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9/30/2011 Chapter 7 Chemical Formula Relationships Elements == > Formula In each chemical formula, one can determine the number of atoms of each element by its corresponding subscripts in each component or formula unit. Exception: If the subscript is 1, it is usually omitted for that component in the formula. Let’s look at some examples!!! EXAMPLES Chemical Formulas Mg(NO3)2 In the formula for magnesium nitrate, this formula is comprised of the following: 1 Mg atom 2 N atoms 6 O atoms HC2H3O2 (aq) In the formula for acetic acid, this formula is comprised of the following: 4 H atoms 2 O atoms 2 C atoms 1 9/30/2011 ATOMIC Mass It is the average mass of the element as they exist in nature. The atomic mass is determined in comparison to the mass of an atom of carbon-12, which by definition is exactly 12 atomic mass units (12u). Calculate ATOMIC Mass CO2 Figure 7.1-Molecular (or formula) mass for carbon dioxide. To calculate the formula mass of a compound, it equals: Formula Mass = Σ atomic masses in the formula unit Rounding Molar Mass When determining the final molar mass of a compound, one must follow these rules: 1. Calculate the formula mass using atomic masses in as many decimal places as they are shown. 2. Round off the answer to the same number of decimal places as the smallest number of decimal places in any atomic mass used in the calculation. 2 9/30/2011 Sample Question Calculate the formula mass of barium nitrate. Solution: FORMULA: Ba(NO3)2 1 Ba 137.3 u 137.3u 2N 2(14.01u) 28.02u 6O 6(16.00u) 96.00u ----------------------------------------------------Molecular Mass = 261.32u The final molar mass of Ba(NO3)2 is 261.3u. Reminders Molar Mass In our textbook, molar mass can be as accurate as the number of significant figures of the atomic mass. In our class, each element has been written to 4 significant figures. For determining the molar mass of a compound, write the molar mass rounded to the element which has the least number of significant figures right of the decimal point. Mole Concept By definition, the “mole” is the amount of any substance that contains the same number of units as the number of atoms in exactly 12 grams of a carbon12 isotope. The definition of a mole refers to a number of particles, but it doesn’t say what that number is. By experiment, it has been found that to 3 significant figures: 1 mol of any substance = 6.02 x 1023 of that substance 3 9/30/2011 Avogadro’s NUMBER (Na) Avogadro’s number (by experiment): 6.02214179 × 1023/mol Avogadro’s number can be written as a conversion factor: 1 mole 6.02 × 1023 items 6.02 × 1023 items 1 mole It is similar as to if you were determining the number of eggs of 1 dozen eggs by analogy as the following manner: 12 eggs dozen 6.02 × 1023 items 1 mole Sample Problem How many moles of NaCl are in 5.66 x 1024 molecules of NaCl? ANSWER: 5.66 ×10 24 molecules NaCl × 1 mole NaCl 6.02 × 10 23 molecules NaCl = 9.40 mole NaCl Molar Mass Molar mass is defined as the mass of one mole of a substance. Molar mass is a bridge between a macroscopic quantity of matter, grams, and a particulate-level quantity of matter, moles. The units of molar mass by definition is in units of grams per mole (g/mol). molar mass = mass g = mole mol 4 9/30/2011 Molar Mass The definitions of atomic mass, the mole, and molar mass are all directly or indirectly related to carbon-12. This leads to two important facts: 1. The mass of one atom of carbon-12—the atomic mass of carbon-12—is exactly 12 atomic mass units. 2. The mass of one mole of carbon-12 atoms is exactly 12 grams; its molar mass is 12 grams per mole. This leads to the conclusion: The molar mass of any substance in grams per mole is numerically equal to the atomic, molecular, or formula mass of that substance in atomic mass units. Illustration Molar Mass Figure 7.2-(a) One mole of elements and (b) One mole of molecular and ionic compounds Molar Mass What is the molar mass of CH4? Express your answer in grams/mole. 1C 12.01 g 12.01g 4H 4(1.008g) 4.032g ----------------------------------------------------Molecular Mass = 16.042g The final molar mass of CH4 is 16.04g. 5 9/30/2011 Moles, Grams, and Particles With knowing that molar mass can be determined by using the periodic table, and Avogadro’s number, one