chem 161-2013 some chapter 2 practice problems with solutions

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CHEM 161-2011 CHAPTER 2 - ATOMS, MOLECULES AND IONS

PRACTICE PROBLEMS DR. ED TAVSS

Fundamental chemical laws Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes Periodic table (Also, see Chapter 7 - Periodic Table Trends) Atoms, ions, moles and molecular wts.

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FUNDAMENTAL CHEMICAL LAWS 3 Chem 161-2012 Final Exam

Burdge and Overby Chapter 2 Atoms and the Periodic Table History (Corrleates to H&P Chapter 2) Which of the following statements of Daltons Atomic Theory are still believed to be correct? X. Elements are composed of atoms which are indivisible. Y. All atoms of a given element are identical. Z. Atoms combine in simple whole number ratios to form compounds. 1. A.Xonly2. B.Yonly

3. C.Zonly4. D.YandZonly5. E.X,Y,andZ

6. X.False.Elementsarecomposedofatoms,buttheyareabletobedivided.(Atomsareneutralincharge.)Anatomcanloseanelectron,aproton,aneutron.Y.False.Atomsexistasisotopes,soonetypeofatom,e.g.,hydrogen,canexistasisotopescontainingnoneutrons(protium),oneneutron(deuterium)ortwoneutrons(tritium).Thereforetherecanbethreedifferenttypesofhydrogenatoms.Z.True;e.g.,CO=1atomC+1atomO;CO2=1atomC+2atomsO;CO3

2=1atomC+3atomsO

C

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19 Chem 161-2008 Exam I Hill, Petrucci et al. CHAPTER 2 - ATOMS, MOLECULES AND IONS

Fundamental chemical laws

Five samples were analyzed for elemental composition. Four of the samples might be the same compound. One could not. Which one of the following is different?

Fe (g) C (g) N (g)A 3.16 4.08 4.76B 1.84 2.38 2.78C 60.60 78.19 91.21D 4.74 6.12 7.14

E 6.40 4.80 4.80 The law of definite proportions states that all samples of a particular compound have the same composition. That is, all samples have the same proportion by mass of the elements present. A: The ratio of N:C:Fe = 1.51:1.29:1.00 B: The ratio of N:C:Fe = 1.51:1.29:1.00 C: The ratio of N:C:Fe = 1.51:1.29:1.00 D: The ratio of N:C:Fe = 1.51:1.29:1.00 E: The ratio of N:C:Fe = 0.75:0.75:1.00

Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 2 Atoms, Molecules and Ions Fundamental chemical laws 17. Which of the following is a hypothesis? A. A beaker weights 157.27g B. Water freezes when the temperature drops because the interactions between molecules becomes greater than the kinetic energy of the molecules. C. Sugar turns black when heated. D. The mixture of two solutions can produce a solid. E. Ice is less dense than liquid water. A hypothesis is a mechanism of action, which means it explains the results mechanistically. A. A beaker weighing 157.27 g is a result, not a hypothesis. B. The mechanism by which water freezes at lower temperatures is that several factors provide a driving force. This is an explanation of what has happened. It is a mechanism. C. Sugar turns black when heated is a result, not a hypothesis. D. The mixture of two solutions producing a solid is a result, not an explanation of what is happening.

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E. Ice being less dense than water is a result of a measurement, not a mechanism of action.

NATURE OF THE ATOM (PROTONS, NEUTRONS, ELECTRONS, SYMBOLS, ETC.)

30 Chem 161-2011 Final exam

B&O CHAPTER 2 ATOMS AND THE PERIODIC TABLE Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes (H&P Chapter 2) How many protons, electrons and neutrons are in the 47Ti2+ ion?

protons electrons neutrons A 22 22 23 B 20 22 25

C 22 20 25 D 22 24 25 E 25 20 22

The atomic number for titanium is 22. Therefore, it has 22 protons. The upper left-hand number, 47, is the mass number, which is the sum of the protons and neutrons. Therefore, this titanium isotope has 25 neutrons. Since it has a charge of 2+, then it is missing two electrons. Therefore, it has 20 electrons.

C

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48 Chem 161-2011 Final exam B&O CHAPTER 2 ATOMS AND THE PERIODIC TABLE Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes (H&P Chapter 2) Lithium forms compounds which are used in dry cells and storage batteries and in high-temperature lubricants. It has two naturally occurring isotopes, 6Li (isotopic mass = 6.015121 u) and 7Li (isotopic mass = 7.016003 u). Lithium has an atomic mass of 6.9409 u. What is the percent abundance 6Li?

A. 92.50% B. 86.66% C. 46.16% D. 7.504% E. 6.080%

(fraction[1] x value[1]) + (fraction[2] x value[2]) = weight average 6Li 7Li = Li 6.015121 u 7.016003 u 6.9409 u Total 6Li + 7Li = 100%, or 1. Let X = fraction abundance of 6Li. (X x 6.015121) + ((1 X) x 7.016003) = 6.9409 X = 0.0750 = 7.50%

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Chem 161-2011 Exam I Burdge and Overby Chapter II Atoms and the Periodic Table Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes Correlates to Hill & Petrucci Chapter 2 24. How many protons, neutrons and electrons are in the nucleus of 89Y3+?

