UNIT 7 CHEMICAL FORMULAS WRITING FORMULAS NOTES. EXAMPLES: 1. carbon tetrachloride 2. calcium oxide. 3. iron (III) bromide 4.

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1 WRITING FORMULAS NOTES EXAMPLES: 1. carbon tetrachloride 2. calcium oxide 3. iron (III) bromide 4. lead (II) nitrate 5. aluminum hydroxide 6. ammonium chromate Notes- HONORS 1

2 NAMING COMPOUNDS NOTES EXAMPLES: 1. P2O5 2. MgSO4 3. CuCl2 4. (NH4)3PO4 5. FeCO3 6. K2O POLYATOMIC IONS (LISTED ALPHABETICALLY) Name Formula Name Formula Name Formula acetate C2H3O2-1 dichromate Cr2O7-2 nitrite NO2-1 ammonium NH4 +1 hydrogen carbonate (or bicarbonate) HCO3-1 perchlorate ClO4-1 bromate BrO3-1 hydrogen sulfate HSO4-1 permanganate MnO4-1 carbonate CO3-2 hydroxide OH -1 phosphate PO4-3 chlorate ClO3-1 hypochlorite ClO -1 sulfate SO4-2 chlorite ClO2-1 iodate IO3-1 sulfite SO3-2 chromate CrO4-2 nitrate NO3-1 thiocyanate SCN -1 cyanide CN -1 Notes- HONORS 2

3 OXIDATION NUMBERS NOTES Any uncombined element (element not in a compound) has an oxidation number of 0. Fluorine always has an oxidation number of -1 in a compound. Oxygen has an oxidation number of -2 in all compounds except when it is part of a binary compound with a halogen. Hydrogen has an oxidation number of +1 except when it is in a binary compound with a metal. The algebraic sum of the oxidation numbers in a compound is zero. The algebraic sum of the oxidation numbers in a polyatomic ion is the charge on the ion. To find the oxidation number of another element in a compound, use this general formula: Σ (# of each element in cmpd. oxidation # of each element) = 0 Let x = unknown oxidation number EXAMPLE: Find the oxidation number of carbon (C) in Na2CO3. x = carbon s oxidation number Na = +1 O = -2 (2. +1) + (1. x) + (3. -2) = 0 Na C O 2 + x 6 = 0 x 4 = 0 x = +4 is carbon s oxid. # in Na2CO3 Find the oxidation number of the underlined element in each compound. 1. KMnO4 2. MnO2 3. LiNO3 4. Ca(NO2)2 5. NaClO 6. Ba(ClO4)2 PERCENT COMPOSITION NOTES PERCENT COMPOSITION: the percentage by mass of each element in a compound FORMULA FOR % COMPOSITION: % composition = mass of element in compound x 100 molar mass of compound EXAMPLE 1: Find the % composition of copper (I) sulfide, Cu2S. ~ Finding % composition means that you have to find the % of each element in the cmpd. molar mass of Cu2S: % Cu = x 100 = % S = x 100 = Cu: x = S: x = MM of Cu2S = EXAMPLE 2: Find the percent of oxygen in calcium phosphate, Ca3(PO4)2. molar mass of Ca3(PO4)2: % O = x 100 = Ca: x = P: x = O: x = MM of Ca3(PO4)2 = Notes- HONORS 3

4 EMPIRICAL FORMULAS NOTES opposite of percent composition use % to find formula for compound EMPIRICAL FORMULA: simplest formula; subscript numbers are reduced to lowest terms MOLECULAR FORMULA: subscripts are multiples of empirical formula subscripts MOLECULAR FORMULA C6H6 C6H12O6 EMPIRICAL FORMULA C12H16O4N TO SOLVE EMPIRICAL FORMULA PROBLEMS: A sample of a compound is found to contain 36.0 % calcium and 64.0 % chlorine. Calculate the empirical formula. Step 1: Rewrite % as grams g Ca 64.0 g Cl Step 2: Find moles of each element. Ca: 36.0 g Ca 1 mole Ca = moles Ca Cl: 64.0 g Cl 1 mole Cl = 1.80 moles Cl 40.1 g Ca 35.5 g Cl Step 3: Find mole ratio. (Divide by smallest number of moles.) Ca: moles = 1 Cl: 1.80 moles = * These whole numbers are subscripts in formula.* Step 4: Write the formula. Ca1Cl2 ====> CaCl2 Example 2: A sample of a compound contains 66.0 % calcium and 34.0 % phosphorus. What is the empirical formula? Ca: 66.0 g Ca 1 mole Ca = 1.65 moles Ca P: 34.0 g P 1 mole P = 1.10 moles P 40.1 g Ca 31.0 g P Ca: 1.65 = 1.5 P: 1.10 = Q: So, what happens now? I can't write Ca1.5P1. And 1.5 is not close enough to round to 2. A: The easiest way to get 1.5 to a whole # is to multiply by 2. Remember to multiply both #'s by 2 to get your answer. Ca: 1.5 x 2 = 3 P: 1 x 2 = 2 So, formula is Ca3P2 PRACTICE - A compound contains 43.4 % sodium, 11.3 % carbon, and 45.3 % oxygen. What is the empirical formula for this compound? Notes- HONORS 4

5 MOLECULAR FORMULAS NOTES To find the molecular formula, one more piece of information must be given - the molar mass (also called molecular mass or formula mass). EX. 1- An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. If the molecular mass is 78, what is the molecular formula? STEP 1: Find the empirical formula. C: g C 1 mole C = 7.69 moles C H: 7.75 g H 1 mole H = 7.75 moles H 12 g C 1 g H 7.69 moles C = moles H = 1 So... empirical formula is CH STEP 2: Find molar mass of the empirical formula. C: 1 x 12.0 = 12.0 H: 1 x 1.0 = MM = 13.0 STEP 3: Find "multiple" number. MM of molecular formula = multiple # 78 = 6 MM of empirical formula 13 STEP 4: Write molecular formula. Multiply "multiple" # by all subscripts in the empirical formula. So... molecular formula is C6H6. PRACTICE - An oxide of nitrogen contains 30.4 % nitrogen and 69.6 % oxygen. If the molar mass of this compound is 92 g/mole, what is the molecular formula? HYDRATES NOTES Hydrates are compounds with a certain number of water molecules attached to them. Their formulas look the same except that there is a. # H2O after it. Example: MgSO4. 7 H2O When determining the empirical formula for a hydrate, generally you will be determining the number in front of the H2O in the formula. In order to determine this number, you will need to find the mole ratio between the moles of the compound and moles of water. EXAMPLE: A hydrated sample of sodium carbonate (Na2CO3. # H2O) has a mass of grams. The sample is then heated and all water is removed. The anhydrous salt that remains has a mass of grams. What is the empirical formula for the hydrated sodium carbonate? STEP 1: Find moles of sodium carbonate g Na2CO3 1 mole Na2CO3 = moles Na2CO3 Na: 2 x 23.0 = g Na2CO3 C: 1 x 12.0 = 12.0 O: 3 x 16.0 = Notes- HONORS 5

6 STEP 2: Find moles of water. First, find grams of water g H2O 1 mole H2O = moles H2O 18 g H2O hydrated sample = grams anhydrous sample = grams mass of water = grams STEP 4: Find mole ratio of water to sodium carbonate moles = 10 So empirical formula for this hydrate is Na2CO3. 10 H2O moles Notes- HONORS 6