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Chem 109 Chapter 3 Notes

by: mkennedy24

Chem 109 Chapter 3 Notes Chem 109

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About this Document

These notes cover chapter 3 using information from both the textbook and form lecture.
General Chemistry
Eric Malina
Class Notes
Chemistry, General Chemistry




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This 5 page Class Notes was uploaded by mkennedy24 on Tuesday February 9, 2016. The Class Notes belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 144 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.


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Date Created: 02/09/16
I. Chapter 3: Molecules, Compounds, and Chemical Equations a. Why do atoms bond? i. Minimum Energy Principal ii. Lowest Potential Energy (most stable) b. How do atoms bond? i. Share electrons  Covalent Bond (using electron configuration) ii. Exchange electrons  Ionic Bonds (electrostatic attraction) c. Objective: Identify type of substance and types of bonds present from a formula i. Types of Substances: 1. Molecular substances have covalent bonds only  Nonmetal + Nonmetal 2. Ionic substances have ionic attractions/bonds present  Nonmetal + Metal ii. Polyatomic ions: A molecule with a charge 1. NO 3- 2. NaNO has no charge because the negative 3 - + from NO a3d the positive from Na cancel out 3. **Table 3.5 from textbook** - 2- HSO 3 Hydrogen Sulfate O Oxide 3- - PO 4 Phosphate N Nitride H 2O 4- Dihydrogen NO 2- Nitrite Phosphate CrO 42- Chromate NO 3- Nitrate - - ClO 4 Perchlorate OH Hydroxide Cr2O 72- Dichromate F - Fluoride O 2- Peroxide MnO - Permanganate 2- 4- CN Cyanide HSO 4 Hydrogen sulfate Cl- Chloride S 22- Disulfide ClO 2- Chlorite SO 2- Sulfite 2- - ClO 3 Chlorate N 3 Azide CO 32- Carbonate S 2- Sulfide - 2- Br Bromide HPO 4 Hydrogen Phospate + - NH 4 Ammonium HCO 3 Bicarbonate C 2 3 2 Acetate SO 42- Sulfate 4. Above are polyatomic ions that should be known! iii. Diatomic elements 1. Nitrogen, Iodine, Fluorine, Oxygen, Bromine, Chlorine, Hydrogen 2. The seven diatomic elements make up a seven in the periodic table using 6 of the elements (Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, and Iodine) and the last element is Hydrogen which is in Group 1A d. Chemical Formulas i. Empirical Formulas: Lowest whole number ratio for a given substance (CH O2is the empirical formula for table sugar) ii. Molecular Formulas: Actual number of atoms of each element in a compound (C H 6 12 t6e molecular formula for table sugar) iii. Formula Units: Empirical formulas for ionic substances only iv. Structural Formula: uses lines to represent covalent bonds and shows how atoms in a molecule are bonded to each other 1. H 2 2 H-O-O-H 2. CO 2 O=C=O e. Formula Mass and the Mole i. Atomic Mass: Mass of one atom of an element ii. Formula Mass: Mass of one of those formula units (Ionic Compounds) iii. Molecular Mass: Mass of one molecule iv. Objective: Determine formula/ molecular mass v. Molar Mass: mass of one mole of ionic or molecular compound 1. Example: 63 Page 132 in textbook a. Determine the number of moles(of molecules or formula units) in each sample b. 25.5g NO 2 i. 25.5g NO x 1molNO2 = .554 2 46.01gNO2 mol NO 2 ii. 38.2g KNO x 1mol KNO3 = . 