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1 The Mole 2 The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x 10-24 g 4 This is an infinitesimal 1.673 x 10-24 g mass 5 • Chemists require a unit for counting which can express large numbers of atoms using simple numbers. • Chemists have chosen a unit for counting atoms. • That unit is the 6 1 mole = 6.02 x 23 10 objects 7 6.02 x 23 10 is a very LARGE number 8 6.02 x 23 10 is Avogadro’s Number number 9 Amadeo (Amedeo) Avogadro 10 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number. 11 1 mole of any element contains 6.02 x 1023 particles of that substance. 12 The atomic weight in grams of any element23 contains 1 mole of atoms. 13 This is the same number of particles 6.02 x 1023 as there are in exactly 12 grams of 12 6 C 14 Examples 15 Species Quantity Number of H atoms H 1 mole 6.02 x 23 10 16 Species Quantity Number of H2 molecules H2 1 mole 6.02 x 23 10 17 Species Quantity Number of Na atoms Na 1 mole 6.02 x 23 10 18 Species Quantity Number of Fe atoms Fe 1 mole 6.02 x 23 10 19 Species C6H6 Quantity 1 mole Number of 23 6.02 x 10 C6H6 molecules 20 1 mol of atoms = 6.02 x 1023 atoms 1 mol of molecules = 6.02 x 1023 molecules 1 mol of ions = 6.02 x 1023 ions 21 • The mole weight of an element is its atomic weight in grams. • It contains 6.02 x 1023 atoms (Avogadro’s number) of the element. 22 H Atomic Number of Mole weight mass atoms 1.008 amu 1.008 g 6.02 x 1023 Mg 24.31 amu 24.31 g 6.02 x 1023 Na 22.99 amu 22.99 g 6.02 x 1023 Element 23 Problems 24 Convert To Moles! 25 How many moles of iron does 25.0 g of iron represent? Atomic weight iron = 55.85 Conversion sequence: grams Fe → moles Fe Set up the calculation using a conversion factor between moles and grams. 1 mol Fe (grams Fe) 55.85 g Fe 1 mol Fe (25.0 g Fe) 0.448 mol Fe 55.85 g Fe 26 How many iron atoms are contained in 40.0 grams of iron? Atomic weight iron = 55.85 Conversion sequence: grams Fe → atoms Fe Set up the calculation using a conversion factor between atoms and grams. 6.02 x 1023 atoms Fe (grams Fe) 55.85 g Fe 6.02 x 1023 atoms Fe 23 (40.0 g Fe) 4.31 x 10 atoms Fe 55.85 g Fe 27 What is the mass of 3.01 x 1023 atoms of sodium (Na)? Mole weight Na = 22.99 g Conversion sequence: atoms Na → grams Na Set up the calculation using a conversion factor between grams and atoms. 22.99 g Na (atoms Na) 23 6.02 x 10 atoms Na 22.99 g Na (3.01 x 10 atoms Na) 11.5 g Na 23 6.02 x 10 atoms Na 28 23 What is the mass of 0.365 moles of tin? Atomic weight tin = 118.7 Conversion sequence: moles Sn → grams Sn Set up the calculation using a conversion factor between grams and atoms. mole weight Sn (moles Sn) 1 mole Sn 118.7 g Sn (0.365 moles Sn) 43.3 g Sn 1 mole Sn 29 How many oxygen atoms are present in 2.00 mol of oxygen molecules? Two conversion factors are needed: 6.02 x 10 molecules O2 1 mol O 2 23 2 atoms O 1 molecule O 2 Conversion sequence: moles O2 → molecules O2 → atoms O 6.02 x 1023 molecules O2 2 atoms O (2.00 mol O2 ) 1 mol O2 1 molecule O2 = 2.41 x1024 atoms O 30 Mole Weight of Compounds 31 The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula. 32 Calculate the mole weight of C2H6O. 2 C = 2(12.01 g) = 24.02 g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = 16.00 g 46.08 g 33 Calculate the mole weight of LiClO4. 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = 35.45 g 4 O = 4(16.00 g) = 64.00 g 106.39 g 34 Calculate the mole weight of (NH4)3PO4 . 3 N = 3(14.01 g) = 42.03 g 12 H = 12(1.01 g) = 12.12 g 1 P = 1(30.97 g) = 30.97 g 4 O = 4(16.00 g) = 64.00 g 149.12 g 35 Calculate the mole weight of NaC2H3O2 · 3 H2O Solids that contain water molecules as part of their structure are called Hydrates. 