Protons Neutrons Electrons A. 39 36 50 B. 36 50 39 C. 42 50 39

D. 39 50 36 E. 50 39 36

Y-89 has an atomic number of 39, and therefore 39 protons. The mass number, which is the sum of the neutrons and protons, is 89. Therefore, it has 50 neutrons. Since it has a +3 charge, it must have only 36 electrons (39 protons + 36 electrons = +3).

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Chem 161-2011 Exam I Burdge and Overby Chapter II Atoms and the Periodic Table Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes Correlates to Hill & Petrucci Chapter 2 9. A hypothetical element Q (atomic mass: 186.207 u) consists of two stable isotopes: 63.201% of isotope A and 36.799% of isotope B. If the isotopic mass of A is 185.924 u, what is the isotopic mass of B? A. 187.3 u B. 186.7 u C. 187.6 u D. 186.1 u E. 187.9 u QA + QB = Mixture of QA + QB 63.201% 36.799% AW=185.924 u ?AW AW=186.207u (fraction[1] x value[1]) + (fraction[2] x value[2]) = weight average (0.63201 x 185.924) + (0.36799 x X) = 186.207 X = 186.693 u

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1 CHEM 161-2010 FINAL EXAM CHAPTER 2 - ATOMS, MOLECULES AND IONS Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes A hypothetical element Q (atomic mass: 75.5492 u) consists of two stable isotopes: A (isotopic mass: 74.6745 u) and B (isotopic mass: 76.1238 u). Calculate the percent natural abundance of isotope A. A. 27.43 % B. 72.84 % C. 39.65 % D. 65.29 % E. 51.73 % A B = Q 74.6745 u 76.1238 u 75.5492 u Fraction: X (1-X) 75.5492 u is the weight average mass of isotopes A and B. We can apply the weight with percentage or fractions. Well use fractions. (74.6745 x X) + (76.1238 x (1-X)) = 75.5492 X = 0.3965 = 39.65% = Isotope A. This answer makes sense because 75.5492 means that there is less of isotope A than isotope B.

Chem 161-2010 Exam I Hill & Petrucci et al. 4th edition Chapter 2 Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 10. How many protons, neutrons and electrons are in the 103Rh atom? A. 45 protons, 45 neutrons and 58 electrons B. 58 protons, 45 neutrons and 58 electrons C. 58protons, 58 neutrons and 45 electrons D. 45 protons, 58 neutrons and 45 electrons E. 45 protons, 58 neutrons and 58 electrons The atomic number of Rh is 45. Therefore, this uncharged atom has 45 protons. Since it is an atom, and not an ion, then the number of protons and electrons must be the same. Therefore, Rh has 45 electrons. The upper left hand corner of the Rh symbol is the mass number, which is the sum of the protons and neutrons. Therefore, Rh has 58 neutrons.

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Chem 161-2010 Exam I Hill & Petrucci et al. 4th edition Chapter 2 Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 25. A hypothetical element P has an atomic mass of 27.857 u and consists of 3 isotopes: Isotope Q (26.9886 u), isotope R (27.9986 u) and isotope S (28. 8769 u). If the percent natural abundance of isotope Q is 45.67%, what is the percent natural abundance of isotope S? A. 45.6 % B. 36.1 % C. 27.3 % D. 52.0 % E. 17.8 % P = 27.857 Q R S 26.9886 27.9986 28.8769 0.4567 X Y X + Y = 1-0.4567 X + Y = 0.5433 Y = 0.5433 X Q R S 26.9886 27.9986 28.8769 0.4567 X 0.5433-X (0.4567 x 26.9886) + (X x 27.9986) + ((0.5433-X) x 28.8769) = 27.857 X = 0.179 0.5433-0.179 = 0.364 = 36.4% S

Chem 161-2009 Final exam Hill, Petrucci et al. Chapter 2 Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 5. Silicon has three major isotopes: 28Si (isotopic mass 27.9769), 29Si (isotopic mass 28.97649) and 30Si (isotopic mass 29.97376, fractional abundance 3.10%). What is the fractional abundance of 29Si? A. 82.0% B. 18.0% C. 3.57% D. 4.72% E. 95.3% (fraction[1] x value[1] ) + (fraction[2] x value[2]) = weight average atomic weight

