Ch. 3 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements Brady & Senese, 5th Ed.
Index 3.1 The mole conveniently links mass to number of atoms or molecules 3.2 Chemical formulas relate amounts of substances in a compound 3.3 Chemical formulas can be determined from experimental mass measurements 3.4 Chemical equations link amounts of substances in a reaction 3.5 The reactant in shortest supply limits the amount of product that can form 3.6 The predicted amount of product is not always obtained experimentally 2
Particles Have Characteristics Masses The same mass may not represent the same number of molecules Suppose one rabbit has a mass of 250 g. What mass in kg would a case of 24 rabbits have? 24 rabbits 6.0 kg 250 g rabbit kg 1000g 3.1 The mole conveniently links mass to number of atoms or molecules 3
Counting Atoms By Their Mass The mass of an atom is called its atomic mass Atomic mass provides a means to count atoms by measuring the mass of a sample The periodic table gives atomic masses of the elements in u per atom to reduce rounding errors, use the most precise values possible 3.1 The mole conveniently links mass to number of atoms or molecules 4
Learning Check How many atoms of C are there in 3.5 10 8 u? 3.5 10 8 u 1atom C 12.0107 u 2.9 10 7 What is the mass (in u) of 2.33 10 16 atoms of H? 2.35 10 16 u 16 1.00794 u 2.33 10 atoms 1atom H atomic masses: C=12.0107 u; H=1.00794 u 3.1 The mole conveniently links mass to number of atoms or molecules 5
Your Turn! Given that the atomic mass of Ba is 137.327u, what is the mass of 23 atoms of Ba? A. 3.2 10 3 u B. 3.2 10-4 u C. 1.37 10 2 u D. none of these 3.1 The mole conveniently links mass to number of atoms or molecules 6
Your Turn! A new element is discovered that has a mass of 3.2 10 2 u for15 atoms. What is the atomic mass? A. 3.2 10 2 B. 0.047 C. 21 D. not enough information E. None of these answers 3.1 The mole conveniently links mass to number of atoms or molecules 7
Relationships 1.66 10-27 kg = 1 u (from the inside back cover of the book) may also be written as: 6.0223 10 23 u = 1 g ( a form you will often use) We can use this as a conversion factor to convert between mass quantities in u, and those in g 1 g 23 6.0223 10 u atomic mass units (u) grams (g) 6.0223 10 1 g 23 u 3.1 The mole conveniently links mass to number of atoms or molecules 8
Relationships Atomic Mass (AM) u = 1 particle We can use this as a conversion factor to convert between these quantities. 1 particle AM u particles mass (u) AM u particle 3.1 The mole conveniently links mass to number of atoms or molecules 9
Learning Check How many u of Na are there in 55.2 kg Na? 3 10 g 55.2kg kg 6.0223 10 g 23 u 3.32 10 28 u How many g Na are there in 3.2 x 10 15 u of Na? 1g 3.2 10 u 23 6.0223 10 u 15 5.3 10-9 g 3.1 The mole conveniently links mass to number of atoms or molecules 10
Your Turn! Which of the following are not equivalent to a sample of 10.5 10 7 u of Cu? A. 1.74 10-16 g B. 1.65 10 6 atoms C. 63.54 u D. None of these 3.1 The mole conveniently links mass to number of atoms or molecules 11
What Is The Formula Mass Of? Ba 3 (PO 4 ) 2 : (NH 4 ) 2 CO 3 : 601.9261 u/fu 96.08603 u/fu atomic masses: Ba: 137.327(7)u; P:30.973761(2)u; O: 15.9994(3)u; H:1.00794u; N:14.00672u; C 12.0107(8)u 3.1 The mole conveniently links mass to number of atoms or molecules 12
Relationships Formula mass (FM) u = 1 particle We can use this as a conversion factor to convert between these quantities. 1 particle FM u particles mass (u) FM u particle 3.1 The mole conveniently links mass to number of atoms or molecules 13
