Course Outline
General Chemistry
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Course Outline General Chemistry |
03/24/07 |
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Before we discuss the concept of the mole, we need to differentiate among several similar terms: Atomic Number--the sum of the protons and neutrons in a particular isotope of an element. For example, carbon has six protons. There are three naturally occurring carbon isotopes: one with six neutrons, one with seven neutrons, and one with eight neutrons. The atomic numbers of theses three isotopes are 12, 13, and 14. Isotopic Mass--the mass of one atom of a particular isotope of an element, which is most often reported in atomic mass units, or amu. The atomic mass scale is set relative to carbon-12 isotope, so that one atomic mass unit, amu, is one twelfth of the mass of one atom of carbon-12. The isotopic mass is different from the mass number for two reasons: because it includes the masses of the electrons AND because the mass of a proton and a neutron is not exactly one amu. Atomic weight—the average mass of the atoms of an element taking into account the masses of the various isotopes and their relative abundance. For example, oxygen has an atomic weight of 15.9994. This means that the average of the atoms’ masses is 15.9994 atomic mass units (amu). Calculating atomic weights—the atomic weight is a weighted average of all of the atomic masses of the naturally occurring isotopes. This means that, instead of each mass counting equally toward the average, the mass is scaled according to each isotope’s abundance. To find the atomic weight multiply each isotope’s atomic mass by its percent abundance in decimal form then find the sum of these weighted masses.
Concept Check: Chromium has four naturally occurring isotopes: 50Cr, 52Cr, 53Cr, and 54Cr. The atomic masses and abundance data are:
What is chromium's atomic weight? Answer: Find the weighted average after converting the percentages to decimals. Atomic mass Cr = 0.04345(49.9460464 amu) + 0.83789(51.9405098 amu) + 0.09501(52.9406513 amu) + 0.02365(53.9388825 amu) = 51.996 amu This answer is rounded to the third decimal place because we apply the multiplication rule and then the addition rule for finding significant figures. Concept Check: There are two naturally occurring isotopes of boron, 10B and 11B. The atomic weight of boron is 10.811 amu, the atomic mass of 11B is 11.0093054 and its abundance is 80.1%. What is the atomic mass of 10B? Answer: Set the atomic weight equal to the weighted average, with 19.9% of the 10B isotope at an unknown mass. Solve for this unknown. 10.811 amu = 0.199(10B) + 0.801(11.0093054 amu) 10B = [10.811 amu - 0.801(11.0093054 amu)] ÷ 0.199 = 10.0 amu The answer is rounded to three significant figures because of the degree of uncertainty of the abundance data. The experimental value for the atomic mass of 10B is 10.0129369 amu.
Atomic mass and abundance data were found at Web Elements (September 16, 2004).
© Copyright 2006, Kelley Whitley, ChemProfessor. All rights reserved. This site was last updated 10/05/06 |
