Compounds 1) Types of Compounds 2) Formula Writing 3) Formula Naming 4) Empirical Formulas 5) Molecular Formulas 6) Types of Chemical Reactions 7) Balancing Chemical Reactions 8) Attractive Forces

Formula Writing The charge of the (+) ion and the charge of the (-) ion must cancel out to make the formula. Use subscripts to indicate how many atoms of each element there are in the compound, no subscript if there is only one atom of that element. Na+1 and Cl-1 = NaCl Ca+2 and Br-1 = CaBr2 Al+3 and O-2 = Al2O3 Zn+2 and PO4-3 = Zn3(PO4)2 Try these problems!

Formulas to Write Ba+2 and N-3 NH4+1 and SO4-2 Li+1 and S-2 Cu+2 and NO3-1 Al+3 and CO3-2 Fe+3 and Cl-1 Pb+4 and O-2 Pb+2 and O-2

Formula Naming Compounds are named from the elements or polyatomic ions that form them. KCl = potassium chloride Na2SO4 = sodium sulfate (NH4)2S = ammonium sulfide AgNO3 = silver nitrate Notice all the metals listed here only have one charge listed? So what do you do if a metal has more than one charge listed? Take a peek!

The Stock System CrCl2 = chromium (II) chloride Try CrCl3 = chromium (III) chloride Co(NO3)2 and CrCl6 = chromium (VI) chloride Co(NO3)3 FeO = iron (II) oxide MnS = manganese (II) sulfide Fe2O3 = iron (III) oxide MnS2 = manganese (IV) sulfide The Roman numeral is the charge of the metal ion!

Formulas, Naming and Properties of Acids Arrhenius Definition of Acids: molecules that dissolve in water to produce H3O+ (hydronium) as the only positively charged ion in solution. HCl (g) + H2O (l)  H3O+ (aq) + Cl- Properties of Acids Naming of Acids Formula Writing of Acids

Properties of Acids Acids react with metals above H2 on Table J to form H2(g) and a salt. Acids have a pH of less than 7. Dilute solutions of acids taste sour. Acids turn phenolphthalein CLEAR, litmus RED and bromthymol blue YELLOW. Acids neutralize bases. Acids are formed when acid anhydrides (NO2, SO2, CO2) react with water for form acids. This is how acid rain forms from auto and industrial emissions.

Naming of Acids (polyatomic ion) -ate +ic acid Binary Acids (H+ and a nonmetal) hydro (nonmetal) -ide + ic acid HCl (aq) = hydrochloric acid Ternary Acids (H+ and a polyatomic ion) (polyatomic ion) -ate +ic acid HNO3 (aq) = nitric acid (polyatomic ion) -ide +ic acid HCN (aq) = cyanic acid (polyatomic ion) -ite +ous acid HNO2 (aq) = nitrous acid

Formula Writing of Acids Acids formulas get written like any other. Write the H+1 first, then figure out what the negative ion is based on the name. Cancel out the charges to write the formula. Don’t forget the (aq) after it…it’s only an acid if it’s in water! Hydrosulfuric acid: H+1 and S-2 = H2S (aq) Carbonic acid: H+1 and CO3-2 = H2CO3 (aq) Chlorous acid: H+1 and ClO2-1 = HClO2 (aq) Hydrobromic acid: H+1 and Br-1 = HBr (aq) Hydronitric acid: Hypochlorous acid: Perchloric acid:

Formulas, Naming and Properties of Bases Arrhenius Definition of Bases: ionic compounds that dissolve in water to produce OH- (hydroxide) as the only negatively charged ion in solution. NaOH (s)  Na+1 (aq) + OH-1 (aq) Properties of Bases Naming of Bases Formula Writing of Bases

Properties of Bases Bases react with fats to form soap and glycerol. This process is called saponification. Bases have a pH of more than 7. Dilute solutions of bases taste bitter. Bases turn phenolphthalein PINK, litmus BLUE and bromthymol blue BLUE. Bases neutralize acids. Bases are formed when alkali metals or alkaline earth metals react with water. The words “alkali” and “alkaline” mean “basic”, as opposed to “acidic”.

