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Chapter 4 Chemical Composition • Mole Quantities • Moles, Masses, and Particles • Determining Empirical and Molecular Formulas • Chemical Composition of Solutions 4-1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Mole • The unit that acts as a bridge between the microscopic world and the macroscopic world • Contains 6.022 x 1023 particles (molecules, atoms, ions, formula units, etc.) – This number is called Avogadro’s number. • The amount of substance that contains as many basic particles (atoms, molecules, or formula units) as there are atoms in exactly 12 g of carbon-12 4-2 Moles of Various Elements and Compounds Figure 4.8 4-3 1 Molar Mass • Describes the mass of 1 mole of a substance • We obtain the Molar Mass (MM) from the periodic table by assigning different units to the atomic mass. – Instead of assigning the atomic mass units of amu, we assign the atomic mass units of grams per 1 mole. • Molar mass is the conversion factor between mass and moles. 4-4 Practice - Molar Mass • Complete the table. Element or Atomic Mass Compound Molar Mass H O Na Cl H2O NaCl 4-5 Practice Solutions – Molar Mass Element or Compound Atomic Mass Molar Mass H 1.008 amu 1.008 g/mol O 16.00 amu 16.00 g/mol Na 22.99 amu 22.99 g/mol Cl 35.45 amu 35.45 g/mol H2O 18.02 amu 18.02 g/mol NaCl 58.44 amu 58.44 g/mol 4-6 2 Percent Composition by Mass • An expression of the portion of the total mass contributed by each element • To find the percent composition of E (E is any element): %E = mass of E x 100% mass of sample 4-7 Practice - Percent Composition 1. What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S? 2. A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone? 4-8 Practice Solutions - Percent Composition 1. What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S? % Fe = %S= 2.54 g Fe 8.33 g sample 2.91 g S 8.33 g sample x 100% = 30.5% Fe in sample x 100% = 34.9% S in sample 4-9 3 Practice Solutions - Percent Composition 2. A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone? % Ca = 1.82 g Ca x 100% = 40.0% Ca in limestone 4.55 g limestone 4 - 10 Practice – Conversions with Molar Mass 1. How many moles of sulfur are in the 1.28 g of sulfur (S) found in a sample of chalcopyrite? 4 - 11 Practice Solutions – Conversions with Molar Mass 1. How many moles of sulfur are in the 1.28 g of sulfur (S) found in a sample of chalcopyrite? The conversion factor needed is the MM of sulfur: 1 mole S = 32.07 g S We can write this equality as a ratio as well: 1 mole S or 32.07 g S 32.07 g S 1 mole S To solve, we need to cancel out the grams of S. Therefore: 1.28 g S x 1 mol S = 0.0399 mol S 32.07 g S 4 - 12 4 Extra Practice – Conversions with Molar Mass 1. How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? 2. If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets? 4 - 13 Extra Practice Solutions – Conversions with Molar Mass 1. How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? 1. We first need to convert from mg to g: 1 g C14H18N2O5 40. mg C14H18N2O5 x = 0.040 g C14H18N2O5 1000 mg C14H18N2O5 2. Next, we need to find the MM of C14H18N2O5: (14 moles C x 12.01 g C) + (18 moles H x 1.01 g H) + 1 mol C 1 mol H (2 moles N x 14.01 g N) + (5 moles O x 16.00 g O) 1 mol N 1 mol O = 294.34 g/mol C14H18N2O5 4 - 14 Extra Practice Solutions Continued – Conversions with Molar Mass 1. How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? 3. Next, convert from grams to moles: 0.040 g C14H18N2O5 x • 1 mol C14H18N2O5 = 1.4 x 10-4 mol C14H18N2O5 294.34 g C14H18N2O5 Finally, we convert from moles to molecules: 1.4 x 10 -4 mol C14H18N2O5 x 6.022 x 10 23 molecules C14H18N2O5 = 8.2 x 1019 molecules C14H18N2O5 1 mol C14H18N2O5 4 - 15 5 Extra Practice Solutions – Conversions with Molar Mass 2. If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets? 0.324 g C9H8O4 1 mol C9H8O4 6.022 x 10 23 molecules C9H8O4 x x 1 aspirin tablet 180.17 g C9H8O4 1 mol C9H8O4 = 1.08 x 1021 molecules C9H8O4 1 aspirin tablet 2 aspirin tablets x 1.08 x 1021 molecules C9H8O4 1 aspirin tablet = 2.16 x 1021 molecules C9H8O4 4 - 16 Summary – Conversions with Molar Mass and Avogadro’s Number • To convert from moles to grams or from grams to moles, use a molar mass (MM) as your conversion factor. • To convert from moles to particles (molecules, atoms, ions, or formula units) or from particles to moles, use Avogadro’s number as your conversion factor. 