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Chapter 6 Chemical Composition

Chapter 6 Chemical Composition. 2006, Prentice Hall. CHAPTER OUTLINE. Why is Knowledge of Composition Important?. everything in nature is either chemically or physically combined with other substances

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Chapter 6 Chemical Composition

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  1. Chapter 6 Chemical Composition 2006, Prentice Hall

  2. CHAPTER OUTLINE

  3. Why is Knowledge of Composition Important? • everything in nature is either chemically or physically combined with other substances • to know the amount of a material in a sample, you need to know what fraction of the sample it is • Some Applications: • the amount of sodium in sodium chloride for diet • the amount of iron in iron ore for steel production • the amount of hydrogen in water for hydrogen fuel • the amount of chlorine in freon to estimate ozone depletion Tro's Introductory Chemistry, Chapter 6

  4. THE MOLECONCEPT • Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. • In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

  5. THE MOLECONCEPT • The number of particles in a mole is called Avogadro’s number and is 6.02x1023. 1 mole 6.02 x 1023 equals to Avogadro’s number (NA)

  6. MOLE THE MOLECONCEPT A mole is a very large quantity 6.02x1023 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

  7. THE MOLECONCEPT 1 mole H atoms = 6.02x1023 H atoms 1 mole H2 molecules = 6.02x1023 H2 molecules = 2 x (6.02x1023) H atoms 1 mole H2O molecules = 6.02x1023 H2O molecules = 2 x (6.02x1023) H atoms = 6.02x1023 O atoms 1 mole Na+ ions = 6.02x1023 Na+ ions

  8. THE MOLECONCEPT • The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams. Mass of 1 H atom = 1.008 amu Mass of 1 Mg atom = 24.31 amu Mass of 1 Cl atom = 35.45 amu Mass of 1 mol Mg atoms = 24.31 g Mass of 1 mol Cl atoms = 35.45 g Mass of 1 mol H atoms = 1.008 grams

  9. MOLEDAY • Chemists and chemistry students celebrate two days in the year in honor of the Mole and call them Mole Days. October 23rd 6:02 a.m. Jun 2nd 10:23 a.m.

  10. MOLECULAR MASS • The sum of atomic masses of all the atoms in one molecule of a substance is called molecular mass, and is measured in amu. Mass of one molecule of H2O 2 H atoms = 2 (1.008 amu) = 2.016 amu 1 O atom = 1 (16.00 amu) = 16.00 amu Molecular mass 18.02 amu

  11. Relationship Between Moles and Mass • The mass of one mole of atoms is called the molar mass • The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu

  12. MOLAR MASS • The mass of one mole of a substance is called molar mass, and is measured in grams. Mass of one mole of H2O 2 mol H atoms = 2 (1.008 g) = 2.016 g 1 mol O atom = 1 (16.00 g) = 16.00 g Molar mass 18.02 g

  13. CALCULATIONSUSING THE MOLE • Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number. Avogadro’s number Molar mass Mass of a substance Moles of a substance Particles of a substance MM NA

  14. Example 1: What is the mass of 5.00 mol of water? 18.02 90.1 1 3 significant figures Molar mass

  15. Example 2: How many Mg atoms are present in 5.00 g of Mg? mass  mol  atoms 1 6.02 x 1023 24.3 1 1.24x1023 atoms Mg Avogadro’s number Molar mass 3 significant figures

  16. Example 3: How many molecules of HCl are present in 25.0 g of HCl? mass  mol  molecules 1 6.02 x 1023 36.45 1 4.13 x 1023 molecules HCl 3 significant figures

  17. Total mass of element Total mass of compound PERCENTCOMPOSITION • The percent composition of a compound is the mass percent of each element in the compound.

