VIEWS: 7 PAGES: 6 POSTED ON: 2/13/2012
The Mole! When you look at the periodic table, the average atomic mass below each element represents the average mass of one atom in amu. (Review: what is an amu?) This is an incredibly small mass and not really reasonable to work with in a lab setting. SO, if we take the average atomic mass and treat it as a quantity of grams, we call this quantity a mole. Example: 1 atom of carbon has the average mass of 12.01 amu. 1 mole of carbon has the mass of 12.01 grams. Question: How many grams in one mole of lead? How many grams in 4.5 moles of lead? A few more details about the mole: ***A guy named Avogadro calculated how many atoms would be in one mole of an element and determined that it would equal 6.022 x 1023 atoms of that element. In this sense, we use the term mole in the same way we use the term "1 dozen". A dozen always equals 12 of something; a mole always represents 6.022 x 1023 of something. ***The mole is also used to describe quantities of compounds and molecules. The "molar mass" (or grams in a mole) of any substance can be determined by adding up the individual masses of the atoms in the compound. For example: Water, H2O, contains two hydrogen atoms and one oxygen atom. H = 1.01 gram per mole H = 1.01 gram per mole O= 16.00 gram per mole H2O = 18.02 grams per mole Question: How many grams would be in 1 mole of ammonia, NH3? How many grams would be in 1 mole of Mg(NO3)2? This quantity (grams in 1 mole) is called: molar mass or molecular mass or molecular weight Percent Composition Percent composition is a way to describe what a substance is made of by listing the percent of each element in the compound. Percent = part/whole x 100 To determine the percent composition of one or all elements in a compound: 1. Determine the molar mass of the compound by adding up the individual masses of all the atoms in the compound. 2. Divide the total individual mass of each element by the overall molar mass and multiply by 100. Example: H3PO4 3 H = 3(1.01) = 3.03 P = 30.97 = 30.97 4 O = 4 (16.00) = 64.00 98.00 H 3.03/98.00 x 100 = %H P 30.97/98.00 x 100 = %P O 64.00/98.00 x 100 = %O Practice: what is the percent composition of glucose, C6H12O6? Empirical Formula So, we have gone from the chemical formula to the percent composition. Now, we learn to go from percent composition back to the chemical formula, in a way. From these percents, we can determine the empirical formula, which is the simplest whole number ratio of atoms in a molecule. The empirical formula may or may not be the same as the molecular formula which is the actual ratio of atoms in a molecule. Example: Molecular formula of glucose: C6H12O6 Empirical formula of glucose: CH2O Determining empirical formula: 1. Treat the percentages as grams. 2. Convert the grams to moles. 3. Divide each mole quantity by the smallest mole quantity to achieve an initial ratio. 4. If this ratio is extremely close to whole numbers, STOP. This is your ratio for your empirical formula. 5. If this ratio is not close to whole numbers, multiply by a whole number to achieve a whole number ratio. (For example, if the ratio is 1:1.5, you will need to multiply by 2 to get to 2:3.) Example: A compound is made of 23.29% Mg, 30.72% S, and 45.99% O. What is its empirical formula?