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Chang, 6th Edition, Chapter 3, Worksheet #1 S. B. Piepho, Fall 1998 Atomic Mass, Moles, and the Periodic Table Atomic Mass and Molar Mass Isotopic masses cannot be obtained by summing the masses of the elementary particles (neutrons, protons, and electrons) from which the isotope is formed. This process would give masses slightly too large, since mass is lost when the neutrons and protons come together to form the nucleus. Atomic masses (also called atomic weights) are thus assigned relative to the mass of a particular carbon isotope, 126 C , which is assigned the mass of 12 amu exactly. Likewise 1 mole of 12 6 C has a mass of exactly 12 g. Atomic masses and molar masses of other isotopes are calculated based on their mass relative of that of Carbon-12. Masses of “average” atoms are found by summing isotopic masses, weighting each isotopic mass by its abundance (see Chang, pg 70). Thus one “average” C atom has a mass of 12.011 amu, and the mass of 1 mole of “average” carbon atoms has a mass of 12.011 g. These average masses are what is given on the periodic chart. What is a Mole? Since atoms and molecules are so tiny, it is convenient to talk about a large number of them at a time. The chemical counting unit is known as the mole. A mole is defined as the amount of substance that contains as many elementary entitites (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the 126 C isotope. It has been found experimentally that 1 mole of particles = 6.022 x 1023 particles This value is known as Avogadro’s number. Just like 1 dozen of anything always contains 12 items, 1 mole of anything always contains 6.022 x 1023 items. Molecular Masses and Compound Masses Molecular masses are found by summing atomic masses (see Chang, pgs 74-75). They are often called molecular weights. Thus the mass of 1 mole of water, H2O, would be 2 x (molar mass of H) plus 1x (molar mass of O) or [(2 x 1.008 g) + (1 x 16.00)] = 18.02 g. Ionic compounds such as NaCl do not contain molecules. Their formulas give the relative numbers of each kind of atom in the sample. What we mean by the molar mass (or the molecular weight) of an ionic compound is really the formula weight. The formula weight is the sum of the atomic masses in the formula. Percent Composition of Compounds The percent composition by mass is the percent by mass of each element in a compound. If there are n moles of an element per mole of compound, the percent by mass of the element is calculated using the equation, n molar mass of element 100% % Composition of Element = molar mass of compound The sum of the % compositions of all elements in a compound is 100%. 1 Chang, 6th Edition, Chapter 3, Worksheet #1 S. B. Piepho, Fall 1998 Exercises 1. The atomic mass scale gives masses in atomic mass units (amu) relative to the mass of carbon-12. (a) What is the mass of one 12C atom in atomic mass units (amu)? ____________ (b) What is the mass of an average C atom in atomic mass units (amu)? ____________ (c) What is the mass of an average Cl atom in amu? ____________ (d) What is the mass of an average Br atom in amu? ____________ 2. The molar mass scale gives masses in grams (g) relative to the mass of 12C. (a) What is the mass in grams of 1 mole (mol) of 12C? ___________ (b) What is the mass in grams of 1 mole (mol) of carbon? ___________ (c) What is the mass in grams of 1 mole (mol) of Cl? ___________ (d) What is the mass in grams of 1 mole (mol) of Na? ___________ 3. The mass of a mole of 12C is 12 g exactly. The conversion factor between atomic mass units (amu) and grams (g) is 1 amu = 1.6605 x 10-24 g. Each atom of 12C has a mass of 12 amu exactly. Use this information to determine the number of 12C atoms in a mole of 12C using the factor-label method. (Your answer should look familiar!) 4. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O. (a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g? Answers: 2 Chang, 6th Edition, Chapter 3, Worksheet #1 1. 2. 3. 4. S. B. Piepho, Fall 1998 (a) 12 amu exactly; (b) 12.10 amu; (c) 35.45 amu; (d) 79.90 amu. (a) 12 g exactly; (b) 12.10 g; (c) 35.45 g; (d) 79.90 g. 6.022 x 1023 atoms of 12C. (a) 80.56% C; 7.51% H; 11.93% O; (b) 2.11 x 1021 molecules of C9H10O. 3