1. Chemical Nomenclature, Formulas, and Equations

2. 2 Formulas and Models

3. 3 A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. An empirical formula shows the simplest whole-number ratio of the atoms in a substance. H2OH2O H2OH2O molecularempirical C 6 H 12 O 6 CH 2 O O3O3 O N2H4N2H4 NH 2

4. 2.2 Write the molecular formula of methylamine, a colorless gas used in the production of pharmaceuticals and pesticides, from its ball-and-stick model, shown below.

5. 2.2 Solution Refer to the labels (also see back end papers). There are five H atoms, one C atom, and one N atom. Therefore, the molecular formula is CH 5 N. However, the standard way of writing the molecular formula for methylamine is CH 3 NH 2 because it shows how the atoms are joined in the molecule.

6. 2.3 Write the empirical formulas for the following molecules: (a)biborane (B 2 H 6 ), which is used in rocket propellants (b)glucose (C 6 H 12 O 6 ), a substance known as blood sugar (c)nitrous oxide (N 2 O), a gas that is used as an anesthetic gas (“laughing gas”) and as an aerosol propellant for whipped creams.

7. 2.3 Strategy Recall that to write the empirical formula, the subscripts in the molecular formula must be converted to the smallest possible whole numbers.

8. 2.3 Solution (a)There are two boron atoms and six hydrogen atoms in diborane. Dividing the subscripts by 2, we obtain the empirical formula BH 3. (b) In glucose there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Dividing the subscripts by 6, we obtain the empirical formula CH 2 O. Note that if we had divided the subscripts by 3, we would have obtained the formula C 2 H 4 O 2. Although the ratio of carbon to hydrogen to oxygen atoms in C 2 H 4 O 2 is the same as that in C 6 H 12 O 6 (1:2:1), C 2 H 4 O 2 is not the simplest formula because its subscripts are not in the smallest whole-number ratio.

9. 2.3 (c) Because the subscripts in N 2 O are already the smallest possible whole numbers, the empirical formula for nitrous oxide is the same as its molecular formula.

10. 10 Ionic compounds consist of a combination of cations and anions. The formula is usually the same as the empirical formula. The sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero. The ionic compound NaCl

11. 11 The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds.

12. 12 Formulas of Ionic Compounds Al 2 O 3 2 x +3 = +63 x -2 = -6 Al 3+ O 2- CaBr 2 1 x +2 = +22 x -1 = -2 Ca 2+ Br - Na 2 CO 3 2 x +1 = +21 x -2 = -2 Na + CO 3 2-

13. 2.4 Write the formula of magnesium nitride, containing the Mg 2+ and N 3− ions. When magnesium burns in air, it forms both magnesium oxide and magnesium nitride.

14. 2.4 Strategy Our guide for writing formulas for ionic compounds is electrical neutrality; that is, the total charge on the cation(s) must be equal to the total charge on the anion(s). Because the charges on the Mg 2+ and N 3− ions are not equal, we know the formula cannot be MgN. Instead, we write the formula as Mg x N y, where x and y are subscripts to be determined.

15. 2.4 Solution To satisfy electrical neutrality, the following relationship must hold: (+2)x + (−3)y = 0 Solving, we obtain x/y = 3/2. Setting x = 3 and y = 2, we write Check The subscripts are reduced to the smallest whole- number ratio of the atoms because the chemical formula of an ionic compound is usually its empirical formula.

16. 16 Chemical Nomenclature Ionic Compounds –Often a metal + nonmetal –Anion (nonmetal), add “-ide” to element name BaCl 2 barium chloride K2OK2O potassium oxide Mg(OH) 2 magnesium hydroxide KNO 3 potassium nitrate

17. 17 Transition metal ionic compounds –indicate charge on metal with Roman numerals FeCl 2 2 Cl - -2 so Fe is +2iron(II) chloride FeCl 3 3 Cl - -3 so Fe is +3iron(III) chloride Cr 2 S 3 3 S -2 -6 so Cr is +3 (6/2)chromium(III) sulfide

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20. 2.5 Name the following compounds: (a)Fe(NO 3 ) 2 (b)Na 2 HPO 4 (c)(NH 4 ) 2 SO 3

21. 2.5 Strategy Our reference for the names of cations and anions is Table 2.3. Keep in mind that if a metal can form cations of different charges (see Figure 2.10), need to use the Stock system.

