Hydrates and Organic Compounds: Chemical Equations and Mole Relationships

1. Midterm WEDNESDAYCh 8-10.5Final Exam will be the last day of class. SL 130, 1-4 pm.

2. Hydrate A compound that contains water molecules weakly bound in the crystals. The formula of a hydrate is written with a dot before the water molecule(s) included. For example: CuSO45H2O

3. Hydrates are named using the anhydrous (without water) compound name followed by the prefix for the number of water molecules included and the word “hydrate.” For example: CuSO45H2O is named copper(II) sulfate pentahydrate.

4. A compound whose common name is green vitriol has the chemical formula FeSO47H2O. What is the chemical name of this compound? FeSO47H2O is iron(II) sulfate heptahydrate.

5. Calcium chloride hexahydrate is used to melt snow on roads. What is the chemical formula of the compound? The chemical formula for calcium chloride hexahydrate is CaCl26H2O.

6. Organic Compounds An important class of molecular substances; they contain carbon combined with other elements – notably hydrogen, oxygen, and nitrogen. Hydrocarbons contain only carbon and hydrogen.

9. A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas. For example: 2Na + Cl2 2NaCl Reactants are the starting materials; they are written on the left of the equation. Products are the materials at the end of the reaction; they are written on the right of the equation.

10. Because a reaction must accurately describe the chemical reaction, it must be consistent with the law of conservation of mass. When this is not the case, after correct formulas are written for each reactant and product, the coefficients are adjusted so that the same number of each atom is present in both the reactants and the products. This is called balancingthe equation.

11. Balance the following equation: NH3 + O2 NO + H2O Tally the number of each atom on each side: N 1 on reactant side; 1 on product side H 3 on reactant side; 2 on product side O 2 on reactant side; 2 on product side Begin by inserting the coefficient “2” before NH3 on the reactant side and the coefficient “3” before H2O on the product side. We leave O2 until later because it is an element.

12. 2NH3 + O2 NO + 3H2O • Tally the atoms again: • N 2 on reactant side; 1 on product side • H 6 on reactant side; 6 on product side • O 2 on reactant side; 4 on product side • To balance N, insert a “2” before NO: • 2NH3 + O2 2NO + 3H2O

13. 2NH3 + O2 2NO + 3H2O • Tally the atoms again: • N 2 on reactant side; 2 on product side • H 6 on reactant side; 6 on product side • O 2 on reactant side; 5 on product side • Since this gives us an odd number oxygens, we double the coefficients on NH3, NO, and H2O and to balance O, insert a “5” before O2.

14. Tally the atoms again to double check: • 4NH3 + 5O2 4NO + 6H2O • N 4 on reactant side; 4 on product side • H 12 on reactant side; 12 on product side • O 10 on reactant side; 10 on product side • The reaction is now balanced!

15. Balance the following equation: C2H5OH + O2 CO2 + H2O Tally the number of each atom on each side: C 2 on reactant side; 1 on product side H 6 on reactant side; 2 on product side O 3 on reactant side; 3 on product side Begin by balancing H. Insert the coefficient “3” before H2O on the product side. We leave O2 until later because it is an element.

16. C2H5OH + O2 CO2 + 3H2O • Tally the number of each atom on each side: • C 2 on reactant side; 1 on product side • H 6 on reactant side; 6 on product side • O 3 on reactant side; 5 on product side • To balance C, insert a “2” before CO2.

17. 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

18. The Mole is based upon the 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 The Mole

19. Mole, mol The quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12 Avogadro’s Number, NA The number of atoms in exactly 12 g of carbon-12 NA = 6.02 × 1023 (to three significant figures)

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 = 256.52 amu 1 mole of S8 = 256.52 g = 6.022 x 1023 molecules

21. One Mole of Common Substances CaCO3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g

22. Figure 3.2: One mole each of various substances.Photo courtesy of American Color.

23. Molar Mass The mass of one mole of substance For example: Carbon-12 has a molar mass of 12 g or 12 g/mol

24. Molecular Mass The sum of the atomic masses of all the atoms in a molecule of the substance Formula Mass The sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not

25. 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 = 18.02 g 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

26. Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures. calcium hydroxide, Ca(OH)2 methylamine, CH3NH2

27. Ca(OH)2 1 Ca 1(40.08) = 40.08 amu 2 O 2(16.00) = 32.00 amu 2 H 2(1.008) = 2.016 amu 3 significant figures 74.1 amu Total 74.095 CH3NH2 1 C 1(12.01) = 12.01 amu 1 N 1(14.01) = 14.01 amu 5 H 5(1.008) = 5.040 amu Total 31.060 3 significant figures 31.1 amu

28. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/moleculeof compound Mass/mole of compound 6 atoms 12 atoms 6 atoms 6 moles of 12 moles of 6 moles of atoms atoms atoms 6(6.022 x 1023) 12(6.022 x 1023) 6(6.022 x 1023) atoms atoms atoms 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) =72.06 amu =12.10 amu =96.00 amu 72.06 g 12.10 g 96.00 g

30. What is the mass in grams of the nitric acid molecule, HNO3? First, find the molar mass of HNO3: 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 (2 decimal places) 63.02 g/mol 63.018

31. Next, convert this mass of one mole to one molecule using Avogadro’s number:

32. A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this? First, find the molar mass of HNO3: 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 (2 decimal places) 63.02 g/mol 63.018

33. Next, using the molar mass, find the mass of 0.253 mole: = 15.94406 g

34. Chapter 4 Chemical Reactions: An Introduction

35. Figure 4.1: Reaction of potassium iodide solution and lead (II) nitrate solution. Photo courtesy of James Scherer.

36. Electrolytes

37. Figure 4.2: Motion of ions in solution.

38. Figure 4.3: Testing the electrical conductivity of a solution: water.Photo courtesy of American Color.

39. Figure 4.3: Testing the electrical conductivity of a solution: sodium chloride.Photo courtesy of American Color.

40. Figure 4.4: Comparing strong and weak electrolytes: HCl. Photo courtesy of American Color.

41. Figure 4.4: Comparing strong and weak electrolytes: NH3. Photo courtesy of American Color.

42. Methanol Li

43. The Role of Water as a Solvent: The Solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called strong Eeectrolytes. NaCl(s) + H2O(l) Na+(aq) + Cl -(aq) When sodium chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

44. Fig. 4.2

46. Molarity (Concentration of Solutions)= M Moles of Solute Moles Liters of Solution L M = = solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

47. Making a solution of known molarity Desire 250.0 mL of 0.3128 M CuSO4 • Calculate mass of solute needed • moles = M × V • moles →mass • 159.61 g CuSO4/mole 2. Accurately weigh solute

48. Do we add exactly 250.00 mL water? • Use volumetric glassware to dissolve solute • Add water until level matches mark on flask 0.07821 moles in 0.2500 L solution

49. Fig. 3.11

50. Preparing a Solution - I • Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l ! • What is the Molarity of the salt and each of the ions? • Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)