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28Si 29Si 30Si WtavgSi mass=27.9769 mass=28.97649 mass=29.97376 mass=28.0855 Let X = fractional abundance of 29Si. 0.0310 = fractional abundance of 30Si. Therefore, 0.9690 X = fractional abundance of 28Si. 28Si 29Si 30Si WtavgSi mass=27.9769 mass=28.97649 mass=29.97376 mass=28.0855 0.9690-X X 0.0310 ((0.9690-X) x 27.9769) + (X x 28.97649) + (0.0310 x 29.97376) = 28.0855 X = 0.0467 = 4.67% This low percentage is consistent with the weight average atomic mass being so close to the atomic mass of 28Si, and so far from the atomic mass of 29Si. Chem 161-2009 Final exam Hill, Petrucci et al. Chapter 2 Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 42. How many protons, neutrons and electrons are there in 119Sn2+? p n e-

A 50 69 48 B 52 50 69 C 50 48 69 D 48 69 50 E 50 69 52 The atomic number of Sn is 50. That means it has 50 protons. The mass number, 119, is the sum of the protons and neutrons. Therefore this substance has 69 neutrons. The charge is 2+; therefore, two electrons are missing; therefore, there are 48 electrons.

Chem 161-2009 Exam I Hill, Petrucci et al. Chapter 2 Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 2. How many protons, neutrons and electrons are in the 85Rb atom? A. 48 protons, 48 neutrons and 37 electrons B. 37 protons, 37 neutrons and 48 electrons C. 48 protons, 37 neutrons and 48 electrons

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D. 37 protons, 48 neutrons and 37 electrons E. 37 protons, 48 neutrons and 48 electrons The atomic number is 37; therefore, it has 37 protons. Since the atom is uncharged, it also has 37 electrons. Since the mass number is the sum of the protons and neutrons, then the number of neutrons is (85-37=) 48. Therefore, this atom has 37 protons, 48 neutrons and 37 electrons. Chem 161-2009 Exam I Hill, Petrucci et al. Chapter 2 Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 16. Europium has two stable isotopes: 151Eu with isotopic mass 150.92 u and 153Eu with mass 152.92. What is the natural abundance of 153Eu? A. 25.0% B. 47.5% C. 37.5% D. 52.0% E. 65.0% Weighted atomic mass of Eu = 151.96 g/mol (fraction x mass) + (fraction x mass) = weighted average atomic mass ((1-X) x (150.92)) + ((X) x (152.92)) = 151.96 X = 0.520 = 52.0%

Chem 161-2008 Final exam Hill, Petrucci et al. Chapter 2 - Atoms, Molecules, and Ions Nature of the atom (protons, neutrons, electons, symbols, isotopes, etc.)

2. How many protons, electrons and neutrons are in the 47Ti2+ion? protons electrons neutrons A 22 22 23 B 20 22 25

C 22 20 25 D 22 24 25 E 25 20 22 The atomic number for titanium is 22. Therefore, it has 22 protons. 47 is the mass number; therefore this titanium isotope contains 25 neutrons. Since the

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charge is 2+, then titanium has two more protons than electrons; therefore it has 20 electrons. C. Protons: 22; electrons: 20; neutrons: 25. ---------------------------------------------------------------------------------------------

Chem 161-2008 Final exam Hill, Petrucci et al. Chapter 2 - Atoms, Molecules, and Ions Nature of the atom (protons, neutrons, electons, symbols, isotopes, etc.) 9. Lithium forms compounds which are used in dry cells and storage batteries and in high-temperature lubricants. It has two naturally occurring isotopes, 6Li (isotopic mass = 6.015121 u) and 7Li (isotopic mass = 7.016003 u). Lithium has an atomic mass of 6.9409 u. What is the percent abundance 6Li? A. 92.50% B. 86.66% C. 46.16% D. 7.503% E. 6.080% 6Li 7Li WtAvgLi AW = 6.015121 u AW = 7.016003 u 6.9409 u ?% Let X = fraction of 6Li. Let 1-X = fraction of 7Li (X x 6.015121) + ((1-X) x (7.016003)) = 6.9409 ((((X x 6.015121) + ((1-X) x (7.016003))) = (6.9409)),X) X = 0.0750 = 7.50% D. 7.503% ---------------------------------------------------------------------------------------------

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Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes

Which of the following contains the fewest neutrons?

A. +24120Ca B. Ge7032

C. +Ti4322 D. Sc4121 E. Ca4120 The number of neutrons is the difference between the mass number (number of protons + neutrons) and the atomic number (number of protons). A. 41-20 = 21 neutrons B. 70-32 = 38 neutrons C. 43-22 = 21 neutrons D. 41-21 = 20 neutrons E. 41-20 = 21 neutrons

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Which of the following correctly matches the number of protons, neutrons and electrons in the atom or ion?

p n e-

A. 122Sb3- 51 71 54 B. 27Al3+ 13 14 16 C. 40Ca 40 20 20 D. 40Ar 18 40 18 E. 56Fe2+ 28 30 26

A. 122Sb3-: 51protons; 122-51 = 71 neutrons; 51+3 = 54 electrons B. 27Al3+: 13 protons; 27-13=14 neutrons; 27-3 = 24 electrons C. 40Ca: 20 protons; 40-20=20 neutrons; 20 electrons D. 40Ar: 18 protons; 40-18=22 neutrons; 18 electrons E. 56Fe2+: 26 protons; 56-26=30 neutrons; 26-2 = 24 electrons

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Gallium has two naturally occurring isotopes, one of which is 69Ga with atomic mass 68.93, and natural abundance 60.3%. What is the mass of the second isotope?

