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PERCENT COMPOSITION Percent composition is the percent by mass of each element in a compound. To determine the percent composition, first find the molar mass of the compound. Then find the total mass of each element in the compound. Divide the mass of the element by the mass of the compound and then multiply by 100 and that is the percent composition of that element in the compound. EX: Na2CO3 Molar Mass = 105.99g C: 12.01g/ 105.99g x 100 = Mass of Na = 45.98g Mass of C = 12.01g Mass of O = 48.00g DETERMINING AN EMPIRICAL FORMULA The empirical formula of a compound is the smallest whole number ratio of elements in the compound. It is based on percent composition. In a sample of a compound, the ratio of the moles of each element present in the sample will form a small whole number ratio. These ratios can help determine the subscripts of the empirical formula of the compound. EX. A sample of an unknown Carbon-Hydrogen compound contains 18 grams of Carbon and 6 grams of Hydrogen. What is the empirical formula of the compound? First, take your values and covert them into moles. 18 g C x 1 mole C = 1.5 moles C 12.01 g 6 g H x _1 mole H = 6 moles H 1.01 g Then, determine the simplest whole number ratio between the two elements. To do this, divide both molar values by the smallest molar value. *If you get a ratio that does not give you a whole number but rather a half (.5), then double all values to get the simplest whole number ratio. 1.5 moles C = 1 6 moles H = 4 1.5 moles 1.5 moles The ratio is the subscript of that element in the formula, CH 4. More often, you will be given the percent composition of the compound. When this happens, assume you have a 100g sample of the compound. And then logically, you would have the percent’s value in grams. Then continue on as normal. 1) Assume 100g. 2) Convert grams into moles. 3) Determine the simplest ratio by dividing all values by the smallest number. 4) Ratio is the subscript (within 0.05). *If you get a 0.5 value – double all ratios! EX. An unknown compound contains 74.2% Sodium and 25.8% Oxygen. What is the empirical formula of the compound? 74.2% of a 100g sample is Sodium, therefore 74.2 g of the sample is Na. 25.8% of a 100g sample is Oxygen, therefore 25.8 g of the sample is O. 74.2g Na x 1 mole Na = 3.23 moles Na = 2 22.99g 1.61 25.8g O x 1 mole O = 1.61 moles O = 1 16.00g 1.61 Formula = Na2O HYDRATES A hydrate is a compound that has water molecules surrounding each compound molecule. It has the same number of water molecules around each compound molecule. Since this is a set situation, the compound must be named for its water molecules also. The formula is written normally for the compound, then there is a dot, then a number, and then the formula for water. The compound is named the same but then the prefix and the word “hydrate” is added to the end of it. The prefix is for the number of water molecules there are. For example, a common hydrate is BaCl 2 6H 2O, this is named Barium Chloride Hexahydrate. Hydrates are compounds too. So you can calculate the percent composition for them by element or as a percent of water. For the above example, Barium Chloride Hexahydrate, BaCl2 6H2O, you can determine the percent Barium, Chlorine, Hydrogen, and Oxygen. You could also determine the percent of Barium, Chlorine, and Water, or you could find the percent of Barium Chloride and the percent water. EX. Barium Chloride can exist as 2 possible hydrates. One of them is a hexahydrate, the other has a 25.74% water. What is the formula of the hydrate? 25.74 g H2O x 1 mole H2O = 1.428 moles H2O = 3.99 ~ 4 Therefore the 18.02g 0.3572 hydrate is a tetrahydrate, 74.26 g BaCl2 x 1 mole BaCl2 = 0.3572 moles BaCl2 = 1 BaCl2 4H2O 207.90g 0.3572 DETERMINING A MOLECULAR FORMULA A molecular formula is the actual formula of a compound. The empirical is the simplest. Therefore the molecular compound is a whole number ratio to the empirical. Once the empirical formula is determined, you must then compare the molar mass of the empirical to the given molar mass of the molecular compound. Divide them to get the ratio and multiply all subscripts by the ratio. You may have to rearrange the formula to acknowledge polyatomic ions. EX. An unknown compound has 79.85% carbon and 20.15% Hydrogen and a molar mass of 30.08g. What is the Empirical formula? Molecular formula? PART 1 79.85 g C x 1 mole C = 6.649 moles C = 1 12.01 g 6.649 20.15 g H x 1 mole H = 19.95 moles H = 3.000 The empirical formula is CH 3 1.01 g 6.649 PART 2 The Molecular Formula Mass given is 30.08g. The molar mass of the empirical formula is 15.04g. So, 30.08g = 2 So, 2xs all the subscripts giving you a molecular formula of C2H6 15.04g