1. Chemists describe a molecule (or the smallest unit of an infinite material) using formulae that list what elements the molecule contains and how many atoms of each element. Chapter 3 Molecules, Ions and Their Compounds Formulae

2. Three main types of formulae are used, depending on how much structural information is available or necessary: A molecular formula gives no further information A structural formula shows all of the bonding in a molecule, and is essentially a Lewis dot structure without the lone pairs. A condensed formula indicates which atoms are grouped together but does not show any bonds. C2H6OC2H6OEx) Ex) CH 3 CH 2 OH Ethanol Ex) CH 3 OCH 3 Dimethyl ether C C O H H HH H H C C O H H H H H H Types of Formulae

3. Molecular formulae tend to be used to describe simple molecules and ions that cannot be broken into stable multi-atom pieces. Condensed formulae are used to describe more complex molecules that can be broken into stable multi-atom pieces Structural formula are used when we want to be very clear about which atoms are bonded to which or the overall shape of the molecule. Ex) H 2 O, CO 2, NaCl Ex) Cu(NO 3 ) 2 can be broken into a Cu 2+ and two NO 3- units Organic chemists also use condensed formulae to describe what is bonded to each carbon atom in a chain Ex) CH 3 CHBrCH 3 is different from CH 3 CH 2 CH 2 Br H C C C H H H H Br H H H C C C H H H H H H Isomers Molecules which have the same molecular formula but different structural formulae

4. A structural formula in some cases is the only way to distinguishe between two isomers Isomers Molecules which have the same molecular formula but different structural formulae H C C C H H H Br H H C C C H H H H Consider: CH 3 CHCHBr Consider: CH 3 CHClBr H C C Cl H H Br H H C C Cl H H Br H Structural Formulas and Isomers

5. Naming Compounds In order to communicate conveniently compounds need to be named unambiguously. Naming Ionic Compounds To name an ionic compound, name the cation then the anion. Monoatomic Cations Most cations are single elements and are given the same name as the element. Ex). Na + = sodiumMg 2+ = magnesium I f an element can give rise to different cations use a Roman numeral to indicate the charge. Ex)Fe 2+ = iron (II)Fe 3+ = iron (III) Au + = gold (I)Au 3+ = gold (III) Pb 2+ = lead (II)Pb 4+ = lead (IV) This is necessary for most transition metals as well as tin and lead

6. Monatomic cations and the Periodic Table

7. Monoatomic Anions To indicate that an element has formed an anion, we change its suffix to -ide. Ex) F = fluorine F - = Fluoride N = nitrogenN 3- = Nitride O = oxygenO 2- = Oxide Se = selenium Se 2- = Selenide Tend to form exclusively from the non-metals and one metalliod Charge = group # -18 Free ions do not exist with charges higher than 3 units. carbides have little true ionic character. But the valence of carbon is still often 4- in many of its compounds

8. Polyatomic Ions Not all ions consist of just one atom. Some groups of covalently bonded atoms have overall charges. Because they are charged, they are ions (not molecules). Polyatomic cations: NH 4 + Ammonium Best known H 3 O + Hydronium Polyatomic Anions: ClO - Hypochlorite (Bleach) HCO 3 - Bicarbonate(baking Soda) Examples CN - Cyanide CrO 4 2- Chromate OH - HydroxideCr 2 O 7 2- Dichromate CO 3 2- Carbonate MnO 4 1- Permanganate When H + is added to a polyatomic atom the prefix “bi” is placed before the ‘old’ name. HCO 3 - BicarbonateHS - Bisulfide HSO 4 - Bisulfate

9. ClNSP Ohypochlorite (ClO - ) O2O2 chlorite (ClO 2 - ) nitrite (NO 2 - ) O3O3 chlorate (ClO 3 - ) nitrate (NO 3 - ) sulfite (SO 3 2- ) phosphite (PO 3 3- ) O4O4 perchlorate (ClO 4 - ) sulfate (SO 4 2- ) phosphate (PO 4 3- ) The polyatomic ions ending with ‘ate’ all consist of an atom surrounded by a number of oxygen atoms. Those ending in “ite” have one fewer O

10. Ionic Compounds To name an ionic compound, name the cation then the anion. The charge of each ion indicates how many are needed to make a neutral compound so no prefixes are necessary. MgCl 2 CuBr 2 NaNO 3 (NH 4 ) 2 SO 4 Sn(CO 3 ) 2 Sn(HCO 3 ) 2 NaNO 2 Magnesium Chloride Copper(II) Bromide Sodium Nitrate Ammonium Sulfate Tin (IV) Carbonate Exercise Tin (II) Bicarbonate Sodium Nitrite

