Chemical Formulas and Chemical Compounds Chapter 7 Chemical Formulas and Chemical Compounds

7.1 Chemical Names and Formulas Chemical formulas show the relative #’s of atoms in a chemical compound Examples C12H22O11 C = 12 H = 22 O = 11 Pb(NO3)4 Pb = 1 N = 4 O = 12 (NH4)2CrO4 N = 2 H = 8 Cr = 1 O = 4

Chemical Names and Formulas Naming monatomic ions End in -ide Look up the charges on the periodic table F vs F-1 F F-1 S vs S-2 S S-2 Fluorine Fluoride Sulfur Sulfide

Writing Formulas for Ionic Compounds Cations – positive ions (metal) metal Anions – negative ions nonmetal Charge on compound = 0

Writing Formulas for Ionic Compounds Binary ionic compounds Rules if the charges on the ions are the same, drop ‘em if the charges are different, criss-cross Same charges – Na+1 Cl-1 - Mg+2 O-2+- Different charges- Na+1 S-2- Mg+2 Cl-1- NaCl Sodium Chloride MgO Magnesium Oxide Na2S Sodium Sulfide MgCl2 Magnesium Chloride

Writing Formulas for Ionic Compounds Ternary Ionic compounds Metal + (Polyatomic ion) When naming, do not use the ending –ide Sodium Nitrate Sodium Carbonate Aluminum Nitrate Na+1 NO3-1 Na(NO3) Na+1 CO3-2 Na2(CO3) Al+3 NO3-1 Al(NO3)3

Writing Formulas for Ionic Compounds Aluminum Phosphate Aluminum Bihypophosphite Aluminum Carbonate Al+3 PO4-3 Al(PO4) Al+3 HPO2-2 Al2 (HPO2)3 Al+3 CO3-2 Al2 (CO3)3

Writing Names for Ionic Compounds Front name – positive (cation – metal) Back name – negative (anion – nonmetal) Binary ionic compounds – composed of only 2 types of elements ( M + NM) – end in -ide NaCl MgCl2 Al2O3 NaH Sodium Chloride Magnesium Chloride Aluminum Oxide Sodium Hydride

The BIG Lie Stock System – use Roman Numerals for naming compounds with metals that have multiple charges (the transitions!) Normal Name Symbol Charge Stock Name Copper Cu +1 Cu (I) +2 Cu (II) Mercury Hg Hg (I) Hg (II) Lead Pb Pb (II) +4 Pb (IV) Tin Sn Sn (II) Sn (IV) Iron Fe Fe (II) +3 Fe (III) Cory Matthews Loves Topanga Intensely

The BIG Lie More exceptions to the Lie! Ag is always = +1 charge DON’T write Ag I Zn always = +2 charge DON’T write Zn II

Practice! Sn3N2 AgOH PbCO3 Zn(OH)2 Fe2(SO4)3 Tin (II) Nitride Silver Hydroxide Lead (II) Carbonate Zinc Hydroxide Iron (III) Sulfate

More Practice!! CuSO2 CuHSO2 Cu(HSO2)2 Copper (II) Bihyposulfite Copper (II) Hyposulfite Copper (I) Bihyposulfite

Practice! LiClO3 LiClO2 CaCO3 Ca(HCO2)2 Fe(NO3)3 Lithium Chlorate Lithium Chlorite Calcium Carbonate Calcium Bicarbonite Iron (III) Nitrate

Writing Names for Molecular Compounds Molecular Compounds – covalent compounds 2 nonmetals To name, we use prefixes # Prefix 1 mono 6 hexa 2 di 7 hepta 3 tri 8 octa 4 tetra 9 nona 5 penta 10 deca don’t use the prefix Mono on the first atom

Writing Names for Molecular Compounds Prefix-name prefix-name-ide CO CO2 PCl3 CBr4 N2O5 SF6 Carbon Monoxide Carbon Dioxide Phosphorous trichloride Calcium tetrabromide Dinitrogen pentoxide Sulfur hexafluoride

Writing Formulas for Molecular Compounds The prefixes = the subscripts. Do NOT look at the charges. Sulfur Dioxide Disulfur Trioxide Dinitrogen pentoxide SO2 S2O3 N2O5

Naming Acids Acid - when a solution yields H+ ions in solution 2 types Binary H and one other type of atom ternary (sometimes called oxy) acids that have H with a polyatomic ion

Naming Binary Acids Rules Hydro__(begninng of name)__ic acid Ex. HCl Hydrochloric acid HBr HF H2S H3P Hydrobromic acid Hydrofluoric acid Hydrosulfuric acid Hydrophosphoric acid

Writing Formulas from Names for Acids Do the criss-cross Ex. Hydronitric acid H+1 N-3 H3N Hydroiodic acid H+1 I-1 HI Hydrosulfuric acid H+1 S-2 H2S

Naming Ternary Acids H + polyatomic Rules Do NOT start with hydro- If the ending of polyatomic is –ate -ic + acid If the ending of polyatomic is –ite -ous + acid Ate/ite ic/ous Example: H2SO4 H+ and SO4-2 – sulfate sulfuric acid

Naming Ternary Acids H2SO3 HClO4 HClO3 HClO2 HClO

Formulas for Ternary Acids Use the criss-cross method Nitric acid Phosphorous acid

7.2 Oxidation Numbers Since electrons are shared, there is no definite charge - we assign the more electronegative element the “apparent” negative charge - this is known as the oxidation # oxidation numbers can also be positive. oxidation # - a number assigned to an atom to show the distribution of elements