A. 69.93u B. 69.33u C. 71.72u D. 70.53u E. 70.92u

The other isotope must be 39.7%. Look up the weight average atomic weight of gallium in the attached periodic table. 69Ga xGa Ga 68.93g/mol ?g/mol 69.723g/mol = wt. average AW 0.603 0.397 (fraction x AW) + (fraction x AW) = weight average AW (0.603 x 68.93) + (0.397 x X) = 69.723 X = 70.93g/mol Hence, 60.3% of the 68.93g/mol Ga isotope mixed with 39.7% of the 70.93g/mol Ga isotope provides a weight average of the two isotopes of 69.723 g/mol.

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37. Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 2 - ATOMS, MOLECULES AND IONS Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes Chlorine has 2 natural isotopes: 3517Cl and 3717Cl. What is the percentage of 3517Cl in naturally occurring chlorine? A. 10% B. 22% C. 50 % D. 78 % E. 90% The weight average atomic weight of Cl is 35.453. (AW[1]x percentage[1]) + (AW[2] x percentage[2] = weight average atomic weight) X = percentage of Cl AW 35; 1-X = percentage of Cl AW 37. Note X + (1 X) = 1 = 100% (35 x X) + (37 x (1 X)) = 35.453 X = 77%

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28. Chem 161-2007 Final exam Hill, Petrucci et al. CHAPTER 2 - ATOMS, MOLECULES AND IONS Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes How many protons and neutrons are in the nucleus of 88Sr ? A. 19 protons and 69 neutrons B. 24 protons and 64 neutrons C. 38 protons and 50 neutrons D. 33 protons and 55 neutrons E. 44 protons and 44 neutrons Sr has the atomic number, 38, which is the number of protons in the Sr nucleus. 88 is the mass number, which is the sum of the protons and neutrons in the Sr nucleus. Therefore, there are 50 neutrons in the Sr nucleus.

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Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 2 Atoms, molecules and ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes

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12. Antimony consist of two stable isotopes: 57.25% of one isotope and 42.75% 123Sb. If the isotopic mass of 123Sb is 122.904u what is the isotopic mass of the other natural isotope? A. 121.0 u B. 120.8 u C. 123.0 u D. 121.9 u E. 120.3 u Answer: 120.9 u. Possible answers: B. 120.8u; A. 121.0u Sb 123Sb Total: 121.760g 0.5725g 0.4275g X 122.904u (0.5725 x X) + (0.4275 x 122.904) = 121.760 X = 120.906 u One answer option of 120.8; one answer option of 121.0. Hence, the correct answer is not provided. Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 2 Atoms, molecules and ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 14. All atoms of the same element have the same A. number of protons B. mass number C. number of neutrons D. number of protons and neutrons E. number of neutrons and electrons A. The atom and the number of its protons are inextricably intertwined; hence, every atom has one specific number of protons. B. The mass number is the sum of the neutrons and protons. The neutrons can vary in any atom. C. The neutrons can vary in any atom. D. Since the neutrons can vary in any atom, then the number of protons and neutrons can vary. E. Both the neutrons and electrons can vary in any atom. Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 2 Atoms, molecules and ions

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Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes 18. How many neutrons are in a single atom of carbon-14? A. 8 B. 7 C. 6 D. 9 E. 14 Carbon -14 may be written as: 146C. 14 is the mass number, which means the sum of the protons and neutrons. 6 is the number of protons. This means that C-14 has 6 protons and 8 neutrons.

PERIODIC TABLE (Also, see Chapter 7 - Periodic Table Trends)

Chem 161-2011 Exam I Burdge and Overby Chapter II Atoms and the Periodic Table Periodic table Correlates to Hill & Petrucci Chapter 2 25. Which of the following elements is an alkaline earth metal? A. Cu B. Rb C. Ce D. Sn E. Ba Barium (Ba) is in group II, which is the alkaline earth metals.

Chem 161-2010 Exam I Hill & Petrucci et al. 4th edition Chapter 2 Periodic Table 2. Which element is a transition metal? A. Ba B. Sm

C. Pb D. Sb E. Ag

All of the elements between groups IIa and IIIa are transition metals. This includes Ag.

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Periodic table (Also, see Chapter 7 - Periodic Table Trends)

Which of the following represents a metal, a nonmetal and a metalloid?