4 - 17 Determining Empirical and Molecular Formulas • Empirical formula – Expresses the simplest ratios of atoms in a compound – Written with the smallest whole-number subscripts • Molecular formula – Expresses the actual number of atoms in a compound – Can have the same subscripts as the empirical formula or some multiple of them 4 - 18 6 Determining Empirical and Molecular Formulas 4 - 19 Practice – Empirical and Molecular Formulas • For which of these substances is the empirical formula the same as the molecular formula? 4 - 20 Empirical or Molecular Formulas Table 4.1 Some Empirical and Molecular Formulas Substance Molecular Formula Empirical Formula cyclopentane C5H10 CH2 cyclohexane C6H12 CH2 ethylene C2H4 CH2 hydrogen sulfide H2S H2S calcium chloride This compound does not have a molecular formula CaCl2 4 - 21 7 Empirical Formulas from Percent Composition 4 - 22 Finding Empirical Formulas • To find the empirical formula: 1. If starting with a percent composition, find the mass of the element by assigning the percent composition (which has no units but a % instead) the units of grams. – If starting with another set of units, then convert the units to masses if necessary. 2. Convert from mass to moles using the MM of the element. 4 - 23 Finding Empirical Formulas • To find the empirical formula cont’d: 3. Repeat for all elements in the compound. 4. Find whole number subscripts by: 1. Dividing the moles of the each element by the smallest number of moles. The quotients will give whole numbers which are now the subscripts for the empirical formula. 2. If #1 does not give whole numbers, then multiply all numbers by a multiplier that will resolve the quotients into whole numbers. 4 - 24 8 Practice – Finding Empirical Formulas 1. Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. 2. Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. 4 - 25 Practice Solutions – Finding Empirical Formulas 1. Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. First, reassign the percentages units of grams: 66.5 g Cu and 33.5 g S. Then, convert to moles and divide both numbers by the lowest number. 66.5 g Cu x 1 mol Cu = 1.05 mol Cu 1.05 mol Cu = 1 63.55 g Cu 1.05 mol 33.5 g S x 1 mol S = 1.05 mol S 1.05 mol S = 1 32.07 g S 1.05 mol The whole numbers in purple then become our subscripts. The empirical formula is therefore: CuS 4 - 26 Practice Solutions Continued – Finding Empirical Formulas 2. Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. 48.43 g Cu x 1 mol Cu = 0.7621 mol Cu = 2.5 mol Cu x 2 = 5 mol Cu 63.55 g Cu 0.3069 17.12 g Si x 1 mol Si = 0.6095 mol Si = 2 mol Si x 2 = 4 mol Si 28.09 g Si 0.3069 34.14 g O x 1 mol O = 2.134 mol O = 7 mol O x 2 = 14 mol O 16.00 g O 0.3069 0.31 g H x 1 mol H = 0.3069 mol H = 1 mol H x 2 = 2 mol H 1.01 g H 0.3069 Therefore, the empirical formula is: Cu5Si4O14H2 4 - 27 9 Molecular Formulas • • To determine a molecular formula, the problem must give a piece of experimental data, such as a molar mass, MM. To find the molecular formula: 1. Find the empirical formula first. 2. Divide the empirical formula’s molar mass by the experimental molar mass (which is given). 4 - 28 Practice - Molecular Formulas 1. Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? 4 - 29 Practice Solutions - Molecular Formulas 1. Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? First, find the empirical formula: 28.93 g K x 1 mol K = 0.7399 mol K = 1 mol K 39.10 g K 0.7396 23.72 g S x 1 mol S = 0.7396 mol S = 1 mol S 32.07 g S 0.7396 47.35 g O x 1 mol O = 2.959 mol O = 4 mol O 16.00 g O 0.7396 Thus, the empirical formula is KSO4. 4 - 30 10 Practice Solutions Continued – Molecular Formulas 1. Potassium persulfate is a strong bleaching agent. It has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate? To find the molecular formula: Calculate the MM of KSO4 = (1 mol K x 39.10 g/mol K) + (1 mol S x 32.07 g/mol S) + (4 mol O x 16.00 g/mol O) = 135.17 g/mol KSO4 270.0 g/mol (Experimental MM) = 2 135.17 g/mol (Empirical Formula’s MM) Thus, the molecular formula is K2S2O8. 