22. 2.5 Solution (a)The nitrate ion (NO 3 − ) bears one negative charge, so the iron ion must have two positive charges. Because iron forms both Fe + and Fe 2+ ions, we need to use the Stock system and call the compound iron(II) nitrate. (b)The cation is Na + and the anion is HPO 4 2− (hydrogen phosphate). Because sodium only forms one type of ion (Na + ), there is no need to use sodium(I) in the name. The compound is sodium hydrogen phosphate. (c) The cation is NH 4 + (ammonium ion) and the anion is SO 3 2− (sulfite ion). The compound is ammonium sulfite.

23. 2.6 Write chemical formulas for the following compounds: (a)mercury(I) nitrite (b)cesium sulfide (c)calcium phosphate

24. 2.6 Strategy We refer to Table 2.3 for the formulas of cations and anions. Recall that the Roman numerals in the Stock system provide useful information about the charges of the cation.

25. 2.6 Solution (a)The Roman numeral shows that the mercury ion bears a +1 charge. According to Table 2.3, however, the mercury(I) ion is diatomic (that is, ) and the nitrite ion is. Therefore, the formula is Hg 2 (NO 2 ) 2. (b)Each sulfide ion bears two negative charges, and each cesium ion bears one positive charge (cesium is in Group 1A, as is sodium). Therefore, the formula is Cs 2 S.

26. 2.6 (c) Each calcium ion (Ca 2+ ) bears two positive charges, and each phosphate ion ( ) bears three negative charges. To make the sum of the charges equal zero, we must adjust the numbers of cations and anions: 3(+2) + 2(−3) = 0 Thus, the formula is Ca 3 (PO 4 ) 2.

27. 27 Molecular compounds −Nonmetals or nonmetals + metalloids −Common names −H 2 O, NH 3, CH 4 −Element furthest to the left in a period and closest to the bottom of a group on periodic table is placed first in formula −If more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom −Last element name ends in -ide

28. 28 HIhydrogen iodide NF 3 nitrogen trifluoride SO 2 sulfur dioxide N 2 Cl 4 dinitrogen tetrachloride NO 2 nitrogen dioxide N2ON2Odinitrogen monoxide Molecular Compounds

29. 2.7 Name the following molecular compounds: (a)SiCl 4 (b)P 4 O 10

30. 2.7 Strategy We refer to Table 2.4 for prefixes. In (a) there is only one Si atom so we do not use the prefix “mono.” Solution (a)Because there are four chlorine atoms present, the compound is silicon tetrachloride. (b)There are four phosphorus atoms and ten oxygen atoms present, so the compound is tetraphosphorus decoxide. Note that the “a” is omitted in “deca.”

31. 2.8 Write chemical formulas for the following molecular compounds: (a)carbon disulfide (b) disilicon hexabromide

32. 2.8 Strategy Here we need to convert prefixes to numbers of atoms (see Table 2.4). Because there is no prefix for carbon in (a), it means that there is only one carbon atom present. Solution (a)Because there are two sulfur atoms and one carbon atom present, the formula is CS 2. (b) There are two silicon atoms and six bromine atoms present, so the formula is Si 2 Br 6.