Counting Molecules By Their Masses The molecular mass allows counting of molecules by mass The molecular mass is the sum of atomic masses of the atoms in the compound s formula Strictly speaking, ionic compounds do not have a molecular mass, we describe an analogous quantity- the formula mass - to cover all possibilities 3.1 The mole conveniently links mass to number of atoms or molecules 14
Learning Check: How many molecules of CO 2 are there in 3.5 10 8 u? 3.5 10 8 1 molecule u 44.0095 6 CO u 2 8.0 10 6 u What is the mass (in u) of 2.33 10 16 molecules of H 2? 2.33 10 16 atoms 2.01588u 1moleculeH 4.70 10 16 u atomic masses: C=12.0107 u; H=1.00794 u; O=15.99943 u 2 3.1 The mole conveniently links mass to number of atoms or molecules 15
What Is a Mole? One mole of any substance contains the same number of units, called Avogadro s number, N 1 mole formula units = 6.0223 x 10 23 formula units It is a large quantity of particles because the particles described are so small. 3.1 The mole conveniently links mass to number of atoms or molecules 16
Why is Molar Mass the Same as Formula Mass? suppose we start with 12.0107g of C. How many atoms of C are there? given that the atomic mass of C is 12.0107 u 12.0107gC 6.0223 10 g 12.0107 g C = 6.0223 x 10 23 atoms thus for any substance, the formula mass (in g) corresponds to the same number of atoms, N 23 u 1 atom 12.0107 u 3.1 The mole conveniently links mass to number of atoms or molecules 17
Molar Mass One mole contains the same number of particles as the number of atoms in exactly 12 g of carbon-12 The molar mass of a substance has the same numeric value as the formula mass The value is different because the units are different Thus if the formula mass of Ba 3 (PO 4 ) 2 is 610.332 u/fu, the molar mass of Ba 3 (PO 4 ) 2 is 610.332 g/mol 3.1 The mole conveniently links mass to number of atoms or molecules 18
Relationships MM g = 1 mole Use this as a conversion factor to convert between these quantities 1mole MMg Mass (g) mole MMg 1mole 3.1 The mole conveniently links mass to number of atoms or molecules 19
Learning Check: Converting Between Mass And Moles Given that the molar mass of CO 2 is 44.0098 g/mol What mass of CO 2 is found in 1.55 moles? 1.55 mol 44.0098 g mol 68.2 g How many moles of CO 2 are there in 10 g? mol g 44.0098 g 10 0.2 mol 3.1 The mole conveniently links mass to number of atoms or molecules 20
Your Turn! What is the molar mass of Ca 3 (PO 4 ) 2 in g/mol? Ca: 40.078 ; P: 30.973761 ; O:15.9994 A. 279.203 B. 215.205 C. 310.177 D. none of these 3.1 The mole conveniently links mass to number of atoms or molecules 21
Your Turn! What mass in g, of Ca 3 (PO 4 ) 2 (MM=310.1767) would a 3.2 mole sample have? A. 1.0 10-3 g B. 9.9 10 2 g C. 6.0 10 26 g D. 1.6 10-21 g E. None of these 3.1 The mole conveniently links mass to number of atoms or molecules 22
Using Avogadro s Number, N Counting formula units by moles is no different than counting eggs by the dozen (12 eggs) or pens by the gross (144 pens) Since the individual particle is very small, the mole is a more practical quantity It is a group, in which 6.0223 10 23 individuals comprise 1 mole The quantity, N, is Avogadro's number and is measured as 6.0223 10 23 3.1 The mole conveniently links mass to number of atoms or molecules 23