Metal Nonmetal Metalloid

A. V Cl Ge B. B Ca Si C. Ra S In D. Hf Pb Sb E. Ge Po I

A. Vanadium is a metal; Cl is a non-metal; Ge is a metalloid. B. Boron is a metalloid; Ca is a metal; Si is a metalloid. C. Radium is a metal; S is a non-metal; In is a metal. D. Hf is a metal; Pb is a metal; Sb is a metalloid. E. Ge is a metalloid; Po is a metalloid; I is a non-metal.

Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 2 Atoms, molecules and ions Periodic table 1. Which element is a transition metal? A. Na B. Al C. Pb D. Be

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E. Fe The transition metals are the b group metals between groups IIa and IIIa. Na is in group Ia; Al is in group IIIa; Pb is in group IVa; Be is in group IIa; Fe is in group VIIIb; therefore it is a transition metal.

ATOMS, IONS, MOLES AND MOLECULAR WTS.

21 Chem 161-2011 Final exam B&O CHAPTER 2 ATOMS AND THE PERIODIC TABLE Moles, mass and molar mass (H&P Chapter 3A) If a 100 gram sample of a compound is found to contain 8 percent hydrogen by mass, which one of the following statements is correct?

A. A. 50 gram sample of the same compound would contain 4 percent hydrogen by mass B. A 10 gram sample of the same compound would contain 8 grams of hydrogen. C. 80 grams of the compound would contain 10 grams of hydrogen. D. A 40 gram sample of the compound would contain 8 percent hydrogen by mass. E. 40 grams of this compound would contain 4 grams of hydrogen.

Sample H 100g 8g A. False. A 50g sample would contain 4gH, which is still 8% H. B. False. A 10g sample would contain 0.8gH. C. False. 80g of sample would contain 6.4 g H. D. True. A 40 g sample would contain 3.2g H, which is still 8 percent. E. False. A 40 g sample would contain 3.2 g H.

D

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45 Chem 161-2011 Final exam B&O CHAPTER 2 ATOMS AND THE PERIODIC TABLE Moles, mass and molar mass (H&P Chapter 3A) What mass of calcium citrate, Ca3(C6 H5 O7 )2, (Molar Mass = 498.44 g/mol) is required to furnish 1.25 g of calcium? A. 5.18 g B. 3.02 g C. 15.5 g D. 8.75 g E. 3.63 g Ca3(C6 H5 O7 )2 3Ca2+ + 2(C6 H5 O7 )3- Xg 1.25g MW = 498.44gmol-1 Plan: gca molca molcacit gcacit (1.25gCa2+/40.08gmol-1) x (1molcacit/3molca2+) x 498.44gcacitmol-1cacit = 5.18g calcium citrate

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11 Chem 161-2011 Final exam

B&O CHAPTER 2 ATOMS AND THE PERIODIC TABLE Moles, mass and molar mass (H&P Chapter 3A)

To produce Mn(ClO4)x, 3.62 g of ClO4- are needed for every 1.00 g of Mn. What is x?

A. 1 B. 2 C. 3 D. 4 E. 5 Mn(ClO4)x Mnx+ + XClO41- 1.00g 3.62g AW Mn = 54.95g/mol mol = 1.00g/54.95gmol-1 = 0.0182 mol MW ClO41- = 35.5 + 64 = 99.5 g/mol mol ClO4- = 3.62g/99.5gmol-1 = 0.0364 mol Ratio of moles of Mnx+ to ClO4- = 0.0182:0.0364 = 2.0000 Hence, X = 2

B

Chem 161-2011 Exam II

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Burdge and Overby, Chapter V Ionic and Covalent Compounds Avogadros number and molar masses (H&P chapter 3A)

10. Which one of the following statements correctly describes CO2? A. The mass of one mole of CO2 is 44.0 kg. B. The mass of one mole of CO2 is 44.0 u. C. 44.0 grams of CO2 contains 16.0 grams of oxygen. D. The mass of one molecule of CO2 is 7.31 x 10-23 grams. E. The mass of one molecule of CO2 is 44.0 grams. A. False. The mass of one mole of CO2 is 44.0 g. B. False. The mass of one mole of CO2 is 44.0 g, not 44.0 amus. 44.0 amus = 7.3 x 10-23g C. False. 44.0 grams of CO2 = 1 mol of CO2.

CO2 C + 2O 2O = 32.0g

D. True. Plan: Moleculeco2 molco2 gco2 1moleculeco2 x (1molco2/(6.022x1023moleculeco2) x (44.0gco2/molco2) = 7.31 x 10-23 g CO2 E. False. The mass of one mole of CO2 is 44.0 grams. The mass of one molecule of CO2 is 7.31 x 10-23 g. Chem 161-2011 Exam II Burdge and Overby, Chapter V Ionic and Covalent Compounds