can make the relationship between each unit for an atom or molecule. Figure 7.3-Conversion among grams, moles, and number of particles. Sample Problem Determine the number of water molecules in 1.5kg of water. Formula of water: H2O Molar Mass (H2O) = 18.02g Avogadro’s Number = 6.02 x 1023 molecules/mol 1 mole H2O 6.02 × 1023 molecules H2O 1.5kg × 1000g × × = 5.01×1025 molecules H 2O 1kg 18.02g H 2O 1 mol H 2O Percent Composition For percent composition, one can determine the percent of one component of a mixture by the following component using the following equation: %A = parts of A ×100% total parts 6 9/30/2011 Sample Problem What percentage of oxygen is in calcium oxide? Formula of calcium oxide (CaO) Molar Mass = 40.08 g/mol Ca + 16.00g/mol O = 56.08g %O= 16.00g ×100% = 28.53% 56.08g Sample Problem Calculate the percentage composition of sodium nitrate. FORMULA: NaNO3 Molar Mass = 22.99g/mol Na 14.01g/mol N +3(16.00g/mol O) ------------------85.00 g/mol NaNO3 Sample Problem (cont’d) Now, to find the percentage of each element. Na: % Na = 22.99g ×100% = 27.05% 85.00g N: %N = 14.01g × 100% = 16.48% 85.00g O: %O = 48.00g × 100% = 56.47% 85.00g TOTAL Percentage: 27.05% + 16.48% + 56.47% = 100.00% 7 9/30/2011 Empirical Formula An empirical formula is the lowest wholenumber ratio between the atoms in a compound. EXAMPLES: For C2H4, the empirical formula is CH2. For C3H6, the empirical formula is CH2. Therefore, the empirical formula of CH2 will have the same empirical formula and same percent composition. Practice Problems Determine the empirical formula (EF) for each of the following substances. a) b) c) d) e) C3H3 P4O10 B2H6 C6H14 Ag2O2 EF: EF: EF: EF: EF: Determining an Empirical Formula 1) Find the masses of different elements in a sample of the compound. 2) Convert the masses into moles of atoms of the different elements. 3) Determine the ratio of moles of atoms. 4) Express the moles of atoms as the smallest possible ratio of integers. 5) Write the empirical formula, using the number for each atom in the integer ratio as the subscript in the formula. 8 9/30/2011 Practice Problem Hydroquinone, used as a photographic developer is 65.4% C, 5.5% H, and 29.1% O by mass. What is the empirical formula of hydroquinone? Solution: It is usually helpful in an empirical formula problem to organize the calculations in a table with the following headings: Element Element C H O Grams Moles Mole Ratio Grams Moles Mole Ratio 65.4 65.4g -------------12.01g/mol 5.45 --------1.82 = 5.45 mol C =2.99 5.5g -------------1.008g/mol 5.5 --------1.82 = 5.5 mol H =3.02 29.1g -------------16.00g/mol 1.82 --------1.82 = 1.82 mol O =1 5.5 29.1 Formula Ratio Formula Ratio Empirical Formula Empirical Formula 3 3 C3H3O 1 Molecular Formula A molecular formula illustrates the total number of atoms present of each element in the compound. This formula can be determined by knowing how many empirical formula units are present in the substance. For example, if the empirical formula is CH, the molecule could be the following: CH, C2H2, C3H3, C4H4, C5H5, C6H6 empirical formula units in 1 molecule = molar mass of compound molar mass of empirical formula 9 9/30/2011 Determining an Molecular Formula 1) Determine the empirical formula of the compound. 2) Calculate the molar mass of the empirical formula unit. 3) Determine the molar mass of the compound (which will be given in this book). 4) Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n (the number of empirical formula units per molecule). 5) Write the molecular formula. Practice Problem Sorbic acid is added to food as a mold inhibitor. Its composition is 64.3% C, 7.2% H, and 28.5% O, and its molecular mass is 112 g/mol. What is its molecular formula? Element Grams Moles Mole Ratio Element Grams Moles Mole Ratio C 64.3 64.3g -------------12.01g/mol 5.35 --------1.78 = 5.35 mol C =3.01 7.2g -------------1.008g/mol 7.1 --------1.78 = 7.1 mol H =3.99 28.5g -------------16.00g/mol 1.78 --------1.78 = 1.78 mol O =1 H O 7.2 28.5 Formula Ratio Formula Ratio Empirical Formula Empirical Formula 3 4 C3H4O 1 10 9/30/2011 Solution (cont’d) Finding the molar mass of C3H4O. Molar Mass (EF) = n= 3(12.01 g/mol C) 4(1.008 g/mol H) +16.00 g/mol O ------------56.06 g/mol C3H4O 112 g/mol = 2.00 = 2 56.06 g/mol (C H O) = C H O 3 4 2 6 8 2 11