Relationships N particles = 1 mole We can use this as a conversion factor to convert between these quantities 1mole N particles particle Moles N particles mole 3.1 The mole conveniently links mass to number of atoms or molecules 24
Learning Check: Mole Conversions Calculate the formula units of Na 2 CO 3 in 1.29 moles of Na 2 CO 3 23 6.0223 10 fu 1.29 mol mol How many moles of Na 2 CO 3 are there in 1.15 x 10 5 formula units of Na 2 CO 3? mol 1.15 10 fu 23 6.0223 10 fu 5 1.91 10-19 7.77 10 23 fu mol 3.1 The mole conveniently links mass to number of atoms or molecules 25
Relationships Between Quantities # particles N moles FM MM mass (u) N mass (g) 3.1 The mole conveniently links mass to number of atoms or molecules 26
Your Turn! Which of the following is not a relationship, but is a sample size? A. molar mass B. Avogadro s number C. formula mass D. Mass in u E. None of these 3.1 The mole conveniently links mass to number of atoms or molecules 27
Your Turn! Given that you have a sample of 5.5 g Na 2 CO 3 how many formula units are present? Na: 22.989770 ; C: 12.011; O:15.9994 # particles moles N A. 6.0 10 23 B. 5.2 10-2 C. 3.2 10-23 D. 3.3 10 24 E. None of these 3.1 10 22 FM mass (u) N mass (g) MM 3.1 The mole conveniently links mass to number of atoms or molecules 28
Using The Chemical Formula To relate components of a compound to the compound quantity we look at the chemical formula In Na 2 CO 3 there are 3 relationships: 2 mol Na: 1 mol Na 2 CO 3 1 mol C: 1 mol Na 2 CO 3 3 mol O: 1 mol Na 2 CO 3 We can also use these on the atomic scale,e.g.: 1 atom C:1 fu Na 2 CO 3 3.2 Chemical formulas relate amounts of substances in a compound 29
Learning Check: Calculate the number of moles of sodium in 2.53 moles of sodium carbonate 2 molna mol 1mol Na 2 CO 3 2.53 5.06 mol Na Calculate the number of atoms of sodium in 2.53 moles of sodium carbonate 2 mol Na 2.53 mol 1mol Na CO 2 3 6.0223 10 atoms 1mol Na 3.05 10 24 atoms Na 23 Na 3.2 Chemical formulas relate amounts of substances in a compound 30
Your Turn! How many atoms of iron are in a 15.0 g sample of iron(iii) oxide (MM 159.6885 9 g/mol)? A. 1.13 10 23 B. 9.39 10-2 C. 5.66 10 22 D. 1.88 10-1 E. None of these 3.2 Chemical formulas relate amounts of substances in a compound 31
Percent Composition Percent composition is a list of the mass percent of each element in a compound Na 2 CO 3 is 43.38% Na 11.33% C 45.29% O What is the sum of the percent composition of a compound? 3.3 Chemical formulas can be determined from experimental mass measurements 32
Percent Composition: How Is It Calculated? What is the % C in CO 2? Determine the molar mass of the compound MM=44.0095 6 g/mol Multiply the ratio of the mass of the element to the molar mass of the compound by 100 (12.0107/44.00965 6 ) 100= 27.2911 %C MM g/mol C:12.0107; O:15.99943 3.3 Chemical formulas can be determined from experimental mass measurements 33
Learning Check A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound? 0.1417 g N 0.1417g + 0.4045 g total 25.94% N 100 74.06% O 3.3 Chemical formulas can be determined from experimental mass measurements 34
Your Turn! A 35.5 g sample is analyzed and found to contain 23.5% Si. What mass of Si is present in the sample? A. 6.62 10-1 g B. 8.88 10 1 g C. 1.51 10 2 g D. 8.34 g E. None of these 3.3 Chemical formulas can be determined from experimental mass measurements 35
