Chapter 4 Chemical Composition. Moles of Various Elements and Compounds Figure 4.8

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1 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 x 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 Moles of Various Elements and Compounds Figure

2 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 Compound H O Na Cl H 2 O NaCl Atomic Mass Molar Mass 4-5 Practice Solutions Molar Mass Element or Compound Atomic Mass Molar Mass H O Na Cl H 2 O NaCl amu amu amu amu amu amu g/mol g/mol g/mol g/mol g/mol g/mol 4-6 2

3 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 mass of sample x 100% 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 (CaCO 3 ) 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? 2.54 g Fe % Fe = x 100% = 30.5% Fe in sample 8.33 g sample 2.91g S % S = x 100% = 34.9% S in sample 8.33 g sample 4-9 3

4 Practice Solutions - Percent Composition 2. A 4.55-g sample of limestone (CaCO 3 ) contains 1.82 g of calcium. What is the percent Ca in limestone? 1.82 g Ca % 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 = g S We can write this equality as a ratio as well: 1 mole S or g S g S 1 mole S To solve, we need to cancel out the grams of S. Therefore: 1mol S 1.28 g S x = g S mol S

5 Extra Practice Conversions with Molar Mass 1. How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? 2. If one aspirin tablet contains g of acetylsalicylic acid (C 9 H 8 O 4 ), 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 (C 14 H 18 N 2 O 5 ) 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: 1g C14H18N2O5 40. mg C14 H18N2O5 x = 0.040g C14H18N2O mg C14H18N2O5 2. Next, we need to find the MM of C 14 H 18 N 2 O 5 : (14 moles C x g C) + (18 moles H x 1.01 g H) + 1 mol C 1 mol H (2 moles N x g N) + (5 moles O x g O) 1 mol N 1 mol O = g/mol C 14 H 18 N 2 O Extra Practice Solutions Continued Conversions with Molar Mass 1. How many moles of aspartame (C 14 H 18 N 2 O 5 ) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? 3. Next, convert from grams to moles: 1mol C14H18N2O g C14H18N2O5 x = 1.4 x 10 mol C14H18N2O g C14H18N2O5 Finally, we convert from moles to molecules: x 10 molecules C14H18N2O5 1.4 x 10 mol C14H18N2O5 x = 1mol C14H18N2O x 10 molecules C14H 18N2O

6 Extra Practice Solutions Conversions with Molar Mass 2. If one aspirin tablet contains g of acetylsalicylic acid (C 9 H 8 O 4 ), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets? g C9H8O4 1mol C9H8O x 10 molecules C9H8O4 x x 1aspirin tablet g C9H8O4 1mol C9H8O4 = 1.08 x molecules C 9 H 8 O 4 1 aspirin tablet x 10 molecules C9H8O 2 aspirin tablets x 1aspirin tablet = 2.16 x molecules C 9 H 8 O 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 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

7 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 cyclopentane cyclohexane Molecular Formula C 5 H 10 C 6 H 12 Empirical Formula CH 2 CH 2 ethylene C 2 H 4 CH 2 hydrogen sulfide calcium chloride H 2 S This compound does not have a molecular formula H 2 S CaCl

8 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 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

9 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 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 g Cu x 1 mol Cu = 1.05 mol Cu 1.05 mol Cu = g Cu 1.05 mol 33.5 g S x 1 mol S = 1.05 mol S 1.05 mol S = 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 g Cu x 1 mol Cu = mol Cu = 2.5 mol Cu x 2 = 5 mol Cu g Cu g Si x 1 mol Si = mol Si = 2 mol Si x 2 = 4 mol Si g Si g O x 1 mol O = mol O = 7 mol O x 2 = 14 mol O g O g H x 1 mol H = mol H = 1 mol H x 2 = 2 mol H 1.01 g H Therefore, the empirical formula is: Cu 5 Si 4 O 14 H

10 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) 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 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 g/mol. What are the empirical and molecular formulas of potassium persulfate? First, find the empirical formula: g K x 1 mol K = mol K = 1 mol K g K g S x 1 mol S = mol S = 1 mol S g S g O x 1 mol O = mol O = 4 mol O g O Thus, the empirical formula is KSO

