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Chapter 6 – Chemical Composition Counting Atoms

Chapter 6 – Chemical Composition Counting Atoms Atoms are too small and too numerous to count individually. A unit of measure called the mole has been established for use in counting atoms. One mole of something consists of 6.022 X 10 23 units of that substance: 1 mol = 6.022 X 10 23.

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Chapter 6 – Chemical Composition Counting Atoms

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  1. Chapter 6 – Chemical Composition Counting Atoms • Atoms are too small and too numerous to count individually. • A unit of measure called the mole has been established for use in counting atoms. • One mole of something consists of 6.022 X 1023 units of that substance: 1 mol = 6.022 X 1023

  2. Moles • A mole is often referred to as “the chemist’s dozen” because it gives the number of “things” in one mole of anything • 1 mole eggs = 6.022 x 1023 eggs • 1 mole pencils = 6.022 x 1023 pencils • 1 mole of an element = 6.022 x 1023 atoms • 1 mole carbon = 6.022 x 1023 carbon atoms • 1 mole of a compound = 6.022 x 1023 molecules • 1 mole water = 6.022 x 1023 water molecules

  3. Example #1: How many sodium atoms are in 4.5 mol of sodium? Given: 4.5 mol of Na Find: number of sodium atoms Conversion Factor: 1 mol Na = 6.022 X 1023 Na atoms Solution Map: mol of Na -> Na atoms Solution: 4.5 mol of Na X 6.022 X 1023 Na atoms 1 mol Na = 2.7 X 1024 Na atoms

  4. Example #2: How many moles of copper are in 1.38 X 1025 copper atoms? Given: 1.38 X 1025 Cu atoms Find: mol of Cu atoms Conversion Factor: 1 mol Cu = 6.022 X 1023 Cu atoms Solution Map: Cu atoms -> mol of Cu Solution: 1.38 X 1025 Cu atoms X 1 mol Cu (6.022 X 1023 Cu atoms) = 22.9 mol Cu

  5. Moles and Molar Mass • The SI definition of the mole is the amount of a substance that contains as many entities as there are in exactly 12 grams of carbon-12. 1 mol carbon = 6.022 x 1023 C atoms = 12.01 g C • The definition of a mole provides a relationship between mass (grams of a substance) and the number of atoms (Avogadro’s number) • The molar mass of any substance is the mass (in grams) of 1 mol of the substance. • The atomic mass gives the number of grams in one mole of an element. • The formula mass gives the number of grams in one mole of a compound.

  6. Example #3: Calculate the number of moles of gold in 26.2 g of gold. Given: 26.2 g of Au Find: mol Au Conversion Factor: 196.97 g Au = 1 mol of Au Solution Map: g Au -> mol Au Solution: 26.2 g Au X 1 mol Au 196.97 g Au = 0.133 mol Au

  7. Example #4: Calculate the number of atoms present in 2.34 g of calcium. Given: 2.34 g of Ca Find: Ca atoms Conversion Factors: 40.08 g Ca = 1 mol of Ca 1 mol Ca = 6.022 X 1023 atoms of Ca Solution Map: g Ca -> mol Ca -> # of Ca atoms Solution: 2.34 g Ca X 1 mol Ca X 6.022 X 1023 atoms Ca 40.08 g Ca 1 mol Ca = 3.52 X 1022 atoms of Ca

  8. Example #5: Calculate the mass (in grams) of 7.9 X 1021 uranium atoms. Given: 7.9 X 1021 U atoms Find: g U Conversion Factors: 1 mol U = 6.022 X 1023 atoms of U 238 g U = 1 mol of U Solution Map: # of U atoms -> mol U -> g U Solution: 7.9 X 1021 U atoms X 1 mol U X 238 g U 6.022 X 1023 U atoms 1 mol U = 3.1 g U

