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.
- 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.
- 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.
- 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.
- 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
- 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.
- 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
- 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
- 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%
- 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
- 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
- Fill in the following table:
Symbol 56Fe3+ Protons 29 55 Neutrons 34 78 7 Electrons 28 10 Net Charge 0 2- 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- - 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- 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 - 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
- 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
- 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
- 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
- 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%
- 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.
- Calculate the following:
- How many grams are in 1.5 moles of atomic nitrogen? Answer
- How many grams of nitrogen are in 1.5 moles of gaseous nitrogen (N2)? Answer
- How many moles of Br are in 79.9 g of Br2? Answer
- How many moles of O2 are there in 32 g O2? Answer
( 1.5 mole N ) ( ) = 21 g N ( 1.5 mole N2 ) ( ) ( ) = 42 g N ( 79.9 g Br2 ) ( ) ( ) = 1.0 mole Br ( 32 g O2 ) ( ) = 1 mole O2 - 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
- 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
- What is the empirical formula of an aluminum oxide sample if it is 52.92% aluminum and 47.08% oxygen? Answer
- 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
( 62.07 g C ) ( ) = 5.17 mole C ( 27.59 g O ) ( ) = 1.72 mole O ( 10.34 g H ) ( ) = 10.34 mole H ( ) = 3 ( ) = 1
⇒ C, 3; O, 1; H, 6 ⇒ C3H6O( ) = 6 ( 0.32 g H ) ( ) = 0.32 mole H ( 0.85 g O ) ( ) = 0.053 mole O ( 3.83 g C ) ( ) = 0.319 mole C ( 0.320 mole H 0.053 mole O ) = 6 ( 0.053 mole O 0.053 mole O ) = 1
⇒ C, 6; H, 6; O, 1 ⇒ C6H6O( 0.319 mole C 0.053 mole O ) = 6 ( 52.92 g Al ) ( ) = 1.96 mole Al ( 47.08 g O ) ( ) = 2.94 mole O ( 1.96 mole Al 1.96 mole Al ) = 1
⇒ Al, 1; O, 1.5 ⇒ Al, 2; O, 3 ⇒ Al2O3( ) = 1.5 ( 1.913 g CO2 ) ( ) ( ) = 0.0435 mole C ( 1.174 g H2O ) ( ) ( ) = 0.130 mole H ( 0.0435 mole C ) ( ) = 0.522 g C ( 0.130 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 ) ( ) = 0.0218 mole O ( 0.0435 mole C 0.0218 mole O ) = 2 ( 0.130 mole H 0.0218 mole O ) = 6
⇒ C, 2; H, 6; O, 1 ⇒ C2H6O (ethanol or ethyl alcohol)( 0.218 mole O 0.0218 mole O ) = 1 - 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
- How many moles of CO2 are produced when when 1.5 moles of C2H5OH are burned? Answer
- How many moles of oxygen are needed to produce 3.3 moles of carbon dioxide? Answer
- How many grams of C2H5OH are needed to produce 36 g H2O? Answer
- How many grams of CO2 are produced when 15 g O2 are reacted? Answer
- 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.
- How many grams of H2O are produced when 256 g of O2 are reacted with 138 g of C2H5OH? Answer
( 1.5 mole C2H5OH ) ( ) = 4.5 mole O2 ( 1.5 mole C2H5OH ) ( ) = 3 mole CO2 ( 3.3 mole CO2 ) ( ) = 2.2 mole O2 ( 36 g H2O ) ( ) = 2 mole H2O ( 2 mole H2O ) ( ) ( 46 g C2H5OH 1 mole C2H5OH ) = 30.67 g C2H5OH ( 15 g O2 ) ( ) ( ) ( ) = 13.75 g CO2 ( 4 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 ) ( ) = 2.67 mole CO2 Determine the moles of each reactant that we are given:
( 256 g O2 ) ( ) = 8 mole O2
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.( 138 g C2H5OH ) ( 1 mole C2H5OH 46 g C2H5OH ) = 3 mole C2H5OH ( 3 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 ) ( ) ( ) = 144 g H2O - 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)
AnswerDetermine the moles of each of the reactants that we are given.
( 32 g O2 ) ( ) = 1 mole O2
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.( 56 g Fe ) ( ) = 1 mole Fe ( 1 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 ) ( ) ( ) ( 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 ) ( ) = 7.94 g O2 left over - 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)
AnswerDetermine the moles of each of the reactants that we are given.
( 32 g O2 ) ( ) = 1 mole O2
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.( 56 g Fe ) ( ) = 1 mole Fe ( 1 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 ) ( ) ( ) ( 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 ) ( ) = 7.94 g O2 left over -
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!