A) Calculate the % composition for CuSO 4 •5H 2 O ) = 249.5 g ( ( ( ( ) + 1(32) + 4 16 ) + 1(32) + 4 16 ) + 1(32) ) ( ) + 5 18 ( ) ( ) MM = 1 63.5 MM = 1 63.5 MM = 1 MM = 1 63.5 MM = 1 63.5 mol Empirical and molecular formulas 63.5 % Cu = 249.5 � 100 100 = 25.25% 32 32 % S = % S = % S = 249.5 � 100 = 12.83% 249.5 � 100 249.5 249.5 % composition = part whole � 100 ( ) ( ) 9 16 9 16 % O = % O = % O = 249.5 � 100 = 57.72% 249.5 � 100 ( ) ( ) 10 1 10 1 % H = % H = % H = 249.5 � 100 249.5 � 100 = 4.01% B) How many g O are in 25.0 g of CuSO 4 •5H 2 O Two different types of chemical formulas g O g total � 100 = mass % Empirical formula Molecular formula g O g total = mass fraction smallest whole # ratio of actual # of atoms in molecule atoms in comp’d g O = = mass fraction � g total Used for ionic comp’d Used for covalent comp’d = 0.5772 � 0.5772 � 25.0 g = 14.43 g O Determining empirical formula for a comp’d Covalent compound 1. Obtain mass of each element Molecular formula Empirical formula 2. Determine the # moles of each C 6 H 12 O 6 CH 2 O 3. Divide the # moles of each by the smallest # of moles to convert the smallest to 1: gives mole ratio C 6 H 16 N 2 C 3 H 8 N 4. If one or more of the numbers are not integers, multiply each by the smallest integer that will convert all into C 12 H 4 Cl 4 O 2 C 6 H 2 Cl 2 O whole numbers . 2, 3, 4, 5 H 2 O H 2 O 5. These #s are the subscripts in empirical formula 6. To determine molecular formula, compare ratio of H 2 O 2 HO molecular to empirical formula masses 1
What is the empirical formula for a compound that has 30.4% N and 69.6% O? Step 3: get mole ratio by dividing each number by the smaller of the two Step 1: calculate mass of each element: 2.17 Assume 100g 2.17 = 1 mole N 30.4 g N and 69.6 g O 4.35 Step 2: convert to moles = 2 mole O � � � � 2.17 30.4 g N 1 mole N 30.4 g N � = 2.17 mole N � � � 14 g N 2 O for every 1 N � NO 2 � � � � � � � � 69.6 g O 1 mole O 69.6 g O � = 4.35 mole O � � � 16 g O � � � � Step 6 What is the empirical formula for a compound that contains 65.2% arsenic and 34.8% oxygen? Find MM of empirical formula unit and Step 1: calculate mass compare to molecular Assume 100g For above problem, MM of molecular formula = 92 g/mol 65.2 g As and 34.8 g O Step 2: convert to moles • Find MM of empirical formula � � � � 65.2 g As 1 mole As NO 2 = 46 g/mol 65.2 g As � = 0.870 mole As 92 � � � 46 = 2 74.9 g As � � � � • Compare the two MM: 34.8 g O 1 mole O � � � � Comp’d has 2 x MM of emp. form.: 34.8 g O � = 2.18 mole O � � � N 2 O 4 16 g O � � � � For the previous problem, if MM is 230 g/mol, what is its molecular formula? Step 3: get mole ratio by dividing by smallest Step 6: Find MM of empirical formula unit and 0.87 87 87 = 1 mole As compare to molecular 0.87 2 � 75 = 150 g X 2 2.18 0.87 = 2.5 mole O 5 � 16 = 80 g 230 g/mol Step 4: Multiply each by small integer to get whole # 230 230 = 1 1 empirical formula unit in molecular 2 arsenic for every 5 oxygen As 2 O 5 As 2 O 5 = molecular formula 2
Step 1: mass % to grams: A compound contains 73.14 % C and 7.37% H. The rest is 100 g = 73.14 g C 73.14 g C + 7.37 g H 7.37 g H + mass O O. A 10.0 g sample was found to contain .06165 moles. Mass O = 19.49 g What are the empirical and molecular formulas? Step 2: Convert mass to moles: � � � 1 mole C � 73.14 g C 73.14 g C � = 6.089 mole C � � � 12.011 g C � � � � � � � � 7.37 g H 1 mole H 7.37 g H � = 7.31 mole H � � � 1.0079 g H � � � � � � � 19.49 g O 1 mole O � � � � 19.49 g O � = 1.218 mole O � � � 16.0 g O � � � � A 10.0 g sample was found to contain 0.06165 moles. Step 3: Find mole ratio and subscripts What are the empirical and molecular formulas? Step 6: calculate molecular mass � 6.089 mole C � = 5 mole C � � 1.218 mole O 10.0 g � � 0.06165 mol = 162.2 g/mol � � � 7.31 mole H � = 6 mole H � � 1.218 mole O � � � 162.2 = ~ 2 � 82 Empirical formula = C 5 H 6 O Molecular formula = C 10 H 12 O 2 MM = 5(12)+6(1)+1(16) = 82 8) A nonmetal oxide compound is 74.1% by mass oxygen 7) A nonmetal oxide compound is 50.0% by mass oxygen annd has an empirical formula of X 2 O 5 . Identify X. and has an empirical formula of XO 2 . Identify X. X 2 O 5 74.1% by mass O XO 2 50% by mass O assume 1 mole of compound : assume 1 mole of compound : 5 mol O 2 mol O 2 mol X 1 mol X let x = MM of element X let x = MM of element X 2(16) 5(16) x = 0.5 2x = 0.741 5(16) + 2x = 2(16) + x 2(16) 5(16) 32 = 16 + 0.5 x 80 = 59.28 .28 + 1.482 x S SO 2 N N 2 O 5 x = 32 g X X = S SO x = 13.98 g X X = N N 3
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