A Getting-It-On Review and Self-Test Formulas and Equations

Transcription

1 A Getting-It-On Review and Self-Test Formulas and Equations The mole is defined as a (1) of things. To three significant figures, this number has the value (2). Atomic weights are based on a mole of (3). A mole of atoms of a given element is called (4). A mole of molecules is called (5).A whole number ratio of the number of atoms of each element in a compound is called a (6). The simplest or reduced ratio of atoms is called the (7).The (8) is a (9) number multiple of this formula. In the case of salts, the (10) formula and the (11) formula are often identical because of the simple relationship between (12). An (13) weight may be found by adding the mole weight of the elements in an (14) formula. This weight can be related to the mole weight of the compound by the equation (15). The percent composition by weight may be determined by comparing the weight of an (16) to the total (17) weight. Given the percentage composition by weight, the (18) but not the (19) formula may be determined. The (20) weight must be known before the (21) formula is determined. Compounds may be generally classified as (22) or (). Covalent compounds contain (24) while ionic compounds contain (25). Binary compounds containing oxygen are called (26). Non-metal oxides usually indicate the relative number of atoms by using (27) whereas metal oxides are usually named based on the (28) of the (29). Covalent ternary compounds containing oxygen and hydrogen are called (30). An acid usually contains (31) which (32) in water solutions. Ionic ternary compounds containing oxygen and hydrogen are generally called (33). These compounds generally contain (34) ions. Naming oxy-acids and the salts of oxy-acids is usually based on the (35). An ic acid becomes a (36) salt and an ous acid becomes a (37) salt. Chemical equations relate (38) to (39) during a chemical change. The balanced chemical equation has an equal number of (40) in the (41) and the (42). In balancing chemical equations, the (43) must not be altered. A number preceding a formula modifies (44) in the formula. For compounds the sum of the (45) is (46). Common values for charges are: elements (47), fluorine (48), oxygen (49), hydrogen (50), IA metals (51), IIA metals (52).

2 2 53. Determine the mole weights of: a. He b. H2O c. BaCl2 2 H2O d. C6H12O6 54. Calculate the number of moles in the following amounts: a grams Ne b grams H2O c grams BaCl2 2 H2O d grams C6H12O6 55. Calculate the weights of the following amounts: a. 3 moles Ne b molecules H2O c mole BaCl2 2 H2O d molecules C6H12O6 56. Calculate the number of atoms, molecules, or formula units from the given information: a mole Ne = atoms Ne b grams H2O = molecules H2O c grams BaCl2 2 H2O = atoms of H d moles C6H12O6 = atoms of C, H, & O 57. Balance the chemical equations and indicate the general type: a. Na + Br2 NaBr b. Pb(NO3)2 + HI PbI2 + HNO3 c. Fe(ClO3)3 FeCl3 + O2 d. Al + Cu(NO3)2 Al(NO3)3 + Cu e. NH3 + H3PO4 (NH4)3PO4 f. C + H2O C12H22O11 g. Ba(HCO3)2 + H3PO4 Ba3(PO4)2 + H2CO3 h. FeI3 + Br2 FeBr3 + I2

3 3 58. Given the reactants and type of reaction, write the formula(s) of the product(s): reactants type a HCl and NaOH double replacement b S and H2 combination c NH3 and H2SO4 combination d Ca and H2O single replacement e Ba(OH)2 and H2SO4 double replacement f KI and F2 single replacement 59. Compute the percent composition by weight of each of the elements in: a. BaCl2 b. CuSO4 12 H2O 60. Given the mole formula or percent composition by weight, determine the empirical formula: a. C2H4O2 b. % Fe = 46.6 % S = 53.4 c. Na3S9O12 d. % Hg = 70.6 % Cl = 12.5 % O = Determine the chemical formulas given the mole weight and percent by weight composition: a. % C = 42.1 % H = 5.3 % N = 24.6 % O = 28.1 b. % Na = % P = % O = mole weight = 228 mole weight = 375.8

4 4 ANSWERS 1. Avogadro s number carbon gram atomic weight 5. gram molecular weight 6. formula 7. empirical formula 8. molecular formula 9. whole 10. Empirical 11. compound 12. charges 13. empirical 14. empirical 15. mole weight empirical weight = n 16. element 17. mole 18. empirical 19. molecular 20. mole 21. molecular 22. covalent. ionic 24. molecules 25. ions 26. oxides 27. prefixes 28. Charge 29. metal 30. oxy-acids 31. hydrogen 32. ionizes 33. bases 34. hydroxide 35. charge 36. ate 37. ite 38. reactants 39. products 40. elements 41. reactants 42. products 43. formulas 44. everything 45. charges 46. zero or a grams b grams c grams d grams

5 5 54. a g Ne 1 mole Ne = mole g Ne b g H2O 1 mole H2O = 20.0 mole g H2O c g BaCl2-2H 2O 1mole = mole g d. -5 1mole g C6H12O 6 = mole 180. g 55. a. b. 20 g 3 mole Ne = 60 g Ne 1 mole 24 1 mole 18 g molec = 90. g molec 1 mole 244 g mole BaCl -2H O = 12.2 g 1 mole 21 1 mole 180. g = g mole c. 2 2 d. 56. a. b atoms mole = atoms 1 mole 1 mole molec g = molec 18 g 1 mole c. 1 mole 4 mole H atom g BaCl 2-2H2O = atom 244 g 1 mole BaCl 2-2H2O 1 mole H d molec 24 atoms mole C6H12O 6 = atoms 1 mole 1 molecule type 57. a. 2, 1 2 combination b. 1, 2 1, 2 double replacement c. 2 2, 9 decomposition d. 2, 3 2, 3 single replacement e. 3, 1 1 combination f. 12, 11 1 combination g. 3, 2 1, 6 double replacement h. 2, 3 2, 3 single replacement

6 6 58. a. H2O + NaCl b. H2S c. (NH4)2SO4 d. CaO + H2 e. BaSO4 + H2O f. KF + I2 59. a. BaCl2 = % Ba = % % Cl = = 34.05% b. CuSO4 12 H2O = Cu + S + 16 O + 24 H = % Cu = % S = 8.50 % O = % H = a. ½ (C2H4O2) = CH2O b Fe S FeS2 c. 1 3 (Na3S9O12) = NaS3O4 d Hg 1 Cl O 3 HgClO First, the old fashioned way: 61. a C H N O C2H3N1O1

7 7 empirical weight = 2 C + 3 H + 1 N + 1 O= 57.1 mole weight empirical weight 57.1 molecular formula = 4 (C2H3N1O1) = C8H12N4O4 Second, the new and improved faster method: 61. a g C 1 mole C 228 g 100 g 12.0 g C = 8 mole C 5.3 g H 1 mole H 228 g 100 g 1.0 g H = 12 mole H 24.6 g N 1 mole N 228 g 100 g 14.0 g N = 4 mole N 28.1 g O 1 mole O 228 g 100 g 16.0 g O = 4 mole O C8H12N4O4 First, the old fashioned way: b. Na P O Na1P1O2.5 = Na2P2O5 empirical weight = 2 Na + 2 P + 5 O = mole weight empirical weight 187.9

8 8 molecular formula = 2 (Na2P2O5) = Na4P4O10 Second, the new and improved faster method: 61. b g Na 1 mole Na g 100 g g Na = 4 mole Na g P 1 mole P g 100 g g P = 4 mole P g O 1 mole O g 100 g g O = 10 mole O Na4P4O10