Avogadros number and molar masses (H&P chapter 3A) 11. If 5.00 g of compound X2CO3 contains 0.0181 mol, what element is X? A. Na B. K C. Ag D. Rb E. Cu mol = g/MW MW = g/mol = 5.00g/0.0181mol = 276.24g = MW of X2CO3 Mass of CO32- = 12.01 + (3 x 16.00) = 60.01g 2X + 60.01 = 276.24g X = 108.11 g Atomic mass Na = 22.99 K = 39.10 Rb = 85.47 Ag = 107.9 Cu = 63.55 X = Ag

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Chem 161-2011 Exam II Burdge and Overby, Chapter V Ionic and Covalent Compounds

Avogadros number and molar masses (H&P chapter 3A) 13. How many oxygen atoms are in 50.0 g of magnesium nitrate? A. 1.22 x 1024 B. 2.03 x 1023 C. 1.02 x 1023 D. 4.06 x 1023 E. 2.44 x 1024 The molar mass of Mg(NO3)2 is 24.31 + (2 x (14.01 + 48.00)) = 148.33 Mg(NO3)2 Mg + 2N + 6O Plan: gmg(no3)2 molmg(no3)2 moleculesmg(no3)2 atomso 50.0g Mg(NO3)2/(148.33g Mg(NO3)2/mol Mg(NO3)2) x (6.022 x 1023molecules Mg(NO3)2/mol Mg(NO3)2) x (6 atoms Mg(NO3)2/molecule Mg(NO3)2) = 1.22 x 1024 oxygen atoms Chem 161-2011 Exam II Burdge and Overby, Chapter V Ionic and Covalent Compounds

Avogadros number and molar masses (H&P chapter 3A) 15. 1.0 mol of CH4 and 0.50 mol of C2H6 have the same: A. mass B. number of molecules C. number of carbon atoms D. number of hydrogen atoms E. number of total atoms A. False. The molar mass of CH4 is 16. The molar mass of C2H6 is 30. Therefore, 1 mole of CH4 is 16g, and 0.50 mol of C2H6 is 15g. B. False. There are 6.022 x 1023 molecules per mole. Therefore, 1.0 mol of CH4 is 6.022 x 1023 molecules; 0.50 mol of C2H6 is 3.011 x 1023 molecules. C. True. 1.0 mol of CH4 has 6.022 x 1023 molecules, which contain 6.022 x 1023 carbon atoms. 0.50 mol of C2H6 has 3.011 x 1023 molecules. Since each molecule has two carbon atoms, then 0.50 mol of C2H6 has 6.022 x 1023 carbon atoms. D. False. 1.0 mol of CH4 has 6.022 x 1023 molecules, and correspondingly 24.088 x 1023 hydrogen atoms. 0.50 mol of C2H6 has 3.011 x 1023 molecules, and correspondingly 18.066 x 1023 hydrogen atoms.

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E. False. CH4 1C + 4H C2H6 2C + 6H CH4: 1.0 mol x (6.022x1023molecules/mole) x (5atom/molecule) = 3.01 x 1024 atoms C2H6: 0.50 mol x (6.022x1023molecules/mole) x (8atom/molecule) = 2.41x 1024 atoms Chem 161-2011 Exam II Burdge and Overby, Chapter V Ionic and Covalent Compounds

Avogadros number and molar masses (H&P chapter 3A) 22. What is the mass of aluminum in 60.0 g of aluminum sulfide? A. 10.8 g B. 13.4 g C. 17.8 g D. 26.7 g E. 21.6 g Aluminum sulfide = Al2S3 MW Al2S3 = (2 x 26.98) + (3 x 32.07) = 150.17 Al2S3 2Al + 3S Plan: gal2s3 molal2s3 molal gal 60.0g Al2S3/(150.17gAl2S3/molAl2S3) x (2molAl/1molAl2S3) x (26.98gAl/molAl) = 21.6g Al

Chem 161-2011 Exam I Burdge and Overby, Chapter V Ionic and Covalent Compounds Atoms, ions, moles and molecular weights Correlates to Hill & Petrucci Chapter 3A 19. How many magnesium atoms can be contained in a magnesium cube with an edge measuring 0.103 feet? Density of magnesium = 1.738 g/cm3. 1 inch = 2.54 cm. A. 1.33 x 1024 atoms B. 9.25 x 1021 atoms C. 1.43 x 1021 atoms D. 2.06 x 1023 atoms E. 5.24 x 1023 atoms Edge = 0.103 feet Density = 1.738g/cm3 ?Mg atoms Plan: feet inches centimeters cm3 grams moles atoms

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0.103ft x (12in/ft) x (2.54cm/in) = 3.139 cm (3.139 cm)3 x 1.738g/cm3 x (1mol/24.305g) x (6.022x1023atom/mol) = 1.33 x 1024 atoms

Chem 161-2011 Exam I Burdge and Overby, Chapter V Ionic and Covalent Compounds Atoms, ions, moles and molecular weights Correlates to Hill & Petrucci Chapter 3A 11. What is the mass (in mg) of 7.32 x 1019 atoms of Ga? A. 42.5 mg B. 85.0 mg C. 0.122 mg D. 8.47 mg E. 7.70 mg Plan: atoms mol g mg

