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1 The mass of a single atom is too small to measure on a balance. mass of hydrogen atom = 1.673 x 10-24 g 2 This is an infinitesimal 1.673 x 10-24 g mass 3 The standard to which the masses of all other atoms are compared to was originally chosen to be the most abundant isotope of hydrogen. 1 1 H 4 A mass of exactly 1 atomic mass units (amu) was assigned to 1 1 H 5 Electric charge and mass of electrons, protons, & neutrons Particle symbol Actual charge (coulombs) Relative charge Actual mass (g) Relative mass (amu) Electron e– –1.602 10–19 –1 9.11 10–28 0 Proton p +1.602 10–19 +1 1.67 10–24 1 Neutron n 0 1.68 10–24 1 0 6 • Most elements occur as mixtures of isotopes. • Isotopes of the same element have different numbers of neutrons. • The listed atomic mass of an element is the average relative mass of the isotopes of that element compared originally to the mass of hydrogen: 1amu (now carbon). 7 • Balances in the lab room weigh in metric ,i.e., grams not in amu’s. So how do we change the number on the periodic table from amu’s to grams? • We need to know how many atoms of each element we need to obtain the amu weight in grams. • This amount (a big pile) would convert amu’s to grams for each element. 8 1 pile = 1 mole (in Greek) 1 mole = 6.02 x 23 10 objects 9 6.02 x 23 10 is a very LARGE number 10 6.02 x 23 10 is Avogadro’s Number number 11 Amadeo (Amedeo) Avogadro 12 1 mole of any element contains 6.02 x 1023 particles of that substance. 13 The atomic weight in grams of any element23 contains 1 mole of atoms. 14 How many seconds in 1 minute? How many minutes in 1 hour? How many hours in 1 day? So how many seconds in 1 day? 60 × 60 × 24 = 86400 15 Suppose you count at a rate of 1 number per second? How long to count to 6.02 × 1023? 6.02 1023 6.97 1018 days 86400 How many years is this? 6.97 1018 1.91 1016 years 365 16 How many years? 19,100,000,000,000,000 How old is the universe? 14 billion years 1.4 × 1010 14,000,000,000 ǃ 17 Examples 18 Species Quantity Number of H atoms H 1 mole 6.02 x 23 10 19 Species Quantity Number of H2 molecules H2 1 mole 6.02 x 23 10 20 Species Quantity Number of Na atoms Na 1 mole 6.02 x 23 10 21 Species Quantity Number of Fe atoms Fe 1 mole 6.02 x 23 10 22 Species C6H6 Quantity 1 mole Number of 23 6.02 x 10 C6H6 molecules 23 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 24 • 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. 25 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 26 Problems 27 Convert To Moles! 28 Weight (mass) Numbers Moles Periodic Table Avogadro’s Number 6.02 × 1023 29 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 mole Fe (grams Fe) 55.85 g Fe 1 mole Fe (25.0 g Fe) 0.448 mole Fe 55.85 g Fe 30 How many iron atoms are contained in 40.0 grams of iron? Atomic weight iron = 55.85 Conversion sequence is: grams Fe → moles Fe → atoms Fe 1 mole Fe 6.02 x 1023 atoms Fe (40.0 g Fe) 1 mole Fe 55.85 g Fe 4.31 x 1023 atoms Fe 31 What is the mass of 3.01 x 1023 atoms of sodium (Na)? Mole weight Na = 22.99 g Conversion sequence: atoms Na → moles Na → grams Na 1 mole Na 22.99 g Na (3.01 x 10 atoms Na) 23 6.02 x 10 atoms Na 1 mole Na 23 11.5 g Na 32 How many oxygen atoms are present in 2.00 moles of oxygen molecules? Conversion sequence: moles O2 → molecules O2 → atoms O 6.02 x 1023 molecules O2 (2.00 mole O2 ) 1 mole O2 2 atoms O 1 molecule O 2 = 2.41 x1024 atoms O 33 Mole Weight of Compounds 34 The mole weight of a compound can be determined by adding the mole weights of all of the atoms in its formula. 35 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 36 Calculate the mole weight of lithium perchlorate. 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 37 Calculate the mole weight of ammonium phosphate. (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 38 Solids that contain water as part of their crystalline structure are known as hydrates. Water in a hydrate is known as water of hydration or water of crystallization. 