Chem I Homework Page, Exam 1 Material

Homework Page with 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 12th 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:
  2. Convert to or from exponential notation as needed.
  3. Answer the following density questions.
  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 ± 0.1 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 2.54 g/mL to 2.74 g/mL. The known value is between these two numbers.
  5. Use units and dimensional analysis to do the following problems.

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:
  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:
  10. How many protons, neutrons, and electrons are in the following isotopes:
  11. Fill in the following table:
    Symbol56Fe3+
    Protons2955
    Neutrons34787
    Electrons2810
    Net Charge02-
    Answer
    Symbol56Fe3+63Cu+133Cs15O2-
    Protons2629558
    Neutrons3034787
    Electrons23285510
    Net Charge3+1+02-
  12. Predict the compounds formed by combining the cations across the top with the anions down the side.
    IonZn2+Na+Ba2+Al3+
    SO42-
    NO3-
    CO32-
    OH-
    Cl-
    PO43-
    Answer
    IonZn2+Na+Ba2+Al3+
    SO42-ZnSO4Na2SO4BaSO4Al2(SO4)3
    NO3-Zn(NO3)2NaNO3Ba(NO3)2Al(NO3)3
    CO32-ZnCO3Na2CO3BaCO3Al2(CO3)3
    OH-Zn(OH)2NaOHBa(OH)2Al(OH)3
    Cl-ZnCl2NaClBaCl2AlCl3
    PO43-Zn3(PO4)2Na3PO4Ba3(PO4)2AlPO4
  13. Give the name or formula for each of the following compounds.
  14. Give the name or formula for each of the following molecules.

Stoichiometry, Etc. (Ch. 3)

  1. Balance the following chemical equations.
  2. Caculate the formula weights (also called molecular weight or molar mass) of the following substances. See Chapter Three problems for interesting examples.
  3. Caculate the percentage by mass of the indicated element in the following substances.
  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:
  6. Calculate the empirical formulas.
  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)

  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