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Chapter 3 Exam Review KEY

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Chapter 3 Exam Review KEY Chapter 3 Exam Review KEY I. Atomic Theory 1. Explain the law of conservation of mass in your own words. F I V E MOD ELS OF T H E ATOM Matter is neither created nor destroyed in a chemical reaction. 2. What were Rutherford’s two main conclusions about the structure of the atom? List the piece of evidence NUCLEAR MODEL: The atom can be SOthatLID SP supportsHERE MODE EACHL: The conclusion.atom is a divided into a nucleus and electrons. The solid sphere that cannot be divided up into nucleus occupies a small amount of space at smaller particles or pieces. the center of the atom. The nucleus is dense Evidence: Most of the α-particles passed right through the gold foil without deflection. and positively charged. The electrons circle around the nucleus. The electrons are tiny Conclusion: Most of an atom is composed of empty space. and negatively charged. Most of the atom is empty space. Evidence: Some were deflected somewhat, fewer were greatly deflected, and a very few Negative bounced back toward the source. electron Positive Conclusion: There is a dense positive nucleus. nucleus 3. Draw and describeLBCTCM_01_037 the atom model of each of the following scientists. a. Dalton PROTON MODEL: The atom can be divided PLUM PUDDING MODEL: The atomDalton can thought the atom was a solid sphere that was indivisible. into a nucleus and electrons. The nucleus be divided into a fluid (the “pudding”) and occupies a small amount of space at the electrons (the “plums”). The fluid spreads The Atomic Model Through Time center of theLBCTCM_01_036 atom. The nucleus consists of out in the atom and is positively charged. protons that are positively charged. The The electrons are very tiny and negatively electrons circle the nucleus. The electrons are charged.b. Most Thomson of the atom is made of fluid. tiny and negatively charged. Most of the atom Thomson discovered the electron and, consequently, the proton. Thomson’s is empty space. Positive model of the atom is called the plum pudding model. He though that most of the Solid sphere model fluidPlum pudding model Nuclear model Dalton, 1803 Thomson,atom 1897 wasRutherford, made 1911 of positively charged fluid. He thought that electrons were tiny Negative specs of negatively charged particles throughout the positive fluid. electron c. Rutherford Solar system model RutherfordProton model discovered the nucleus of the atom. His model is called the nuclear Bohr, 1913 model.Rutherford, He1918 broke the atom into two regions the nucleus and the electron cloud. SOLAR SYSTEM MODEL: The atom can He didn’t reallyElectrons know what he meant by the electron cloud; he just knew that the be divided into a nucleusLBCTCM_01_039 and electrons. electrons make up little mass of the atom, but most the atom’s volume. His The nucleus is at the center of the atom and model mainlyNucleus focuses on the nucleus of the atom. He describes the nucleus of LBCTCM_01_038 is positively charged. The electrons circle the atom as a positively charged dense region that contains most of the atoms around the nucleus in specified orbits. The electrons are tiny and negatively charged. mass but hardly any of the atom’s volume. Different electrons are in orbits at different Cloud Model, 1932 Simple Atomic Model Living By Chemistry Teaching and Classroom Resources Unit 1 Alchemy 45 distances from the nucleus.II. Atoms, Most of the Ions, atom¥7( &REEMAN IsotopesANDLBCTCM_01_033#OMPANY"&7 LBCTCM_01_032 LBCTCM_01_031Lesson 11 s4RANSPARENCY is empty space. LBCTCM_01_035LBCTCM_01_035aLBCTCM_01_034 LBCTCM_01_BFW1st12.indd 45 2/25/12 12:30:06 PM 4. Based onLBCTCM_01_035b the information in the table determine if it is describing an atom and ion. Negative electron Number of Number of Number of Element Atom or Ion Positive Protons Electrons Neutrons nucleus Helium 2 2 2 Atom Helium 2 3 3 Ion Living By Chemistry Teaching and Classroom Resources Cobalt Unit27 1 Alchemy 2543 33 Ion ¥7(&REEMANAND#OMPANY"&7 