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Chapter 31 Molecules, Ions, and Their Compounds Chapter 3.

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1 Chapter 31 Molecules, Ions, and Their Compounds Chapter 3

2 2 Molecular and Empirical Formulas Molecules and Molecular Compounds Molecular formula – A formula which gives the actual number and type of atoms in a molecule. Examples: H 2 O, CO 2, CO, CH 4, H 2 O 2, O 2, O 3, and C 2 H 4. Empirical formula – A formula which gives the lowest whole number ratio of atoms in a molecule. Examples: SubstanceMol. formulaEmpirical Formula Ethane C 2 H 6 CH 3 Water H 2 OH 2 O

3 Chapter 33 Molecular and Empirical Formulas Condensed formula – A formula which indicates how atoms are grouped together in a molecule. Name Molecular FormulaCondensed formula Ethane C 2 H 6 CH 3 CH 3 Diethyl etherC 4 H 10 O CH 3 CH 2 OCH 2 CH 3 Molecules and Molecular Compounds

4 Chapter 34 Picturing Molecules Structural Formula – A formula which shows how the atoms of a molecule are joined. Structural formulas do not necessarily show the three dimensional shape of the molecule. Molecules and Molecular Compounds

5 Chapter 35 Molecular Models These are three-dimensional representations of molecules. Molecules and Molecular Compounds

6 Chapter 36 If an electron is removed or added to a neutral atom a charged particle or ion is formed. A positively charged ion is called a cation. Ions and Ionic Compounds

7 Chapter 37 If an electron is removed or added to a neutral atom a charged particle or ion is formed. A positively charged ion is called a cation. A negatively charged ion is called an anion. Ions and Ionic Compounds

8 Chapter 38 Predicting Ionic Charge Metals tend to form cations Non-metals tend to form anions. Ions and Ionic Compounds

9 Chapter 39 Ions and Ionic Compounds Molecules can also gain or lose electrons and form ions, They are called polyatomic ions.

10 Chapter 310 Ions and Ionic Compounds Ion NameFormulaIon NameFormula PeroxideO 2 2- SulfateSO 4 2- TriiodideI3-I3- SulfiteSO 3 2- AmmoniumNH 4 + PhosphatePO 4 3- NitrateNO 3 - AcetateCH 3 CO 2 - NitriteNO 2 - PerchlorateClO 4 - HydroxideOH - PermanganateMnO 4 - CarbonateCO 3 2- DichromateCr 2 O 7 2-

11 Chapter 311 Ionic Compounds Ionic Compound – A compound that contains positively charged ions and negatively charged ions. Ions and Ionic Compounds

12 Chapter 312 Predicting Formulas Let’s consider a compound containing Mg and N. The common charge on Mg is +2 (or Mg 2+ ). The common charge on N is –3 (or N 3- ). Since we want to make a neutral (uncharged) compound, the total charges from the cations and anions must cancel-out (or sum to zero). Therefore, Mg needs to lose 6 electrons (3  2 + ) and N gain those 6 electrons (2  3 - ). The resulting formula is: Mg 3 N 2. Ions and Ionic Compounds

13 Chapter 313 Names and Formulas of Ionic Compounds Naming of compounds (nomenclature) is divided into: organic compounds (those containing C) inorganic compounds (the rest of the periodic table). We will consider the naming rules of the Inorganic compounds. Naming Inorganic Compounds

14 Chapter 314 Naming Ionic Compounds 1.Cations a)Cations from metal atoms have the same name as the metal. b)If the cation can form more than one ion, the positive charge is indicated by a roman numeral in parenthesis. c)Cations of nonmetals end in –ium. P +3 phosphorium Names and Formulas of Ionic Compounds

