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Chapter 7 Chemical Quantities Masuk High School Mr. J. K. Stoelzel 1 Section 7.1 The Mole: A Measurement of Matter OBJECTIVES: –Describe how Avogadro’s number is related to a mole of any substance. –Calculate the mass of a mole of any substance. 2 What is a Mole? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES. 3 Moles (abbreviated: mol) 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 Avogadro’s number. 4 What is the Mole? 1 mole of hockey pucks would equal the mass of the moon! 1 mole of basketballs would fill a bag the size of the earth! 1 mole of pennies would cover 5 the Earth 1/4 mile deep! Representative particles The smallest pieces of a substance. –For a molecular compound: it is the molecule. –For an ionic compound: it is the formula unit (ions). –For an element: it is the atom. »Remember the 7 diatomic elements (made of molecules) 6 Types of questions How many oxygen atoms in the following? –CaCO3 –Al2(SO4)3 How many ions in the following? –CaCl2 –NaOH –Al2(SO4)3 7 Types of questions How many molecules of CO2 are there in 4.56 moles of CO2 ? = (4.56 moles)(6.02 x 1023 molecules/mole) = 2.75 x 1024 molecules How many moles of water is 5.87 x 1022 molecules? = (5.87 x 1022 molecules)/(6.02 x 1023 molecules/mole) = 0.0975 moles 8 Types of questions How many atoms of carbon are there in 1.23 moles of C6H12O6 ? = (1.23 moles)(6 atoms of C/mole) = 7.38 atoms of C Howmany moles is 7.78 x 1024 formula units of MgCl2? = (7.78 x 1024 units)/(6.02 x 1023 units/mole) = 12.9 moles 9 Measuring Moles 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. 10 Gram Atomic Mass (gam) Equals the mass of 1 mole of an element in grams 12.01 grams of C has the same number of pieces as 1.008 grams of H and 55.85 grams of iron. We can write this as 12.01 g C = 1 mole C We can count things by weighing 11 them. Examples How much would 2.34 moles of carbon weigh? = (2.34 moles C)(12.01 g/mole) = 28.1 g C How many moles of magnesium is 24.31 g of Mg? = (24.31 g Mg)/(24.31 g/mole) = 1.000 moles Mg 12 Examples How many atoms of lithium is 1.00 g of Li? (1.00 g Li)(6.02 x 1023 atoms Li/mol) (6.94 g Li/mol) 8.67 x 1022 atoms Li How much would 3.45 x 1022 atoms of U weigh? (231.04 g U/mol)(3.45 x 10 22 atoms U) (6.02 x 1023 atoms U/mol) 38.38 g U 13 What about compounds? 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 14 – add them up What about compounds? What is the mass of one mole of CH4? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH4 = 12.01 + 4.04 = 16.05g The Gram Molecular Mass (gmm) of CH4 is 16.05g – this is the mass of one mole of a molecular compound. 15 Gram Formula Mass (gfm) The mass of one mole of an ionic compound. Calculated the same way as gmm. What is the GFM of Fe2O3? 2 moles of Fe x 55.85 g = 111.70 g 3 moles of O x 16.00 g = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70 g 16 Section 7.2 Mole-Mass and Mole-Volume Relationships OBJECTIVES: – Use the molar mass to convert between mass and moles of a substance. – Use the mole to convert among measurements of mass, volume, and number of particles 17 Molar Mass Molar mass is the generic term for the mass of one mole of any substance (in grams) The same as: 1) gram molecular mass, 2) gram formula mass, and 3) gram atomic mass- just a much broader term. 18 Molar Mass The number of grams of 1 mole of atoms, ions, or molecules. We can make conversion factors from these. –To change grams of a compound to moles of a compound. 