  9. Example #6: Calculate the number of moles in 8.91 g of potassium permanganate. Given: 8.91 g of KMnO4 Find: mol KMnO4 Conversion Factor: formula mass of KMnO4 = 1(Atomic mass of K) + 1(Atomic mass of Mn) + 4(Atomic mass of O) = 1(39.10 g/mol) + 1(54.94 g/mol) + 4(16.00 g/mol) = 158.04 g/mol or 158.04 g mol-1 Solution Map: g KMnO4 -> mol KMnO4 Solution: 8.91 g KMnO4 X 1 mol KMnO4 158.04 g KMnO4 = 5.64 X 10-2 mol KMnO4

  10. Example #7: Calculate the mass (in grams) of 1.27 mmol of carbon dioxide. Given: 1.27 mmol of CO2 Find: g CO2 Conversion Factor: formula mass of CO2 = 1(Atomic mass of C) + 2(Atomic mass of O) = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol or 44.01 g mol-1 Solution Map: CO2 mmol -> mol CO2 -> g CO2 Solution: 1.27 CO2 mmol X 0.001 mol CO2 X 44.01 g CO2 1 mmol CO2 1 mol CO2 = 0.0559 or 5.59 X 10-2 g CO2

  11. Example #8: Calculate the mass (in grams) of 4.03 X 1022 molecules of benzene, C6H6. Given: 4.03 X 1022 C6H6 molecules Find: g C6H6 Conversion Factors: 1 mol C6H6 = 6.022 X 1023 molecules of C6H6 formula mass of C6H6 = 6(Atomic mass of C) + 6(Atomic mass of H) = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.12 g/mol or 78.12 g mol-1 Solution Map: C6H6 molecules -> mol C6H6 -> g C6H6 Solution: 4.03 X 1022 C6H6 molecules X 1 mol C6H6 X 6.022 X 1023 C6H6 molecules 78.12 g C6H6 1 mol C6H6 = 5.23 g C6H6

  12. Counting Moles Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound. For example, 1 mol of CH4 molecules consists of 1 mol of carbon atoms and 4 mol of hydrogen atoms. Moles of Compound Moles of Constituents 1 mol NaCl 1 mole Na, 1 mole Cl 1 mol H2O 2 mol H, 1 mole O 1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O 1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

  13. Example #9: Determine the number of moles of oxygen in 3.5 mol of Ca(OH)2. Given: 3.5 mol of Ca(OH)2 Find: mol O Conversion Factor: 1 mol of Ca(OH)2 molecules contains 2 mol of O atoms Solution Map: mol Ca(OH) 2 -> mol of O Solution: 3.5 mol Ca(OH)2 X 2 mol of O 1 mol of Ca(OH)2 = 7.0 mol O

  14. Example #10: A 15.5 gram sample of diphosphorous pentoxide contains how many grams of phosphorous? Given: 15.5 g of P2O5 Find: g of P Conversion Factors: formula mass of P2O5 = 2(30.97 g/mol) + 5(16.00 g/mol) = 141.94 g/mol 1 mol of P2O5 molecules contains 2 mol of P atoms 1 mol of P = 30.97 g of P Solution Map: g P2O5 -> mol P2O5 -> mol P -> g P Solution: 15.5 g P2O5 X 1 mol P2O5 X 2 mol P X 30.97 g P 141.94 g P2O5 1 mol P2O5 1 mol of P = 6.76 g P

  15. Example #11: Find the grams of iron in 79.2 g of iron (III) oxide. Given: 79.2 g of Fe2O3 Find: g of Fe Conversion Factors: formula mass of Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol 1 mol of Fe2O3 molecules contains 2 mol of Fe atoms 1 mol of Fe = 55.85 g of Fe Solution Map: g Fe2O3 -> mol Fe2O3 -> mol Fe -> g Fe Solution: 79.2 g Fe2O3 X 1 mol Fe2O3 X 2 mol Fe X 55.85 g Fe 159.70 g Fe2O3 1 mol Fe2O3 1 mol of Fe = 55.4 g Fe