15 Chapter 315 Names and Formulas of Ionic Compounds Naming Inorganic Compounds

16 Chapter 316 Naming Inorganic Compounds 2.Anions a)Monoatomic anions have names formed by dropping the ending of the name of the element and adding –ide. b)Polyatomic anions containing oxygen have names ending in –ate or –ite. c)Anions derived by adding H + to an oxyanion are named by adding as a prefix the word hydrogen- or dihydrogen-. HSO 4 - Hydrogensulfate H 2 PO 4 - Dihydrogenphsophate Names and Formulas of Ionic Compounds

17 Chapter 317 Naming Inorganic Compounds 2.Anions d) Oxyanions There are rules for these, but they are confusing. Ion Name ClO 4 - perchlorate ion ClO 3 - chlorate ion ClO 2 - chlorite ion ClO - hypochlorite ion Names and Formulas of Ionic Compounds

18 Chapter 318 Naming Inorganic Compounds 3.Ionic Compounds Name the compound by naming the cation followed by the anion. Names and Formulas of Ionic Compounds

19 Chapter 319 Naming Binary Molecular Compounds Binary molecular compounds have two elements. 1.The name of the left-most element is written first. 2.If the elements are in the same group the lower element is written first. 3.The name of the second element ends in –ide. 4.Prefixes are used to indicate the number of atoms of each element. Naming Inorganic Compounds

20 Chapter 320 Naming Binary Molecular Compounds Naming Inorganic Compounds

21 Chapter 321 Naming Binary Molecular Compounds Binary molecular compounds have two elements. 1.The name of the left-most element is written first. 2.If the elements are in the same group the lower element is written first. 3.The name of the second element ends in –ide. 4.Prefixes are used to indicate the number of atoms of each element. Mono is never used in the first element. Naming Inorganic Compounds

22 Chapter 322 Naming Inorganic Compounds N2ON2O This is a molecular compound. The first element (N), just takes its name, Nitrogen. The second compound takes its name, ending in -ide, Oxide. Now we must consider how to show that there are two nitrogen atoms, use di- as a prefix. Dinitrogen Oxide

23 Chapter 323 Naming Inorganic Compounds N2O5N2O5 This is a molecular compound. The first element (N), just takes its name, Nitrogen. The second compound takes its name, ending in -ide, Oxide. Now we must consider how to show that there are two nitrogen atoms, use di- as a prefix. Finally, we must consider how to show that there are five oxygen atoms, use penta- as a prefix. Dinitrogen Pentoxide

24 Chapter 324 Formula and Molecular Weights Molecular weight The sum of the atomic weights of each atom in the molecular formula. Formula weight is the general term, molecule weight refers specifically to molecular compounds. Formulas, Compounds, and the Mole

25 Chapter 325 Formula and Molecular Weights Formula weight (FW) The sum of the atomic weights of each atom in the chemical formula. Example: CO 2 Formula Weight = 1(AW, carbon) + 2(AW, oxygen) Formula Weight = 1(12.011amu) + 2(16.0amu) Formula Weight = 44.0 amu Formulas, Compounds, and the Mole

26 Chapter 326 Formula and Molecular Weights Chemistry “trick” The masses of the atoms are on a “gram equivalent scale”. 1 atom (average)1 mole C 12.01 amu12.01 g H 1.008 amu1.008 g So, the mass of a single atom or a mole is numerically equvalent. Formulas, Compounds, and the Mole

27 Chapter 327 Converting Between Mass, Moles, Molecules and Atoms Formulas, Compounds, and the Mole

28 Chapter 328 Moles Numbers of Particles Moles  Numbers of Particles The Mole

29 Chapter 329 Mass  Moles The Mole

30 Chapter 330 Moles  Mass The Mole

31 Chapter 331 A sample of hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many atoms of carbon does it contain? The Mole

32 Chapter 332 A sample of hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many atoms of carbon does it contain? The Mole

33 Chapter 333 A sample of hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many atoms of carbon does it contain? The Mole

34 Chapter 334 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many atoms of carbon does it contain? The Mole

35 Chapter 335 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many atoms of carbon does it contain? The Mole

36 Chapter 336 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many molecules of estradiol does it contain? The Mole

