Back on Sept 15<\/a> I opened the discussion about what a mole is. Whether or not students ever get comfortable with it, they need to use the concept. Calculations in chemistry center around moles; here’s an example that shows why:<\/p>\n 2C2<\/sub>H6<\/sub> + 7O2<\/sub> \u2192 4CO2<\/sub> + 6H2<\/sub>O<\/p>\n The equation above means “2 moles C2<\/sub>H6<\/sub> with 7 moles O2<\/sub> reacts to produce 4 moles carbon dioxide and 6 moles water.” Reactions happen based on ratios of moles. Therefore, when predicting the output (or required input) of a reaction, moles are central.<\/p>\n Concerning moles, a primary skill a chemistry student needs to develop is finding the molar mass. Let’s see how it’s done in the following example:<\/p>\n Example<\/strong>: Find the molar mass of (NH4<\/sub>)3<\/sub>PO4<\/sub><\/p>\n Solution<\/strong>: From the periodic table, we find the mass of each atom present, then multiply by its frequency in the compound. Afterward, we sum up the contributions to arrive at the molar mass:<\/p>\n So, apparently, the molar mass (aka M.M.) of ammonium phosphate (aka (NH4<\/sub>)3<\/sub>PO4<\/sub>) is 149.1g.<\/p>\n Even when finding M.M. on paper, using a tabulated format helps you keep track of more involved calculations:)<\/p>\n Source:<\/p>\n Mortimer, Charles E. Chemistry<\/em>. Belmont: Wadsworth, Inc., 1986.<\/p>\n\n
\n atom<\/td>\n mass(g)<\/td>\n mulitplicity<\/td>\n mass contribution(g)<\/td>\n<\/tr>\n \n N<\/td>\n 14.0<\/td>\n 3<\/td>\n 42.0<\/td>\n<\/tr>\n \n H<\/td>\n 1.01<\/td>\n 12<\/td>\n 12.12<\/td>\n<\/tr>\n \n P<\/td>\n 31.0<\/td>\n 1<\/td>\n 31.0<\/td>\n<\/tr>\n \n O<\/td>\n 16.0<\/td>\n 4<\/td>\n 64.0<\/td>\n<\/tr>\n \n MOLAR MASS(M.M.)<\/td>\n <\/td>\n <\/td>\n 149.12<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n