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34. 34 An acid can be defined as a substance that yields hydrogen ions (H + ) when dissolved in water. For example: HCl gas and HCl in water Pure substance, hydrogen chloride Dissolved in water (H 3 O + and Cl − ), hydrochloric acid

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36. 36 An oxoacid is an acid that contains hydrogen, oxygen, and another element. HNO 3 nitric acid H 2 CO 3 carbonic acid H 3 PO 4 phosphoric acid

37. 37 Naming Oxoacids and Oxoanions

38. 38 The rules for naming oxoanions, anions of oxoacids, are as follows: 1. When all the H ions are removed from the “-ic” acid, the anion’s name ends with “-ate.” 2. When all the H ions are removed from the “-ous” acid, the anion’s name ends with “-ite.” 3. The names of anions in which one or more but not all the hydrogen ions have been removed must indicate the number of H ions present. For example: –H 2 PO 4 - dihydrogen phosphate –HPO 4 2- hydrogen phosphate –PO 4 3- phosphate

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40. 2.9 Name the following oxoacid and oxoanion: (a)H 2 SO 3, a very unstable acid formed when SO 2 (g) reacts with water (sulfurous acid  sulfite +2H+) (b)H 2 AsO 4 −, once used to control ticks and lice on livestock (dihydrogen arsenate) (c)SeO 3 2−, used to manufacture colorless glass. (selenite) H 3 AsO 4 is arsenic acid, and H 2 SeO 4 is selenic acid. H 2 SeO 3 is selenous acid

41. 2.9 Strategy We refer to Figure 2.14 and Table 2.6 for the conventions used in naming oxoacids and oxoanions. Solution (a)We start with our reference acid, sulfuric acid (H 2 SO 4 ). Because H 2 SO 3 has one fewer O atom, it is called sulfurous acid. (b)Because H 3 AsO 4 is arsenic acid, the AsO 4 3− is named arsenate. The H 2 AsO 4 − anion is formed by adding two H + ions to AsO 4 3−, so H 2 AsO 4 − is called dihydrogen arsenate. (c)The parent acid is H 2 SeO 3. Because the acid has one fewer O atom than selenic acid (H 2 SeO 4 ), it is called selenous acid. Therefore, the SeO 3 2− anion derived from H 2 SeO 3 is called selenite.

42. 42 A base can be defined as a substance that yields hydroxide ions (OH - ) when dissolved in water. NaOH sodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide

43. 43 Hydrates are compounds that have a specific number of water molecules attached to them. BaCl 2 2H 2 O LiClH 2 O MgSO 4 7H 2 O Sr(NO 3 ) 2 4H 2 O barium chloride dihydrate lithium chloride monohydrate magnesium sulfate heptahydrate strontium nitrate tetrahydrate CuSO 4 5H 2 O CuSO 4

44. 44 Organic chemistry is the branch of chemistry that deals with carbon compounds. C H H H OH C H H H NH 2 C H H H COH O methanolmethylamineacetic acid Functional Groups:

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46. 46 Chemical Equations Symbolic representation of a chemical reaction that shows: 1.reactants on left side of reaction 2.products on right side of equation 3.relative amounts of each using stoichiometric coefficients

47. 47 Chemical Equations Attempt to show on paper what is happening at the laboratory and molecular levels.

48. 48 Chemical Equations Look at the information an equation provides: ReactantsYieldsProducts

49. 49 Chemical Equations Look at the information an equation provides: ReactantsYieldsProducts 1 formula unit 3 molecules 2 atoms3 molecules

50. 50 Chemical Equations Look at the information an equation provides: ReactantsYieldsProducts 1 formula unit 3 molecules 2 atoms3 molecules 1 mole3 moles2 moles3 moles

51. 51 Chemical Equations Look at the information an equation provides: ReactantsYieldsProducts 1 formula unit 3 molecules 2 atoms3 molecules 1 mole3 moles2 moles3 moles 159.7 g84.0 g111.7 g132 g

52. 52 Chemical Equations Law of Conservation of Matter – There is no detectable change in quantity of matter in an ordinary chemical reaction. – Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. – This law was determined by Antoine Lavoisier.

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