Chem I Homework Page, Exam 1 Material

Homework Page Without Visible Answers

This page has all of the required homework for the material covered in the first exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12^th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Matter, Conversion Factors and Dimensional Analysis (Ch. 1)

1. Label each of the following as either a chemical or a physical process:
   • Burning paper

   Answer

   Chemical change, paper is changed to CO₂ and H₂O

   • A nail that is rusting.

   Answer

   Chemical change, iron is changing to iron oxide.

   • Dissolving sugar in water.

   Answer

   Physical change, the sugar can be recovered by evaporating the water.

   • Acid rain reacting with limestone headstones.

   Answer

   Chemical change, the limestone reacts with the acid to form CO₂, H₂, and water.

   • Dissolving table salt in water.

   Answer

   Physical change, the salt can be recovered by evaporating the water.

2. Convert to or from exponential notation as needed.
   • 23,465

   Answer

   2.3 x 10⁴

   • 0.005316

   Answer

   5.3 x 10^-3

   • 6.74 x 10⁵

   Answer

   674,000

   • 4.54 x 10^-4

   Answer

   0.000454

   • 78.632 x 10⁴

   Answer

   7.8632 x 10⁵ and 786,320

3. Answer the following density questions.
   • Under normal conditions aluminum has a density of 2.70 g/ml. What is the volume of 1.85 g of Aluminum?

   Answer

   (

   1.85 g Al

   )

   (

   1 mL Al   2.70 g Al

   )

   = 0.685 mL of Al

   • One way to liquify carbon dioxide is to put it under a lot of pressure. At a certain pressure the density of the liquid is 0.466 g/cm³. What is the mass of a 15 mL sample of this liquid CO₂?

   Answer

   Note: 1 cm³ = 1 mL

   (

   15 mL CO2

   )

   (

   0.466 g CO2   1 mL CO2

   )

   = 6.99 g CO2

   • A student was given the task of determining if the pure metal she had was nickel (d = 8.90 g/mL) or zinc (d = 7.14 g/mL). She used a balance to determine a mass of 9.7 g. She then used a displacement technique to determine a volume of 1.36 mL. What would she report?

   Answer

   Since 9.7 g divided by 1.36 mL is 7.13 g/mL it is more likely to be zinc.

4. A reported value with its uncertainty (+/- or ±) should encompass the correct value for the measurement. A group of students reported a value of 2.64 ± 0.08 g/mL for the density of aluminum (known value is 2.70 g/mL). What is the range of their reported value and does it encompass the known value for the density of aluminum?

   Answer

   The range goes from 2.56 g/mL to 2.72 g/mL. The known value is between these two numbers.

5. Use units and dimensional analysis to do the following problems.
   • The periodic table provides conversion factors between grams and moles. We call these numbers the atomic weight or the molar mass. Atomic nitrogen has a molar mass of 14 g/mole. How many grams are in 1.5 moles of atomic nitrogen?

   Answer

   (

   1.5 mole N

   )

   (

   14 g N   1 mole N

   )

   = 21 g N

   • How many grams are in 1.5 moles of gaseous nitrogen (N₂)? See the last question for data.

   Answer

   (

   1.5 mole N2

   )

   (

   2 mole N   1 mole N2

   )

   (

   14 g N   1 mole N

   )

   = 42 g N

   • How many liters (L) are there in 500 mL?

   Answer

   (

   500 mL

   )

   (

   1 L   1000 mL

   )

   = 0.5 L

   • How many centimeters (cm) are there in 1.2 m?

   Answer

   (

   1.2 m

   )

   (

   100 cm   1 m

   )

   = 120 cm

Atoms, Molecules, and Ions (Ch. 2)

1. What is the difference between an atom and an element?

   Answer

   An atom is the smallest piece of an element that retains the properties of that element.

2. Define a molecule two different ways.

   Answer

   A molecule is two or more atoms that are hooked together, are stuck together, and act as one unit. A molecule is the smallest piece of a substance that retains the properties of that substance.

3. Do ionic compounds have molecules? Explain.

   Answer

   Ionic compounds do not have molecules. There is no single unit that retains the properties of a compound.

