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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas. How you measure how much?. 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. Moles.

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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

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  1. ChemicalQuantities The Mole, % Composition, Empirical and Molecular Formulas

  2. How you measure how much? • 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 • 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.

  4. Representative particles • 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.

  5. 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 3 12 3 2 5

  6. Types of questions using the equality;1 mole = 6.02 x 1023 • How many molecules of CO2 are the in 4.56 moles of CO2? • 4.56 mole x 6.02x1023 mc = 1 1 mole • How many moles of water is 5.87 x 1022 molecules? • 5.87 x 1022 mc x 1 mole = 1 6.02x1023 mc 2.75x1024mc 0.0975 mole

  7. Types of questions using the equality;1 mole = 6.02 x 1023 4.44x1024 atoms • How many atoms of carbon are there in 1.23 moles of C6H12O6? • 1.23 moles x 6.02x1023 mc x 6 atoms = 1 1 mole 1 mc • How many moles is 7.78 x 1024 formula units of MgCl2? 7.78x1024 FU x 1 mole = 12.9 mole 1 6.02x1024 FU

  8. 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.

  9. Gram Atomic Mass • 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.

  10. Examples • How much would 2.34 moles of carbon weigh? • 2.34 moles C x 12 g C 1 1mole • How many moles of magnesium in 24.31 g of Mg? 24.31 g Mg x 1 mole = 1 24g Mg = 28.08 g 1.013 mole

  11. 8.60x1022 atoms 13.6 g How many atoms of lithium in 1.00 g of Li? 1.00 g Li x 1 mole x 6.02x1023 atoms 1 7 g Li 1 mole How much would 3.45 x 1022 atoms of U weigh? 3.45x1022 atoms U x 1 mole x 238 g U 1 6.02x1023atoms 1 mole

  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 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.

  14. Gram Formula Mass • 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

  15. Molar Mass • The generic term for the mass of one mole. • The same as gram molecular mass, gram formula mass, and gram atomic mass.

  16. Examples • Calculate the molar mass of the following and indicate what type it is. • Na2S • 2 (23) + 32 = • N2O4 • 2(14) + 4(16) = • C 46 + 32 = 78g 1mole Na2S = 78g Gram Formula Mass 28 + 64 = 92g 1 mole N2O4 = 92g Gram Molecular Mass 12g 1 mole C = 12 g Gram Atomic Mass

  17. Molar Mass Cont. Gram Formula Mass • Ca(NO3)2 • 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g • 1 mole Ca(NO3)2 = 164g • C6H12O6 • 6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g • 1 mole C6H12)6 = 180g Gram Molecular Mass • (NH4)3PO4 • 3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g • 1 mole (NH4)3PO4 = 149g Gram Formula Mass

  18. Using Molar Mass Finding moles of compounds Counting pieces by weighing

  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 = 23g 1 mol O = 16.00 g 1 mole of H = 1 g

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

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

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

  28. Examples • How many moles is 4.56 g of CO2? • 4.56g CO2 x 1 mole 1 44gCO2 • How many grams is 9.87 moles of H2O? • 9.87 moles x 18g H2O 1 1 mole = 0.104 moles = 178g

  29. Examples = 2.56x1023 mc • How many molecules in 6.8 g of CH4? • 6.8g CH4 x 1 mole x 6.02x1023 mc 1 16g CH4 1mole • 49 molecules of C6H12O6 weighs how much? • 49 mc x 1 mole x 180g C6H12O6 = 1 6.02x1023 mc 1 mole • 8820 g____= 1.5x10-20 g 6.02x1023

  30. Gases and the Mole

  31. 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 • Scientists compare gases at Standard Temperature and Pressure

  32. 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 • Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

  33. Examples • What is the volume of 4.59 mole of CO2 gas at STP? • 4.59mole x 22.4L 1 1mole • How many moles is 5.67 L of O2 at STP? • 5.67L x 1 mole 1 22.4L • What is the volume of 8.80g of CH4 gas at STP? • 8.80g CH4x 1mole x 22.4L 1 16g CH4 1mole = 102.816 = 103L = .523moles = 12.32 = 12.3L

  34. We have learned how to • change moles to grams • moles to atoms • moles to formula units • moles to molecules • moles to liters • molecules to atoms • formula units to atoms • formula units to ions

  35. Mass Periodic Table Moles

  36. Mass PeriodicTable Volume Moles

  37. Mass Periodic Table Volume 22.4 L Moles

  38. Mass Periodic Table Volume 22.4 L Moles Representative Particles

  39. Mass Periodic Table Volume 22.4 L Moles 6.02 x 1023 Representative Particles

  40. Mass Periodic Table Volume 22.4 L Moles 6.02 x 1023 Representative Particles Atoms

  41. Mass Periodic Table Volume 22.4 L Moles 6.02 x 1023 Representative Particles Ions Atoms

  42. Percent Composition • Like all percents • Part x 100 % whole • Find the mass of each component, • divide by the total mass.

  43. Example • 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 • S + 4.30g • 33.3g /33.3 = .8709 x 100 = 87.09% = .1291 x 100 = 12.91% /33.3

  44. Getting % from the formula • If we know the formula, assume you have 1 mole. • Then you know the pieces and the whole.

  45. Examples • Calculate the percent composition of C2H4? • C 2(12g)=24 • H 4(1g) = +4 • 28g /28 = .8571 x 100 = 85.71% /28 = .1429 x 100 = 14.29%

  46. Example /234 = .2308 x 100 = 23.08% /234 = .15.38 x 100 = 15.38% /234 = .6154 x 100 = 61.54% Calculate the percent composition of Aluminum carbonate. Al2(CO3)3 Al 2(27g)= 54 C 3(12g)= 36 O 9(16)= 144 234g

  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 94.1% x 3.54g = 3.33g S

  48. Calculate the mass of nitrogen in 25g of (NH4)2CO3. Calculate the mass of nitrogen in 25g of (NH4)2CO3. N 2(14g) = 28 H 8(1g) = 8 C 1(12g) = 12 O 8(16g) = +128 176g (NH4)2CO3 %N = 28/176 x 100 = 15.91% 15.91% x 25g = 4.0g N

  49. Calculate the mass of magnesium in 97.4g of Mg(OH)2. 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

  50. Empirical Formula From percentage to formula

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