Unit 2 Review Answers
Student Textbook pages 276 –279
Answers to Knowledge/Understanding Questions
True/False
1. False — The molecular formula of a compound is an exact formula of a molecule,
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
giving the types of atoms and the number of each type. The empirical formula is the
simplest whole number ratio of atoms in a compound.
True
False — The average atomic mass of an element is the weighted average mass of all
the isotopes of a naturally occurring atom.
True
False — The fundamental unit for chemical quantity is the mole.
True
True
False — The Avogadro constant is a precise number that does not vary.
True
True
False — The theoretical yield is predicted by stoichiometry.
False — Stoichiometric calculations are used to predict the amount of products
expected from a known amount of reactants or to calculate the necessary amounts of
reactants required to yield a specific amount of product.
Multiple Choice
13. (c)
14. (a)
15. (c)
16. (d)
17. (c)
18. (b)
19. (b)
20. (d)
21. (e)
22. (d)
23. (d)
24. (e)
25. (b)
26. (b)
27. (e)
28. (d)
Answers to Short Answer Questions
29. (a) 1.00 mol N2 × (6.02)(1023) = (6.02)(1023) molecules N2
# atoms N2 = 2 × (6.02)(1023) = (1.20)(1024) atoms
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24
(b) # PO3−
4 ions in 2.5 mol Ca3(PO4)2 = (3.01)(10 )
24
(c) # O atoms in 0.47 mol Ca3(PO4)2 = (2.26)(10 )
30. A balanced chemical equation has the same number of each type of atom on both
sides.
C6H14 + 9.5O2 → 7H2O + 6CO2
In the above balanced equation, there are 6 carbon atoms on both sides and each
carbon atom has a mass of 12.01g. The law of conservation of mass is followed
because the mass of carbon on the reactant side of the equation is the same as the
mass of carbon on the product side of the equation.
31. – The number of atoms, molecules, and moles of the products and reactants
– The mass of the products and reactants.
– The amount of products that are expected from a known amount of reactants
– The amount of reactants required to produce a desired amount of products
– The amount of one reactant needed to react completely with another reactant
32. (a) A reactant that is completely used up in a reaction is a limiting reactant. When
the limiting reactant is used up, the reaction stops. The following analogy explains
this concept. A hand-knitted sweater requires 9 balls of blue yarn, 2 balls of white
yarn, and 1 ball of red yarn. Suppose you have 50 balls of blue yarn, 15 balls of
white yarn, and 3 balls of red yarn. You can only knit three sweaters because each
sweater requires 1 ball of red yarn. The amount of red yarn you have limits the
amount of sweaters you can knit.
(b) The opposite of a limiting reactant is the excess reactant.
(c) Reactants are often not present in stoichiometric amounts in chemical reactions.
In industrial settings, it may be inefficient in terms of time and money to ensure
that reactants are present in 1:1 ratios. If one of the reactants is inexpensive and
readily available it is generally used in excess. In nature, reactions involving
oxygen almost never have reactants in stoichiometric amounts because of the
abundance of oxygen found in the air.
33. (a) mol C3H8 = 0.167 mol
(b) molecules of C3H8 = (1.00)(1023)
(c) atoms of C = (3.00)(1023)
34. mol P4 = (8.07)(10−2)
atoms of P = (4.86)(1022)
35. One molecule has 3 uranium atoms and 2(3) + 4(2) + 12(1) = 26 oxygen atoms.
amount of mineral
# of U atoms
# of O atoms
1 molecule
3
26
2.00 g
x
26
(1.38)(1021)
3
21
(1.38)(10 )
x
=
or x = (1.196)(1022)
Thus there are (1.20)(1022) oxygen atoms present.
36. Mass percent of hydrogen in C2H5OH = 13.1%
Mass percent of hydrogen in C32H84O2 = 16.9%
Cetyl palmitate, C32H84O2 has the greatest mass percent of hydrogen.
