Week of Notes 10/12-10/16
Week of Notes 10/12-10/16 CH 101
Popular in General Chemistry
Popular in Chemistry
verified elite notetaker
This 9 page Class Notes was uploaded by Rani Vance on Monday October 19, 2015. The Class Notes belongs to CH 101 at University of Alabama - Tuscaloosa taught by a professor in Fall 2015. Since its upload, it has received 13 views. For similar materials see General Chemistry in Chemistry at University of Alabama - Tuscaloosa.
Reviews for Week of Notes 10/12-10/16
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/19/15
Chem 101 Week of Notes 10121016 10 12 Molar mass of compounds contribution chemical formula What is a mole 6022x 1023 Molar mass of a compound is equal to the sums of atomic masses of its constituent elements Formula mass of atoms of 1st element atomic mass 1st element of atoms of 2rld element atomic mass of 2nd element Example NaCl formula MMNa MMCI 2299 gmol Na 3545 gmol Cl 5844 gmol Water 11008 gmol H 2 1600 gmol O 18016 gmol 9 1802 gmol Mass of moles X molar mass What is the mass of 25 moles of NaCl Nmol Nac1 25 mol MMNaCl 5844gmol 25mol NaCl 5844 gmol NaCl 1461 g Clicllter Question Find the molar mass of CH4 Answer 1201 gmol C X 1 gmol H4 1601 g How many formula units of NaCl are there in 1 mol 6022 X1023 How many formula units are there in 8766 g of NaCl Mass moles of molecules 5844 gmol of NaCl 8766 g 1 mol X 6022 X 1023 molecules 903 X1023 molecules 5844 gmol 1 mol How many atoms of 02 are present in 32 g of water Mass of water 32g MM of water 18g Mass of water moles of water moles 02 32g X 1 mol 1778 mol 18g mol How many moles in 453 of A12 SO43 A12 227 gmol 54 S 33206 gmol 96 O 121600 gmol 192 342 453g lmol 0132mols 342 gmol Chemical formula Empirical formula gives the relative of atoms think of it as a ratio Molecular gives the actual of atoms eX A12 SO43 EX Molecular Empirical Glucose C6H1206 CHzO Hydrogen Peroxide H202 HO Carbon Dioxide C02 C02 Water H20 H20 Composition Chemical formulas give the element ratio Within a compound EX C6H1206 Ratio 1 C 2 H 1 0 Mass ratio different from atom ratio How to calculate composition mass of 1 element mass of element in moles Mass of molecule mass of 1 mole of molecule Find the C by mass in C02 Mass of C mass C in moles C02 Mass of C02 in mole C02 MM C02 MMC 2 MMO 1201gmol 21601gmol 4401 gmol 9 1201gmol 273 4401 gmol How do chemists know What the composition of something is Chemists are able to very precisely measure the gases generated When something is combusted Works well Wcompounds containing H amp C What comes out C02 H20 Chapter 5 Calculating the Empirical formula 1 Convert the to grams assume you start With 100g of the compound 2 Convert g9 moles 3 write pseudoformula using mol transcripts 4 divide all by smallest of moles if result is Within 01 round to the nearest Whole 5 Multiply all mol ratios by to make Whole a if ratio is 5 X by 2 if 33 or 67 X by 3 if 2575 X by 4 MUST KNOW HOW TO DO FOR EXAM Composition Empirical Formula EX l C 4000 H 671 O 5328 1 C9 4000g H9 671g 09 5328 g 2 g9 moles C 4000g X liol 333 moles C 1201g H 671g X 1 mol 664 moles H l01g O 5328g X 1mol 333 moles O 1600g 3 C333 H664 0333 4 Divide by the smallest 333 9 C1 H199 01 9 round if it is Within 01 C1 H2 01 5 this step is not necessary for this example because they are easily rounded to Whole numbers EX 2 1 C 6000 9 6000 g C H 448 9 448 g H 0 3553 9 3553 g o 2 C 60g X lmol 4996 moles 1201g H 448 g X 1 mol 443 moles l01g O 3553g X 1 mol 2221 moles 1601g 3 C4996 H443 02221 4 smallest number C2249 H1997 01 5 multiply to make a Whole ratio multiply by 4 because 025 9 C9 H8 04 empirical formula Chapter 6 Polar Covalent Bonding Covalent bonding between unlike atoms results in unequal sharing of the electrons one atom may pull an electron in the bond closer to itself bond becomes polar shifts to one side Result is a polar covalent bond atom With larger bond electron density gets a partial negative charge atom With less bond electron density gets a partial positive charge Electronegativity the ability of an atom to attract bonding electrons to itself is called electronegativity Most EN element is F least EN element is Fr the larger the difference in EN the more polar the bond F is good at attracting electrons because of the effective nuclear charge F more EN than H9 bond shifted toward F H9 partial positive 9 dipole partial and partial Bond Polarity most bonds have some degree of sharing and some degree of ion formation to them bonds classified as covalent if the amount of energy transfer in insufficient for the material to display the classic properties of ionic compounds if sharing is unequal enough to produce dipole Know polar ionic spectrum Determine What bond NO ENN 30 ENC 35 ENdelta ENoENN 05 biggestsmallest 05 on scale polar covalent Bond Polarity ClCl pure covalent HCl 3021 09 polar covalent Bond Dipole Moments Dipole moment mu is a measure of bond polarity material With positive and negative end directly proportional to size and charges Generally the more electrons two atoms share and the larger the atoms are the larger the dipole moment 1016 Exceptions to Octet Rule LeWis theory says that atoms most stable When they have their octet of valence electrons Some violate this rule Be 2 bonds With no lone pairs B9 3 bonds With no lone pairs 3rd row elements often ignore the octet rule and more than eight electrons odd electron species unpaired electrons freeradicals Lewis Structure Memorize common bonding pattern Picture Writing Lewis Structures from Molecular Compounds in 5 STEPS 1 Draw correct skeleton structure for molecule 2 calculate total of valence electrons add an extra electron for each negative charge subtract an electron for positive charge 3 Place two bonding electrons between each bonding atom Then try to fill the valence shell of atoms by adding electrons Start with the terminal atoms first 4 If any atoms lack an octet for double or triple bonds do this by moving electron pairs from terminal atoms into the bonding regions 5 check your work total number of electrons added to the Lewis structure should number of electrons in step 2 EX O 5 F0 g i 1 6V 4 6 l FOZHWOa lI50 flv39ol p 09 H 0 4 39t l mm mm ammt 3 53 H A C 1 D39L l 39 OU 7 0 z w 0 quotl W2 0 WM I 0 04601 0 l 3 Z I l 2 N971 an 0 75000 0 L1 1 0 g g 3 N 0 7 V quot39 F6 2 v 4 9 0 t N 5 wifgfwl o 7 W T Fiji 10 mm 1 olt501m 4 V might v 5mg cHCW format qu at I O39x Miw v39 5 Formal Charge During bonding atoms may end With more or fewer electrons than the valence electrons have brought in order to fulfill octets results in atoms having a formal charge FC starting With valence electrons nonbonding electrons 12 bonded electrons