39 Formulas of hydrates are written by first writing the formula for the anhydrous compound and then adding a dot followed by the number of water molecules present. CoCl2 6H2O 40 Calculate the mole weight of NaC2H3O2 · 3 H2O Sodium acetate trihydrate 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 41 Avogadro’s Number of Particles 6.02 x 1023 Particles 1 MOLE Mole Weight 42 Avogadro’s Number of Ca atoms 6.02 x 1023 Ca atoms 1 MOLE Ca 40.078 g Ca 43 Avogadro’s Number of H2O molecules 6.02 x 1023 H2O molecules 1 MOLE H2O 18.02 g H2O 44 In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules. 45 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 46 Problems 47 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 = 4.992 moles C6 H 6 (390.0 g C6 H 6 ) 78.12 g C6 H 6 48 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 )3PO4 (2.52 mole (NH 4 )3PO 4 ) 1 mole (NH 4 )3PO4 = 376g (NH 4 )3 PO4 49 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 mole N 2 28.02 g N 2 6.02 x 1023 molecules N 2 1 mole N 2 1 mole N 2 6.02 x 10 molecules N 2 (56.04 g N 2 ) 1 mole N 2 28.02 g N 2 23 = 1.20 x 1024 molecules50N 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 mole N 2 6.02 x 1023 molecules N 2 2 atoms N 1 molecule N 2 28.02 g N 2 1 mole N 2 1 mole N 2 6.02 x 1023 molecules N 2 (56.04 g N 2 ) 1 mole N 28.02 g N 2 2 2 atoms N 1 molecule N 2 = 2.41 x 1024 atoms N 51 Percent Composition of Compounds 52 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 53 Percent Composition From Formula 54 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. 55 total mass of the element x 100 = percent of the element mole weight 56 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 57 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% 58 Percent Composition From Experimental Data 59 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. 60 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 61 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 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% 62 Empirical Formula versus Molecular Formula 63 • 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. 64 • 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. 65 Examples 66 Molecular Formula C2H4 Empirical Formula CH2 Smallest Whole Number Ratio C:H 1:2 67 Molecular Formula C6H6 Empirical Formula CH Smallest Whole Number Ratio C:H 1:1 68 Molecular Formula H2O2 Empirical Formula HO Smallest Whole Number Ratio H:O 1:1 69 70 Two compounds can have identical empirical formulas and different molecular formulas. 71 72 Calculating Empirical Formulas 73 The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance. C___O___ 74 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. 75 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. 76 The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance. C___O___ 27g 73g 12.0 g/mole 16.0 g/mole C___O___ 2.3 4.6 moles 2.3moles moles 2.3moles 77 The analysis of a compound shows that it contains 27% carbon and 73% oxygen. Calculate the empirical formula for this substance. C___O___ 1 2 78 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 79 • 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. 80 Problems 81 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 82 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 mole K atoms K: 56.58 g K 1.447 mole K atoms 39.10 g K 1 mole C atoms 0.723 mole C atoms mole C atoms C: 8.68 g C 0.723 C has the smallest number 12.01 g C of moles 1 mole O atoms O: 34.73 g O 2.171 mole O atoms 83 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 mole 0.723 mole K= = 2.00 C: = 1.00 0.723 mole 0.723 mole 0.723 mole C atoms 2.171 mole O= = 3.00 C has the smallest number 0.723 mole of moles The simplest ratio of K:C:O is 2:1:3 Empirical formula K2CO3 84 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 85 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 mole N atoms N: 25.94 g N 1.852 mole N atoms 14.01 g N 1 mole O atoms O: 74.06 g O 4.629 mole O atoms 16.00 g O 86 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 mole 4.629 mole N= = 1.000 O: = 2.500 1.852 mole 1.852 mole This is not a ratio of whole numbers. 87 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 88 Calculating the Molecular Formula from the Empirical Formula 89 • 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 90 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 91 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 92 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: 1mole N 138gcomp'd 30.4 gN 3mole N 14.0 g 1mole comp'd 100 gcomp'd 138gcomp'd 69.6gO 1mole O 6mole O 1mole comp'd 100 gcomp'd 16.0 g N3O6 93 94