37. Mass of C = 60.00 g
Mass of H = 12.12 g
Mass of O = 32.00 g
Mass of 1 mol of C5H12O = 104.12 g
% C = 57.6
% H = 11.6
% O = 30.7
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38.
2NH4Cl(s)
+Ca(OH)2(s)
CaCl 2(s)
+2NH3 (g)
+2H2O(g)
Mole ratio
2
1
1
2
2
Molar mass
53.49 g/mol
74.1 g/mol
17.03 g/mol
Given
8.93 g
7.48 g
m
Mol NH4Cl = 0.167 mol
Mol Ca(OH)2 = 0.101 mol
Mol NH3 produced based on:
NH4Cl → n = 0.167 mol
Ca(OH)2 → n = 0.202 mol
The limiting reagent is NH4Cl.
m = 2.84 g NH3 produced.
Answers to Inquiry Questions
39. Assuming that this hydrate will become anhydrous, but not decompose, when heated,
here is a procedure.
Measure and record the mass of a clean evaporating dish. Put a shallow layer of the
hydrate in the dish. Measure and record total mass. Heat the dish and contents
strongly for a good period of time. Let dish and contents cool then measure and
record total mass again. The required observations are the three measured masses. By
subtracting, find the mass of; the original hydrate, the anhydrous solid, the water of
hydration given off. Then calculate the number of moles of: anhydrous solid, water of
hydration. Finally determine x by dividing the number of moles of water by the number of moles of anhydrous solid. Then you can write your conclusion, the
formula of the hydrate Na2S2O3.xHsO using your value of x.
40. From the balanced equation the mole-to-mole ratio between potassium iodide and
lead(II) nitrate is 2:1 Thus the volumes of the equal concentration solutions
(provided) should be in this same ratio. Here is a procedure.
Place 10 ml of lead(II) nitrate solution in a graduated cylinder and 20 ml of
potassium iodide solution in a second cylinder. Empty both cylinders into a beaker.
Rinse both with small volumes of distilled water and add the rinses to the beaker.
Mix well with a stirring rod and let settle. Rinse rod into beaker. Heat, but not to
boiling, to help the precipitate to form larger particles. Measure the mass of a filter
paper and then fold it and place in a funnel. Filter the contents of the beaker. Rinse
all the precipitate into the filter cone. Dry the filter paper and precipitate in an oven
and find the total mass of paper and precipitate. By subtraction, find the mass of the
precipitate. From the volumes of solutions used, find the number of moles of
precipitate expected. Convert the number of moles to mass and compare to the actual
mass of precipitate. If the measured and predicted masses are the same, the predicted
mole-to-mole ratio is confirmed.
41. The reaction is unlikely to produce a 100% yield in warm water because lead(II)
chloride is moderately soluble under such conditions. A competing reaction would be
the dissolution of the solid PbCl2 produced in the reaction. The lead and chloride
ions are then free to recombine with the sodium and nitrate ions also formed in the
reaction.
42. Given m = 0.40 g, l = 10.0 cm, w = 10.0 cm
3
(a) Given D = 2.70 g/cm find the thickness, h.
The volume, V, may be expressed to 2 ways: V = lwh and V = m/D
Equating these 100.00 h = 0.40 g/2.70 = 0.14814 cm3 which gives
h = (1.48)(10−3) cm
Thus the foil is (1.48)(10−2) mm thick
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(b) For the Al, m = 0.40 g and M = 26.98 g/mol so n = m/M = 0.0148 mol
Thus there are (0.0148)(6.02)(1023) or (8.9246)(1021) Al atoms in the foil.
Assume each atom is a cube, x cm to a side. State the total volume in two ways.
V = 0.14814 = (8.9246)(1021)(x3) and this gives
x = (2.55)(10−8) cm = 0.255 nm
Thus the “radius” would be 0.1275 nm
(c) For a spherical atom, V = (4/3)πr 3 which leads to
0.14814 = (8.9246)(1021)(4/3)πr 3 and r = (1.58)(10−8) cm = 0.158 nm
Thus the radius is 0.158 nm.
