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Empirical and Molecular Formulas Empirical Formula: The smallest whole number mole ratio of elements in a compound Assume that each percent by mass represents the mass of the element in a 100.00-g sample How to Calculate Empirical Formula 1 : Calculate the % composition of each element (If not given) and change % into grams 2nd: Calculate Molar Mass of each element 3rd: Determine simplest whole # ratio 4th: Write Empirical Formula st Example #1 The mass of C is 48.64g, the mass of H is 8.16g, and the mass of O is 43.20g. Find the Empirical Formula (EF). Step 1: Find Molar mass of each element 48.64 g of C X 1 mol of C/12.01g of C (atomic mass) =4.050 mol of C 8.16 g of H X 1 mol of H/1.008g of H (atomic mass) =8.10 mol of H 43.20 g of O X 1 mol of O/16.00g of O (atomic mass) =2.7 mol of O Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 4.050/2.7 = 1.5 mol of C 8.10/2.7 = 3 mol of H 2.7/2.7 = 1 mol of O Then look at the three numbers of moles and determine the lowest number they can be multiplied by to get all whole numbers. In this case the number is 2. 4.050/2.7 = 1.5 mol of C x2=3 mol of C 8.10/2.7 = 3 mol of H X 2=6 mol of H 2.7/2.7 = 1 mol of O X2=2 mol of O Step 3: Create Empirical Formula from moles in Step two C3H6O2 1 Example #2 Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% C, 5.08% H, and 54.24% oxygen and has a molar mass of 118.1g/mol. Determine the empirical formula for succinic acid. Step 1: Determine molar mass. 1st: Convert percentages into grams of elements. 2nd: Use molar mass formula to find moles. 40.68 g of C X 1mol C/12.01g (atomic mass) of C= 3.390 mol of C 5.08 g of H X 1mol H/1.008g (atomic mass) of H= 5.04 mol of H 54.24g of O X 1mol O/16.00g (atomic mass) of O= 3.390 mol of O Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 3.390/3.390 = 1 mol of C 5.04/3.390 = 1.5 mol of H 3.390/3.390 = 1 mol of O Then look at the three numbers of moles and determine the lowest number they can be multiplied by to get all whole numbers. In this case the number is 2. 3.390/3.390 = 1 mol of C X 2 =2 mol of C 5.04/3.390 = 1.5 mol of H X 2=3 mol of H 3.390/3.390 = 1 mol of O X 2 =2 mol of O Step 3: Create Empirical Formula from moles in Step two C2H3O2 Practice Problems for Empirical Formula Directions: What is the empirical formula of the compounds below? 1. 75% C, 25% H 2. 52.7% K, 47.3% Cl 3. 22.1% Al, 25.4% P, and 52.5% O 4. 13% Mg, 87%Br 5. 32.4% Na, 22.5% S, and 45.1% O 2 Answers to Practice Problems for Empirical Formula Directions: What is the empirical formula of the compounds below? 1. 75% C, 25% H Step 1: Find Molar mass of each element 75 g of C X 1 mol of C/12.01g of C (atomic mass) =6.25 mol of C 25 g of H X 1 mol of H/1.008g of H (atomic mass) =24.8 mol of H Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 6.25/6.25 = 1 mol of C 24.8/6.25 = 4 mol of H They are already in whole numbers so move to Step 3 1 mol of C 4 mol of H Step 3: Create Empirical Formula from moles in Step two CH4 2. 52.7% K, 47.3% Cl Step 1: Find Molar mass of each element 52.7 g of K X 1 mol of K/39.098g of K (atomic mass) =1.35 mol of K 47.3g of Cl X 1 mol of Cl/35.453 of Cl (atomic mass) =1.33 mol of Cl Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 In this case both are basically the same number so both are 1 1 mol of K 1 mol of Cl Already in whole numbers so move to Step 3 Step 3: Create Empirical Formula from moles in Step two KCl 3. 22.1% Al, 25.4% P, and 52.5% O Step 1: Find Molar mass of each element 22.1 g of Al X 1 mol of Al/26.9815g of Al (atomic mass) =.82 mol of Al 25.4 g of P X 1 mol of P/30.9738 of P (atomic mass) =.82 mol of P 52.5 g of O X 1 mol of O/16.00g of O (atomic mass) =3.29 mol of O 3 Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 .82/.82= 1 mol of Al .82/.82= 1 = 1 mol of P 3.29/.82 = 4 mol of O Already whole numbers so move to the next step Step 3: Create Empirical Formula from moles in Step two AlPO4 4. 