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Chemical Stoichiometry: Atomic and Molecular Mass, Isotopes, Moles, and Empirical Formulas

This chapter covers topics such as atomic and molecular mass calculations, isotopes, moles of atoms and molecules, percentage composition, derivation of formulas, and solutions.

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Chemical Stoichiometry: Atomic and Molecular Mass, Isotopes, Moles, and Empirical Formulas

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  1. Chapter 3 Chemical Stoichiometry

  2. 3.1: Atomic Mass/Weight The average mass of a sulfur atom is 32.06 amu S The average mass of a sodium atom is 22.99 amu Na

  3. 3.2: Molecular Mass/Weight Formula Mass/Weight • The formula of an Acetylsalicylic Acid (Aspirin) molecule is C9H8O4 • Acetylsalicylic Acid (Aspirin) has molecular mass of 180.159 amu

  4. 3.2: Molecular Mass/Weight Formula Mass/Weight • Aluminum sulfate is an ionic compound with the formula Al2(SO4)3 • Aluminum sulfate has a formula mass of 342.153 amu.

  5. Isotopes are atoms of an element that differ only in the number of neutrons in the nucleus of the atom. Chlorine has two isotopes: chlorine – 35 35Cl chlorine – 37 37Cl 3.3: Isotopes

  6. In any random sample of chlorine about 1 in 4 atoms is chlorine – 37. 3.3: Isotopes The atomic mass of chlorine is therefore about 35.5 amu.

  7. How do atoms of chlorine – 35 and chlorine – 37 differ? 3.3: Isotopes

  8. 3.4: Moles of Atoms and Avogadro’s Number • What is a mole of atoms?

  9. 3.4: Moles of Atoms and Avogadro’s Number • What is a mole of atoms? One mole of magnesium atoms contains 6.02 x 1023 Mg atoms.

  10. 3.4: Moles of Atoms and Avogadro’s Number • What is the mass of one mole of Mg atoms? • The atomic mass of Mg is 24.305 amu. • The molar mass of Mg is 24.305 grams.

  11. 3.5: Moles of Molecules • What is a mole of molecules?

  12. 3.5: Moles of Molecules • What is a mole of molecules? • One mole of CHCl3 (chloroform) contains 6.02 x 1023 molecules of CHCl3.

  13. 3.5: Moles of Molecules • What is a mole of molecules? • One mole of CHCl3 (chloroform) contains 6.02 x 1023 molecules of CHCl3. • What else does it contain?

  14. 3.5: Moles of Molecules • How many H atoms are in one mole of CHCl3?

  15. 3.5: Moles of Molecules • How many H atoms are in one mole of CHCl3? 6.02 x 1023 H atoms

  16. 3.5: Moles of Molecules • How many Cl atoms are in one mole of CHCl3?

  17. 3.5: Moles of Molecules • How many Cl atoms are in one mole of CHCl3? 3(6.02 x 1023) Cl atoms

  18. 3.5: Moles of Molecules • What is the molar mass of CHCl3?

  19. 3.5: Moles of Molecules • What is the molar mass of CHCl3? 119.377g

  20. Key Concept • You have one mole of ascorbic acid (Vitamin C) C6H8O6. What else do you have? • 6.02 x 1023 molecules of C6H8O6 • 176.13 grams of C6H8O6 • 6 moles C or 6(6.02 x 1023) C atoms • 8 moles H or 8(6.02 x 1023) H atoms • 6 moles O or 6(6.02 x 1023) O atoms

  21. Key Concept • If you have a certain number of moles of any compound you can always find the moles of each element present? • How many mol C are in 0.80 mol of CO2? • How many mol O are in 0.80 mol of CO2? • How many mol H are in 1.2 mol of C12H22O11? • How many mol S are in 3.8 mol of Cr(SO4)3? • How many mol O are in 3.8 mol of Cr(SO4)3?

  22. 3.6: Percentage Composition or Percent by Mass • What is the percent by mass of hydrogen in water?

