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Dr. G. Pernía - Atomic Structure Review – ANSWER KEY ATOMIC PARTICLES: Define the following terms. PROTON: IDENTIFIES an element NEUTRON: Inside nucleus, stabilizes the nucleus, No charge ELECTRON: negatively charged, outside nucleus MASS NUMBER: number of particles inside nucleus: proton + neutron. ATOMIC NUMBER: The number of PROTONS in an atom ATOM: EQUAL number of electrons and protons ION: A NEGATIVELY charged particle ISOTOPE: Has a DIFFERENT number of electrons and protons therefore DIFFERENT atomic masses Answer the following What would happen to a neutral atom if We changed the amount of PROTONS? ______CREATE A NEW ELEMENT__________________ We changed the amount of NEUTRONS? _____CREATE AN ISOTOPE______________________ We changed the amount of ELECTRONS? ____CREATE AN ION___________________________ AVERAGE ATOMIC MASS 1. What is the atomic mass of hafnium if, out of every 100 atoms, 5 have a mass of 176; 19 have a mass of 177; 27 have a mass of 178; 14 have a mass of 179; and 35 have a mass of 180.0? (176 x 0.05) + (177 x 0.19) + (178 x 0.27) + (179 x 0.14) + (180 x 0.35) = 178.55 amu 2. Calculate the average atomic mass of lithium, which occurs as two isotopes that have the following atomic masses and abundances in nature: 6.017 amu with 7.30%; and 7.018 amu with 92.70%. (6.017 x 0.073) + (7.018 x 0.927) = 6.945 amu 3. Hydrogen is 99% 1H, 0.8% 2H, and 0.2% 3H. Calculate its average atomic mass. (1 x 0.99) + (2 x 0.008) + (3 x 0.002) = 1.012 amu 4. 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) + (24.986 x 0.1013) + (25.983 x 0.1117) = 24.310 amu SCIENTISTS – fill in 4 scientists’ names and complete the table. (Only TWO of the scientists have experiments!) Nuclear Model Gold Foil experiment Discovered electrons Atomic Theory SCIENTIST energy levels “Cookie Dough/Plum Pudding” positively charged nucleus EXPERIMENT Dalton CONCEPTS/IDEAS “Solar System” “Billiard Ball” Cathode Ray Tube MODEL Atomic Theory “Billiard Ball” Thomson Cathode Ray Tube Discovered electrons “Cookie Dough/Plum Pudding” Rutherford Gold Foil experiment positively charged nucleus Nuclear Model energy levels “Solar System” Bohr UNDERSTANDING ISOTOPIC NOTATION DIRECTIONS: Fill in the missing information for the given element symbols. Round Atomic Masses to the nearest whole number. SYMBOL K-40 2 Na+ Al3+ 2 # OF PROTONS ELEMENT, # OF # OF ION, or NEUTRONS ELECTRONS ISOTOPE? 40 19 21 19 isotope 48 113 48 65 48 isotope oxygen 8 18 8 10 10 ion sodium 11 23 11 12 10 ion carbon 6 14 6 8 6 isotope aluminum 13 27 13 14 10 ion sulfur 16 332 16 16 18 ion barium 56 138 56 72 56 isotope NAME ATOMIC NUMBER potassium 19 cadmium ATOMIC MASS Ba-138 MOLAR MASS CONVERSIONS How many ATOMS are in: 1. 6.15 moles of sulfur 6.15 mol S x 6.02 x 1023 atoms 1 mol S 2. 13 g of cadmium 13 g x 3. = 3.70 x 1024 atoms 1 mol Cd 112.411 g x 6.02 x 1023 atoms 1 mol Cd = 6.96 x 1022 atoms x 6.02 x 1023 atoms 1 mol Sn = 8.93 x 1023 atoms 176 g of tin 176 g x 1 mol Sn 118.71 g How many MOLES are in: 1. 891 g of sodium 891 g Na x 1 mol = 38.8 mol 22.99 g 2. 2.03 x 1022 atoms of lead 2.03 x 1022 atoms Pb x 1 mol = 0.0337 mol 23 6.02 x 10 atoms Determine the MASS of: 1. 0.380 moles of potassium 0.380 moles K x 39.098 g 1 mol K 2. 9.64 x 1025 atoms of silver 9.64 x 1025 atoms x = 14.86 g 1 mol Ag 6.02 x 1023 atoms x 107.868 g 1 mol Ag = 17,273 g 3. 46.99 x 1014 atoms of barium - report your answer in regular and scientific notation 46.99 x 1014 atoms x 1 mol Ba x 137.327 g = 1.071 x10-6 g = 6.02 x 1023 atoms 1 mol Ba = 0.000001071 g NUCLEAR REACTION EQUATIONS 1. 234 90ℎ 2. 228 89 4 2 _____0-1β___________ + 228 90ℎ 3. ___23290Th________ 4. 5. 113 48 40 19 + ____23088Ra__________ + 10 + −10 4 2 + 228 88 114 48Cd_________ 40 20Ca_______ + HALF-LIFE 1. How much of a 100.0g sample of 198Au is left after 8.10 days if its half-life is 2.70 days? A = A0 x (½)n 8.1/2.7 = 3 half-lives. A = 100 x (½)3 = 12.5 g 2. A 50.0g sample of 16N decays to 12.5g in 14.4 seconds. What is its half-life? A = A0 x (½)n, so A/ A0 = (½)n So 50/12.5 = (½)2 = 2 half-lives. So 1 half-life = 14.4/2 = 7.2 s 3. The half-life of 42K is 12.4 hours. How much of a 750g sample is left after 62.0 hours? 62/12.4 = 5 half-lives. A = A0 x (½)n So A = 750 x (½)5 = 23.4375 g 4. What is the half-life of 99Tc if a 500g sample decays to 62.5g in 639,000 years? A = A0 x (½)n 500/62.5 = (½)3 so 3 half lives 639,000/3 = 213,000 years = 1 half-life 5. There are 5.0g of 131I left after 40.35 days. How many grams were in the original sample if its half-life is 8.07 days? A = A0 x (½)n, so A0 = A / (½)n 40.35/8.07 = 5 half-lives So A = 5 / (½)5 = 160 g