37 Chapter 337 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many molecules of estradiol does it contain? The Mole

38 Chapter 338 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many molecules of estradiol does it contain? The Mole

39 Chapter 339 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many molecules of estradiol does it contain? The Mole

40 Chapter 340 A sample hormone, estradiol, C 18 H 24 O 2, contains 3.0 x 10 20 atoms of hydrogen. How many molecules of estradiol does it contain? The Mole

41 Chapter 341 Percentage Composition from Formulas Describing Compound Formulas

42 Chapter 342 Percentage Composition from Formulas Example: Calculate the percent oxygen in CH 3 CH 2 OH. Formula weight of ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu Mass of oxygen: 1(16.00amu) = 16.00amu % oxygen Describing Compound Formulas

43 Chapter 343 Percentage Composition from Formulas Example: Calculate the percent oxygen in CH 3 CH 2 OH. Formula weight of ethanol: 2(12.01amu) + 6(1.01amu) + 1(16.00amu) = 46.08amu Mass of oxygen: 1(16.00amu) = 16.00amu % oxygen Describing Compound Formulas

44 Chapter 344 Empirical Formulas from Analyses

45 Chapter 345 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = Cl  26.1g/ 35.45g/mol = Empirical Formulas from Analyses

46 Chapter 346 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = 0.368 mol Cl  26.1g/ 35.45g/mol = 0.736 mol Empirical Formulas from Analyses

47 Chapter 347 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = 0.368 mol Cl  26.1g/ 35.45g/mol = 0.736 mol -determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Empirical Formulas from Analyses

48 Chapter 348 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = 0.367 mol Cl  26.1g/ 35.45g/mol = 0.736 mol -determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Empirical Formulas from Analyses

49 Chapter 349 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = 0.368 mol Cl  26.1g/ 35.45g/mol = 0.736 mol -determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Empirical Formulas from Analyses

50 Chapter 350 Analysis Hg  73.9% Cl  26.1% -assume 100g sample Hg  73.9 g Cl  26.1g -convert grams to moles Hg  73.9g / 200.59g/mol = 0.368 mol Cl  26.1g/ 35.45g/mol = 0.736 mol -determine the empirical formula by using the moles of the elements to get the smallest whole number ratio of the elements. Empirical Formulas from Analyses

51 Chapter 351 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Empirical Formulas from Analyses

52 Chapter 352 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Analysis: C  10.4% S  27.8% Cl  61.7% Empirical Formulas from Analyses

53 Chapter 353 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Assume 100g sample C  10.4g S  27.8g Cl  61.7g Empirical Formulas from Analyses

54 Chapter 354 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Empirical Formulas from Analyses

55 Chapter 355 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Empirical Formulas from Analyses

56 Chapter 356 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Empirical Formulas from Analyses

57 Chapter 357 Determine the empirical formula of the compound with the following compositions by mass: C, 10.4%; S, 27.8%, Cl, 61.7%. Moles of each element C  10.4g/12.011g/mol = 0.866 mol S  27.8g/32.066g/mol = 0.867 mol Cl  61.7g/35.453g/mol = 1.74 mol Empirical Formulas from Analyses

58 Chapter 358 Molecular Formula from Empirical Formula To determine the molecular formula from an empirical formula, you must have the molecular weight of the substance. Empirical Formulas from Analyses

59 Chapter 359 Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Empirical Formulas from Analyses

60 Chapter 360 Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Empirical Formulas from Analyses

61 Chapter 361 Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Empirical Formulas from Analyses

62 Chapter 362 Molecular Formula from Empirical Formula Empirical formula: CH Empirical formula weight: 13.019 g/mol Molecular weight: 78.114g/mol Empirical Formulas from Analyses

63 Chapter 363 End of Chapter Problems 6, 8, 12, 14, 20, 22, 28, 32, 36, 42, 48, 54, 68, 86

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