Empirical Formula of Magnesium Oxide Introduction Chemical formulas are used to describe the atomic ratio of atoms in a compound. In molecular formulas, the ratios relate to the molecule as a whole. Benzene, for example, has a molecular formula of C 6H6, which means that there are 6 carbon atoms and 6 hydrogen atoms in each benzene molecule. In other words, the ratio of carbon to hydrogen is 6:6. Empirical formulas give the simplest whole-number ratio of the atoms in a compound. For benzene, this is CH (carbon to hydrogen ratio of 1:1). For propylene, C 3H6, the empirical formula is CH2. Empirical formulas are an important part of analytical chemistry, because they are simple to determine experimentally (or, empirically). Say you have 13.8 g of a compound that contains only nitrogen and oxygen. By decomposing the compound, you determine that it contains 4.2 g of nitrogen. How would you determine the empirical formula? Remember that atomic ratios and molar ratios are the same. If you find the molar ratio of the nitrogen and oxygen, it will be the same as the atomic ratio of the nitrogen and oxygen. The atomic ratio is your empirical formula. So, first, find the number of moles of nitrogen and oxygen in your sample: 1 mol N 4.2 g N X 14.01 g N (13.8 g - 4.2 g) O X 1 mol O 16.00 g O = 0.30 mol N = 0.60 mol O Divide each of those by the smallest number to get the formula: N: 0.30/0.30 = 1 O: 0.60/0.30 = 2 Thus, the empirical formula of your compound is NO 2. In the following experiment, you will determine the empirical formula of magnesium oxide, using the same thinking as the nitrogen dioxide determination you just read about here. Pre-Lab Questions Heating 1.29 g of gallium gives a gallium-oxygen compound with a mass of 1.73 g. Calculate the empirical formula of the compound. Step 1: Figure out how many grams of oxygen are in the 1.73 grams of gallium-oxygen compound by subtracting. 1.73-1.24=.44 grams of oxygen Step 2: Convert the 1.29 g of gallium to moles: 64 x 723 = 0.0185 Moles Step 3: Convert the grams of oxygen you calculated in step 1 to moles: 0.44 grams of oxygen = 0.44/32 (x2) = 0.01375 (x2) = 0.0275 moles Step 4: Figure out the formula. Oxygen moles = 0.0185 = 1 x 4860 “O” X 2 Gallium moles 0.0275 = 1. “G” X2 = 3 “O” / 2 “G” = Ga2O3 Formula: ___Ga2O3___________ Name: ______Gallium oxide__________

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