Naming of Bases Bases are named like any ionic compound, the name of the metal ion first (with a Roman numeral if necessary) followed by “hydroxide”. Fe(OH)2 (aq) = iron (II) hydroxide Fe(OH)3 (aq) = iron (III) hydroxide Al(OH)3 (aq) = aluminum hydroxide NH3 (aq) is the same thing as NH4OH: NH3 + H2O  NH4OH Also called ammonium hydroxide.

Formula Writing of Bases Formula writing of bases is the same as for any ionic formula writing. The charges of the ions have to cancel out. Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq) Potassium hydroxide = K+1 and OH-1 = KOH (aq) Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq) Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq) Lithium hydroxide = Copper (II) hydroxide = Magnesium hydroxide =

Math of Chemistry 1) Formula Mass 2) Percent Composition 3) Mole Problems 4) Gas Laws 5) Neutralization 6) Concentration 7) Significant Figures and Rounding 8) Metric Conversions 9) Calorimetry

Formula Mass Gram Formula Mass = sum of atomic masses of all elements in the compound Round given atomic masses to the nearest tenth H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = 142.1 g/mole Now you try: BaBr2 CaSO4 Al2(CO3)3

Percent Composition The mass of part is the number of atoms of that element in the compound. The mass of whole is the formula mass of the compound. Don’t forget to take atomic mass to the nearest tenth! This is a problem for you to try.

Practice Percent Composition Problem What is the percent by mass of each element in Li2SO4?

Mole Problems Grams <=> Moles Molecular Formula Stoichiometry

Grams <=> Moles How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh? 3.00 moles X 40.0 g/mol = 120. g How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams? (10.0 g) / (40.0 g/mol) = 0.250 mol

Molecular Formula Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 70.0 grams/mole? 1) Find the Empirical Formula Mass: CH2 = 14.0 2) Divide the MM/EM: 70.0/14.0 = 5 3) Multiply the molecular formula by the result: 5 (CH2) = C5H10

Stoichiometry Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given) Given the balanced equation N2 + 3 H2  2 NH3, How many moles of H2 need to be completely reacted with N2 to yield 20.0 moles of NH3? 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2

Limiting Reactant controls the amount of product formed. CO(g) + 2H2 (g)  Ch3OH If 500 mol of CO react with 750 mol of H2, which is the limiting reactant? Use either given amount to calculate required amount of other. Compare calculated amount to amount given b. How many moles of excess reactant remain unchanged? H2 125 mol CO

Percent yield= (actual yield/ theoretical yield)*100 Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant Actual yield is the measured amount of a product obtained from a reaction Theoretical yield= 117.5 g SnF2 Actual yield = 113. 4g SnF2 Percent yield = 113.4 g SnF2 117.5 g SnF2 *100

Determining empirical formula from combustion data When a compound containing C,H and O undergoes combustion, it forms CO2 and H2O. Then from the mass of CO2 and H2O, we can calculate the mass of C and Hand then find the mass of O by subtracting the sum of masses of C and H from total g present of that substance. From the mass of C,H and O, we can calculate the moles of C,H and O.Then the smallest whole number ratios of these moles will give the empirical formula. Ex. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g of H2O. Determine the empirical formula of the compound. Ans. C7H6O2

Empirical Formulas Ionic formulas: represent the simplest whole number mole ratio of elements in a compound. Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound. Many molecular formulas can be simplified to empirical formulas Ethane (C2H6) can be simplified to CH3. This is the empirical formula…the ratio of C to H in the molecule. All ionic compounds have empirical formulas.

Molecular Formulas The count of the actual number of atoms of each element in a molecule. H2O: a molecule made of two H atoms and one O atom covalently bonded together. C2H6O: A molecule made of two C atoms, six H atoms and one O atom covalently bonded together. Molecular formulas are whole-number multiples of empirical formulas: H2O = 1 X (H2O) C8H16 = 8 X (CH2) Calculating Molecular Formulas

Calculating Empirical Formulas One can calculate the empirical formula from the percent composition. © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.105 mol 0.7288 mol 5.09 mol 1.458 mol © 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2 © 2009, Prentice-Hall, Inc.

Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this. C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. © 2009, Prentice-Hall, Inc.

Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products. © 2009, Prentice-Hall, Inc.

Stoichiometric Calculations Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant). © 2009, Prentice-Hall, Inc.

Stoichiometric Calculations C6H12O6 + 6 O2  6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams. © 2009, Prentice-Hall, Inc.

Molecular Formula Actual ratio of atoms in a compound. Ex. H2O, C6H12O6 To determine the molecular formula, divide the molar mass by empirical formula mass. This will give the number of empirical formula units (n) in actual molecule. n= Molar Mass/ Empirical Formula Mass Ex. Determine the empirical and molecular formula of each of the following: Ethylene glycol, the substance used as antifreeze has 38.70 % C, 9.70 % H and 51.60 % O , mm= 62.10 g Caffeine, a stimulant in coffee has the following percent composition: 49.50 % C, 5.15% H, 28.90 % N and 16.50 % O , molar mass= 195.00g

Types of Chemical Reactions Redox Reactions: driven by the loss (oxidation) and gain (reduction) of electrons. Any species that does not change charge is called the spectator ion. Synthesis Decomposition Single Replacement Ion Exchange Reaction: driven by the formation of an insoluble precipitate. The ions that remain dissolved throughout are the spectator ions. Double Replacement

Synthesis Two elements combine to form a compound 2 Na + O2  Na2O Same reaction, with charges added in: 2 Na0 + O20  Na2+1O-2 Na0 is oxidized (loses electrons), is the reducing agent O20 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Na0 to the O20. No spectator ions, there are only two elements here.

Decomposition A compound breaks down into its original elements. Na2O  2 Na + O2 Same reaction, with charges added in: Na2+1O-2  2 Na0 + O20 O-2 is oxidized (loses electrons), is the reducing agent Na+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the O-2 to the Na+1. No spectator ions, there are only two elements here.

Single Replacement An element replaces the same type of element in a compound. Ca + 2 KCl  CaCl2 + 2 K Same reaction, with charges added in: Ca0 + 2 K+1Cl-1  Ca+2Cl2-1 + 2 K0 Ca0 is oxidized (loses electrons), is the reducing agent K+1 is reduced (gains electrons), is the oxidizing agent Electrons are transferred from the Ca0 to the K+1. Cl-1 is the spectator ion, since it’s charge doesn’t change.

Double Replacement The (+) ion of one compound bonds to the (-) ion of another compound to make an insoluble precipitate. The compounds must both be dissolved in water to break the ionic bonds first. NaCl (aq) + AgNO3 (aq)  NaNO3 (aq) + AgCl (s) The Cl-1 and Ag+1 come together to make the insoluble precipitate, which looks like snow in the test tube. No species change charge, so this is not a redox reaction. Since the Na+1 and NO3-1 ions remain dissolved throughout the reaction, they are the spectator ions. How do identify the precipitate?

Identifying the Precipitate The precipitate is the compound that is insoluble. AgCl is a precipitate because Cl- is a halide. Halides are soluble, except when combined with Ag+ and others.

Balancing Chemical Reactions Balance one element or ion at a time Use a pencil Use coefficients only, never change formulas Revise if necessary The coefficient multiplies everything in the formula by that amount 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O. Examples for you to try!

Reactions to Balance ___NaCl  ___Na + ___Cl2 ___Al + ___O2  ___Al2O3 ___SO3  ___SO2 + ___O2 ___Ca + ___HNO3  ___Ca(NO3)2 + ___H2 __FeCl3 + __Pb(NO3)2  __Fe(NO3)3 + __PbCl2

Writing Net Ionic Equations Cancel all the spectator ions. Dissociate all dissociable ionic compounds (refer to solubility rules) All gases and liquids NEVER dissociate. Write the net ionic equation.