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Chapter 7 Chemical Quantities

Chapter 7 Chemical Quantities. Section 7.1 The Mole: A Measurement of Matter. OBJECTIVES: Describe how Avogadro’s number is related to a mole of any substance. Section 7.1 The Mole: A Measurement of Matter. OBJECTIVES: Calculate the mass of a mole of any substance. What is a Mole?.

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Chapter 7 Chemical Quantities

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  1. Chapter 7Chemical Quantities

  2. Section 7.1The Mole: A Measurement of Matter • OBJECTIVES: • Describe how Avogadro’s number is related to a mole of any substance.

  3. Section 7.1The Mole: A Measurement of Matter • OBJECTIVES: • Calculate the mass of a mole of any substance.

  4. 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 inMOLES.

  5. 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.

  6. 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)

  7. 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

  8. Types of questions • How many molecules of CO2 are there in 4.56 moles of CO2 ? • How many moles of water is 5.87 x 1022 molecules? • How many atoms of carbon are there in 1.23 moles of C6H12O6 ? • How many moles is 7.78 x 1024 formula units of MgCl2?

  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 mass 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 them.

  11. Examples • How much would 2.34 moles of carbon weigh? • How many moles of magnesium is 24.31 g of Mg? • How many atoms of lithium is 1.00 g of Li? • How much would 3.45 x 1022 atoms of U weigh?

  12. 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 • add them up

  13. 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.

  14. 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

  15. Section 7.2Mole-Mass and Mole-Volume Relationships • OBJECTIVES: • Use the molar mass to convert between mass and moles of a substance.

  16. Section 7.2Mole-Mass and Mole-Volume Relationships • OBJECTIVES: • 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. Examples • Calculate the molar mass of the following and tell what type it is: • Na2S • N2O4 • C • Ca(NO3)2 • C6H12O6 • (NH4)3PO4

  19. 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.

  20. For example • How many moles is 5.69 g of NaOH?

  21. For example • How many moles is 5.69 g of NaOH?

  22. For example • How many moles is 5.69 g of NaOH? • need to change grams to moles

  23. For example • How many moles is 5.69 g of NaOH? • need to change grams to moles • for NaOH

  24. For example • How many moles is 5.69 g of NaOH? • 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

  25. For example • How many moles is 5.69 g of NaOH? • 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

  26. For example • How many moles is 5.69 g of NaOH? • 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

  27. For example • How many moles is 5.69 g of NaOH? • 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

  28. Examples • How many moles is 4.56 g of CO2? • How many grams is 9.87 moles of H2O? • How many molecules is 6.8 g of CH4? • 49 molecules of C6H12O6 weighs how much?

  29. 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.

  30. 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

  31. Examples • What is the volume of 4.59 mole of CO2 gas at STP? • How many moles is 5.67 L of O2 at STP? • What is the volume of 8.8 g of CH4 gas at STP?

  32. 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) • At STP the volume is 22.4 L.

  33. Examples • Find the density of CO2 at STP. • Find the density of CH4 at STP.

  34. 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. • What is the molar mass of a gas with a density of 1.964 g/L? • 2.86 g/L?

  35. 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.

  36. Section 7.3Percent Composition and Chemical Formulas • OBJECTIVES: • Calculate the percent composition of a substance from its chemical formula or experimental data.

  37. Section 7.3Percent Composition and Chemical Formulas • OBJECTIVES: • Derive the empirical formula and the molecular formula of a compound from experimental data.

  38. Calculating Percent Composition of a Compound • Like all percent problems: part whole • Find the mass of each component, • then divide by the total mass. x 100 %

  39. Example • Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

  40. Getting it from the formula • If we know the formula, assume you have 1 mole. • Then you know the mass of the pieces and the whole.

  41. Examples • Calculate the percent composittion 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

  42. 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 • H2O is both empirical & molecular

  43. 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 atom of C and 2 atoms of O.

  44. 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.

  45. 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 14.01 gN

  46. 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 • A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

  47. 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?

  48. Example • A compound 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? • Sample Problem 7-14, p.194

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