1 Na = 1(22.99 g) = 22.99 g 2 C = 2(12.01 g) = 24.02 g 9 H = 9(1.01 g) = 9.09 g 5 O = 5(16.00 g) = 80.00 g 136.10 g 36 Avogadro’s Number of Particles 6.02 x 1023 Particles 1 MOLE Mole Weight 37 Avogadro’s Number of Ca atoms 6.02 x 1023 Ca atoms 1 MOLE Ca 40.078 g Ca 38 Avogadro’s Number of H2O molecules 6.02 x 1023 H2O molecules 1 MOLE H2O 18.02 g H2O 39 These relationships are present when hydrogen combines with chlorine. H Cl HCl 6.02 x 1023 H atoms 6.02 x 1023 Cl atoms 6.02 x 1023 HCl molecules 1 mol H atoms 1 mol Cl atoms 1 mol HCl molecules 1.008 g H 35.45 g Cl 36.46 g HCl 1 mole weight H atoms 1 mole weight Cl atoms 1 mole weight 40 HCl molecules In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules. 41 Calculate the mole weight of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the mole weight of 1 mole of H2 molecules. 2 H = 2(1.01 g) = 2.02 g 42 Problems 43 How many moles of benzene, C6H6, are present in 390.0 grams of benzene? The mole weight of C6H6 is 78.12 g. Conversion sequence: grams C6H6 → moles C6H6 78.12 grams C6 H 6 Use the conversion factor: 1 mole C6 H 6 1 mole C6 H 6 = 5.000 moles C6 H 6 (390.0 g C6 H 6 ) 78.12 g C6 H 6 44 How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4? The mole weight of (NH4)3PO4 is 149.12 g. Conversion sequence: moles (NH4)3PO4 → grams (NH4)3PO4 149.12 grams (NH 4 )3PO4 Use the conversion factor: 1 mole (NH 4 )3PO4 149.12 g (NH 4 )3 PO4 (2.52 mol (NH 4 )3PO 4 )) 1 mol (NH 4 )3 PO4 = 376g (NH 4 )3 PO445 56.04 g of N2 contains how many N2 molecules? The mole weight of N2 is 28.02 g. Conversion sequence: g N2 → moles N2 → molecules N2 Use the conversion factors 1 mol N 2 28.02 g N 2 6.02 x 1023 molecules N 2 1 mol N 2 1 mol N 2 6.02 x 10 molecules N 2 (56.04 g N 2 ) 1 mol N 2 28.02 g N 2 23 = 1.20 x 1024 molecules46N 2 56.04 g of N2 contains how many N atoms? The mole weight of N2 is 28.02 g. Conversion sequence: g N2 → moles N2 → molecules N2 → atoms N Use the conversion factors 1 mol N 2 6.02 x 1023 molecules N 2 2 atoms N 28.02 g N 2 1 molecule N 2 1 mol N 2 1 mol N 2 6.02 x 1023 molecules N 2 (56.04 g N 2 ) 28.02 g N 1 mol N 2 2 2 atoms N 1 molecule N 2 = 2.41 x 1024 atoms N 47 Percent Composition of Compounds 48 Percent composition of a compound is the mass percent of each element in the compound. H2O 11.19% H by mass 88.79% O by mass 49 Percent Composition From Formula 50 If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the mole weight of the formula. Step 2 Divide the total mass of each element in the formula by the mole weight and multiply by 100. 51 total mass of the element x 100 = percent of the element mole weight 52 Calculate the percent composition of hydrosulfuric acid H2S(aq) . Step 1 Calculate the mole weight of H2S. 2 H = 2 (1.01 g) = 2.02 g 1 S = 1 (32.07 g) = 32.07 g 34.09 g 53 Calculate the percent composition of hydrosulfuric acid H2S. Step 2 Divide the mass of each element by the mole weight and multiply by 100. 2.02 g H H: (100) = 5.93% 34.09 g 32.07 g S S: 34.09 g (100) 94.07% S H 94.07% 5.93% 54 Percent Composition From Experimental Data 55 Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100. 56 A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of product 57 Calculate the percent composition of hydrosulfuric acid H2S. Step 2 Divide the mass of each element by the total mass of the compound formed. 1.52 g N (100) = 30.5% 4.99 g 3.47 g O 4.99 g (100) = 69.5% O 69.5% N 30.5% 58 Empirical Formula versus Molecular Formula 59 • The empirical formula or simplest formula gives the smallest wholenumber ratio of the atoms present in a compound. • The empirical formula gives the relative number of atoms of each element present in the compound. 