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UNIT 2: ATOMIC STRUCTURE – AVERAGE ATOMIC MASS Agenda: • • • • • • Do Now Wrap up White Board Session Quick Review I do We do You do Homework: GOAL: I WILL BE ABLE TO CALCULATE THE AVERAGE ATOMIC MASS OF AN ELEMENT FROM ITS NATURALLY OCCURRING ISOTOPES? Essential Question: How does the abundance of various isotopes affect an elements atomic mass? DO NOW: If an atom has a mass number of 34 and the atom has 16 neutrons, how many protons does the atom have? None for now Monday Tuesday Wednesday Thursday Friday Average Atomic Mass Isotope Quiz Intro to Electron Configuration Electron Configuration Cont. Identifying Core and Valence Electrons Unit Review ISOTOPES • ISOTOPES-ATOMS THAT HAVE THE SAME NUMBER OF PROTONS BUT DIFFERENT NUMBERS OF NEUTRONS ARE CALLED ISOTOPES. • BECAUSE ISOTOPES OF AN ELEMENT HAVE DIFFERENT NUMBERS OF NEUTRONS, THEY ALSO HAVE DIFFERENT MASS NUMBERS. AVERAGE ATOMIC MASS •MOST ELEMENTS OCCUR NATURALLY AS MIXTURES OF ISOTOPES. •THE MASS NUMBERS ON THE PERIODIC TABLE ARE THE WEIGHTED AVERAGE OF THE MOST ABUNDANT ISOTOPES’ MASS NUMBERS. HOW TO CALCULATE AVG. ATOMIC MASS • TO CALCULATE THE ATOMIC MASS OF AN ELEMENT, MULTIPLY THE MASS OF EACH ISOTOPE BY ITS NATURAL ABUNDANCE, EXPRESSED AS A DECIMAL, AND THEN ADD THE PRODUCTS. EXAMPLE 1 • RUBIDIUM HAS TWO COMMON ISOTOPES, RUBIDIUM-85 AND RUBIDIUM-87. IF THE ABUNDANCE OF RUBIDIUM -85 IS 72.2% AND THE ABUNDANCE OF RUBIDIUM-87 IS 27.8%, WHAT IS THE AVERAGE ATOMIC MASS OF RUBIDIUM? Isotope % abundance Fraction of abundance Rb-85 72.2% 0.722 X85 0.278 X87 Rb-87 27.8% Isotope mass amu amu Isotope mass contribution to element = = 61.37 24.186 +______________ 85.556 amu EXAMPLE 2 EXAMPLE BORON HAS TWO ISOTOPES: B-10 (MASS 10.013 AMU) 19.8% ABUNDANCE B-11 (MASS 11.009 AMU) 80.2% ABUNDANCE CALCULATE THE AVERAGE ATOMIC MASS. Isotope % abundance Fraction of abundance mass Isotope mass contribution to element B-10 19.8% 0.198 x 10.013 amu = 1.983 B-11 80.2% 0.802 x 11.009 amu = + 8.829 10.812 EXAMPLE 3 CALCULATE THE AVERAGE ATOMIC MASS OF MG. ISOTOPE 1 - 23.985 AMU (78.99%) ISOTOPE 2 - 24.986 AMU (10.00%) ISOTOPE 3 – 25.982 AMU (11.01%) EXAMPLE 4 HELIUM HAS TWO NATURALLY OCCURRING ISOTOPES, He-3 AND He-4. THE ATOMIC MASS OF HELIUM IS 4.003 AMU. WHICH ISOTOPE IS MORE ABUNDANT IN NATURE? He-4 IS MORE ABUNDANT IN NATURE BECAUSE THE ATOMIC MASS IS CLOSER TO THE MASS OF He-4 THAN TO THE MASS OF He-3.
