Name:__[KEY]___________________
Date:________________
Period:____
REVIEW Unit 3 Test (Chp 1-3): Matter, Measurement, & Stoichiometry
Multiple Choice (27 questions)
(50% of score)
many will be easy questions from chem I (study these topics in chapters 1 & 2)
classifying pure substance (element/compound) vs. mixture (hetero-/homo-)
heterogeneous vs. homogeneous mixtures
states of matter (solid, liquid, gas)
chemical vs. physical changes and properties
density
Dalton's atomic theory
atomic models and experiments of Thomson & Rutherford
atomic structure (protons, neutrons, electrons, atomic #, mass #, isotopes, average atomic mass)
periodic table (metals, nonmetals, metalloids, groups, periods, etc.)
ionic vs. molecular compounds and naming
12 polyatomic ions on the first ion quiz and their charges
oxidation numbers of elements in compounds, like what is the ox. # of C in oxalate ion C2O42– ?
stoichiometry (limiting reactant, theoretical and percent yield)
Free Response (31 points) (5 questions: 1 short, 3 medium, 1 long)
(50% of score)
1) describe isotopes, calculate average atomic mass, convert atoms/grams of an isotope (6 pts)
2) lab question determining formula of a hydrate (5 pts) (study Lab handout & Chp 3 #53)
3) combustion analysis and empirical/molecular formula (7 pts)
4) write 3 equations from verbal descriptions and balance them (some diatomics) (9 pts)
5) limiting reactant and gram to gram conversions (with mol to mol ratios) (5 pts)
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Section I Multiple Choice
NO CALCULATOR
___1.
A measured mass of an unreactive metal was dropped into a small graduated cylinder half filled
with water. The following measurements were made.
Mass of metal = 19.611 grams
Volume of water before addition of metal = 12.4 milliliters
Volume of water after addition of metal = 14.9 milliliters
The density of the metal should be reported as
(A) 7.8444 grams per mL
14.9 – 12.4 = 2.5 mL (keep one decimal)
(B) 7.844 grams per mL
(C) 7.84 grams per mL
19.611 g = 7.8 g/mL (keep two sig figs)
2.5 mL
(D) 7.8 grams per mL
2–
___2. What is the oxidation number (charge) of chromium in the dichromate ion, Cr2O7 ?
(A) 2–
2(Cr) + 7(O) = –2
(B) 6+
2(Cr) + 7(–2) = –2
(C) 7+
(D) 12+
2(Cr) = +12
Cr = +6
___3.
What number of moles of O2 is needed to produce 14.2 grams of P4O10 from P?
(Molecular weight P4O10 = 284)
(A) 0.0500 mole
4 P + 5 O2  P4O10
(B) 0.250 mole
14.2 g P4O10 x 1 mol P4O10 x 5 mol O2 = 0.250 mol
(C) 0.0625 mole
284 g P4O10 1 mol P4O10
(D) 0.500 mole
142 = 0.5 , therefore 14.2 = 0.05 , then 0.05 x 5 = 0.25
(E) 0.125 mole
284
284
10 HI + 2 KMnO4 + 3 H2SO4  5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
___4.
According to the balanced equation above, how many moles of HI would be necessary to produce
2.5 mol of I2, starting with 4.0 mol of KMnO4 and 3.0 mol of H2SO4?
(A) 20.
2.5 mol I2 x 10 mol HI = 5.0 mol HI
(B) 10.
5 mol I2
(C) 8.0
10 = 2 , then 2 x 2.5 = 5
(D) 5.0
5
(E) 2.5
(no need to find limiting reactant b/c moles of product was
already known.)
2
Section II Free Response
CALCULATOR ALLOWED
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if
you do not. Attention should be paid to significant figures.
1. Analysis of naturally occurring iron by a mass spectrometer results in four peaks at 54, 56, 57, and 58.
These four stable isotopes of iron are listed in the table below with Fe–56 as the reference isotope.
54
Name of Isotope
Mass of
Isotope
Relative
Abundance
Iron-54
Iron-56
Iron-57
Iron-58
53.94 amu
55.93 amu
56.94 amu
57.93 amu
5.90%
91.72%
2.10%
0.280%
56 57 58
(a) Differentiate between the terms “element”, “atom”, and “isotope”. (2)
Elements are composed of all the same type of atoms with the same number of protons and electrons.
Atoms are the smallest particles of an element that maintain their properties and are composed of protons,
neutrons, and electrons.
Isotopes are atoms of the same element with the same number of protons and electrons, but different
numbers of neutrons.
(b) Calculate the average atomic mass of iron. (1)
(53.94)(0.0590) + (55.93)(0.9172) + (56.94)(0.0210) + (57.93)(0.00280) = 55.84 amu
(c) How many grams of naturally occurring iron would contain 1.807 x 1024 atoms of Iron-57 ? (3)
1.807 x 1024 atoms 57Fe x
1 mol 57Fe
x 56.94 g 57Fe = 170.9 g 57Fe
23
57
6.022 x 10 atoms Fe
1mol 57Fe
2.10% of naturally occurring Fe is 57Fe.
(0.0210) x (g Fe) = 170.9 g 57Fe
g Fe = 170.9
0.0210
g Fe = 8140 g Fe
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2. Answer the following questions about BeC2O4(s) and its hydrate.