  16. Percent Composition • Percentage of an element in a compound. • Can be determined from either the formula of the compound or the experimental mass analysis of the compound. The percent composition or mass percent of an element can be determined from experimental data using the following formula: Mass percent of element X = Mass of X in a sample of compound X 100% Mass of the sample of compound X • The mass percent tells you the mass of a constituent element in 100 g of the compound. For example, the fact that Al2O3 is 52.9% aluminum by mass means that 100 g of Al2O3 contains 52.9 g of Al. • The mass percent can be used as a conversion factor. For example, g Al2O3 X 52.9 g Al = g Al 100 g Al2O3

  17. Example #12: A 4.67 g sample of iron completely reacts with oxygen to form 6.67 g of iron (III) oxide. Calculate the mass percent composition of iron in iron (III) oxide. Given: sample of 6.67 g Fe2O3 4.67 g of iron in sample Find: mass percent composition of Fe in Fe2O3 Equation: Mass percent of element X = Mass of Fe in the Fe2O3 sample X 100% Mass of the Fe2O3 sample Solution: Mass % Fe = 4.67 g Fe X 100% 6.67 g Fe2O3 = 70.0 % Fe

  18. Example #13: Copper (I) bromide contains 44.30% copper by mass. Calculate the mass of copper in grams contained in 32.5 g of copper (I) bromide. Given: 32.5g CuBr and mass % Cu in CuBr is 44.30% Find: g Cu Conversion Factor: 44.30 g Cu in every 100 g CuBr Solution Map: g CuBr -> g Cu Solution: 32.5 g CuBr X 44.30 g Cu 100 g CuBr = 14.4 g Cu

  19. Example #14: Calculate the mass percent composition of phosphorous in potassium phosphate. Find: mass percent composition of phosphorous in K3PO4 Equation: Mass percent of element X = Atomic Mass of X X 100% Formula Mass of Compound Solution: Atomic Mass of P = 30.97 g/mol Formula Mass of K3PO4 = 3(39.10 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol) = 212.27 g/mol Mass % P = 30.97 g/mol P 212.27 g/mol K3PO4 X 100% = 14.59 % P

  20. Example #15: Calculate the mass percent of each element in carvone, C10H14O. Equation: Mass percent of element X = Mass of X X 100% Formula Mass of Compound Solution: 1) Determine the masses of the various elements in 1 mol of carvone Mass of C in 1 mol = 10 mol X 12.01 g/mol = 120.1 g Mass of H in 1 mol = 14 mol X 1.01 g/mol = 14.14 g Mass of O in 1 mol = 1 X 16.00 g/mol = 16.00 g 2)Determine the formula mass of carvone Mass of 1 mol of C10H14O = 150.24 g

  21. Example # 15 Continued. 3)Find the fraction of the total mass contributed by each element and convert it to percentage. Mass percent of C = 120.1 g C 150.24 g C10H14O Mass percent of H = 14.14 g H 150.24 g C10H14O Mass percent of O = 16.00 g O 150.24 g C10H14O X 100% = 79.94% C = 9.412% H X 100% = 9.412% O X 100%

  22. Example #16 - Problem 108 in the text: A leak in the air conditioning of an older car releases 55 g CF2CL2 per month. How much Cl is emitted into the atmosphere each year by this car? Given: 55 g of CF2Cl2 Find: g of Cl Solution Map: g CF2Cl2 per month -> g Cl per month -> g Cl per year Conversion factors: Mass percent of Cl in CF2Cl2 = Mass of Cl Formula Mass of CF2Cl2 = 70.9 g Cl 120.91 g CF2Cl2 = 58.638657% 12 months = 1 year Solution: 55 g CF2Cl2 X 58.638657 g Cl X 12 months 1 month 100 g CF2Cl2 1 year X 100% X 100% = 3.9 X 102 g Cl per year

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