4. Use the periodic table to predict the charge of an ion created from each of the following elements:
   • K

   Answer

   K⁺¹

   • Ca

   Answer

   Ca⁺²

   • Br

   Answer

   Br^-1

   • Ba

   Answer

   Ba⁺²

   • F

   Answer

   F^-1

5. Define the word isotope.

   Answer

   Two atoms are isotopes if they have the same number of protons, but a different number of neutrons. Two atoms of the same element that have a different number of neutrons.

6. Silver has two naturally occurring isotopes: ¹⁰⁷Ag (mass = 106.905095 g/mole) has an abundance of 51.83% and ¹⁰⁹Ag (mass = 108.904754 g/mole) has an abundance of 48.17. Calculate the molar mass that is reported on the periodic table.

   Answer

   0.5183(106.905095) + 0.4817(108.904754) = 107.868 g/mole

7. Lithium has two naturally occurring isotopes: ⁶Li (mass = 6.015123 g/mole) has an abundance of 7.5% and ⁷Li has an abundance of 92.5%. Calculate the molar mass of  ⁷Li. The periodic table reports 6.941 g/mole for the molar mass of lithium.

   Answer

   0.075(6.015123) + 0.925(x) = 6.941

   x = [6.941 - (0.075)(6.015123)]/0.925 = 7.016 g/mole

8. Boron has two naturally occurring isotopes: ¹⁰B (mass = 10.012938 g/mole) and ¹¹B (mass = 11.009305 g/mole). The atomic weight on the periodic table for boron is 10.811 g/mole. Calculate the percent abundance for each isotope.

   Answer

   x(10.012938) + (1 - x)(11.009305) = 10.811

   x = [10.811 - 11.009305]/[10.012938 - 11.009305] = 0.199

   ¹⁰B is 19.9% and ¹¹B is 80.1%

9. Identify the following elements and the number of neutrons in the nucleus of the isotope:
   • ³⁷₁₇X

   Answer

   Cl, 20 neutrons

   • ²⁵₁₂X

   Answer

   Mg, 13 neutrons

   • ⁷⁰₃₁X

   Answer

   Ga, 39 neutrons

   • ²⁴³₉₅X

   Answer

   Am, 148

10. How many protons, neutrons, and electrons are in the following isotopes:
    • ³⁵Cl

    Answer

    17 protons, 18 neutrons, and 17 electrons

    • ⁸⁵Sr²⁺

    Answer

    38 protons, 47 neutrons, and 36 electrons

    • ¹³¹I^-

    Answer

    53 protons, 78 neutrons, 54 electrons

    • ¹²⁹Cs

    Answer

    55 protons, 74 neutrons, 55 protons

11. Fill in the following table:

    Symbol	56Fe3+
    Protons		29	55
    Neutrons		34	78	7
    Electrons		28		10
    Net Charge			0	2-

    Answer

    Symbol	56Fe3+	63Cu+	133Cs	15O2-
    Protons	26	29	55	8
    Neutrons	30	34	78	7
    Electrons	23	28	55	10
    Net Charge	3+	1+	0	2-

12. Predict the compounds formed by combining the cations across the top with the anions down the side.

    Ion	Zn2+	Na+	Ba2+	Al3+
    SO42-
    NO3-
    CO32-
    OH-
    Cl-
    PO43-

    Answer

    Ion	Zn2+	Na+	Ba2+	Al3+
    SO42-	ZnSO4	Na2SO4	BaSO4	Al2(SO4)3
    NO3-	Zn(NO3)2	NaNO3	Ba(NO3)2	Al(NO3)3
    CO32-	ZnCO3	Na2CO3	BaCO3	Al2(CO3)3
    OH-	Zn(OH)2	NaOH	Ba(OH)2	Al(OH)3
    Cl-	ZnCl2	NaCl	BaCl2	AlCl3
    PO43-	Zn3(PO4)2	Na3PO4	Ba3(PO4)2	AlPO4

13. Give the name or formula for each of the following compounds.
    • CuCl

    Answer

    Copper(I)chloride

    • HgCO₃

    Answer

    Mercury(II)carbonate

    • FeBr₃

    Answer

    Iron(III)bromide

    • Chromium(III)acetate

    Answer

    Cr(C₂H₃O₂)₃

    • Lead(II)hydroxide

    Answer

    Pb(OH)₂

    • Manganese(II)nitrate

    Answer

    Mn(NO₃)₂

14. Give the name or formula for each of the following molecules.
    • Cl₂O

    Answer

    dichlorine monoxide

    • N₂O₄

    Answer

    dinitrogen tetroxide

    • NF₃

    Answer

    nitrogen trifluoride

    • chlorine pentaflouride

    Answer

    ClF₅

    • Dinitrogen oxide

    Answer

    N₂O

    • Phosphorous trichloride

    Answer

    PCl₃

Stoichiometry, Etc. (Ch. 3)