Empirical vs. Molecular Formulas The empirical formula is the lowest whole number ratio of atoms in a compound Note that the molecular formula is a whole number multiple of the empirical formula. glucose C 6 H 12 O 6 CH 2 O C 1x6 H 2x6 O 1x6 3.3 Chemical formulas can be determined from experimental mass measurements 36
Strategy Convert starting quantities to moles Divide all quantities by the smallest number of moles to get the smallest ratio of moles Convert any non-integers into integers If any number ends in a common decimal equivalent of a fraction, multiply by the least common denominator Otherwise, round the numbers to the nearest integers 3.3 Chemical formulas can be determined from experimental mass measurements 37
Common Ratios And Their Decimal Equivalents decimal Fraction equivalent multiplier For example:.25 ¼ or ¾ 4 or.75.3333 1/3 or 2/3 3 or.6667.50 ½ 2 1.25 x 4 = 5 3.3 Chemical formulas can be determined from experimental mass measurements 38
Learning Check: A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula mass(g) MM mol lowest ratio integer ratio N O 0.522 1.490 14.00674 15.99943 0.0372 68 0.09312 83 1 2.50 2 5 3.3 Chemical formulas can be determined from experimental mass measurements 39
Determining The Multiplier, n Ratio of the molecular mass to the mass predicted by the empirical formula and round to an integer n = molecular formula mass empirical formula mass The actual molecule is larger by this amount If the empirical formula is A x B y, the molecular formula will be A n x B n y 180.1572 g C H 30.0262 g CH O O 6 12 6 glucose n = = 6 2 3.3 Chemical formulas can be determined from experimental mass measurements 40
Example: The empirical formula of hydrazine is NH 2, and its molecular mass is 32.0. What is its molecular formula? n=(32.0/16.02)=2 N 2 H 4 A substance is known to be 35.00% N, 5.05% H and 59.96% O. What is its EF? Determine the Molecular Formula if the MM of the compound is 80.06 g/mol EF: N 2 H 4 O 3 n=(80.06/80.043)=1 N 2 H 4 O 3 MM: N:14.00674; H:1.00794; O:15.99943 3.3 Chemical formulas can be determined from experimental mass measurements 41
Your Turn! Given the composition analysis of lindane (a controversial pesticide ) what is its empirical formula? C H Cl A. C 24 H 2 Cl 73 24.77928% 2.07943% 73.14129% B. C 2 H 2 Cl 2 C. C 142 HCl 126 D. CHCl E. None of these 3.3 Chemical formulas can be determined from experimental mass measurements 42
Your Turn! We found that the empirical formula was CHCl. Given that the MM is 290.8316 g/mol, what is the molecular formula? A. C 6 H 6 Cl 6 B. C 8 H 17 Cl 5 C. C 3 H 5 Cl 7 D. C 5 H 18 Cl 7 E. none of these 3.3 Chemical formulas can be determined from experimental mass measurements 43
Combustion Analysis: Empirical formulas may also be calculated indirectly When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen, only carbon dioxide and water are produced 3.3 Chemical formulas can be determined from experimental mass measurements 44
Combustion Analysis: Empirical formulas may be calculated from the analysis of combustion information grams of C can be derived from amount of CO 2 grams of H can be derived from amount of H 2 O the mass of oxygen is obtained by difference: g O = g sample ( g C + g H ) 3.3 Chemical formulas can be determined from experimental mass measurements 45
Learning Check: The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO 2 and 4.512 g of H 2 O. Calculate the empirical formula of the compound. 12.0107 g C 7.406 g CO2 44.00956 g CO2 2.01588 g H 4.512 g H O 18.01531g H2O 2.021 18 g C 2.5048 84 g H 5.217g- 2.021 18 g C-.5048 84 g H= 2.690 94 g O H: 1.00794; C:12.0107; O: 15.99943 3.3 Chemical formulas can be determined from experimental mass measurements 46