(d) The answer changes from 0.128 nm to 0.158 nm, a 23% increase. The question
arises: “How are the spheres, which seem more likely than cubes, packed
together?”
43. CaCl2(aq) + Na2SO4(aq) → CaCO3(s) + 2NaCl(aq)
(a) Accurately weigh 1.00g of CaCl2 into a beaker and record in your notebook. Add
50 mL of Na2SO4 to the beaker and stir. Allow to sit until the reaction is
complete. Collect the solid precipitate by filtration. Place the CaCO3 in a drying
oven until completely dry. Weigh and record the mass of CaCO3 produced. Using
the mass of CaCl2 you used, calculate the theoretical yield of calcium carbonate.
Using the mass of calcium carbonate actually obtained in the experiment,
calculate the percentage yield of the reaction.
(b) The measurement of masses can affect the outcome. The chemist’s skill can affect
the amount and purity of the precipitate obtained.
Answers to Communication Questions
44. The atomic mass giving for neon, 20.18 u, is based on the mixture of the different
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MHR • Unit 2 Chemical Quantities
isotopes of neon. These isotopes have slightly different masses because of the differing
numbers of neutrons. The atomic mass of neon is the average value based on the
isotopic composition.
The unit of measure of the molecular mass of a compound is based 12C as the
standard. In this system, 12C is assigned a mass of exactly 12 atomic mass units. The
masses of all other atoms are given relative to this standard. The molar mass of a
molecular compound represents the mass in grams of 1 mole of the substance
Na2C4H4O6 is a molecular formula. The numbers of each type of atom are multiples
of two and can be reduced to the simpler whole ratio numbers. The empirical
formula of this molecule would be NaC2H2O3 . The molecule C63H88CN14O14F
is an empirical formula. The numbers representing each type of atom cannot be
reduced to a simpler whole number ratio.
An empirical formula can represent many different molecules because it represents
the simplest whole-number ratio of the various types of atoms in a compound.
Multiples of the simplest ratio may give rise to many different molecules.
Industrial chemists work generally work on very large scales and require large
quantities of reactants. Furthermore, the same reactions are usually carried out over
and over to produce particular products. Knowing the percentage yield of reaction
saves money and time. Industrial chemists choose the reactants and reaction
conditions that optimize the percentage yield of a reaction. Even small deviations in
product yield can lead to large financial repercussions for the industry.
2A + 3B → 4C + D
mol of A = mass A (g) ÷ molar mass A (g/mol)
mol of B = mass B (g) ÷ molar mass B (g/mol)
Determine the limiting reagent:
Mol C produced based on:
A → n = mol of A × 4/2 = mol CA
B → n = mol of B × 4/3 = mol CB
Assume A is the limiting reagent:
Mol C produced = mol CA (mol)
Mass C expected (g) = mol CA (mol) × molar mass C (g/mol)
Answers to Making Connections Questions
50. (a) To help students approach this question, suggest they check out the following web
site set up by a company involved with drug testing: http://www.cannamm.com/
The site is set up to aid students with projects, in addition to providing information to prospective clients.
(b) In addition to the site mentioned above, students should be referred to sites for
human rights commissions (Canadian: http://www.chrc-ccdp.ca, Ontario:
http://www.ohrc.on.ca)
Student answers will hopefully reflect a balance between the concerns of the
employers and the rights of the employees.
51. 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
(a) Oxygen is the limiting reactant. Even well tuned engines form some carbon
monoxide, which indicates insufficient oxygen for complete combustion.
(b) A car will run poorly at higher altitudes because the density of oxygen in the air is
smaller. Since oxygen is the limiting reactant in gasoline-powered vehicles, a
smaller proportion of oxygen leads to less efficient combustion.
(c) The oxygen-nitrogen reaction removes oxygen that could otherwise react with the
gasoline. The adjustment needed is to increase the proportion of air to fuel to
allow for more oxygen to ensure as complete combustion of the fuel as possible.
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