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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas 1 How you measure how much? 2 n You can measure mass, or volume, or you can count pieces. n n We measure mass in grams. We measure volume in liters. n We count pieces in MOLES. Moles n n n n 3 Defined as the number of carbon atoms in exactly 12 grams of carbon-12. 1 mole is 6.02 x 1023 particles. Treat it like a very large dozen 6.02 x 1023 is called Avagadro’s number. Representative particles n n n n 4 The smallest pieces of a substance. For a molecular compound it is a molecule. For an ionic compound it is a formula unit. For an element it is an atom. Types of questions n n How many oxygen atoms in the following? 3 – CaCO3 – Al2(SO4)3 12 How many ions in the following? – CaCl2 – NaOH 3 2 – Al2(SO4)3 5 5 Types of questions using the equality; 1 mole = 6.02 x 1023 n n n n 6 How many molecules of CO2 are the in 4.56 moles of CO2 ? 4.56 mole x 6.02x1023 mc = 2.75x1024mc 1 1 mole How many moles of water is 5.87 x 1022 molecules? 5.87 x 1022 mc x 1 mole = 0.0975 mole 1 6.02x1023 mc Types of questions using the equality; 1 mole = 6.02 x 1023 n How many atoms of carbon are there in 1.23 moles of C6H12O6 ? n 1.23 moles x 6.02x1023 mc x 6 atoms = 4.44x1024 atoms 1 1 mole 1 mc n How many moles is 7.78 x 1024 formula units of MgCl2? 7.78x1024 FU x 1 mole 1 6.02x1024 FU 7 = 12.9 mole Measuring Moles n n n n 8 Remember relative atomic mass? The amu was one twelfth the mass of a carbon 12 atom. Since the mole is the number of atoms in 12 grams of carbon-12, the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams. Gram Atomic Mass n n n n 9 The mass of 1 mole of an element in grams. 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron. We can right this as 12.01 g C = 1 mole We can count things by weighing them. Examples How much would 2.34 moles of carbon weigh? n 2.34 moles C x 12 g C = 28.08 g 1 1mole n How many moles of magnesium in 24.31 g of Mg? 24.31 g Mg x 1 mole = 1.013 mole 1 24g Mg n 10 11 n How many atoms of lithium in 1.00 g of Li? n 1.00 g Li x 1 mole x 6.02x1023 atoms 8.60x1022 atoms 1 7 g Li 1 mole n How much would 3.45 x 1022 atoms of U weigh? n 3.45x1022 atoms U x 1 mole x 238 g U 1 6.02x1023atoms 1 mole 13.6 g What about compounds? n n n n n 12 in 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound determine the moles of the elements they have Find out how much they would weigh add them up What about compounds? n n n n n n 13 What is the mass of one mole of CH4? 1 mole of C = 12 g 4 mole of H x 1 g = 4g 1 mole CH4 = 12 + 4 = 16g The Gram Molecular mass of CH4 is 16.05g The mass of one mole of a molecular compound. Gram Formula Mass n n n n n n 14 The mass of one mole of an ionic compound. Calculated the same way. What is the GFM of Fe2O3? 2 moles of Fe x 56 g = 112 g 3 moles of O x 16 g = 48 g The GFM = 112 g + 48 g = 160g Molar Mass n n 15 The generic term for the mass of one mole. The same as gram molecular mass, gram formula mass, and gram atomic mass. Examples n n n Calculate the molar mass of the following and indicate what type it is. Na2S 2 (23) + 32 = 46 + 32 = 78g 1mole Na2S = 78g Gram Formula Mass n N2O4 n 2(14) + 4(16) = 28 + 64 = 92g 1 mole N2O4 = 92g Gram Molecular Mass n C 16 12g 1 mole C = 12 g Gram Atomic Mass Molar Mass Cont. n n n Ca(NO3)2 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g 1 mole Ca(NO3)2 = 164g Gram Formula Mass C6H12O6 n 6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g n 1 mole C6H12)6 = 180g Gram Molecular Mass n n n n 17 (NH4)3PO4 3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g 