1. Balance the following chemical equations.
   • CH₄ + O₂ → H₂O + CO₂

   Answer

   CH₄ + 2O₂ → 2H₂O + CO₂

   • HgCO₃ + HCl → H₂O + CO₂ + HgCl₂

   Answer

   HgCO₃ + 2HCl → H₂O + CO₂ + HgCl₂

   • FeBr₃ + AgNO₃ → Fe(NO₃)₃ + AgBr

   Answer

   FeBr₃ + 3AgNO₃ → Fe(NO₃)₃ + 3AgBr

   • H₂ + O₂ → H₂O

   Answer

   H₂ + ½O₂ → H₂O   or   2H₂ + O₂ → 2H₂O

   • K + N₂ → K₃N

   Answer

   6K + N₂ → 2K₃N

   • Al(OH)₃ + H₂SO₄ → Al₂(SO₄)₃ + H₂O

   Answer

   2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

2. Calculate the formula weights (also called molecular weight or molar mass) of the following substances. See Chapter Three problems for interesting examples.
   • Ammonium hydroxide

   Answer

   NH₄OH ⇒   35 g/mole

   • HgCO₃

   Answer

   260.59 g/mole

   • H₂SO₄

   Answer

   98 g/mole

   • Copper(II)chloride

   Answer

   CuCl₂ ⇒ 134.452 g/mole

3. Calculate the percentage by mass of the indicated element in the following substances.
   • Oxygen in ammonium hydroxide

   Answer

   NH₄OH ⇒   16/35 = 0.457 ⇒ 45.7%

   • Hg in HgCO₃

   Answer

   200.59/260.59 = .767 ⇒ 76.7%

   • N in Fe(NO₃)₃

   Answer

   (3*14)/241.845 = .174 ⇒ 17.4%

   • Chlorine in copper(II)chloride

   Answer

   CuCl₂ ⇒ (2*35.453)/134.452 = 0.527 52.7%

4. What is a mole? Where are the conversion factors between grams and moles found?

   Answer

   1 mole = 6.02 x 10²³ objects. The atomic mass on the periodic table is the number of grams associated with one mole of that element.

5. Calculate the following:
   • How many grams are in 1.5 moles of atomic nitrogen?

   Answer

   (

   1.5 mole N

   )

   (

   14 g N   1 mole N

   )

   = 21 g N

   • How many grams of nitrogen are in 1.5 moles of gaseous nitrogen (N₂)?

   Answer

   (

   1.5 mole N2

   )

   (

   2 mole N   1 mole N2

   )

   (

   14 g N   1 mole N

   )

   = 42 g N

   • How many moles of Br are in 79.9 g of Br₂?

   Answer

   (

   79.9 g Br2

   )

   (

   1 mole Br2   159.8 g Br2

   )

   (

   2 mole Br   1 mole Br2

   )

   = 1.0 mole Br

   • How many moles of O₂ are there in 32 g O₂?

   Answer

   (

   32 g O2

   )

   (

   1 mole O2   32 g O2

   )

   = 1 mole O2

6. Calculate the empirical formulas.
   • What is the empirical formula for acetone, a common solvent, if it is 62.07% carbon, 27.59% oxygen, and 10.34% hydrogen?

   Answer

   (

   62.07 g C

   )

   (

   1 mole C   12.01 g C

   )

   = 5.17 mole C

   (

   27.59 g O

   )

   (

   1 mole O   16 g O

   )

   = 1.72 mole O

   (

   10.34 g H

   )

   (

   1 mole H   1 g H

   )

   = 10.34 mole H

   (

   5.17 mole C   1.72 mole O

   )

   = 3

   (

   1.72 mole O   1.72 mole O

   )

   = 1

   (

   10.34 mole H   1.72 mole O

   )

   = 6

   ⇒ C, 3; O, 1; H, 6 ⇒ C₃H₆O

   • What is the empirical formula for phenol if a sample analysis yielded 0.32 g hydrogen, 0.85 g oxygen, and 3.83 g carbon?

