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Lecture 10 Notes

by: nichl

Lecture 10 Notes 01:160:161


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General Chemistry Chapter 6
General Chemistry I
Class Notes
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This 4 page Class Notes was uploaded by nichl on Saturday October 15, 2016. The Class Notes belongs to 01:160:161 at Rutgers University taught by in Fall 2016. Since its upload, it has received 4 views. For similar materials see General Chemistry I in Chemistry at Rutgers University.


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Date Created: 10/15/16
General Chemistry Lecture 10 10/6/16 Molecular Formula Determination 1. Obtain empirical formula 2. Compute mass corresponding to empirical formula Molar Mass 3. Calculate ratio: EmpiricalFormula Mass 4. Multiply subscript for each element in the empirical formula by the ratio to obtain the molecular formula Questions: You have a compound that is: 71.65% Cl, 24.27% C, 4.07% H. What is the empirical formula? 1. For 100g sample, you have 71.65g Cl, 24.27g C, 4.07g H 2. Determine number of moles of each substance: 71.65∗1mole =2.021 a. Moles Cl: 35.45gCl moles Cl 24.27∗1mole b. Moles C: 12.01gC =2.021 moles C 4.07∗1 mole =4.04 c. Moles H: 1.008H moles H 3. Divide all moles with smallest number of moles (in this case 2.021): a. Cl: 1, C: 1, H: 2 4. Empirical formula: ClCH 2 Molar mass is 98.46 g/mol What is the molecular form? 1. ClCH 2empirical form 2. Mass for empirical form: 49.48 g/mol Molar Mass = 98.96 3. Find ratio: Empirical Formula Mass 49.48 4. Multiply each subscript of empirical form by the ratio to get molecular form: Cl2C2H4 Compound contains 43.64% P & 56.36% O What is its empirical form? 1. In 100g of sample: 43.64g P & 56.36g O 43.64∗1mol 56.36∗1mol 2. Moles P: =1.409 Moles O: =1.409 30.97g 16.00g 3. Divide both mole values by 1.409 (smallest value) General Chemistry Lecture 10 10/6/16 a. P = 1 b. O = 2.5 4. Multiply by an integer (in this case 2) until a whole number is reached 5. Empirical form: P 2 5 Combustion Analysis  Experimental determination (by combustion) of an empirical formula. Carbon & Hydrogen containing compounds yield CO & H 2 upo2 combustion (*if N present, forms N 2pon combustion)  Combustion uses oxygen from the air. If compound contains oxygen, the only way to figure out how much oxygen is present in compound is by subtraction (difference)  Question: o 0.255g compound containing C, H, O produces .561 g CO & 0.326 g H 2. What is the empirical formula? 0.561∗1 =0.0127 0.0127∗12.01=.153 o 44.01 moles CO 2 g Carbon 0.306∗1 ∗2molH o 18.016 moles H O 0.0340∗1.008=.0343 g =0.0340 2 1molH O2 Hydrogen o Amount of O in sample is 0.255g− 0.153+0.034 =0.068 g O∗1mol =4.2∗10 −3 ( ) 16.00 −3 o Divide all moles by 4.2∗10 to get formula: C3H 8 Organic Compounds  **Know the first 10 alkanes in the PT  Simplest organic compounds  Hydrocarbons (only contain Carbon and Hydrogen)  Saturated hydrocarbons: maximum number of hydrogens for the number of carbons  General formula: C n 2n  CH 4 Methane  C2H6– Ethane CH 3H 3ondensed structural formula Other Hydrocarbons   alkene Name: ethene   alkyne Name: ethyne General Chemistry Lecture 10 10/6/16 Octet Rule  Atoms will lose, share, or gain electrons to achieve a Noble Gas configuration o Does this to achieve octet – 8 valence electrons Lewis Structures  Shows non-metals combining, sharing electrons to achieve octets o F2: o Exception: Hydrogen can only achieve a duet Polar Covalent Bond & Electronegativity  Related to ionization energy and electron affinity  Measures ability for an atom to compete for electrons with the other atoms it is bonded with  The greater the electronegativity of the atom in a molecule, the more strongly it attracts electrons to itself when bonded with other atoms  Trend: (increases along arrows) Periodic Table Electronegativity Difference & Bond Polarity  No difference in electronegativity between diatomic molecules (i.e. Cl-Cl or O- O)  Chlorine is more electronegative than hydrogen   Cl is more electronegative  Therefore, there is an electronegativity difference and this is a “polar covalent bond”  Electrons are being pulled toward more electronegative atom like a magnet  Na Cl- o Electron completely transferred from metal to non-metal o Very large electronegativity difference – bigger the difference in electronegativity, the more polar the bond  Ex: Rank the following according to increasing polarity of the bond o O---H; H---H; S---H; F---H o H---H < S---H < O---H < F---H General Chemistry Lecture 10 10/6/16 Depicting Polar Covalent Bonds (showing bond dipoles) δ−¿ ¿  δ+¿H−−¿l ❑  If electronegativity differs by less than 0.5, the bond is considered non-polar  If difference between bonds is between 0.5 & 2, it’s considered to be a polar covalent bond  If difference between bonds is greater than 2, it’s an ionic bond Equation for Dipole Moment  μ=Qr μ – dipole moment (in Debye[D]) Q – Product of Charge r – distance between charges −30  1Debye=3.336∗10 Coulomb Meters *Note: Charge Q must be divided by magnitude of 1 electron to give partial charge of each atom in the bond  The more electronegative atom gets the – sign  Question: o Partial charges in C & O in Carbon Monoxide are: +0.020 and -0.020. Calculate dipole moment of CO if distance between partial charges is 113 pm Q=0.020∗1.6022∗10 −1=3.2∗10 −21C  (113∗10−12∗1D  μ= (3.2∗10−2) =0.11D 3.36∗10−30


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