7.32 x 1019 atoms x (1mol/(6.022x1023atom)) x 69.723g/mol x (1000 mg/g) = 8.48 mg

54 CHEM 161-2010 FINAL EXAM

CHAPTER 3A STOICHIOMETRY: CHEMICAL CALCULATIONS Atoms, ions, moles and molecular wts. What is the mass of 0.500 mol of sodium in atomic mass units? A. 6.93 u B. 11.5 u C. 0.500 u D. 6.93 x 1024 u E. 3.01 x 1023 u 1 mole Na = 22.99g Therefore, 1 atom Na = 22.99 amus (which = 22.99 u). 6.022 x 1023 atoms/mole Therefore, 0.5 mol Na = 3.011 x 1023 atoms Na (22.99 amu/1 atom Na) x (3.011 x 1023 atoms Na) = 6.92 x 1024 amu = 6.92 x 1024 u This problem could also be solved knowing the relationship between atomic mass units and grams. One atomic mass unit = the mass of 1 atom of protium (the hydrogen isotope with no neutrons) = 1.66 x 10-24 g. 0.500 mol Na = 11.5g. 11.5g x (1amu/1.66x10-24g) = 6.93x1024u.

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11 CHEM 161-2010 FINAL EXAM CHAPTER 3A STOICHIOMETRY: CHEMICAL CALCULATIONS Atoms, ions, moles and molecular wts. B&O Chapter 5 Calculate the number of oxygen atoms that are present in 34.7 g of magnesium nitrate pentahydrate. A. 1.06 x 1023 B. 5.30 x 1023 C. 7.54 x 1023 D. 1.23 x 1023 E. 9.64 x 1023

Mg(NO3)25H2O 1Mg + 1N + 11O + 10H 34.7g ?Oatoms

Plan: gMg(NO3)25H2O molMg(NO3)25H2O molO atomsO 24.3 + (2 x 14.0) + (11 x 16.0) + (10 x 1.01) = 238.4 gMg(NO3)25H2O/molMg(NO3)25H2O 34.7g Mg(NO3)25H2O /(238.4gMg(NO3)25H2O/1mol Mg(NO3)25H2O) x (11molO/1molMg(NO3)25H2O) x (6.022x1023atomsO/molO) = 9.64x1023 O atoms

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45 CHEM 161-2010 FINAL EXAM CHAPTER 3A STOICHIOMETRY: CHEMICAL CALCULATIONS Atoms, ions, moles and molecular wts. A sample of 0.3127 mole of the compound Li2ZO4 has a mass of 41.53 g. The element Z is: A. Fe B. I C. Cr D. Mn E. Br Li2ZO4 0.3127 mol 41.53g ? Identification of Z MW = g/mol (41.53gLi2ZO4/0.3127molLi2ZO4) = 132.81gLi2ZO4/molLi2ZO4 ((2 x 6.94) + Z + (4 x 16.00)) = 132.81g/mol Z = 54.93g/mol Fe = 55.85 g/mol I = 126.90 g/mol Cr = 52.00 g/mol Mn = 54.94 g/mol Br = 79.90 g/mol

Chem 161-2010 Exam I Hill & Petrucci et al. 4th edition Chapter 3A Atoms, ions, moles and molecular weights 1. How many molecules of propanol are present in 55.8 mL of propanol (density = 0.8034 g/mL)? A. 4.49 x 1023 B. 7.56 x 1023 C. 2.18 x 1023 D. 9.07 x 1023 E. 4.54 x 1023 Plan: mL g mol molecules 55.8mLP x (0.8034gP/mLP) x (1molP/60gP) x (6.022x1023moleculesP/molP)

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= 4.50x1023 molecules P

Chem 161-2010 Exam I Hill & Petrucci et al. 4th edition Chapter 3A Atoms, ions, moles and molecular weights 4. How many moles of Cu(NO3)2 would contain 7.86 x 1021 atoms of O? A. 6.54 x 103 B. 8.73 x 103 C. 2.18 x 103 D. 7.54 x 103 E. 4.36 x 103 Cu(NO3)2 Cu + 2N + 6O atomso molo molcu(no3)2 (7.86 x 1021atomsO/6.022 x 1023atoms/mol) = 0.01305 mol O. 0.01305molO x (1 mol Cu(NO3)2/6mol O) = 0.00218 mol Cu(NO3)2.

Chem 161-2009 Exam I Hill, Petrucci et al. Chapter 3A Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. 13. The mass of 0.1630 mol sample of M(HSO4)2 is 45.91g. Which of the following elements is M? A. Cr B. Ca C. Ni D. Cu E. Sr mol = g/MW MWM(HSO4)2 = g/mol = 45.91g/0.1630mol = 281.7 g/mol The MW of the (HSO4)2 part of the molecule is 194 g/mol. Therefore, M is 281.7 194 = 87.7g/mol, which is close to Sr (87.6g/mol). Chem 161-2009 Exam I Hill, Petrucci et al. Chapter 3A Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. 14. What is the total number of N atoms present in 37.6 g of NH4NO3?