   Answer

   (

   0.32 g H

   )

   (

   1 mole H   1 g H

   )

   = 0.32 mole H

   (

   0.85 g O

   )

   (

   1 mole O   16 g O

   )

   = 0.053 mole O

   (

   3.83 g C

   )

   (

   1 mole C   12 g C

   )

   = 0.319 mole C

   (

   0.320 mole H   0.053 mole O

   )

   = 6

   (

   0.053 mole O   0.053 mole O

   )

   = 1

   (

   0.319 mole C   0.053 mole O

   )

   = 6

   ⇒ C, 6; H, 6; O, 1 ⇒ C₆H₆O

   • What is the empirical formula of an aluminum oxide sample if it is 52.92% aluminum and 47.08% oxygen?

   Answer

   (

   52.92 g Al

   )

   (

   1 mole Al   26.98 g Al

   )

   = 1.96 mole Al

   (

   47.08 g O

   )

   (

   1 mole O   16 g O

   )

   = 2.94 mole O

   (

   1.96 mole Al   1.96 mole Al

   )

   = 1

   (

   2.94 mole O   1.96 mole Al

   )

   = 1.5

   ⇒ Al, 1; O, 1.5 ⇒ Al, 2; O, 3 ⇒ Al₂O₃

   • A common organic solvent only contains carbon, hydrogen, and oxygen. Combustion analysis of 1.000 g of this solvent gives 1.913 g CO₂ and 1.174 g H₂O. What is the empirical formula?

   Answer

   (

   1.913 g CO2

   )

   (

   1 mole CO2   44 g CO2

   )

   (

   1 mole C   1 mole CO2

   )

   = 0.0435 mole C

   (

   1.174 g H2O

   )

   (

   1 mole H2O   18 g H2O

   )

   (

   2 mole H   1 mole H2O

   )

   = 0.130 mole H

   (

   0.0435 mole C

   )

   (

   12 g C   1 mole C

   )

   = 0.522 g C

   (

   0.130 mole H

   )

   (

   1 g H   1 mole H

   )

   = 0.13 g H

   1.000 g sample - 0.522 g C - 0.130 g H = 0.348 g O

   (

   0.348 g O

   )

   (

   1 mole O   16 g O

   )

   = 0.0218 mole O

   (

   0.0435 mole C   0.0218 mole O

   )

   = 2

   (

   0.130 mole H   0.0218 mole O

   )

   = 6

   (

   0.218 mole O   0.0218 mole O

   )

   = 1

   ⇒ C, 2; H, 6; O, 1 ⇒ C₂H₆O (ethanol or ethyl alcohol)

7. In combustion reactions carbon-containing substances react with oxygen to form carbon dioxide and water. Burning ethanol is an example:

   C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

   • How many moles of O₂ are needed to react with 1.5 moles of C₂H₅OH?

   Answer

   (

   1.5 mole C2H5OH

   )

   (

   3 mole O2   1 mole C2H5OH

   )

   = 4.5 mole O2

   • How many moles of CO₂ are produced when when 1.5 moles of C₂H₅OH are burned?

   Answer

   (

   1.5 mole C2H5OH

   )

   (

   2 mole CO2   1 mole C2H5OH

   )

   = 3 mole CO2

   • How many moles of oxygen are needed to produce 3.3 moles of carbon dioxide?

   Answer

   (

   3.3 mole CO2

   )

   (

   2 mole O2   3 mole CO2

   )

   = 2.2 mole O2

   • How many grams of C₂H₅OH are needed to produce 36 g H₂O?

   Answer

   (

   36 g H2O

   )

   (

   1 mole H2O   18 g H2O

   )

   = 2 mole H2O

   (

   2 mole H2O

   )

   (

   1 mole C2H5OH   3 mole H2O

   )

   (

   46 g C2H5OH   1 mole C2H5OH

   )

   = 30.67 g C2H5OH

   • How many grams of CO₂ are produced when 15 g O₂ are reacted?

   Answer

   (

   15 g O2

   )

   (

   1 mole O2   32 g O2

   )

   (

   2 mole CO2   3 mole O2

   )

   (

   44 g CO2   1 mole CO2

   )

   = 13.75 g CO2

   • How many mole of CO₂ are produced when 4 moles of O₂ are reacted with 1.5 moles of C₂H₅OH?