Learning Check (con.): Calculate the empirical formula of the compound. H: 1.00794; C:12.0107; O: 15.99943 C H O mass 2.02118 0.504884 2.69094 MM 12.0107 1.00794 15.99943 mol low ratio.16828 1.500907.16819 2.97 1 CH 3 O integer ratio 1 3 1 3.3 Chemical formulas can be determined from experimental mass measurements 47
Your Turn! Combustion analysis of 3.88 g of a compound containing C, H, and S reveals the following data. What is the empirical formula of the compound? A. C 6 H 5 S B. C 9 H 2 S C. C 5 H 5 S D. C 3 H 9 S 2 E. None of these CO 2 9.377 g H 2 O 1.59 g 3.3 Chemical formulas can be determined from experimental mass measurements 48
What Does The Balanced Equation Mean? 2CO (g) + O 2(g) 2CO 2(g) For every 2 CO reacted, 1 O 2 is also reacted and 2 CO 2 are also reacted 3.4 Chemical equations link amounts of substances in a reaction 49
Using The Balanced Equation: The balanced equation gives the relationship between amounts of reactants used and amounts of products likely to be formed The numeric coefficient tells: how many individual particles are needed in the reaction on the microscopic level how many moles are necessary on the macroscopic level The stoichiometric coefficient 3.4 Chemical equations link amounts of substances in a reaction 50
Stoichiometric Ratios Consider the reaction N 2 + 3H 2 2NH 3 What is the ratio between N 2 and H 2? 1 mole N 2 : 3 mole H 2 N 2 and NH 3? 1mole N 2 : 2 mole NH 3 H 2 and NH 3? 3 mole H 2 : 2 mole NH 3 3.4 Chemical equations link amounts of substances in a reaction 51
Learning Check: For the reaction N 2 + 3 H 2 2NH 3, How many moles of N 2 are used when 2.3 moles of NH 3 are produced? 2.3 mol NH 3 1 mol N2 2 mol NH If 0.575 mole of CO 2 is produced by the combustion of propane, C 3 H 8, how many moles of oxygen are consumed? The balanced equation is C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O 5 mol O 2 mol CO 2 3 mol CO 2 0.575 0.958 mol O 2 3 3.4 Chemical equations link amounts of substances in a reaction 52 1.2 mol N 2
Learning Check How many grams of Al 2 O 3 are produced when 41.5 g Al react? 2Al (s) + Fe 2 O 3(s) Al 2 O 3(s) + 2 Fe (l) 41.5 g Al 1mol Al 26.9815 g Al 78.4 g Al 2 O 3 1mol Al2O 2 mol Al 3 101.9613 gal 1molAl2O 2 3 O 3 MM (g/mol): Al: 26.9815; Al 2 O 3 :101.9613 3.4 Chemical equations link amounts of substances in a reaction 53
Your Turn! Given the reaction: H 2 SO 4 + 2KOH 2H 2 O + K 2 SO 4, How many moles of KOH are required to make 3.0 moles of K 2 SO 4? A. 3.0 moles B. 6.0 moles C. 1.5 moles D. None of these 3.4 Chemical equations link amounts of substances in a reaction 54
Your Turn! Given the reaction: H 2 SO 4 + 2KOH 2H 2 O + K 2 SO 4, How many g of H 2 O (18.0153) would result from the complete reaction of 1.2 g H 2 SO 4 (98.08)? A. 2.4 g B. 1.2 g C. 0.60 g D. 0.44 g E. none of these 3.4 Chemical equations link amounts of substances in a reaction 55
Balancing By Inspection Balance the most complex substance in the equation first Balance elements, H and O last Use coefficients to adjust quantities, not subscripts Some equations may be balanced using fractions, but the most common approach allows only for integer coefficients If polyatomic ions remain intact in a reaction balance them as a group If you have an even/odd problem dilemma, multiply all previously balanced moieties by 2 3.4 Chemical equations link amounts of substances in a reaction 56
