Honors Chemistry In Class Notes Percent Composition and Chemical Formulas 1 Objectives When you complete this presentation, you will be able to: o describe how to calculate the percent by mass of an element in a compound. o interpret an empirical formula. o distinguish between empirical and molecular formulas. Introduction Chemical formulas tell us about the number of atoms in a compound. In general, there are two kinds of chemical formulas. o In ________________ formulas, the ________________ ________________ of atoms in the compound is used. o In ________________ formulas, the ________________ ________________ ________________ ________________ of atoms in the compound is used. In molecular formulas, the total number of atoms in the compound is used. o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each molecule. Therefore, its molecular formula is C6H6. o For example, acetic acid has 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms in each molecule. Therefore, its molecular formula is C2H4O2. o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each molecule. Therefore, its molecular formula is C3H8. In empirical formulas, the lowest whole number ratio of atoms in the compound are used. o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each molecule. Therefore, its molecular formula is C6H6 and its empirical formula is CH (we divide all subscripts by 6). 2 The Mole: A Measurement of Matter o For example, acetic acid has 2 carbon atoms, 6 hydrogen atoms, and 2 oxygen atoms in each molecule. Therefore, its molecular formula is C2H4O2. and its empirical formula is CH2O (we divide all subscripts by 2). o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each molecule. Therefore, its molecular formula is C3H8 and its empirical formula is also C3H8 (there is no common divisor for all subscripts). Percent Composition If we know the ________________ of a compound and the ________________ of one or more of the elements in the compound, then we can find the ________________ ________________ of those atoms in the compound. % mass of an element = mass of atoms 100 % mass of compound For example, o 1.716 g of C, 0.577 g of H, and 2.286 g of O combine together to form 4.579 g of a compound. Sample Problem 10.9 When a 13.60 g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Known: mass of compound = 13.60 g Unknown: % Mg = ?% mass of O = 5.40 g % O = ?% mass of Mg = 13.60 g – 5.40 g = 8.20 g Honors Chemistry In Class Notes Percent Composition and Chemical Formulas If we know the ________________ formula of a compound, then we can find the percent composition of each of the atoms in the compound. We use the ________________ ________________ of the compound and the ________________ ________________ ________________ of the atoms in the compound. % composition = atomic mass of atoms 100 % molar mass of compound For example: o the percent composition of benzene, C6H6, is: o the percent composition of acetic acid, C2H3O2, is: Sample Problem 10.10 Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane Known: molar mass of C3H8 = 44.0 g Unknown: % C = ?% mass of C = 3 12.01 g/mol = 36.03 g/mol % H = ?% mass of H = 8 1.01 g/mol = 8.08 g/mol 3 The Mole: A Measurement of Matter Practice Problems: find the percent composition of … 1. C6H14 2. NaCl 3. KNO3 4. CuSO4 5. FeCO3 Empirical Formulas In ________________ ________________, the lowest whole number ratios of atoms in a compound is used. o Benzene, C6H6, has an empirical formula of ________________. A ratio of 6/6 = 1/1. o Acetic acid, C2H4O2, has an empirical formula of ________________. A ratio of 2/4/2 = 1/2/1 o Propane, C3H8, has an empirical formula of ________________. A ratio of 3/8 =3/8. If we know the percent composition of a compound, then we can find the empirical formula of the compound. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. o Second, we find the number of mols of each atom. o Third, we find the lowest whole number ratio of mols. 4 Honors Chemistry In Class Notes Percent Composition and Chemical Formulas For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. 52.2 g of C 13.1 g of H 34.7 g of O o Next, we we find the number of mols of each atom. nC = mC/MC = 52.2 g/12.0 g/mol = 4.35 mol C nH = mH/MH = 13.1 g/1.01 g/mol = 13.0 mol H nO = mO/MO = 34.7 g/16.0 g/mol = 2.17 mol O o Finally, we find the lowest whole number ratio of mols. nC/nO = 4.35 mol/2.17 mol = 2/1 nH/nO = 13.0 mol/2.17 mol = 6/1 o This means that there is a ratio of 2:6:1 for C:H:O The empirical formula is C2H6O For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. 44.9 g of K 18.4 g of S 36.7 g of O o Next, we we find the number of mols of each atom. nK = mK/MK = 44.9 g/39.1 g/mol = 1.15 mol K nS = mS/MS = 18.4 g/32.1 g/mol = 0.573 mol S nO = mO/MO = 36.7 g/16.0 g/mol = 2.29 mol O o Finally, we find the lowest whole number ratio of mols. nK/nS = 1.15 mol/0.573 mol = 2/1 nO/nS = 2.29 mol/0.573 mol = 4/1 o This means that there is a ratio of 2:1:4 for K:S:O The empirical formula is K2SO4 5 The Mole: A Measurement of Matter Sample Problem 10.11 A compound is analyzed and found to contain25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. Next, we find the number of mols of each atom. Finally, we find the lowest whole number ratio of mols. Practice problems: find the empirical formulas of compounds that have the following percent compositions. 1. 11.21% H and 88.79% O 2. 92.24% C and 7.76% H 3. 39.99% C, 6.73% H, and 53.28% O 4. 24.27% C, 4.08%H, and 71.65% Cl 5. 39.96% N, 14.40% H, and 45.64% O 6 Honors Chemistry In Class Notes Percent Composition and Chemical Formulas 7 Molecular Formulas In ________________ ________________, the total number of atoms in the compound is used. o Benzene has a molecular formula of ________________. o Acetic acid has a molecular formula of ________________. o Propane has a molecular formula of ________________. If we know the ________________ ________________ of a compound, then we can find the molecular formula of the compound, if we know the ________________ ________________ of the compound. o First, we determine the empirical mass of the compound. o Second, we divide the molar mass by the empirical mass to get our multiplier. o Third, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula. For example, we have a compound with a molecular formula of CH2O and a molar mass of 180.16 g/mol. What is the molecular formula of the compound? o First, we determine the empirical mass of the compound. EM = (1×12.01) + (2×1.01) + (1×16.00) = 30.03 o Next, we divide the molar mass by the empirical mass to find the multiplier. M/EM = (180.16)/(30.03) = 5.999333999 ≈ 6 o Finally, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula. CH2O ⇒ C6H12O6 Sample Problem 10.12 Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. The Mole: A Measurement of Matter Practice problems: find the molecular formulas of compounds that have the following empirical formulas and molar masses. 1. CH2O; M = 60.0 g/mol 2. CH2; M = 42.1 g/mol 3. NaCO2; M = 134.0 g/mol 4. CH2Cl; M = 98.96 g/mol 5. NH5O; M = 35.06 g/mol 8