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A. 5.66 x 1023 B. 2.79 x 1023 C. 1.35 x 1023 D. 1.12 x 1024 E. 3.57 x 1023 NH4NO3 2N 37.6g ?atoms gnh4no3 molnh4no3 moln atomsn (37.6g/(80.06g/mol)) x (2 molN/1molNH4NO3) x (6.022 x 1023atomsN/molN) = 5.66x1023 atoms N Chem 161-2009 Exam I Hill, Petrucci et al. Chapter 3A Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. 19. What is the mass % of C in propanoic acid. Propanoic acid is C3H6O2. A. 18.72% B. 24.67% C. 16.21% D. 35.21% E. 48.64% H H : O : | | . . H C C C O H | | . . H H Propanoic acid is C3H6O2 C3H6O2 3C Arbitrarily, assume there is one mole of C3H6O2. One mole of C3H6O2 = 74.09g Plan: gC3H6O2 molC3H6O2 molC gC 74.09gC3H6O2/74.09gmol-1) x (3molC/1molC3H6O2) x (12.01gC/molC) = 36.03gC (36.03gC/74.09gC3H6O2) x 100 = 48.6 mass % C

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12 Chem 161-2008 Exam I Hill, Petrucci et al. CHAPTER 3A STOICHIOMETRY: CHEMICAL CALCULATIONS

Atoms, ions, moles and molecular wts. How many sulfur atoms are there in 0.566g of Na2S2O3?

A. 1.20 1024 atoms B. 2.16 1021 atoms C. 4.32 1021 atoms D. 0.00716 atoms E. 2 atoms

MW Na2S2O3 = 158.12gNa2S2O3/molNa2S2O3 Na2S2O3 2Na + 2S + 3O 0.566g ?atoms Plan: gNa2S2O3 mol Na2S2O3 molS atomsS 0.566gNa2S2O3 x (1molNa2S2O3/158.12gNa2S2O3) x (2molS/1molNa2S2O3) x 6.022x1023atomsS/molS = 4.31x1021atomsS

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Atoms, ions, moles and molecular wts. What is the mass % of N in ammonium nitrate?

A. 17.5% B. 21.2% C. 35.0% D. 53.8% E. 22.2%

MWNH4NO3=80.06g/mol AW N = 14.01g/mol Plan: Begin with any quantity of NH4NO3 and determine what percent of that is nitrogen. Lets arbitrarily begin with 100g NH4NO3. Plan: gNH4NO3 molNH4NO3 molN gN NH4NO3 2N + . . . 100g ?g 100gNH4NO3 x (1molNH4NO3/80.06gNH4NO3) x (2molN/1molNH4NO3) x (14.01gN/molN) = 35.0gN (35.0gN/100gNH4NO3) x 100 = 35.0% N in NH4NO3

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Atoms, ions, moles and molecular wts.

What is the mass of MgSO47H2O (Molar Mass = 246.3 g/mol) that should decompose to yield 54.0 g of water?

A. 125g B. 145g C. 246g D. 75.2g E. 105g

MgSO47H2O 7H2O ?g 54.0g Plan: gH2O molH2O mol MgSO47H2O g MgSO47H2O 54.0g H2O x (1molH2O/18.02gH2O) x (1mol MgSO47H2O/7mol H2O) x (246.3g MgSO47H2O/mol MgSO47H2O) = 105.4g MgSO47H2O Answer: E


Atoms, ions, moles and molecular wts.

Which one of the following is true regarding the mass of one molecule of H2O

Mass (u) Mass (g) A. 18 18 B. 18 1 C. 1.11025 1.110-23

D. 18 310-23 E. 3.010-23 18

The mass of one molecule of water is frequently referred to in atomic mass units (amus), also called Daltons (u). Since there is 18 g H2O/mole H2O, then there is 18 u H2O/molecule H2O. The actual mass of one molecule of H2O may be calculated: 18g/mol x (1mol/6.022x1023 molecules) = 2.99x10-23g/molecule Answer: D

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Chem 161-2007 Exam I Hill, Petrucci et al., 4th edition Chapter 3A Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular weights 22. How many sodium atoms are there in 6.0g of Na3N? A. 3.6 x 1024 atoms B. 1.3 x 1023 atoms C. 0.22 atoms D. 4.6 x 1022 atoms E. 0.072 atoms Na3N 6g MW Na3N: (3 x 22.990) + 14.01 = 82.98g/mol Plan: massna3n molna3n molna atomsna Na3N 3Na 6g Na3N/(82.98g Na3N/mol Na3N) x (3 mol Na/1 mol Na3N) x (6.022 x 1023 atom Na/mol Na) = 1.31 x 1023 atom Na

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