11. Naming Covalent Compounds List the elements in order from least to most electronegative, using a prefix to indicate how many atoms there are of each element. Change the suffix of the final (most electronegative) element to “ide” as if it was an anion. SO 2 NO 2 CO Sulfur dioxide Nitrogen dioxide Carbon monoxide Carbon dioxide Carbon tetrachloride Compounds with hydrogen are often referred to by their common name or as ionic compounds CH 4 MethaneNH 3 Ammonia BH 3 Borane H2OH2O Water If there is only one atom of the first (least electronegative) element, no prefix is used before it. Dinitrogen tetroxide HF HCl H2SH2S Hydrogen FluorideHydrogen Chloride CO 2 CCl 4 N 2 O 4 Hydrogen Sulfide

12. Hydrated Compounds Many ionic compounds incorporate water molecules in the ionic lattice. The electronegative oxygen atoms of the water molecules are attracted to the positively charged metal cations. The water molecules can be removed by heating to “dehydrate” it. Thus, the water is listed at the end of the chemical formula. Ex)Hydrated CuSO 4 is CuSO 4 ·5H 2 O The name of a hydrated compound, is composed of the name of the unhydrated molecule followed by a prefix before “hydrate” indicating the number of water molecules Ex)CuSO 4 ·5H 2 O copper(II) sulfate pentahydrate If an ionic lattice contains no water, the compound is said to be anhydrous.

13. Exercise 2.46 g of a magnesium sulfate hydrate is dehydrated to give 1.20 g magnesium sulfate. What is the chemical formula for the hydrate? MgSO 4 M.W. = 24.3050 + 32.066 + 4*15.9994 g/mol H2OH2O M.W. = 2*1.0079 + 15.9994 g/mol =120.3686 g/mol =18.0152 g/mol Mol. MgSO 4 = (1.20 g)/(120.3686 g/mol)= 0.0100 mol Mol. H 2 O = (2.46g - 1.20 g)/(18.0152 g/mol)= 0.0700 mol 0.0100 mol MgSO 4 : 0.07 mol H 2 O 1 MgSO 4 : 7 H 2 O MgSO 4 ·7H 2 O mangesium sulfate heptahydrate

14. Empirical Formula and Molecular Formula Recall that molecular formula indicates the total number of atoms of each element in a molecule. Empirical formula indicates the ratio of atoms of each element in a molecule. It can be obtained by dividing the subscripts in the molecular formula by the largest common factor. Write empirical formulae for the following molecules: C6H6C6H6 C 6 H 12 O 6 C2H4OC2H4O CH CH 2 O C2H4OC2H4O C 10 H 12 O 4 C5H6O2C5H6O2

15. We can use the empirical formula to calculate the percent mass of each element in a molecule. Ex). What is the percent composition of C, H and O in glucose, C 6 H 12 O 6 ? Percent Composition Empirical formula?CH 2 O mass of emp. form.= 12.011 + 2*1.0079+15.9994 g/mol= 30.0262 g/mol Molecular Mass = (30.0262 g/mol)*6 = 180.1572 g/mol % mass C = 100*mass C/total mass=100*(12.011 g/mol)/(30.1572 g/mol) = 39.8280 % % mass H = 100*mass H/total mass =100*(2*1.0079 g/mol)/(30.1572 g/mol) = 6.6843 % % mass O = 100*mass O/total mass=100*(15.9994 g/mol)/(30.1572 g/mol) = 53.0533 %

16. Percent composition can be used to determine the empirical formula. Ex). A compound was subjected to elemental analysis and found to contain 58.01% C, 16.23% H and 25.76% O. What is its empirical formula? Molecular formula cannot be calculated, since only the relative proportions are known, not their common factor. Percent Composition & Empirical Formula Assume 100 g of material, or any mass of choice Compute mass of each element using % composition Mass C = (% C)*(total mass)/100 = (58.01 %)*(100 g)/(100 %) = 58.01 g Mass H = (% H)*(total mass)/100 = (16.23 %)*(100 g)/(100 %) = 16.23 g Mass O = (% O)*(total mass)/100 = (25.76 %)*(100 g)/(100 %) = 25.76 g

17. Compute the number of moles of each element. Mol. C = (mass C)/ (molar mass C) = (58.01 g)/ (12.011 g/mol) = 4.830 mol Mol. C = (mass H)/ (molar mass H) = (16.23 g)/ (1.0079 g/mol) = 16.10 mol Mol. O = (mass O)/ (molar mass O) = (25.76 g)/ (15.9994 g/mol) =1.610 mol Find common ratio by dividing through by the smallest number 4.830 C : 16.10 H: 1.610 ODivide by 1.610 3.000 C : 10.00 H: 1.000 O Empirical Formula is C 3 H 10 O

18. You are told that the molecular mass is 186.33 g, determine the molecular formula. Recall that the empirical Formula is C 3 H 10 O Mass of emp. form. = 3*12.011 + 10*1.0079 + 15.9994 g/mol = 62.1114 g/mol Ratio of molecular mass to mass of empirical formula gives the common factor we need. Molecular mass/Mass of emp. form = (186.33 g/mol)/ 62.1114 g/mol) = 3.000 Therefore the molecular formula is C 9 H 30 O 3 Molecular Formula

