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Chapter 7 Chemical Formulas & Naming(Nomenclature)

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Presentation on theme: "Chapter 7 Chemical Formulas & Naming(Nomenclature)"— Presentation transcript:

1 Chapter 7 Chemical Formulas & Naming(Nomenclature)

2 Chemical Formulas Chemical formulas indicate the relative # of each type of atom in a compound Ex: C 8 H 18 (octane)Al 2 (SO 4 ) 3 8 Carbons18 Hydrogens 2 Aluminum1 Sulfur & 4 Oxygen 3x everything in ( )

3 Monatomic Ions Monatomic Ions are formed from a single atom (+) positive ions = cations = lose e - (-) negative ions = anions = gain e - Atoms gain or lose electrons in an attempt to achieve a full valence shell (octet)

4 Naming Monatomic Ions Cations(+) K + : potassium cation Mg 2+ : magnesium cation D-Block Cations Cu + : copper(I) Fe 3+ : iron(III) * Use roman # to indicate pos. charge Anions(-) F - : fluoride N 3- : nitride O 2- : oxide * Element name ending in ide

5 Binary Ionic Compounds Binary ionic compounds are made from the combination of cations and anions The charges of the positive and negative ions must balance out to zero(neutral). Ex: Mg 2+ + Br -  MgBr 2

6 Crossing Over Method 1)Write the cation followed by the anion 2) Cross over charges to form subscripts 3) Check to make sure charges are balanced

7 Naming Binary Ionic Compounds Combine the names of both ions Ex: Al 2 O 3 = aluminum oxide AgCl = silver chloride CaBr 2 = calcium bromide NaCl = sodium chloride

8 Stock System of Nomenclature Use Roman #’s to indicate the charge of the cation(+) Ex: CuCl 2 = copper(II) chloride FeO = iron(II) oxide Fe 2 O 3 = iron(III) oxide SnF 4 = tin(IV) fluoride ZnBr = zinc(I) bromide

9 Polyatomic Ions Polyatomic ions are molecules with an overall positive or negative charge Oxyanions- polyatomic ions containing oxygen Ex: NO 2 - = nitrite NO 3 - = nitrate ClO 2 - = chloriteClO - = hypochlorite ClO 3 - = chlorateClO 4 - = hyperchlorate

10 Naming Binary Molecular Compounds (covalent bonds) Use prefixes to indicate the # of each type of element in the molecule 1)Write the least electronegative element 1 st 2)Use a prefix is there is more than 1 of the 1 st element 3)Write the second element always using a prefix to indicate the # of atoms present

11 Prefixes 1 = mono 2 = di 3 = tri 4 = tetra 5 = penta 6 = hexa 7 = hepta 8 = octa 9 = nona 10 = deca

12 Molecular Compounds: Ex: P 4 O 10 = tetraphosphorus decaoxide N 2 O = dinitrogen monoxide SO 3 = sulfur trioxide

13 Acids & Salts Binary Acids: contain H & a halogen(17) Ex: HCL = hydrochloric acid Oxyacids: contain H, O, & a 3 rd element Ex: H 2 SO 4 = sulfuric acid HNO 3 = nitric acid Salts: formed from a cation and an anion of an acid Ex: HCl + NaOH  NaCl + H 2 O (Acid)(Base)(Salt)

14 Oxidation #’s (states) Rules: 1)The oxidation number of a free(single) element is always 0(zero) 2)The oxidation number of a monatomic ion equals the charge of the ion 3)The usual oxidation number of hydrogen(H) is +1 4)The oxidation number of oxygen(O) in compounds is usually -2 5)The oxidation # of other elements is based on the elements group # 6)The sum of the oxidation numbers of all of the atoms in a neutral compound is 0 7)The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion

15 Examples

16 Chemical Formulas Review Formula Mass- sum of the atomic masses of all elements in a compound, units (amu) Ex: H 2 O Molar Mass- mass of one mole of a substance, numerically equivalent to formula mass, different units (g/mol) Ex: H 2 SO 4

17 % Composition The percentage by mass of each element in a compound Mass of Element -------------------------- x 100 = % Composition Mass of Compound Ex: H 2 SO 4

18 Deriving Chemical Formulas Empirical Formula- the symbols of the elements in a compound w/ subscripts showing the smallest whole # ratio between the atoms –Ex: C 6 H 12 O 6  CH 2 O Chemical Formula Empirical Formula

19 Determining Empirical Formulas 1)Assume 100.0 g sample mass (if no sample mass is given) 2) Convert % Composition to Mass for each element 71 % A = 71 g of “A” 29 % B = 29 g of “B”

20 Determining Empirical Formulas 3) Determine the # of Moles of each element using the Molar Mass 71.0 g of A / 10.0 g/mol = 7.10 mol of A 29.0 g of B / 2.01 g/mol = 14.5 mol of B 4) Determine Mole Ratio by dividing both values from step 3 by the lower # A: 7.10 / 7.10 = 1.0 B: 14.5 / 7.10 = 2.0

21 *** If the mole ratio is NOT a whole # ratio after dividing then… Multiply both #’s by the same factor to get a whole # ratio –Ex: A = 1.5 x 2 = 3.0 B = 1.0 x 2 = 2.0 A = 2.25 x 4 = 9.0 B = 1.0 x 4 = 4.0

22 Practice Empirical Formulas 1)32.38% Sodium(Na) 22.65% Sulfur(S) 44.97% Oxygen(O) 2) 10.15 gram sample Contains 4.43 g of Phosphorus(P)

23 Determining Empirical Formulas 1)Convert % of each Element to Grams(g) 2)Convert Grams(g) to Moles(mol) 3)Divide Moles by Lowest # 4)Determine Whole # Ratio Ex: 47.3% Carbon 10.6% Hydrogen 42.1% Oxygen

24 Calculating Molecular Formulas 1)Divide: Actual Mass / Empirical Formula Mass 2) Multiply: Multiply the # from step 1 by the subscript of each element in the empirical formula Ex: BH 3 Actual Mass = 27.67 g/mol

25 END OF CHAPTER 7 NOTES


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