13% Mg, 87% B Step 1: Find Molar mass of each element 13 g of Mg X 1 mol of Mg/24.305g of Mg (atomic mass) =.535 mol of Mg 87 of B X 1 mol of B/10.811g of B (atomic mass) = 8.05 mol of B Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 In this case both are basically the same number so both are 1 .535/.535 = 1 mol of Mg 8.05/.535 =5 mol of B Already in whole numbers so move to Step 3 Step 3: Create Empirical Formula from moles in Step two MgB5 5. 32.4% Na, 22.5% S, and 45.1% O Step 1: Find Molar mass of each element 32.4 g of Na X 1 mol of Na/22.9898 g of Na (atomic mass) =1.4 mol of Na 22.5 g of S X 1 mol of S/32.064g of S (atomic mass) =.70 mol of S 45.1 g of O X 1 mol of O/16.00g O (atomic mass) =1.4 mol of O Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 1.40/.70= 2 mol of Na .70/.70= 1 = 1 mol of S 1.4/.70 = 2 mol of O Already whole numbers so move to the next step Step 3: Create Empirical Formula from moles in Step two Na2SO3 4 Molecular Formula Molecular Formula: A formula that states that actual number of atoms of each element in one compound or formula unit of the substance. How to Calculate Molecular Formula: 1st: 2nd: n= Experimentally determined Molar Mass of the compound (MF) Mass of Empirical formula of the compound (EF) Multiple the subscripts in the empirical formula by n Example #1 Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% C, 5.08 H, and 54.24% oxygen and has a molar mass of 118.1g/mol. Determine the Molecular formula for succinic acid. Empirical Formula for Succinic Acid is: C2H3O2 and the molar mass of Succinic acid is 59.04 g/mol Step 1: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n n= Experimental Molar mass of succinic acid Molar mass of C2H3O2 n= 118.1g/mol 59.04 g/mol n= 2.000 Step 2: Multiple subscripts by 2 (n) C4H6O4 Example #2 The empirical formula of a compound is NO2. Its molecular mass is 92g/mol. What is the molecular formula? Step 1: Determine the empirical formula molar mass N has 1 atom 1 X 14.0067= 14.0067g/mol O has 2 atoms 2X 16.00= 32.00g/mol 46.0067 g/mol 5 Step 2: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n n= Experimental Molar mass of NO2 Molar mass of NO2 n= 92g/mol 46.0067 g/mol n= 2.000 Step 2: Multiple subscripts by 2 (n) N2O4 Practice problems for Molecular Formula 1) The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is the molecular formula? 2) A compound is found to be 40% C, 6.7% H and 53.5%O. Its Molecular mass is 60 g/mol. What is its molecular formula? 6 Answers from Molecular Formula Problems 1) The empirical formula of a compound is CH2. Its molecular mass is 70g/mol. What is the molecular formula? Step 1: Determine the empirical formula molar mass C has 1 atom 1 X 12.011= 12.011g/mol H has 2 atoms 2X 1.008= 2.016g/mol 14.027 g/mol Step 2: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n n= Experimental Molar mass of CH2 Molar mass of CH2 n= 70g/mol 14.027 g/mol n= 5.000 Step 2: Multiple subscripts by 5 (n) C5H10 2) A compound is found to be 40% C, 6.7% H and 53.5% O. Its Molecular mass is 60 g/mol. What is its molecular formula? Step 1: Determine the empirical formula and then its molar mass 40 g of C X 1 mol of C/12.01g of C (atomic mass) =3.33 mol of C 6.7 g of H X 1 mol of H/1.008g of H (atomic mass) =6.648 mol of H 53.6 g of O X 1 mol of O/16.00g of O (atomic mass) =3.35 mol of O 3.33 mol of C/3.33 =1 mol of C 6.648 mol of H/3.33 =2 mol of H 3.35 mol of O/3.33 = 1 mol of O So E.F: CH2O C has 1 atom 1 X 12.011= H has 2 atoms 2X 1.008= O has 1 atom 1X 16.00= 12.011g/mol 2.016g/mol 16.00g/mol 30.021 g/mol Step 2: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n n= Experimental Molar mass of CH2O Molar mass of CH2O 7 n= 60g/mol 30.021 g/mol n= 2.000 Step 2: Multiple subscripts by 2 (n) C2H4O2 8 9