19 B. Molar Mass Examples water – H 2O – 2(1.01) + 16.00 = 18.02 g/mol sodium chloride – NaCl – 22.99 + 35.45 = 58.44 g/mol 20 B. Molar Mass Examples sodium bicarbonate – NaHCO3 – 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol sucrose – C12H22O11 – 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mol 21 Examples Calculate the molar mass of the following and tell what type it is: Na2S 2 mole of Na x 22.99 g = 45.98g 1 mole of S = 32.07 g 1 mole Na2S = 45.98 + 32.07 = 78.05 g/mol 22 Examples Calculate the molar mass of the following and tell what type it is: N2O4 2 mole of N x 14.01 g = 28.02 g 4 mole of O x 16.00 g = 64.00 g 1 mole N2O4 = 28.02 + 64.00 = 92.02 g/mol 23 Examples Calculate the molar mass of the following and tell what type it is: Ca(NO3)2 1 mole of Ca = 40.08 g 2 mole of N x 14.01 g = 28.02 g 6 mole of O x 16.00 g = 96.00 g 1 mole Ca(NO3)2 = 40.08 + 28.02 + 96.00 = 164.10 g/mol; ionic compound 24 For example How many moles is 5.69 g of NaOH? 25 For example How many moles is 5.69 g of NaOH? 5.69 g 26 For example How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles 27 For example How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH 28 For example How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 29 For example How many moles is 5.69 g of NaOH? mole 5.69 g g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g 30 For example 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 = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g 31 For example How many moles is 5.69 g of NaOH? 1 mole 5.69 g = 0.142 mol NaOH 40.00 g need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g 1 mole NaOH = 40.00 g 32 Examples Howmany moles is 4.56 g of CO2? 1 mole 12.01 2 *16.00 g = 0.104 mol CO 2 4.56 g Howmany grams is 9.87 moles of H2O? 1 mole 16.00 2 *1.01 g = 0.548 mol H 2 O 9.87 g 33 Examples Howmany molecules is 6.8 g of CH4? 1 mole 12.01 4 *1.01 g = 0.399 mol CH 4 6.8 g 6.02 x 10 molecules 23 0.399 mol CH 4 1 mole 23 = 2.40 x 10 molecules CH 4 34 Examples 49molecules of C6H12O6 weighs how much? 1 mole 6 *12.01 12 *1.01 6 *16.00 g 49 molecules 6.02 x 10 molecules 23 1 mole 1 180.18 49 23 g 6.02 x 10 1 = 1.47 x 10 -20 g C 6 H12 O 6 35 Gases Many of the chemicals we deal with are gases. –They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas –Temperature and pressure We need to compare them at the same temperature and pressure. 36 Standard Temperature and Pressure 0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume 1 mole = 22.4 L of any gas at STP 37 Examples Whatis the volume of 4.59 mole of CO2 gas at STP? 22.4 L 4.59 mole 103 L CO 2 1 mole How many moles is 5.67 L of O2 at STP? 1 mole 5.67 L 0.253 mole O 2 22.4 L 38 Examples Whatis the volume of 8.8 g of CH4 gas at STP? 1 mole 22.4 L 12.01 4 *1.01 g 1 mole 12 L 8.8 g 39 Density of a gas D=m/V –for a gas the units will be g / L We can determine the density of any gas at STP if we know its formula. To find the density we need the mass and the volume. If you assume you have 1 mole, then the mass is the molar mass (from PT) 40 At STP the volume is 22.4 L. Examples Find the density of CO2 at STP 12.01 2 * 16.00 g 1 mol 1L 1 mol 22.4 L 1000 ml 1.96 x 10 -3 g/ml Find the density of CH4 at STP. 12.01 4 * 1.01 g 1 mol 1L 1 mol 22.4 L 1000 ml 7.17 x 10 -4 g/ml 41 The other way Given the density, we can find the molar mass of the gas. Again, pretend you have 1 mole at STP, so V = 22.4 L. m = D x V m is the mass of 1 mole, since you have 22.4 L of the stuff. 42 Examples What is the molar mass of a gas with a density of 1.964 g/L? 22.4 L 1.964 g/L 44.0 g/mol 1 mol 2.86 g/L? 22.4 L 2.86 g/L 64.1 g/mol 1 mol 43 Summary These four items are all equal: a) 1 mole b) molar mass (in grams) c) 6.02 x 1023 representative particles d) 22.4 L at STP Thus, we can make conversion factors from them. 44 Section 7.3 Percent Composition and Chemical Formulas OBJECTIVES: – Calculate the percent composition of a substance from its chemical formula or experimental data. – Derive the empirical formula and the molecular formula of a compound from experimental data. 45 Calculating Percent Composition of a Compound Like all percent problems: Part x 100 % whole Find the mass of each component, then divide by the total mass. 46 Example Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. 