   Answer

   Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen O₂ below.

   (

   4 mole O2

   )

   (

   1 mole C2H5OH   3 mole O2

   )

   = 1.33 mole C2H5OH needed

   We need 1.33 moles, but we have 1.5 moles according to the problem statement. We have extra C₂H₅OH which means that O₂ is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

   (

   4 mole O2

   )

   (

   2 mole CO2   3 mole O2

   )

   = 2.67 mole CO2

   • How many grams of H₂O are produced when 256 g of O₂ are reacted with 138 g of C₂H₅OH?

   Answer

   Determine the moles of each reactant that we are given:

   (

   256 g O2

   )

   (

   1 mole O2   32 g O2

   )

   = 8 mole O2

   (

   138 g C2H5OH

   )

   (

   1 mole C2H5OH   46 g C2H5OH

   )

   = 3 mole C2H5OH

   Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen C₂H₅OH below. Get the mole ratios from the balanced equation.

   (

   3 mole C2H5OH

   )

   (

   3 mole O2   1 mole C2H5OH

   )

   = 9 mole O2 needed

   We need 9 mole O₂, but have 8 moles. We don't have enough O₂ which means that O₂ is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

   (

   8 mole O2

   )

   (

   3 mole H2O   3 mole O2

   )

   (

   18 g H2O   1 mole H2O

   )

   = 144 g H2O

8. How many grams of Fe₂O₃ are produced when 56 g of Fe are reacted with 32 g of O₂? How many grams of the excess reactant will remain after the reaction is complete?

   4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

   Answer

   Determine the moles of each of the reactants that we are given.

   (

   32 g O2

   )

   (

   1 mole O2   32 g O2

   )

   = 1 mole O2

   (

   56 g Fe

   )

   (

   1 mole Fe   55.845 g Fe

   )

   = 1 mole Fe

   Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen Fe below. Get the mole ratios from the balanced chemical equation.

   (

   1 mole Fe

   )

   (

   3 mole O2   4 mole Fe

   )

   = 0.752 mole O2 needed

   We need 0.752 mole O₂, but have 1 mole. We have extra O₂ which means that Fe is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

   (

   56 g Fe

   )

   (

   1 mole Fe   55.845 g Fe

   )

   (

   2 mole Fe2O3   4 mole Fe

   )

   (

   159.69 g Fe2O3   1 mole Fe2O3

   )

   = 80.07 g Fe2O3

   We have 1 mole - 0.752 mole = 0.248 moles of extra O₂

   (

   0.248 mole O2

   )

   (

   32 g O2   1 mole O2

   )

   = 7.94 g O2 left over

9. How many grams of Fe₂O₃ are produced when 56 g of Fe are reacted with 32 g of O₂? How many grams of the excess reactant will remain after the reaction is complete?

   4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

   Answer

   Determine the moles of each of the reactants that we are given.

   (

   32 g O2

   )

   (

   1 mole O2   32 g O2

   )

   = 1 mole O2

   (

   56 g Fe

   )

   (

   1 mole Fe   55.845 g Fe

   )

   = 1 mole Fe

   Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen Fe below. Get the mole ratios from the balanced chemical equation.

   (

   1 mole Fe

   )

   (

   3 mole O2   4 mole Fe

   )

   = 0.752 mole O2 needed

   We need 0.752 mole O₂, but have 1 mole. We have extra O₂ which means that Fe is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

   (

   56 g Fe

   )

   (

   1 mole Fe   55.845 g Fe

   )

   (

   2 mole Fe2O3   4 mole Fe

   )

   (

   159.69 g Fe2O3   1 mole Fe2O3

   )

   = 80.07 g Fe2O3

   We have 1 mole - 0.752 mole = 0.248 moles of extra O₂

   (

   0.248 mole O2

   )

   (

   32 g O2   1 mole O2

   )

   = 7.94 g O2 left over

10. Problem for limiting reactant lab: Aluminum hydroxide reacts with sulfuric acid as follows:

    2Al(OH)₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6H₂O(l)

    Determine the limiting reactant when 0.600 mole Al(OH)₃ and 0.600 mole H₂SO₄ are allowed to react? How many moles of Al₂(SO₄)₃ can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction? Show all work!