1 mole (NH4)3PO4 = 149g Gram Formula Mass Using Molar Mass Finding moles of compounds Counting pieces by weighing 18 Molar Mass 19 n The number of grams of 1 mole of atoms, ions, or molecules. n We can make conversion factors from these. To change grams of a compound to moles of a compound. For example n 20 How many moles is 5.69 g of NaOH? For example n How many moles is 5.69 g of NaOH? 5.69 g 21 For example n l 22 How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles For example n l l 23 How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH For example n l l l 24 How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16.00 g 1 mole of H = 1 g For example n l l l l 25 How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g 1 mole NaOH = 40 g For example n l l l l 26 How many moles is 5.69 g of NaOH? 1 mole 5.69 g 40.00 g need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g 1 mole NaOH = 40 g For example n How many moles is 5.69 g of NaOH? 5.69 g l l l l 27 1 mole = 0.142 mol NaOH 40.00 g need to change grams to moles for NaOH 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g 1 mole NaOH = 40 g Examples n n n n 28 How many moles is 4.56 g of CO2? 4.56g CO2 x 1 mole = 0.104 moles 1 44gCO2 How many grams is 9.87 moles of H2O? 9.87 moles x 18g H2O = 178g 1 1 mole Examples n How many molecules in 6.8 g of CH4? n 6.8g CH4 x 1 mole x 6.02x1023 mc 1 16g CH4 1mole n 49 molecules of C6H12O6 weighs how much? n 49 mc x 1 mole x 180g C6H12O6 = 1 6.02x1023 mc 1 mole n 29 8820 g____= 1.5x10-20 g 6.02x1023 = 2.56x1023 mc Gases and the Mole 30 Gases Many of the chemicals we deal with are gases. n They are difficult to weigh. n Need to know how many moles of gas we have. n Two things effect the volume of a gas n Temperature and pressure n Scientists compare gases at Standard Temperature and Pressure n 31 Standard Temperature and Pressure 0ºC and 1 atm pressure n abbreviated STP n At STP 1 mole of gas occupies 22.4 L n Called the molar volume n Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. n 32 Examples n n n n n n 33 What is the volume of 4.59 mole of CO2 gas at STP? 4.59mole x 22.4L = 102.816 = 103L 1 1mole How many moles is 5.67 L of O2 at STP? 5.67L x 1 mole = .523moles 1 22.4L What is the volume of 8.80g of CH4 gas at STP? 8.80g CH4 x 1mole x 22.4L = 12.32 = 12.3L 1 16g CH4 1mole We have learned how to change moles to grams n moles to atoms n moles to formula units n moles to molecules n moles to liters n molecules to atoms n formula units to atoms n formula units to ions n 34 Periodic Table Moles 35 Mass Volume Periodic Table Moles 36 Mass Volume 22.4 L Periodic Table Moles 37 Mass Volume 22.4 L Periodic Table Moles Representative Particles 38 Mass Volume 22.4 L Periodic Table Moles 6.02 x 23 10 Representative Particles 39 Mass Volume 22.4 L Periodic Table Moles 6.02 x 23 10 Representative Particles Atoms 40 Mass Volume 22.4 L Periodic Table Mass Moles 6.02 x 23 10 Representative Particles Atoms 41 Ions Percent Composition n n n n 42 Like all percents Part x 100 % whole Find the mass of each component, divide by the total mass. Example n Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S. Ag 29.0g /33.3 = .8709 x 100 = 87.09% n S n n 43 + 4.30g /33.3 = .1291 x 100 = 12.91% 33.3g Getting % from the formula n n 44 If we know the formula, assume you have 1 mole. Then you know the pieces and the whole. Examples n Calculate the percent composition of C2H4? n C 2(12g)=24 /28 = .8571 x 100 = 85.71% n H 4(1g) = +4 /28 = .1429 x 100 = 14.29% 28g n 45 Example n Calculate the percent composition