Lesson 11 s(ANDOUT Cobalt 27 27 33 Atom Chlorine 17 18 18 Ion LBCTCM_01_BFW1st12.inddLBCTCM_01_040 43 2/25/12 12:30:03 PM Chlorine 17 18 18 Ion 5. From the table above. List each group of isotopes and write the names in hyphen notation. Helium-4 Helium-5 (remember you find the mass # by adding the neutrons and protons) 1 6. Write the nuclear symbol for krypton atom with the atomic mass of 84 amu. �� ���� III. Subatomic Particles of Neutral Atoms Nuclear Isotope Name Atomic # (Z) Mass # (A) Protons Neutrons Electrons Symbol 17 8 O Oxygen-17 8 17 8 9 8 75 33 As Arsenic - 75 33 75 33 42 33 209 83 Bi Bismuth - 209 83 209 83 126 83 24 12 Mg Magnesium - 24 12 24 12 12 12 80 35 Br Bromine - 80 35 80 35 45 35 50 23 V Vanadium-50 23 50 23 27 23 70 31 Ga Gallium-70 31 70 31 39 31 119 50 Sn Tin-119 50 119 50 69 50 7 3 Li Lithium-7 3 7 3 4 3 14 7 N Nitrogen-14 7 14 7 7 7 40 19 K Potassium - 40 19 40 19 21 19 31 P 15 P Phosphorous-31 15 31 15 16 15 237 93 Np Neptunium-237 93 237 93 144 93 73 34 Se Selenium-73 34 73 34 39 34 58 26 Fe Iron-58 26 58 26 32 26 201 80 Hg Mercury-201 80 201 80 121 80 243 95 Am Americium-243 95 243 95 148 95 94 41 Nb Niobium-94 41 94 41 53 41 263 106 Sg Seaborgium-263 106 263 106 157 106 281 110 Ds Darmstadtium-281 110 281 110 171 110 222 86 Rn Radon-222 86 222 86 136 86 89 39 Y Yttrium-89 39 89 39 50 39 247 97 Bk Berkelium-247 97 247 97 150 97 2 IV. Average Atomic Mass Calculate the average atomic masses. Round all answers to two decimal places. 7. Calculate the average atomic mass of magnesium using the following data for three magnesium isotopes. Isotope mass (u) relative abundance Mg-24 23.985 0.7870 Mg-25 24.986 0.1013 Mg-26 25.983 0.1117 23.985 x 0.7870 = 18.87 24.986 x 0.1013 = 2.531 25.983 x 0.1117 = 2.902 18.87 + 2.531 + 2.902 = 24.31 amu 8. Calculate the average atomic mass of iridium using the following data for two iridium isotopes. Isotope mass (u) relative abundance Ir-191 191.0 0.3758 Ir-193 193.0 0.6242 191.0 x 0.3758 = 71.78 193.0 x 0.6242 = 120.5 71.78 + 120.5 = 192.3 amu . 9. Rubidium has two common isotopes, 85Rb and 87Rb. If the abundance of 85Rb is 72.2% and the abundance of 87Rb is 27.8%, what is the average atomic mass of rubidium? 85 x 0.722 = 61.4 87 x 0.278 = 24.2 61.4+ 24.2= 85.6 amu 3 10. The two naturally occurring isotopes of chlorine are chlorine-35 with a mass of 34.9689 amu and chlorine-37 with a mass of 36.9659 amu. The atomic mass of elemental chlorine on earth is found to be 35.46 amu. Calculate the percent abundance of each of the two chlorine isotopes. 4 11. Calculate the atomic mass of the unknown element. Then identify the element. Isotope Mass (amu) Percent abundance 185X 184.953 37.40 187X 186.956 62.60 184.953 x 0.3740 = 69.17 186.956 x 0.6260 = 117.0 69.17 + 117.0 = 186.2 amu Re 12. A sample of naturally occurring silicon consists Si-28 (amu = 27.9769), Si-29 (amu = 28.9765) and Si-30 (amu = 29.9738). If the atomic mass of silicon is 28.0855 and the natural abundance of Si-29 is 4.67%, what are the natural abundances of Si-28 and Si-30? SKIP THIS PROBLEM!!! 13. Uranium has three common isotopes. If the abundance of 234U is 0.01%, the abundance of 235U is 0.71%, and the abundance of 238U is 99.28%, what is the average atomic mass of uranium? 234 x 0.0001 = 0.02 235 x 0.0071 = 1.7 238 x 0.9928 = 236.3 0.02 + 1.7 + 236.3= 238.0 amu 14. Naturally occurring boron is 80.20% boron-11 (atomic mass is 11.01 amu) and 19.80% of some other isotopic form of boron. What must the atomic mass of the second isotope be if the average atomic mass of born is 10.81 amu? SKIP THIS PROBLEM!!! 15. Calculate the atomic mass of the unknown element. Then identify the element. Isotope Mass (amu) Percent abundance 113X 112.904 4.30 115X 114.904 95.70 112.904 x 0.0430 = 4.85 114.904 x 0.9570 = 110.0 4.85 + 110.0 = 114.9 amu Element is In The following equation will be provided on the exam: Calculating Average Atomic Mass % ��������� ���� % ��������� ���� + … … = ������� ������ ���� �� ������� � �� ������� � �� ������� � �� ������� � 5 .
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