfterVAfter
(2.00 M)(2.50 L) = (MAfter)(4.00 L)
MAfter = 1.25 M
Abbreviated Solubility Rules (see Table 4.1—Page 128 when working ChemSkill Builder):
Rule 1: All nitrates, group 1A metal salts and ammonium salts are soluble.
Rule 2: All carbonates, hydroxides, phosphates and sulfides are insoluble.
Rule 3: Rule 1 always takes precedent. Example: NaOH is soluble.
1.
A student places 15.00 grams of solid sodium sulfate into a 500.00-mL volumetric flask
and fills to the mark with water. Draw a picture of the ions in solution (similar to Figure
4.2—Page 125; but rather than label the ions as (+) and (-), label the ions with the
chemical formula. Calculate the number of sodium ions present. Calculate the number
of sulfate ions present. Write a balanced equation for the dissociation of the solid ionic
compound.
Na2SO4 (s) → 2 Na+ (aq) + SO42- (aq)
Note that two sodium ions and one sulfate polyatomic ion are generated for each sodium
sulfate unit that dissociates in water.
 1mol 
 0.1056 mol Na2SO4
15.00 grams Na2SO4 
 142.06 g 
 6.022 x10 23 Na2 SO4 units 
 6.359 x 1022 Na2SO4 units
0.1056 mol Na2SO4 
1
mol


2.
 2 Na  ions
6.359x 1022 Na2SO4 units 
 1Na 2SO 4 unit

 1.272 x 1023 Na+ ions

 1SO4 2 ions
6.359 x 1022 Na2SO4 units 
 1Na 2 SO 4 unit

 6.359 x 1022 SO42- ions


List three strong acids and show how they exist in water (in other words, show a reaction
for each that shows them dissociate).
Hydrochloric acid
HCl → H+ + Cl(because HCl is a strong acid, HCl dissociates 100%).
Nitric acid
HNO3 → H+ + NO3(because HNO3 is a strong acid, HNO3 dissociates 100%).
Sulfuric acid
H2SO4 → H+ + HSO4(because H2SO4 is a strong acid, H2SO4 dissociates 100%).
3.
Draw the structure of propanoic acid (CH3CH2COOH) and carefully show how it exists
in water (in other words, show a reaction that shows it dissociate).
The above reaction does not go 100%. Propanoic acid dissociates less than 2%. So, in
water, there is a considerable amount of CH3CH2COOH and very little CH3CH2COOand H+.
Why does propanoic acid dissociate less than 2%? Because it is a weak acid (it is a
carboxylic acid; it contains the –COOH group). Weak acids dissociate less than 100%
(usually less than 2%).
The double arrows shown (one forward and one backward) indicate that at any given
moment there are reactant (CH3CH2COOH) and products (CH3CH2COO- and H+)
present. As a matter of fact, the forward and backward arrows indicate that the system is
not static, but dynamic. At any given moment reactant is changing to products and
products are changing into reactant (species are changing "positions.")
Warning! Because of cellular phones the following analogy is outdated and is therefore
disclosing the age of chemistry faculty.
Think of it like two pay phones in an entire dorm. At any given moment two residents
are on the phones (these residents are the products), but 198 residents are not on the
phones (these residents are the reactants). When one resident concedes the phone,
another resident starts to use the phone. So, there are always two residents on the phones
and 198 residents not on the phones, but the residents are changing places (this is
dynamic—ever changing). The same two residents are not always on the phones (this
would be considered static—fixed).