of Aluminum carbonate. Al2(CO3)3 n Al 2(27g)= 54 /234 n C 3(12g)= 36 /234 = .15.38 x 100 = 15.38% n O 9(16)= 144 /234 = .6154 x 100 = 61.54% n n 46 234g = .2308 x 100 = 23.08% n n n n n n n n 47 You can also calculate the mass of an element in a given amount of a compound using % composition. • Step 1: calculate the % comp. only of the element you want to find the mass of. • Step 2: Multiply the elements %, by the mass of the compound given. Example: Calculate the mass of sulfur in 3.54g of H2S. MM of H2S = H 2 (1) = 2 S 1(32) = +32 34g H2S % S = 32/34 x 100 = 94.1% S n 94.1% x 3.54g = 3.33g S Calculate the mass of nitrogen in 25g of (NH ) CO . 42 3 n Calculate the mass of nitrogen in 25g of (NH4)2CO3. n n n n n N 2(14g) H 8(1g) C 1(12g) O 8(16g) = 28 = 8 = 12 = +128 176g (NH ) CO 42 3 n %N = 28/176 x 100 = 15.91% n 15.91% x 25g = 4.0g N 48 Calculate the mass of magnesium in 97.4g of Mg(OH)2. n n n n n n 49 Mg 1 (24g) = 24 O 2(16g) = 32 H 2 ( 1g) = +2 58g Mg(OH)2 %Mg = 24/58 x 100 = 41.38% 41.38% x 97.4g = 40.3g Mg Empirical Formula From percentage to formula 50 The Empirical Formula The lowest whole number ratio of elements in a compound. n The molecular formula the actual ration of elements in a compound. n The two can be the same. n CH2 empirical formula n C2H4 molecular formula n C3H6 molecular formula n H2O both n 51 Calculating Empirical Just find the lowest whole number ratio n C6H12O6 n CH4N n It is not just the ratio of atoms, it is also the ratio of moles of atoms. n In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. n In one molecule of CO2 there is 1 atom of C and 2 atoms of O. n 52 Calculating Empirical Means we can get ratio from percent composition. n Assume you have a 100 g. n The percentages become grams. n Can turn grams to moles. n Find lowest whole number ratio by dividing by the smallest moles. n 53 Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. n Assume 100 g so n 38.67 g C x 1mol C = 3.223 mole C 12 g C n 16.22 g H x 1mol H = 16.22 mole H 1gH n 45.11 g N x 1mol N = 3.222 mole N 14 g N n 54 Example The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N n The ratio is 16.00 mol H = 5 mol H 3.222 mol N 1 mol N n C1H5N1 n 55 n n n n n n 56 A compound is 43.64 % P and 56.36 % O. What is the empirical formula? 43.64 g P x 1mol P = 1.408 mole P 31 g P 56.36 g O x 1mol O = 3.523 mole O 16 g O The ratio is 3.523 mol O = 2.5 mol O 1.408 mol P 1 mol P Can not have 2.5 atoms! Double P2O5 n n n n n 57 Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? 49.48 g C x 1mol C = 4.123 mole C 12 g C 5.15 g H x 1mol H = 5.15 mole H 1gH 28.87 g N x 1mol N = 2.062 mole N 14 g N 16.49 g O x 1mol O = 1.031 mole O 16 g O n n n n 58 The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O The ratio is 5.15 mol H = 5 mol H 1.031 mol O 1 mol O The ratio is 2.062 mol N = 2 mol N 1.031 mol O 1 mol O C 4H 5N 2O 1 Empirical to molecular n n n n n n 59 n Since the empirical formula is the lowest ratio the actual molecule would weigh more. By a whole number multiple. Divide the actual molar mass by the the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular formula? C4H5N2O1 = 97g 194/97 = 2 Molecular Formula = C8H10N4O2 Example n A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula? 71.65 g Cl x 1mol Cl = 2.047 mole Cl 35 g Cl n 24.27 g C x 1mol C = 2.023 mole C 12 g C n 4.07 g H x 1mol H = 4.07 mole H 1gH n 60 n n n n n n 61 The ratio is 2.047 mol Cl = 1 mol Cl 2.023 mol C 1 mol C The ratio is 4.07 mol H = 4 mol H 1.031 mol C 1 mol C Empirical Formula = CClH4 12+35+4 = 51g 98.96/51 = 2 Molecular Formula = C2Cl2H8