19. Ex)2.472 g of Manganese metal was completely reacted with 2.564 g of fluorine gas to produce the metal fluoride, MnxFy. Information from a chemical reaction can be used to determine the empirical formula Chemical Reaction a) Calculate % composition of Mn and F in the product. Complete reaction means that all the F and Mn have combined. Mass of products = mass of reactants = 2.472 g Mn + 2.564 g F = 5.036 g of product % Mn =100*(mass Mn)/(mass product) = 100*(2.472 g)/(5.036 g) = 49.09 % % F =100*(mass F)/(mass product) = 100*(2.564 g)/(5.036 g) = 50.91 %

20. b) Determine the empirical formula We need to find the ratio of Mn atoms to F Compute number of moles of Mn and F. Mol Mn = mass Mn/ molar mass Mn =(2.472 g)/(54.938 g/mol) = 0.04500 mol Mn Mol F = mass F/ molar mass F =(2.564 g)/(18.9984 g/mol) = 0.1350 mol Mn Ratio of Mn to F: 0.04500 mol Mn : 0.135 mol F Divide by 0.04500 3.000 mol Mn : 1.000 mol F Empirical formula is MnF 3

21. Concepts from Chapter 3 molecular vs. condensed vs. structural formula naming ionic compounds (including polyatomic ions) naming covalent compounds hydrated vs. anhydrous compounds empirical vs. molecular formula calculating empirical formula from percent composition calculating percent composition from empirical formula ionic and covalent bonding (see chapter 9 notes) the mole (see chapter 2 notes)

22. Problem #114 Ch 3 A sample of CaCl 2 2H 2 O was weighed at 0.832 g. After heating it was weighed again giving a mass of 0.739 g. Have they dehydrated the sample? Assuming it was pure determine how many moles of CaCl 2 2H 2 O there is in the sample before heating Moles CaCl 2 2H 2 O = mass/molar mass = (0.832 g)/(40.0780 + 2*35.453 +2*(2*1.0079+15.9994)) = (0.832 g)/(147.0144 g/mol) = 0.00566 mol Moles of water lost would have to be twice as many as the hydrate What would happen if the sample were heated again? Moles water = 0.01132 mol Mass water = (0.01132 mol)(2*1.0079+15.9994) = 0.204 g Therefore final weight should have been 0.832 -0.204 = 0.628 g Was the sample pure?

23. Moles CaCl 2 = mass/molar mass = (0.739 g)/(40.078 + 2*35.453 g/mol) Assuming the heating removed all the water compute moles of CaCl 2 = (0.739 g)/(110.984 g/mol) If the sample was pure then twice as many moles of water would be lost mass water lost = (0.0133 mol)(18.0152 g/mol) = 0.240 g Moles water lost = 0.0133 mol = 0.00666 mol Therefore the initial weight should have been 0.739 g + 0.240 g = 0.979 g Both calculations tell us the sample is not dehydrated. Compute how many moles of water was removed Moles H 2 O = weight change/ molar mass = (0.832 -0.739 g)/(2*1.0079+15.9994) = 0.093 g/18.0152 g/mol = 0.0052 mol If we had pure hydrate in the beginning half as many moles as water lost would have been in the sample

24. Moles hydrate = (0.0052 mol)/2 = 0.0026 mol Mass of hydrate = (0.0026 mol)((147.0144 g/mol) = 0.38 g The number of moles of hydrate before heating would be the same as the number of moles of CaCl 2 after heating Moles CaCl 2 = 0.0026 mol Mass CaCl 2 = (0.0026 mol) (110.984 g/mol) = 0.28 g Both disagree with observed weigh therefore 1) sample is not dehydrated or 2) was impure to begin with Only by reheating the sample repeatedly until it stops losing weight can one determine the total water lost. If half the moles of water lost corresponds to total moles of CaCl 2 2H 2 O before heating or number of moles of CaCl 2 after heating we know the sample was pure. If this is not the case the composition of the sample can be determined from the amount of water lost.

25. Concepts from Chapter 9 DRAWING LEWIS ELECTRON DOT DIAGRAMS Octet rule Resonance structures Bond polarity (ionic, polar covalent and covalent bonds) Ionic vs. covalent compounds Electronegativity Dipole vectors Calculating formal charges and partial charges Bond order Bond lengths VSEPR and predicting shapes of molecules Polarity of molecules

26. Concepts from Chapter 10 MOLECULAR ORBITALS LCAO theory Correlation diagrams Bonding, nonbonding and antibonding interactions Calculating bond order using MO theory hybridization (sp, sp2 and sp3 orbitals) Sigma (σ) vs. pi (π) bonds Composition of single, double and triple bonds Resonance according to MO theory