  18. Example 1: Calculate the percent composition of sodium chloride (NaCl). Step 1: determine molar mass of NaCl 1 mol Na atoms = 1 (22.99 g) = 22.99 g 1 (35.45 g) = 35.45 g 1 mol Cl atom = Molar mass 58.44 g/mol

  19. Sum = 100% Example 1: Step 2: calculate the mass % of each element % Na = % Cl =

  20. Example 2: 1.63 g of zinc combines with 0.40 g of oxygen to form zinc oxide. Determine the % composition of the compound formed. Step 1: determine total mass of sample mass of sample = 1.63 g + 0.40 g = 2.03 g

  21. Sum = 100% Example 2: Step 2: calculate the mass % of each element % Zn = % O =

  22. Example 3: Calculate the percent composition of sodium hydroxide (NaOH). Step 1: determine molar mass of NaOH 1 mol Na atoms = 1 (23.0 g) = 23.0 g 1 (16.0 g) = 16.0 g 1 mol O atom = 1 (1.01 g) = 1.01 g 1 mol H atoms = Molar mass 40.0 g/mol

  23. Example 3: Calculate the percent composition of sodium hydroxide (NaOH). Step 1: determine molar mass of NaOH 1 mol Na atoms = 1 (23.0 g) = 23.0 g 1 (16.0 g) = 16.0 g 1 mol O atom = 1 (1.01 g) = 1.01 g 1 mol H atoms = Molar mass 40.0 g/mol

  24. CHEMICALFORMULAS Molecular formula Empirical formula Shows the actual number of atoms in a compound Can be written for molecular compounds only Shows the simplest ratio of atoms in a compound Can be written for molecular and ionic compounds

  25. CHEMICALFORMULAS HO 2 H2O 1 CH2O 6 6 CH CH 2 CH2 6

  26. CHEMICALFORMULAS • Several compounds may possess the same percent composition and empirical formula, but different molecular formulas. 13.02 26.04 (2x13.02) 78.12 (6x13.02)

  27. Finding an Empirical Formula • convert the percentages to grams • skip if already grams • convert grams to moles • use molar mass of each element • write a pseudoformula using moles as subscripts • divide all by smallest number of moles • multiply all mole ratios by number to make all whole numbers • if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. • skip if already whole numbers

  28. If, after dividing by the smallest number of moles, the subscripts are not whole numbers, multiply all the subscripts by a small whole number to arrive at whole-number subscripts.

  29. Assume 100 g CALCULATINGEMPIRICAL FORMULAS Arsenic (As) reacts with oxygen (O) to form a compound that is 75.7% As and 24.3% oxygen by mass. What is the empirical formula for this compound? Step 1: Percent to mass 75.7% As 75.7 g As 24.3% O 24.3 g O

  30. Use atomic mass of oxygen CALCULATINGEMPIRICAL FORMULAS Step 2: Mass to mole 75.7 g As 24.3 g O

  31. CALCULATINGEMPIRICAL FORMULAS Step 3: Divide by small As = O =

  32. CALCULATINGEMPIRICAL FORMULAS Step 4: Multiply till Whole As1.00O1.50 As2O3 x 2 = 2 x 2 = 3

  33. Example 1: Determine the empirical formula for a compound containing 11.2% H and 88.8% O. mol H = mol O = H2O

  34. Example 2: Determine the empirical formula for a compound with the following percent composition: 52.14% C, 13.12% H, 34.73% O. C2H6O mol C = mol H = mol O =

  35. MOLECULARFORMULAS • Molecular formula can be calculated from empirical formula if molar mass is known. Molecular formula = (empirical formula) n

  36. Example 1: A compound of N and O with a molar mass of 92.0 g, has the empirical formula of NO2. What is its molecular formula? Mass of empirical formula = 14.0 + 2(16.0) = 46.0 Molecular formula = 2 x (NO2) = N2O4

  37. Example 2: Calculate the empirical and molecular formulas of a compound that contains 80.0% C and 20.0% H, and has a molar mass of 30.00 g. mol C = mol H = Empirical formula CH3

  38. Example 2: Empirical formula = CH3 Mass of empirical formula = 12.0 + 3(1.01) = 15.0 Molecular formula = 2 x (CH3) = C2H6

  39. THE END

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