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Chapter 5 Lecture Notes

by: Caroline Smith

Chapter 5 Lecture Notes CHEM 1030 - 002

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Caroline Smith

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These notes cover chapter 5 lecture. If it would be helpful, I can make online flashcards for the items needed to be memorized, but they are listed in the textbook.
Fundamentals Chemistry I
Dr. Rik Blumenthal
Class Notes
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This 6 page Class Notes was uploaded by Caroline Smith on Tuesday October 18, 2016. The Class Notes belongs to CHEM 1030 - 002 at Auburn University taught by Dr. Rik Blumenthal in Fall 2016. Since its upload, it has received 103 views. For similar materials see Fundamentals Chemistry I in Chemistry at Auburn University.


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Date Created: 10/18/16
Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal  Ionic Compounds and Ionic Bonding o Ionic compounds form when cations (atoms with positive charge) and anions (atoms with negative charge) combine.  Ex. NaCl Na (sodium) is a cation Cl (chlorine) is an anion When combined, the single electron from sodium completely transfers to chlorine, a process called ionization o Ionic compounds denoted in chemical formulas (show elements in ration in which they combine)  Ex. Li + O  L2 O Notice that Li has one valence electron and O has two valence electrons. SO To be electrically neutral, Li must appear in two. o The structural arrangement of ionic compounds are appear in a lattice structure  Ionization, the addition of electrons to an atom, requires energy. That energy is called lattice energy (amount of energy required to convert a mole of an ionic solid to the gas phase). o Lattice energy is a measure of a compound’s stability  Lattice energy increases with electronegativity Stability increases with electronegativity Higher Lattice Energy = More Stable Compound  Naming Ions and Ionic Compounds o Nomenclature is the process of naming compounds o For naming m+natomic ions, add the word “ion” to the element name  Ex. K is potassium ion  Ex. Na sodium ion o Transition metals sometimes have multiple charges. For those, we use the stock system which uses roman numerals  Ex. Iron Fe 2+= iron (II) ion Fe3+= iron (III) ion o List of ions with multiple charges (because these have multiple charges, use roman numerals to notate number of atoms of each element) 1. Scandium Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal 2. Titanium 3. Vanadium 4. Chromium Manganese 5. Iron 6. Cobalt 7. Nickel 8. Copper 9. Zinc o For naming monatomic cations, change the end of the element name to “-ide” and add the word “ion.”  Ex. Cl = chloride ion -  Ex. O = oxide ion  Formulas of Ionic Compounds o To be electrically neutral, the sum of the charges must equal zero. If the charges are different, write the subscript for the cation numerically equal to the anion (and vice versa)  Ex. Potassium bromide + - K + Br KBr  Ex. Zinc iodide Zn 2+ + I  ZnI2  Ex. Aluminum oxide 3+ 2+ Al + O  Al2O3  Naming Ionic Compounds o Name the cation followed by the anion with the ending “-ide” but without the word “ion.”  SEE ABOVE EXAMPLES  Covalent Bonding and Molecules o Electrons are shared instead of transferred o Occurs between atoms of similar electronegativity o Molecular Formulas: show exact number of atoms of each element in the molecule o Empirical Formulas: tells what whole number ratio elements combine in  Ex. Hydrogen peroxide: H O 2 2 Empirical formula: HO  Naming Molecular Compounds o Name the first element o Name the second element and change the ending to “-ide”  Ex. HCl = hydrogen chloride o If need be, specify the numbers of the atoms using Greek suffixes  Mono – 1  Di – 2  Tri – 3  Tetra – 4  Penta – 5  Hexa – 6  Hepta – 7  Octa – 8 Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal  Nona – 9  Deca – 10  Ex. SO = sulfur dioxide 2 (Notice that “mono” is ignored if it is for the first element, but it is used if for the second element)  Naming Compounds Containing Hydrogen o Put “hydrogen” first and change it to “hydro” o Change “-ide” ending in the second element to “-ic”  Ex. HCl = hydrochloric acid  Covalent Bonding in Ionic Species o Many ionic substances contain polyatomic ions (ions containing multiple ions) held together by covalent bonding o The chart on the next page is a list of all of the polyatomic ions we will need to MEMORIZE for the test. The same table is on page 165 (Table 5.10) in our book. (I have no idea if he will actually ask for the names of all of them, but he did say we would need to memorize them.) Common Polyatomic Ions Name Formula/Charge Cations + ammonium NH 4 hydronium H3O + mercury (I) Hg 22+ Anions acetate C 2 3 2 azide N - 32- carbonate CO 3 chlorate ClO3- 2- chromate CrO 4 Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal cyanide CN - - dichromate Cr 2 7 dihydrogen phosphate H PO - 2 - hydrogen carbonate of HCO 3 bicarbonate hydrogen phosphate HPO 22- - hydrogen sulfate HSO 4 hydroxide OH - - hypochlorite ClO nitrate NO 3- - nitrite NO 2 oxalate C 2 42- - perchlorate ClO 4 permanganate MnO 4- 2- peroxide O 2 phosphate PO 43- 3- phosphite PO 3 sulfate SO 42- 2- sulfite SO 3 thiocyanate SCN - o Here are the rules to help you memorize the oxoanions (polyatomic anions with one or more oxygen atoms) above polyatomic ions: 1. The ion with one more oxygen atom then the “- ate” ion is “per…ate” ion.  Ex. ClO i3 chlorate, so ClO is 4erchlorate 2. The ion with one less oxygen atom than the “- ate” ion is chlorite. - -  Ex. ClO i3 chlorate, so ClO is 2hlorite 3. The ion with two fewer oxygen atoms than the “- ate” ion is “hypo…ite” ion. - -  Ex. ClO i3 chlorate, so ClO is hypochlorite o Here are some rules to help you memorize oxoacids (oxoanions with a hydrogen attached): 1. An acid based on an “- ate” ion is “…ic acid.”  Ex. ClO is chlorate, so HClO is chloric acid 3 3 I remember this with the phrase “I ate something icky.” 2. An acid based on “- ite” ion is called “…ous” acid. -  Ex. ClO i2 chlorite, so HClO is c2lorous acid 3. Prefixes in oxoanions retain their name in the oxoacids. Ex. ClO is hypochlorite, so HClO is hypochlorous acid Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal  Organic Compounds: compounds containing carbon o Organic compounds are named based on the number of carbon atoms they contain. o Here is the chart from the page 163 (Table 5.8) Formulas and Names of Alkanes (simple hydrocarbons) Name Formula CH 4 Methane C2H6 Ethane C3H8 Propane C 4 10 Butane C 5 12 Pentane C 6 14 Hexane C 7 16 Heptane C 8 18 Octane C 9 20 Nonane C10 22 Decane  Molecular and Formula Masses o We can use a compound’s molecular formula and atomic masses (from periodic table) to determine the molecular mass of a molecule.  Formula: Molecular Mass = ∑(number of atoms)(atomic mass)  Ex. Water H O 2 MM = 2(1.001) + 1(15.999) = 18.001 o Percent Composition: we can also use the molecular formula to calculate the percentage one element contributes to the total mass.  Formula: %mass of element = (mass of element)/(mass of compound) x 100  Ex. Water H 2 Let’s find the percent of hydrogen’s mass %H = ( (2 x 1.001)/(2 x 1.001 + 15.999) ) x 100 = 5.926 %H o We can use the percent composition of an element to determine the empirical formula (least common multiple of the ratios of atoms)and the molecular formula (actual ratio that atoms appear in).  Ex. Determine the empirical formula of a compound that is 30.45 %N and 69.55 %O. 1. Assume the mass percent equals the element masses. 30.45 %N = 30.45 g N 69.55 %O = 69.55 g O 2. Now, convert grams to moles 30.45 g N x 1 mol/14.01 g N = 2.173 mol N 69.55 g O x 1 mol/15.999 g O = 4.347 mol O Chapter 5 Lecture Notes CHEM 1030 Rik Blumenthal 3. Now, divide the mole amount by the smallest of the two values to get the smallest possible whole numbers (these will end up being the subscripts in the empirical formula.) 2.173/2.173 = 1 mol 4.347/2.173 = 2 mol Solution: NO i2 the empirical formula It is important to note that we can’t find the actual molecular formula unless we are given the molar mass of the compound. Dr. Blumenthal didn’t do an example of this in class, here is how you find the molecular formula: Given that the molar mass of a compound is 92 g/mol, determine the molecular formula of a compound. 1. Divide the given molar mass by the empirical formula mass Empirical Formula Mass = 14.001 + (2 x 15.999) = 46.01. 92/46.01 = about 2. 2. Now, multiply the subscripts in the empirical formula by 2 N O is the molecular formula 2 4


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