  23. 3.6: Percentage Composition or Percent by Mass A glass of water contains 126g of water. How many grams of hydrogen are present?

  24. 3.7: Derivation of Formulas • To calculate a formula of a substance we often use a mole ratio.

  25. 3.7: Derivation of Formulas • To calculate a formula of a substance we often use a mole ratio. • Analysis of a gas shows that it is composed of 0.090 mol carbon and 0.36 mol hydrogen. What is the empirical formula gas?

  26. 3.7: Derivation of Formulas What is the empirical formula of a compound that is 27.29% carbon and 72.71% oxygen.

  27. 3.7: Derivation of Formulas A sample of hematite contains 34.97g of iron and 15.03g of oxygen. What is the empirical formula of hematite?

  28. Determine the empirical and molecular formulas for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.

  29. Determine the empirical and molecular formulas for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol.

  30. Determine the empirical and molecular formulas for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol. CH2O

  31. Determine the empirical and molecular formulas for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O. The molar mass of the compound is 180. g/mol. CH2O • 12.01 + 2(1.008) + 1 (16.00) = 30.03 • 180./30.03 ≈ 6 • 6(CH2O) = C6H12O6

  32. 3.7: Derivation of Formulas 3.22g of a compound decomposes when heated into 1.96g of KCl and oxygen. What is the empirical formula of the compound?

  33. 3.8: Solutions • Solute: The substance that dissolves (the minor component of a solution). KMnO4

  34. 3.8: Solutions • Solvent: The substance in which the solute dissolves (the major component of a solution).

  35. Solution: A homogeneous mixture of the solute and solvent. KMnO4 solution

  36. 3.8: Solutions Dilute Concentrated

  37. Concentration of a solution Molarity(M) is the moles of solute per liter of solution. moles of solute Molarity(M) = Liters of solution

  38. Preparation of Solutions 28.3 grams of nickel(II) chloride Add water to make the desired volume (500.0 ml)

  39. Molarity • Calculate the molarity of an aqueous nickel (II) chloride solution containing 28.3 grams of nickel (II) chloride in 500.0 mL of solution.

  40. Molarity The maximum solubility of lead (II) chromate, PbCrO4, is 4.3 x 10-5 g/L. What is the molarity of a saturated solution of PbCrO4?

  41. Molarity How many moles of sulfuric acid, H2SO4, are contained in 0.80L of a 0.050M solution of sulfuric acid? How many grams of H2SO4 would be needed to make this solution?

  42. Dilution • If we take a more concentrated solution we can dilute it to a lower concentration by adding water to it. • We determine the concentration (molarity) of the diluted solution by using the following formula. M1V1 = M2V2

  43. Dilution • Calculate the concentration of the resulting solution when enough water is added to 250.0mL of 0.60M NaOH to make 300.0mL of solution?

  44. Dilution • How much water would need to be added to 125mL of a 1.50M solution of HCl to dilute the solution to a concentration of 0.570M?

  45. Density • An understanding of density is often necessary in solving various problems. • Density is a ratio of mass to volume and therefore can be used in the conversion of mass to volume or volume to mass.

  46. Mass Percent • An understanding of mass percent is often necessary in solving various problems. • Mass percent indicates the percentage of a particular substance in a mixture.

  47. Aniline, C6H5NH2, a key ingredient in the preparation of dyes for fabrics, is produced by the reaction of C6H5Cl with a solution containing 28.2% NH3 by mass. If the density of the NH3 solution is 0.899g/cm3. What mass of NH3 is needed to prepare 125mL of the ammonia solution? aniline

  48. Aniline, C6H5NH2, a key ingredient in the preparation of dyes for fabrics, is produced by the reaction of C6H5Cl with a solution containing 28.2% NH3 by mass. If the density of the NH3 solution is 0.899g/cm3. What mass of NH3 is needed to prepare 125mL of the ammonia solution?

  49. Stoichiometry: Calculations based on balanced equations. • We use the balanced equation to determine a mole ratio. • Then we use “Moles to Move”.

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