29.0 g 29.0 4.30 g 100 % 87.1 % Ag 4.30 g 29.0 4.30 g 100 % 12.9 % S 47 Getting it from the formula Ifwe know the formula, assume you have 1 mole. Then you know the mass of the pieces and the whole. 48 Examples Calculate the percent composition of C2H4? How about Aluminum carbonate? –Sample Problem 7-11, p.191 We can also use the percent as a conversion factor –Sample Problem 7-12, p.191 49 Examples Calculate the percent composition of C2H4? 2 * 12.01 g 100 % 83.99 % C 2 * 12.01 4 * 1.01 g 4 * 1.01 g 2 * 12.01 4 * 1.01 g 100 % 16.01 % H 50 Examples the percent composition Calculate of Aluminum carbonate, Al(CO3 ) 3 100 % 13.87 % Al 28.98 g 28.98 3 *12.01 9 *16.00 g 100 % 17.24 % C 3 *12.01 g 28.98 3 *12.01 9 *16.00 g 100 % 68.90 % O 9 *16.00 g 28.98 3 *12.01 9 *16.00 g 51 The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula = the actual ratio of elements in a compound. The two can be the same. CH2 is an empirical formula C2H4 is a molecular formula C3H6 is a molecular formula 52 H2O is both empirical & molecular Calculating Empirical Just find the lowest whole number ratio C6H12O6 CH4N It is not just the ratio of atoms, it is also the ratio of moles of atoms. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. In one molecule of CO2 there is 1 53 atom of C and 2 atoms of O. Calculating Empirical We can get a ratio from the percent composition. Assume you have a 100 g. The percentages become grams. Convert grams to moles. Find lowest whole number ratio by dividing by the smallest. 54 Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 55 14.01 gN Example The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N = C1H5N1 56 Example A compound is 43.64 % P and 56.36 % O. What is the empirical formula? Assume 100 g so 43.64 g P x 1mol P = 1.409 mole P 30.97 g P 56.36 g O x 1mol O = 3.523 mole O 16.00 g O The ratio is 1.409 mol P = 2 mol P 3.523 mol O 5 mol O = P2O5 57 Example Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? Assume 100 g so 49.48 g C x 1mol C = 4.120 mole C 12.01 gC 5.15 g H x 1mol H = 5.010 mole H 1.01 gH 28.87 g N x 1mol N = 2.061 mole N 14.01 gN 16.49 g N x 1mol N = 1.031 mole O 16.00 gN 58 Example The ratio is 4.120 mol C = 4 mol C 1.031 mol O 1 mol O The ratio is 5.010 mol H = 5 mol H 1.031 mol O 1 mol O The ratio is 2.061 mol N = 2 mol N 1.031 mol O 1 mol O = C4 H 5 N 2 O 1 59 Empirical to molecular 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 empirical formula mass. Caffeine has a molar mass of 194 g. what is its molecular formula? 60 Empirical to molecular Caffeine has a molar mass of 194 g. what is its molecular formula? C4H5N2O1 empirical formula mass is = 4*12.01 + 5*1.01 + 2*14.01 + 16.00 = 97.11 g/mol Molarmass/empirical formula mass = 194/97.11 = 2.00 So, actual formula is C8H10N4O2 61 Example Acompound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What is its molecular formula? 62 Example Assume 100 g so 71.65 g Cl x 1mol Cl = 2.021 mole Cl 35.45 g Cl 24.27 g C x 1mol C = 2.021 mole C 12.01 g C 4.07 g H x 1mol H = 4.030 mole H 1.01 g H 63 Example The ratio is 2.021 mol C = 1 mol C 2.021 mol Cl 1 mol Cl The ratio is 4.030 mol H = 2 mol H 2.021 mol Cl 1 mol Cl = C1H2Cl1 C1H2Cl1 empirical formula mass is = 12.01 + 4*1.01 + 35.45 = 51.50 g/mol Molar mass/empirical formula mass = 98.96/51.5 = 1.92 = 2 So, actual formula is C2H4Cl2 64

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