(a) Calculate the mass percent of carbon in the hydrated form of the solid that has the formula
BeC2O4•3H2O. (1)
24.02 g C
x 100 = 15.90 % C
[(9.01) + 2(12.01) + 4(16.00) + 3(18.02)]
(b)When heated to 220.oC, BeC2O4•3H2O(s) dehydrates completely as represented below.
BeC2O4•3H2O(s)  BeC2O4(s) + 3 H2O(g)
If 3.21 g of BeC2O4•3H2O(s) is heated to 220.oC, calculate the mass of BeC2O4(s) formed. (1)
3.21 g BeC2O4•3H2O x 1 mol BeC2O4•3H2O x
1 mol BeC2O4
x 97.03 g BeC2O4 = 2.06 g BeC2O4
151.09 g BeC2O4•3H2O 1 mol BeC2O4•3H2O
1 mol BeC2O4
(c) Calculate the moles of H2O(g) formed. (1)
3.21 g BeC2O4•3H2O x 1 mol BeC2O4•3H2O x
3 mol H2O
= 0.0637 mol H2O
151.09 g BeC2O4•3H2O 1 mol BeC2O4•3H2O
(d) When a different student heated the hydrate, the following data were obtained.
Mass of empty container
32.155 g
Initial mass of sample and container
36.382 g
Mass of sample and container after first heating
34.011 g
Mass of sample and container after second heating
33.962 g
The student then calculated the formula of the hydrate to be BeC2O4•2H2O.
Explain why the student’s calculations resulted in a formula with too few waters of hydrate.
Your explanation should make reference to the data. (2)
The student did not heat the hydrated salt to a constant mass. There may have been more water left in the salt
after the second heating.
Failing to remove all of the water would cause the calculated amount of water by difference to be smaller than it
should have been, therefore causing the ratio of waters in the formula of the hydrated salt to be too small.
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3. Answer the following questions about a pure compound that contains only carbon, hydrogen, and
oxygen.
(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of H2O(g).
(i) Calculate the individual masses of C, H, and O in the 0.7549 g sample. (3)
1.9061 g CO2 x 1 mol CO2 x 1 mol C x 12.01 g C = 0.5202 g C
44.01 g CO2 1 mol CO2
1 mol C
0.3370 g H2O x 1 mol H2O x 2 mol H x 1.008 g H = 0.03770 g H
18.02 g H2O 1 mol H2O
1 mol H
0.7549 g sample – (0.5202 g C) – (0.03770 g H) = 0.1970 g O
(ii) Determine the empirical formula for the compound. (2)
0.5202 g C x 1 mol C = 0.04331 mol C
12.01 g C
÷ 0.01231 = 3.5 C x 2 = 7 C
0.03770 g H x 1 mol H = 0.03740 mol H ÷ 0.01231 = 3 H
1.008 g H
0.1970 g O x 1 mol O =
16.00 g O
x 2 = 6 H
0.01231 mol O ÷ 0.01231 = 1 O x 2 = 2 O
C7H6O2
(iii) Determine the molecular formula for the compound if the molecular weight is 244 g∙mol–1. (1)
C7H6O2 = 122 g/mol
244 = 2
122
2(C7H6O2) = C14H12O4
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4. For each of the following three reactions, in part (i) write a balanced equation for the reaction. In part (i),
coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless
otherwise indicated. You may use the empty space at the bottom of the next page for scratch work, but
only equations that are written in the answer boxes provided will be graded.
(a) Solid potassium chlorate is heated strongly forming potassium chloride and oxygen gas.
(i) Balanced equation: (3)
2 KClO3  2 KCl + 3 O2
(b) Solid chromium is oxidized by potassium permangate to form potassium chromate and
manganese(IV) oxide.
(i) Balanced equation: (3)
Cr + 2 KMnO4  K2CrO4 + 2 MnO2
(c) Sulfur trioxide gas is bubbled through water producing sulfuric acid.
(i) Balanced equation: (3)
SO3 + H2O  H2SO4
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5. A 0.150 g sample of solid lead(II) nitrate is added to 0.905 g of sodium iodide in solution.
Assume no change in volume of the solution. The chemical reaction that takes place is represented by
the following equation.
Pb(NO3)2(aq) + 2 NaI(aq)  PbI2(s) + 2 NaNO3(aq)
(a) List an appropriate observation that provides evidence of a chemical reaction between the two
compounds. (1)
A solid precipitate forms (yellow color).
(b) Calculate the number of moles of each reactant. (2)
0.150 g Pb(NO3)2 x 1 mol Pb(NO3)2 = 0.000453 mol Pb(NO3)2
331.22 g Pb(NO3)2
0.905 g NaI x 1 mol NaI = 0.00604 mol NaI
149.89 g NaI
(c) Identify the limiting reactant. Show calculations to support your identification. (2)
0.000453 mol Pb(NO3)2 x
1 mol PbI2 = 0.000453 mol PbI2
1 mol Pb(NO3)2
0.00604 mol NaI x 1 mol PbI2 = 0.00302 mol PbI2
2 mol NaI
Pb(NO3)2 is limiting
(d) What is the theoretical yield of PbI2(s) ? (1)
0.000453 mol PbI2 x
461 g PbI2
1 mol PbI2
= 0.209 g PbI2
(e) If 0.174 g of PbI2(s) was obtained by filtration and weighing, what is the percent yield of
PbI2(s) for this reaction? (1)
0.174 g PbI2
0.209 g PbI2
x 100 = 83.3%
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Answer KEY
FR
1.
2.
3.
4.
5.
created
2000 #3
2005 #2
created
2008B #3(a)(b)(c)
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