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(Campbell / Callis C142B) Chapter #3 : Stoichiometry - Mole - Mass

(Campbell / Callis C142B) Chapter #3 : Stoichiometry - Mole - Mass Relationships in Chemical Systems. 3.1: The Mole 3.2: Determining the Formula of an Unknown Compound 3.3: Writing and Balancing Chemical Equations 3.4: Calculating the amounts of Reactant and Product

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(Campbell / Callis C142B) Chapter #3 : Stoichiometry - Mole - Mass

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  1. (Campbell / Callis C142B) Chapter #3 : Stoichiometry -Mole - Mass Relationships in Chemical Systems 3.1: The Mole 3.2: Determining the Formula of an Unknown Compound 3.3: Writing and Balancing Chemical Equations 3.4: Calculating the amounts of Reactant and Product 3.5: Fundamentals of Solution Stoichiometry

  2. MOLE • The Mole is based upon the definition: The amount of substance that contains as many elementary particles (atoms, molecules, ions, or other?) as there are atoms in exactly 12 grams of carbon -12. 1 Mole = 6.022045 x 1023 particles (atoms, molecules, ions, electrons, or…) = NA particles ~100 million x 100 million x 100 million

  3. Avogadro’s Number (NA)NA = 6.022045 x 1023 = # of particles (atoms, molecules, ions, electrons, or…) in one mole of that thing.

  4. Fig 3.1 (P 90) Counting objects of fixed relative mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S

  5. Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023atoms 1 atom of S = amu 1 mole of S = g = atoms 1 atom of O = amu 1 mole of O = g = atoms Molecular mass: 1 molecule of O2 = amu 1 mole of O2 = g = molecules 1 molecule of S8 = amu 1 mole of S8 = g = molecules

  6. Mole - Mass Relationships of Elements Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms Molecular mass: 1 molecule of O2 = 16.00 x 2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule 1 molecule of S8 =32.07 x 8 = 256.56 amu 1 mole of S8 = 256.56 g = 6.022 x 1023 molecules

  7. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams, called itsmolar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) + amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) + g = g g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  8. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called itsmolar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  9. Fig 3.2 (P 87) One mole of common sbustances CaCO3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g (This balloon volume is not really big enough. Need ~10-20 liters, depending on pressure inside.)

  10. Fig. 3.3

  11. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = No. of W atoms =

  12. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

  13. Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Trisodium phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = Converting mass to moles: # Formula units =

  14. Calculating the Moles and Number of Formula Units in a given Mass of Cpd. Problem: Trisodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4 = 1.46 x 1023 formula units

  15. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Multiply by M (g / mol of X) Mass (g) of X in one mole of compound Divide by mass (g) of one mole of compound Mass fraction of X Multiply by 100 % Mass % of X

  16. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a)Determining the mass percent of each element: mass of C per mole sucrose = mass of H / mol = mass of O / mol = total mass per mole = Finding the mass fraction of C in Sucrose & % C : mass of C per mole mass of 1 mole sucrose = To find mass % of C = Mass Fraction of C = =

  17. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a)Determining the mass percent of each element: mass of C per mole sucrose = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H / mol = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O / mol = 11 x 16.00 g O/mol = 176.00 g O/mol total mass per mole = 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : mass of C per mole 144.12 g C/mol mass of 1 mole sucrose 342.30 g Cpd/mol = 0.4210 To find mass % of C = 0.4210 x 100% = 42.10% Mass Fraction of C = =

  18. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = Mass % of O = x 100% = (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = mol H x M of H mass of 1 mol sucrose mol O x M of O mass of 1 mol sucrose

  19. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose x = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

  20. Calculate M and % composition of NH4NO3. • 2 mol N x • 4 mol H x • 3 mol O x Molar mass = M = 28.02g N2 80.05g %N = x 100% = 35.00% 4.032g H2 80.05g %H = x 100% = 5.037% 48.00g O2 80.05g %O = x 100% = 59.96% 99.997%

  21. Calculate M and % composition of NH4NO3. • 2 mol N x 14.01 g/mol = 28.02 g N • 4 mol H x 1.008 g/mol = 4.032 g H • 3 mol O x 15.999 g/mol = 48.00 g O 80.05 g/mol 28.02g N2 80.05g %N = x 100% = 35.00% 4.032g H2 80.05g %H = x 100% = 5.037% 48.00g O2 80.05g %O = x 100% = 59.96% 99.997%

  22. Calculate the Percent Composition of Sulfuric Acid H2SO4 Molar Mass of Sulfuric acid = 2(1.008g) + 1(32.07g) + 4(16.00g) = 98.09 g/mol 2(1.008g H2) 98.09g %H = x 100% = 2.06% H 1(32.07g S) 98.09g %S = x 100% = 32.69% S 4(16.00g O) 98.09 g %O = x 100% = 65.25% O Check = 100.00%

  23. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

  24. Some Examples of Compounds with the same Elemental Ratio’s Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6

  25. Steps to Determine Empirical Formulas Mass (g) of Element ÷ M (g/mol ) Moles of Element Use no. of moles as subscripts. Preliminary Formula Change to integer subscripts: ÷ smallest, conv. to whole #. Empirical Formula

  26. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = Moles of Cr = Moles of O =

  27. Determining Empirical Formulas from Masses of Elements - I Problem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the empirical formula and name of the compound? Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound. Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  28. Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Converting to integer subscripts (dividing all by smallest subscript): Rounding off to whole numbers:

  29. Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na0.2469 Cr0.1235 O0.4939 Converting to integer subscripts (dividing all by smallest subscript): Na1.99 Cr1.00 O4.02 Rounding off to whole numbers: Na2CrO4 Sodium Chromate

  30. Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = Mass Hydrogen = Mass Oxygen =

  31. Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is Glucose (M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the Molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements. Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g Cpd

  32. Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Moles of H = Moles of O = Constructing the preliminary formula: Converting to integer subscripts, ÷ all subscripts by the smallest:

  33. Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, ÷ all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H 1.008 g H 1 mol O 16.00 g O

  34. Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: M of Glucose empirical formula mass Whole-number multiple = = Therefore the Molecular Formula is:

  35. Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the Molecular Formula: The formula weight of the empirical formula is: 1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 g/mol M of Glucose empirical formula mass Whole-number multiple = = = = 6.00 = 6 180.16 30.03 Therefore the Molecular Formula is: C 1 x 6 H 2 x 6 O 1 x 6 =C6H12O6

  36. Adrenaline is a very Important Compound in the Body - I • Analysis gives : • C = 56.8 % • H = 6.50 % • O = 28.4 % • N = 8.28 % • Calculate the Empirical Formula !

  37. Adrenaline - II • Assume 100g! • C = • H = • O = • N =Divide by smallest (0.591) => • C = • H = • O = • N =

  38. Adrenaline - II • Assume 100g! • C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C • H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H • O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O • N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N • Divide by smallest (0.591) => • C = 8.00 mol C = 8.0 mol C or • H = 10.9 mol H = 11.0 mol H • O = 3.01 mol O = 3.0 mol O C8H11O3N • N = 1.00 mol N = 1.0 mol N

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