3 101.11 gKNO3 378 mol KNO 3 vi. Objective: Calculate Molar Mass vii. Empirical Mass: Mass of one mole of the empirical formula viii. Objective: Determine number of atoms and/or molecules and/or formula units based on mass of a substance f. Composition of Compounds Massof element i. Mass Percent= x 100 Mass of compound ii. Objective: Calculate mass percent based on formula g. Chemical Formulas from Experimental Data i. Objective: Produce chemical formulas provided by data 1. Example: Rust is 69.94% Fe and 30.06% O. By mass, what is the formula unit of rust. a. Assume 100g total (69.94 + 30.06 =100) 1moleFe i. 69.94g Fe x 55.85gFe = 1.252 mole Fe 1moleO ii. 30.06g O x 16.00 gO = 1.879 mole O b. Simple whole numbers are wanted NOT decimals i. Divide by smallest mole amount 1.252moleFe 1. = 1.000 mole 1.252 Fe 1.879 moleO 2. 1.252 = 1.5 mole O ii. If after dividing by smallest mole amount, the number comes out to be a decimal that does not have a 9 in the tenths place (.9), then multiply each mole by a number that results in a whole number iii. HINT: 0.5’s multiply by 2; 0.33’s, 0.66’s multiply by 3; 0.25’s, 0.75’s multiply by 4; 0.20’s, 0.40’s, 0.60’s, 0.80’s multiply by 5 c. Multiply everything by simple whole numbers (meaning final mole numbers : 1.000 mole Fe and 1.500 mole O) to get ride of the fraction and get simple whole numbers i. 1.000 mole Fe x 2= 2.000 mole Fe ii. 1.500 mole O x 2= 3.000 mole O d. Final Answer: Fe2O 3 ii. Objective: Determine molecular formulas form empirical formulas and molar masses 1. Lactic Acid: a. Empirical Formula: CH O2 b. Molar Mass: 90.08 g/mol c. What is the molecular formula? i. Empirical mass: (12.01 x 1) + (1.008 x 2) + (16.00 x 1) = 30.03 g/mol Molar Mass 90.08g/mol ii. Empirical Mass  30.03g/mol = 3 iii. Multiply everything in empirical formula by 3  C H3O6 3 iii. Combustion Analysis 1. Example: Unknown compound containing only C,H,N; exactly 1 gram of unknown is burned and produces 2.836g CO and2.6773 H O. Mo2ar mass is 186.24 g/mol . What is the molecular formula? 1 molCO2 a. Mass of C = 2.836g CO x 2 x 44.01gCO 2 1molC 12.01gC x =.7739 g C 1molCO2 1molC 1 mol H 2O b. Mass of O = .6773g H O 2 18.02 g H 2O x 2molH x 1.008gH = .7577g H 1molH 2O 1mol H c. 1.000g total - .7739g C -.7577g O= . 1503g N 1mol C d. .7739g C x 12.01 gC = .06444 mole C 1molH e. .7577g H x = .07517 mole H 16.00gH 1mol N f. .1503g N x = .01073 mole N 14.01 g N g. Divide by smallest mole amount .06444molC h. = 6.006 Empirical Formula: .01073 C 6 7  Empirical Mass: mol C 93.13 g/mol .07517mol H i. .01073 = 7.006 mol H .01073mol N j. .01073 = 1.000 mol N k. Molar Mass  186.24 = 2  C H N Empirical Mass 93.13 12 14 2 h. Write and balance chemical equations i. Example: Combustion of gaseous methane (CH ) with 4 oxygen gas produces gaseous carbon dioxide and water vapor 1. CH (4) + O  2O (g) +2H O(g) 2 2. BALANCE: Have the same amount of atoms on each side 3. CH (4) + 2O (g)2 CO (g) 2 2H O(g) 2 ii. When balancing, if a fraction is the only way to balance, multiply everything by two to get rid of the fraction 1. C 2 +6O  C2 + H O2 2 2. (C 2 +63.5O  22O + 3H 2 ) x 22= 2C H + 2 6 7O 2 4CO + 2H O 2 iii. Polyatomic Ions 1. **Treat the polyatomic ions as a whole instead of the separate elements** iv. Strategies: 1. Start with understood “1” on the compound with the most elements (atoms) 2. Balance those elements with coefficients 3. Balance the rest with coefficients 4. Scale to lowest whole number ratio v. States: 1. (g): gas 2. (l): liquid 3. (s): solid 4. (aq): aqueous


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