1 / 96

Chapter 3

Chapter 3. Stoichiometry. Chapter 3 - Stoichiometry. 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations

sheehan
Download Presentation

Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 Stoichiometry

  2. Chapter 3 - Stoichiometry 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant

  3. Reaction of zinc and sulfur.

  4. Figure 3.1: Mass spectrometer

  5. Chemists using a mass spectrometer to analyze for copper in blood plasma. Source: USDA Agricultural Research Service

  6. A herd of savanna-dwelling elephants Source: Corbis

  7. Figure 3.2: Relative intensities of the signals recorded when natural neon is injected into a mass spectrometer.

  8. Figure 3.3: Mass spectrum of natural copper

  9. Atomic Definitions II: AMU, Dalton, 12C Std. Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom on this scale Hydrogen has a mass of 1.008 AMU. Dalton (D) = The new name for the Atomic Mass Unit, one dalton = one Atomic Mass Unit on this scale, 12C has a mass of 12.00 daltons. Isotopic Mass = The relative mass of an Isotope relative to the Isotope 12C the chosen standard. Atomic Mass = “Atomic Weight” of an element is the average of the masses of its naturally occurring isotopes weighted according to their abundances.

  10. 11H 1 Proton 0 Neutrons 99.985 % 1.00782503 amu 21H (D) 1 Proton 1 Neutron 0.015 % 2.01410178 amu 31H (T) 1 Proton 2 Neutrons -------- ---------- The average mass of Hydrogen is 1.008 amu 3H is Radioactive with a half life of 12 years. H2O Normal water “light water “ mass = 18.0 g/mole , BP = 100.000000C D2O Heavy water mass = 20.0 g/mole , BP = 101.42 0C Isotopes of Hydrogen

  11. 168O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 178O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 188O8 Protons 10 Neutrons 0.204 % 17.999160 amu Element #8 : Oxygen, Isotopes

  12. Calculating the “Average” Atomic Mass of an Element Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%). 24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amux 0.102 = 2.548556 amu 26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu ___________ amu With Significant Digits = __________ amu

  13. Calculate the Average Atomic Mass of Zirconium, Element #40 Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr. Isotope (% abd.) Mass (amu) (%) Fractional Mass 90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu 91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu 92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu 94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu 96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu ____________ amu With Significant Digits = ___________ amu

  14. Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol. Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0 Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br

  15. LIKE SAMPLE PROBLEM 3.2 During a perplexing dream one evening you come across 200 atoms of Einsteinium. What would be the total mass of this substance in grams? SOLUTION: 200 Atoms of Es X 252 AMU/Atom = 5.04 x 104 AMU 5.04 x 104 AMU x (1g / 6.022 x 1023 AMU) = ______________ g of Einsteinium

  16. MOLE • The Mole is based upon the following definition: • The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly • 12 grams of carbon -12. • 1 Mole = 6.022045 x 1023 particles

  17. Figure 3.4: One-mole samples of copper, sulfur, mercury, and carbon

  18. One mole of common Substances CaCO3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g

  19. Molecular Formula Atoms Molecules Avogadro’s Number 6.022 x 1023 Moles Moles

  20. Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole 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 1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule 1 molecule of S8 = 256.56 amu 1 mole of S8 = _______ g = 6.022 x 1023 molecules

  21. 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. 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 atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = __________ g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  22. LIKE SAMPLE PROBLEM 3.4 How many carbon atoms are present in a 2.0 g tablet of Sildenafil citrate (C28H38N6O11S) ? SOLUTION: MW of Sildenafil citrate = 28 X 12 amu (C) + 38 X 1 amu (H) + 6 X 14 amu (N) + 11 X 16 amu (O) + 1 X 32 amu (S) = 666 AMU

  23. LIKE SAMPLE PROBLEM 3.4 (cont…) 2.0 g (C28H38N6O11S) X 1 mol/666g = 3.0 X 10-3 mol (C28H38N6O11S) 3.0 X 10-3 mol (C28H38N6O11S) X 6.022 X 1023 molecules / 1 mol (C28H38N6O11S) = 1.8 X 1021 molecules of C28H38N6O11S 1.8 X 1021 molecules of C28H38N6O11S X 28 atoms of C / 1 molecules of C28H38N6O11S = Carbon Atoms

  24. 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 = = __________________ atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

  25. 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 = ______________ formula units

  26. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound 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

  27. 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 = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose& % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = ______% Mass Fraction of C = =

  28. 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 = 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

  29. Mol wt 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% N 4.032g H2 80.05g %H = x 100% = 5.037% H 48.00g O2 80.05g %O = x 100% = 59.96% O 99.997%

  30. 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%

  31. Penicillin is isolated from a mold Source: Getty Images

  32. 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.

  33. Figure 3.6: Examples of substances whose empirical and molecular formulas differ.

  34. 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 Empirical Formula

  35. 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

  36. Figure 3.7: Structural Formula of P4O10

  37. 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 = _________ mol Na Moles of Cr = 6.420 g Cr x = ___________ mol Cr Moles of O = 7.902 g O x = ____________ mol O 1 mol Na 22.99 g Na 1 mol Cr 52.00 g Cr 1 mol O 16.00 g O

  38. 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

  39. 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

  40. 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, divide 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

  41. 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 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

  42. 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 !

  43. 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 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

  44. Figure 3.5: Combustion device

  45. Ascorbic acid ( Vitamin C ) - I contains C , H , and O • Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O • Calculate it’s Empirical formula! • C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C • H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H • Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg O

  46. Vitamin C combustion - II • C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C • H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H • O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O • Divide each by 2.21 x 10-4 • C = 1.00 Multiply each by 3 = 3.00 = 3.0 • H = 1.32 = 3.96 = 4.0 • O = 1.00= 3.00 = 3.0 C3H4O3

More Related