Learning Check: Balance The Following: Ba(OH) 1 2(aq) + 1 Na 2 SO 4(aq) BaSO 1 4(s) + NaOH 2 (aq) KClO 2 3(S) KCl 2 (s) + 3 O 2(g) H 2 3 PO 4(aq) + 3 Ba(OH) 2(aq) Ba 1 3 (PO 4 ) 2(s) + H 6 2 O (l) 3.4 Chemical equations link amounts of substances in a reaction 57
Your Turn! Given the following reaction: KCl + Hg 2 (NO 3 ) 2 KNO 3 + Hg 2 Cl 2, when it is balanced, what is the coefficient for KCl? A. 1 B. 2 C. 3 D. 4 E. none of these
Limiting Reagent Consider the reaction of N 2 with H 2 to form NH 3: N 2(g) + 3H 2(g) 2NH 3(g) The stoichiometry suggests that for every mole of N 2 we will need 3 moles of H 2 to form 2 moles of NH 3. So what happens if these proportions are not met? The reaction proceeds, to use up one of the reactants (the limiting reagent) and will not use all of the other reactant (it is in excess) 3.5 The reactant in shortest supply limits the amount of product that can form 59
Limiting Reagents Note that in this reaction, some of the O 2 is not consumed. This is because there is not enough CO to continue consuming the O 2. Thus, CO is the limiting reagent. 3.5 The reactant in shortest supply limits the amount of product that can form 60
Determining The Limiting Reagent (LR) There are several approaches to this. One method is to compare the quantities available to the quantities required. Any substance present in excess of the requirement cannot be limiting. 3.5 The reactant in shortest supply limits the amount of product that can form 61
Learning Check: Ca(OH) 2(aq) + 2HCl (aq) 2 H 2 O (l) + CaCl 2(s) when 1.00 g of each reactant is combined: What is the theoretical yield of H 2 O? The limiting reagent? Ca(OH) 2 HCl 1.00 74.09468 1.00 36.46094 0.013496 0.027427 0.0134 96 mol Ca(OH) 1 2 mol H2O 1mol Ca(OH) 0.027427 mol HCl 2 mol H2O = 0.0274 1 2 mol HCl 2 2 H 2 O 0.486 277 = 0.0269 27 18.01528 0.0269 92 92 mol H O 2 mol H2O mass (g) MM (g/mol) mol TY H 2 O (mol) Ca(OH) 2: 74.09468; HCl: 36.46094; H 2 O: 18.01528 3.5 The reactant in shortest supply limits the amount of product that can form 62
Learning Check: How many grams of NO can form when 30.0 g NH 3 and 40.0 g O 2 react according to: 4 NH 3 + 5 O 2 4 NO + 6 H 2 O NH 3 30.0 17.03052 O 2 40.0 31.9988 NO 30.0 30.0061 1.76 15 1.25 00 1.00 00 1.76 1.25 15 00 mol NH3 1 mol O2 1 mass (g) MM (g/mol) mol 4 mol NO TY NO (mol) = 1.7615mol NO 4 mol NH3 4 mol NO = 1.0000mol NO 5 mol O2 NH 3 : 17.03052; O 2 =31.9988; NO: 30.0061 3.5 The reactant in shortest supply limits the amount of product that can form 63
Your Turn! Given 1.0 g each of KCl and Hg 2 (NO 3 ) 2, what is the expected mass of Hg 2 Cl 2? A. 1.0 g B. 2.0 g C. 0.90 g D. 3.2 g E. none of these KCl Hg 2 (NO 3 ) 2 Hg 2 Cl 2 74.5513 525.1899 472.086 MM (g/mol)
Actual Yield Often we do not obtain the quantity expected This may be due to errors, mistakes, side reactions, contamination or a host of other events Thus we describe the actual yield, the amount obtained experimentally 3.6 The predicted amount of product is not always obtained experimentally 65
Percent Yield The amount of product, predicted by the limiting reagent is termed the theoretical yield Percent yield relates the actual yield to the theoretical yield It is calculated as: actual yield % = x100 theoretical yield If a cookie recipe predicts a yield of 36 cookies and yet only 24 are obtained, what is the % yield? 24 % = x100 36 3.6 The predicted amount of product is not always obtained experimentally 66