60 • The molecular formula is the true formula of a compound. • The molecular formula represents the total number of atoms of each element present in one molecule of a compound. 61 Examples 62 Molecular Formula C2H4 Empirical Formula CH2 Smallest Whole Number Ratio C:H 1:2 63 Molecular Formula C6H6 Empirical Formula CH Smallest Whole Number Ratio C:H 1:1 64 Molecular Formula H2O2 Empirical Formula HO Smallest Whole Number Ratio H:O 1:1 65 66 Two compounds can have identical empirical formulas and different molecular formulas. 67 68 Calculating Empirical Formulas 69 Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element’s mole weight. 70 Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value – If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. – If the numbers obtained are not whole numbers, go on to step 4. 71 Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. FeO1.5 Fe1 x 2O1.5 x 2 Fe2O3 72 • The results of calculations may differ from a whole number. – If they differ ±0.1 round off to the next nearest whole number. 2.9 3 – Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number. 73 Problems 75 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. K = 56.58 g C = 8.68 g O = 34.73 g 76 The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 2 Convert the grams of each element to moles. 1 mol K atoms 1.447 mol K atoms K: 56.58 g K 39.10 g K 1 mol C atoms C atoms atoms C: 8.68 g C 0.723 mol C 12.01 g C C has the smallest number of moles 1 mol O atoms 2.171 mol O atoms O: 34.73 g O 77 16.00 g O The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance. Step 3 Divide each number of moles by the smallest value. 1.447 mol 0.723 mol K= = 2.00 C: = 1.00 0.723 mol 0.723 mol 0.723 mol C atoms 2.171 mol O= = 3.00 C has the smallest number 0.723 mol of moles The simplest ratio of K:C:O is 2:1:3 Empirical formula K2CO3 78 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound. N = 25.94 g O = 74.06 g 79 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 2 Convert the grams of each element to moles. 1 mol N atoms 1.852 mol N atoms N: 25.94 g N 14.01 g N 1 mol O atoms O: 74.06 g O 4.629 mol C atoms 16.00 g O 80 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 3 Divide each number of moles by the smallest value. 1.852 mol N= = 1.000 1.852 mol 4.629 mol O: = 2.500 1.852 mol This is not a ratio of whole numbers. 81 The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 4 Multiply each of the values by 2. N: (1.000)2 = 2.000 O: (2.500)2 = 5.000 Empirical formula N2O5 82 Calculating the Molecular Formula from the Empirical Formula 83 • The molecular formula can be calculated from the empirical formula if the mole weight is known. • The molecular formula will be equal to the empirical formula or some multiple n of it. • To determine the molecular formula evaluate n. • n is the number of units of the empirical formula contained in the molecular formula. mole weight n= = empirical weight number of empirical formula units 84 What is the molecular formula of a compound which has an empirical formula of CH2 and a mole weight of 126.2 g? Let n = the number of formula units of CH2. Calculate the mass of each CH2 unit 1 C = 1(12.01 g) = 12.01g 2 H = 2(1.01 g) = 2.02g 14.03g 126.2 g n 9 (empirical formula units) 14.03 g The molecular formula is (CH2)9 = C9H18 85 A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula. Old Way: 30.4 g 1mole 2.17 mole of N 14.0 g 1mole 69.6g 4.35mole of O 16.0 g 138 3units 46 N2.17 O4.35 NO2 46g/mole 2.17 2.17 3(NO2 ) N3O6 86 A compound made of 30.4% N and 69.6% O has a mole weight of 138 grams/mole. Find the molecular formula. New Way: 138gcomp'd 30.4 gN 1mole N 3mole N 1mole comp'd 100 gcomp'd 14.0 g 138gcomp'd 69.6gO 1mole O 6mole O 1mole comp'd 100 gcomp'd 16.0 g N3O6 87 88