If we know a compound's percent composition we can figure out the formula!
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1 If we know a compound's percent composition we can figure out the formula! Empirical formula the formula with the smallest wholenumber mole ratio of the elements. This may be the same as the molecular formula, but it may not be. *If different, the molecular formula will always be a simple multiple of the empirical formula. Use mass % to find empirical formula Find the mole ratio from masses of elements in the compound must be whole numbers! Example: HO is the empirical formula for hydrogen peroxide (H 2 O 2 molecular formula) Example: Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% Carbon, 8.16% Hydrogen, 43.20% Oxygen. STEP 1: Assume you have a 100g sample If you have 100g then you have 43.20g O, 48.64g C, 8.16g H. STEP 2: Convert these into moles using the conversion factor of molar mass molc = mol C 12.01g C 8.16g H 1 mol H = 8.10 mol H 1.008g H 43.20g O 1 mol O = mol O 16.00g O 1
2 STEP 3: Mole ratio mol C : 8.10 mol H : mol O *the smallest amount is O, use this to get a whole # ratio mol C = 1.500mol C mol 8.10 mol H = 3.00 mol H mol mol = 1 mol O 2.7 mol STEP 4: Note: If all of these were whole numbers you'd be done BUT they're not so... you must multiply each piece by a number that will turn them all to whole #s. In this case it would be x 2 = 3 C 3 x 2 = 6 H 1 x 2 = 2 O To check your Answer: Calculate % composition represented by the formula should be the same as what you were given in the problem. Empirical Formula shows simplest ratio but doesn't always show the actual ratio. Molecular formula specifies the actual number of atoms of each element in one formula unit of a substance. To determine the molecular formula for a compound, the molar mass of the compound must be determined and compared with the mass represented by the empirical formula. Ex: Acetylene is 26.04g/mol and mass of the empirical formula (CH) is 13.02g/mol. 1. Divide actual molar mass by the empirical = This tells us that acetylene is 2 times the mass of the empirical so it must have 2x the amount of elements in the compound. Acetylene C 2 H 2 Note: We can calculate molecular formulas using the same setup as we did to find the empirical formula and then add this last piece on to find the actual equation. 2
3 Homework: pg 346 #58-60 Formulas for HYDRATES In the Christmas ornaments that we made IF you were able to watch the crystals forming from the water solution you would have seen water molecules sometimes sticking to the crystals and becoming part of the structure. We call these waters of hydration. Hydrate a compound that has a specific number of water molecules bound to its atoms. Example: Opal When a hydrate is heated, water molecules are driven off leaving an anhydrous compound, one without water. 3
4 USES OF HYDRATES Hydrates are commonly found in skin care products such as moisturizer, shampoo and lip balm. Hydrates replace the skin's moisture and repair tissue damaged by cold and dryness. Also used in keeping electronics dry in moist climates. NAMING HYDRATES AND WRITING FORMULAS *We use dots to show hydration in the molecular formula. *To name them: we name the compound and then use a prefix to state the number of water molecules and then include the word hydrate after. Example: Na 2 CO 3 *10H 2 O Sodium carbonate decahydrate *To determine the formula of a hydrate: find the number of moles of water with 1 mol of hydrate. Example: A mass of 2.50g of blue, hydrated CuSO 4 *xh 2 O is placed in a crucible and heated. After heating, 1.59g of white anhydrous copper sulfate (CuSO 4 ) remains. What is the formula for the hydrate. Name the hydrate. 4
5 EXAMPLE Here's what we know: The mass of hydrated compound = 2.50g The mass of anhydrous compound = 1.59g Molar mass of water = 18.02g/mol Molar mass of CuSO 4 = 159.6g/mol 1st determine the amount of water lost: 2.50g 1.59g = 0.91g 2nd convert the known masses of water and anhydrous CuSO 4 into moles 1.59g CuSO 4 1mol CuSO g CuSO 4 = mol CuSO g H 2 O 1 mol H 2 O 18.02g H 2 O =.050 mol H 2 O x = moles of H 2 O =.050 mol H 2 O = 5 moles of CuSO mol CuSO 4 Note: X represents the moles of a compound. The ratio of water to cupric sulfate is 5:1 This means CuSO 4 * 5H 2 O The name is Copper(II) sulfate pentahydrate 5
6 Homework: 1. READ Everyday Chemistry pg very interesting 2. Practice Problems: pg and pg 353 6
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