Oxidation Numbers Rules Free elements = 0 Ex. Mg = O Ions = charges Ex. F = -1 S = -2 Oxygen (0) = -2 except in peroxides (H2O2) O = -1 H = +1 except in metal hydrides (MgH2, NaH) H = -1

Oxidation Numbers ….Rules More electronegative atom gets a (-) charge Ox #’s add up to 0 in compounds Ox #’s = the charge in polyatomic ions

Oxidation # Practice FeO (Iron II Oxide) O = -2 Fe = ? Fe2O3 (Iron III oxide) O = -2 Fe = ? -2 + x = 0 x = 2 3(-2) + 2x = 0 x = 3

Oxidation # Practice H2SO4 (Hydrogen Sulfate or Sulfuric Acid) O = -2

Oxidation # Practice H2SO3 (Hydrogen sulfite or sulfurous acid) H = +1 +4 -2 SO3-2 X + 3(-2) = -2 X = 4

Oxidation # Practice H2Cr2O7 (Hydrogen Dichromate or Dichromic Acid NO3-1 (Nitrate) N = O = -2 MgH2 (Magnesium hydride) Mg = H = -1 +6 +4 +2

7.3 Using Chemical Formulas Step 1 – be able to calculate molar mass (aka – formula mass, molecular weight, atomic weight, atomic mass, gram formula weight, etc.) Add atomic weights from the periodic table round to the nearest 10th place Examples CH4 MgSO4· 7H2O 12.0 + 4(1.0) = 16.0 g/mol 24.3 + 32.0 + 4(16.0) + 14(1.0) + 7(16.0) = 246.3 g/mol

7.3 Using Chemical Formulas Step 2 – be able to convert between grams, moles, particles, and liters Liters Atoms, molecules, particles Grams Mole 22.4 L 6.022 x1023 Molar Mass

Using Chemical Formulas Convert 32.0 g of CH4 to moles, liters, molecules, total atoms, atoms of H Moles Liters Molecules Atoms Atoms H

Conversions Moles CH4 Liters 32.0 g 1 mole 1 16.0g 32.0g 1 mole 22.4L

Conversions Molecules Atoms Atoms H 32.0g 1 mole 6.022 x 1023molecules 32.0g 1 mole 6.022 x 1023atoms 5 Atoms H 32.0g 1 mole 6.022 x 1023atoms 4 1.20 x 1024molecules 6.00 x 1024molecules 4.80 x 1024molecules

Percent Composition Percentage Composition - every compound has a certain percentage of each type of atom (we measure it by mass) Formula % composition = mass element mass compound X 100 =

Practice - % Composition Calculate % composition if a compound contains 24 g of Carbon and 64 g of Oxygen % composition = mass element mass compound Total mass compound = 24 + 64 = 88 g % Composition C = 24 x 100 = 27 % C 88 % Composition O = 64 x 100 = 73% O X 100 =

Practice - % Composition What is the % composition of Ba(OH)2? Ba = 137.3 g O = 2 (16.0) = 32.0g H = 2 (1.0) = 2.0g Ba(OH)2 = 171.3 g %Ba = 137.3 171.3 = 80.2% %O = 32.0 171.3 =18.7% %H = 2.0 171.3 =1.1%

Practice - % Composition What is the % composition of C6H12O6 C = 6 (12.0) = 72.0g H = 12 (1.0) = 12.0g O = 6 (16.0) = 96.0g C6H12O6 = 180.0g 40.0% 6.7% 53.3%

7.4 Determining a compound’s empirical and molecular formula Empirical formula - the lowest whole number ratio of atoms in a compound (simplest formula) 4 rules to find empirical formula Cross out the % → change to grams Divide each by own molar mass Divide by the smallest number If needed, multiply by 2 or 3 ONLY if a whole number ratio isn’t the result of step 3

Determining a compound’s empirical formula Ex1 Calculate the empirical formula if there is 52.17 % C, 13.04% H, and 34.78 % O 52.17 g C 1 mole = 4.3 = 1 12.0 g 13.04 g H 1 mole = 13.04 = 1 1.0 g 34.78 g O 1 mole = 2.17 = 1 16.0 g 2 2.17 6 2.17 1 2.17 C2H6O

Determining a compound’s empirical formula Ex2 Calculate the empirical formula is there is 26.56 % K, 35.41 % Cr, and 38.03 % O 26.56 g K 1 mole = 39.1 g 35.41 g Cr 1 mole = 52.o g 38.03 g O 1 mole = 16.0 g .68 = 1 x2 .68 .68 = 1 x2 .68 2.38 = 3.5 x2 .68 K2Cr2O7

Determining a compound’s empirical formula Ex3 Find the empirical formula if a sample contains 5.6 g N and 12.8 g O

Determining a compound’s molecular formula Rules Same steps as empirical formula + 3 more Find the mass of the empirical compound Divide this mass by the given molecular weight Multiply the empirical formula by this number

Determining a compound’s molecular formula Find the molecular formula of a compound (MW = 144.0 g) with 66.67 % C, 11.11 % H, and 22.22 % O Empirical = C4H8O Mass empirical = 72.0 g 144.0 g/72.0 g = 2 Molecular Formula = C8H16O2

Determining a compound’s molecular formula Find the molecular formula of that compound that contains 19.80% C, 2.50% H, 66.10% O, 11.60% N and MW = 242.0 Empirical Formula = C2H3O5N Mass Empirical = 121 g 242 g / 121 g = 2 C4H3O10N2