11 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 g/mol. What are the empirical and molecular formulas of potassium persulfate? To find the molecular formula: Calculate the MM of KSO 4 = (1 mol K x g/mol K) + (1 mol S x g/mol S) + (4 mol O x g/mol O) = g/mol KSO g/mol (Experimental MM) = g/mol (Empirical Formula s MM) Thus, the molecular formula is K 2 S 2 O Determining Percent Composition Using Molar Mass To determine the percent composition of an element (E) in a compound using molar mass (MM): [ MM (E) x # of moles E in compound] % E = 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 1 st ] 1. iron(ii) chloride 2. dinitrogen tetroxide 3. sodium phosphate

12 Practice Solutions Percent Composition Calculate the percent composition of each element in the following compounds: 1. iron(ii) chloride FeCl 2 Find the MM of the compound 1 st: (1 mol Fe x g/mol Fe) + (2 mol Cl x g/mol Cl) = g/mol FeCl 2 % Fe = (1 mol Fe x g/mol Fe) = 44.06% Fe in FeCl g/mol FeCl 2 % Cl = (2 mol Cl x g/mol Cl) = 55.94% Cl in FeCl g/mol FeCl 2 NOTE: % Cl = % FeCl % 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 N 2 O 4 Find the MM of the compound 1 st: (2 mol N x g/mol N) + (4 mol O x g/mol O) = g/mol % N = (2 mol N x g/mol N) = 30.45% N in N 2 O g/mol N 2 O 4 % O = (4 mol O x g/mol O) = 69.55% O in N 2 O g/mol N 2 O 4 NOTE: % O = % N 2 O % 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 Na 2 PO 4 Find the MM of the compound 1 st: (2 mol Na x g/mol Na) + (1 mol P x g/mol P) + (4 mol O x g/mol O) = g/mol Na 2 PO 4 % Na = (2 mol Na x g/mol Na) = 32.62% Na in Na 2 PO g/mol Na 2 PO 4 % P = (1 mol P x g/mol P) = 21.97% P in Na 2 PO g/mol Na 2 PO 4 % O = (4 mol O x g/mol O) = 45.41% O in Na 2 PO g/mol Na 2 PO

13 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

14 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 H 2 S dissolved in sufficient water to give ml of solution. What is the molarity of the solution? 2. Bluestone is copper(ii) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of g/mol. A sample of pond water was found to have a concentration of 6.2 x 10 5 M copper(ii) sulfate. If the pond has a volume of 1.8 x 10 7 L, then what mass of bluestone did the farmer add to the pond?

15 Practice Solutions Solution Concentration 1. A solution is prepared from 22.5 g of H 2 S dissolved in sufficient water to give ml of solution. What is the molarity of the solution? M = moles of solute L solution H 2 S is the solute, so Moles of H 2 S = 22.5 g H 2 S x 1 mole H 2 S = moles H 2 S g H 2 S To find L of solution: ml x 1 L = L 1000 ml M = moles H 2 S = 2.64 M H 2 S solution L solution 4-43 Practice Solutions Continued Solution Concentration 2. Bluestone is copper(ii) sulfate pentahydrate, CuSO 4 5H 2 O, with a molar mass of g/mol. A sample of pond water was found to have a concentration of 6.2 x 10 5 M copper(ii) sulfate. If the pond has a volume of 1.8 x 10 7 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 10 7 L x 6.2 x 10 5 mol x g = 2.8 x g CuSO 4 5H 2 O 1 L 1 mol 4-44 Dilution and the Dilution Equation Dilution The process of adding more solvent to solution The dilution equation: Moles con = M con V con Moles 1 = M 1 V 1 M con V con = M dil V dil M 1 V 1 = M 2 V

16 Diluting a More Concentrated Solution 4-46 Practice Dilution 1. If 42.8 ml of 3.02 M H 2 SO 4 solution is diluted to a final volume ml, what is the molarity of the diluted solution of H 2 SO 4? 2. What is the concentration of a solution prepared by diluting 35.0 ml of M KBr to ml? 4-47 Practice Solutions Dilution 1. If 42.8 ml of 3.02 M H 2 SO 4 solution is diluted to a final volume ml, what is the molarity of the diluted solution of H 2 SO 4? M con V con = M dil V dil (3.02 M)(42.8 ml) = M dil (500.0 ml) M dil = (3.02 M)(42.8 ml) = M ml

17 Practice Solutions Dilution 2. What is the concentration of a solution prepared by diluting 35.0 ml of M KBr to ml? M con V con = M dil V dil (0.150 M)(35.0 ml) = M dil (250.0 ml) M dil = (0.150 M)(35.0 ml) = M ml

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