4 - 31 Determining Percent Composition Using Molar Mass • To determine the percent composition of an element (E) in a compound using molar mass (MM): %E= [ MM (E) x # of moles E in compound] Total MM of compound 4 - 32 Practice – Percent Composition • Calculate the percent composition of each element in the following compounds: [HINT: you must determine the compound formula 1st] 1. iron(II) chloride 2. dinitrogen tetroxide 3. sodium phosphate 4 - 33 11 Practice Solutions – Percent Composition Calculate the percent composition of each element in the following compounds: 1. iron(II) chloride – FeCl2 Find the MM of the compound 1st: (1 mol Fe x 55.85 g/mol Fe) + (2 mol Cl x 35.45 g/mol Cl) = 126.75 g/mol FeCl2 • % Fe = (1 mol Fe x 55.85 g/mol Fe) = 44.06% Fe in FeCl2 126.75 g/mol FeCl2 % Cl = (2 mol Cl x 35.45 g/mol Cl) = 55.94% Cl in FeCl2 126.75 g/mol FeCl2 NOTE: % Cl = 100.00% FeCl2 - 44.06% Fe = 55.94% Cl 4 - 34 Practice Solutions Continued Percent Composition Calculate the percent composition of each element in the following compounds: 2. Dinitrogen tetroxide – N2O4 Find the MM of the compound 1st: (2 mol N x 14.01 g/mol N) + (4 mol O x 16.00 g/mol O) = 92.02 g/mol • % N = (2 mol N x 14.01 g/mol N) = 30.45% N in N2O4 92.02 g/mol N2O4 % O = (4 mol O x 16.00 g/mol O) = 69.55% O in N2O4 92.02 g/mol N2O4 NOTE: % O = 100.00% N2O4 – 30.45% N = 69.55% O 4 - 35 Practice Solutions Continued Percent Composition Calculate the percent composition of each element in the following compounds: 3. Sodium phosphate – Na2PO4 Find the MM of the compound 1st: (2 mol Na x 22.99 g/mol Na) + (1 mol P x 30.97 g/mol P) + (4 mol O x 16.00 g/mol O) = 140.95 g/mol Na2PO4 • % Na = (2 mol Na x 22.99 g/mol Na) = 32.62% Na in Na2PO4 140.95 g/mol Na2PO4 % P = (1 mol P x 30.97 g/mol P) = 21.97% P in Na2PO4 140.95 g/mol Na2PO4 % O = (4 mol O x 16.00 g/mol O) = 45.41% O in Na2PO4 140.95 g/mol Na2PO4 4 - 36 12 Chemical Composition of Solutions • Solutions – are any homogeneous mixture at the molecular or ionic scale – are composed of solutes and solvents • Solutes – Are present in a lesser amount – The substances that are dissolved (can be either wet or dry) • Solvents – Are present in the larger amount – The substances that dissolve 4 - 37 Making a Solution 4 - 38 Solution Concentration • Concentration – Is the relative amounts of solute and solvent in a solution – When compared with one another, solutions are classified as dilute or concentrated. • Dilute solution – A solution that contains a relatively small amount of solute • Concentrated solution – A solution that contains a relatively large amount of solute 4 - 39 13 Concentration of Solutions 4 - 40 Determining Concentration Insert diagram at bottom of pg. 138 (or pg. 140) • Percent by Mass – Expresses concentration via percentage % mass = mass solute x 100% mass solution • Molarity (M) – The moles of solute dissolved in 1 L of solution – The most common units of concentration M = moles solute L solution 4 - 41 Practice – Solution Concentration 1. A solution is prepared from 22.5 g of H2S dissolved in sufficient water to give 250.0 mL of solution. What is the molarity of the solution? 2. Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2 x 105 M copper(II) sulfate. If the pond has a volume of 1.8 x 107 L, then what mass of bluestone did the farmer add to the pond? 4 - 42 14 Practice Solutions – Solution Concentration 1. A solution is prepared from 22.5 g of H2S dissolved in sufficient water to give 250.0 mL of solution. What is the molarity of the solution? M = moles of solute L solution H2S is the solute, so Moles of H2S = 22.5 g H2S x 1 mole H2S = 0.660 moles H2S 34.09 g H2S To find L of solution: = 0.2500 L 250.0 mL x 1 L 1000 mL M = 0.660 moles H2S = 2.64 M H2S solution 0.2500 L solution 4 - 43 Practice Solutions Continued – Solution Concentration 2. Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2 x 105 M copper(II) sulfate. If the pond has a volume of 1.8 x 107 L, then what mass of bluestone did the farmer add to the pond? Start with the volume as it only has one set of units: 1.8 x 107 L x 6.2 x 105 mol x 249.7 g = 2.8 x 1015 g CuSO4•5H2O 1L 1 mol 4 - 44 Dilution and the Dilution Equation • Dilution – The process of adding more solvent to solution • The dilution equation: Molescon = MconVcon MconVcon = MdilVdil Moles1 = M1V1 M1V1 = M2V2 4 - 45 15 Diluting a More Concentrated Solution 4 - 46 Practice – Dilution 1. If 42.8 mL of 3.02 M H2SO4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H2SO4? 2. What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL? 4 - 47 Practice Solutions – Dilution 1. If 42.8 mL of 3.02 M H2SO4 solution is diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H2SO4? MconVcon = MdilVdil (3.02 M)(42.8 mL) = Mdil(500.0 mL) Mdil = (3.02 M)(42.8 mL) = 0.259 M 500.0 mL 4 - 48 16 Practice Solutions – Dilution 2. What is the concentration of a solution prepared by diluting 35.0 mL of 0.150 M KBr to 250.0 mL? MconVcon = MdilVdil (0.150 M)(35.0 mL) = Mdil(250.0 mL) Mdil = (0.150 M)(35.0 mL) = 0.0210 M 250.0 mL 4 - 49 17