General Chemistry Chem 1314
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This 5 page Class Notes was uploaded by Morgan Walker on Sunday February 7, 2016. The Class Notes belongs to Chem 1314 at Oklahoma State University taught by Dr. Jimmie Weaver in Winter 2016. Since its upload, it has received 20 views. For similar materials see General Chemistry in Chemistry at Oklahoma State University.
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Date Created: 02/07/16
Week four chapter 3 Naming oxyacids cont. Some acids have more than one ratio of hydrogen o Name according to the number of hydrogen present Ex: PO4-phosphate -2 HPO 4 hydrogen phosphate H2PO 4-dihydrogen phosphate H PO phosphoric acid 3 4 Molecular Mass or molecular weight, is the sum of the atomic masses in g/mol in a molecule o Mass of one SO 2olecule 1S 32.07 ₊ 20 16.00(2) ------------------- SO264.07 Chemical Formula Empirical- simplest Molecular- actual Ex: Formaldehyde- CH O Formula Mass 30g/mol 2 Finding Formula Ex: C17 21N4 C= 67.31% H=6.98% O=21.20% N= 4.62% (if given in percent, assume 100 grams) Step 1- convert grams to moles (by dividing given number by atomic mass) ???? 67.31 =5.5998 H 6.98=6.9109 O 21.20=1.3250 N 4.62=.3298 12.02 1.01 16.00 14.01 Step 2- Divide by smallest mol C 5.5998=16.9 H 6.9109=20.95 O 1.325=4.02 N .3298= 1 .3298 .3298 .3298 .3298 Step 3- round to whole number C 16.9= 17 H 20.95=21 O 4.02=4 N 1=1 Percent Composition Percent by mass of each element in a compound Ex: C13 18 2 Step 1 multiply subscripts by elements mass C 13X12=156 H 18X1=18 O 2X16=32 Add the totals together 156+18+32=206 Step 2 mass of element divided by total mass all times 100 %C 156 ????100= 76% H 18 ????100= 9% O 32 ????100= 15% 206 206 206 Organic Compounds Compounds from living things, nonorganic is nonliving Organic are easily decomposed and cannot be made in a lab Mainly made of Carbon and Hydrogen, can be made from Oxygen, Nitrogen, Phosphorous or Sulfur Main element of focus of organic chemistry is Carbon Carbon atoms bond almost exclusively covalently Compounds with ionic bonding Carbon are generally inorganic When Carbon bonds it forms four covalent bonds 4 single bonds, 2 double bonds, 1 triple + 1 single Carbon is able to form straight chains of Carbon atoms as well as rings of Carbon atoms Hydrocarbons Organic compounds that contain only Hydrogen and Carbon Compose common fuels like oil, gasoline, liquid propane gas and natural gas Naming Hydrocarbons containing only one single bonds are called alkanes With double or triple bonds are bonds are alkenes(double) and alkynes(triple) Consist of a base name and a suffix Base names- 1-meth 2-eth 3-prop 4-but 5-pent 6-hex 7-hept 8-oct 9-non 10-dec Suffix- alkane- (-ane) alkene- (-ene) alkyne- (-yne) Functionalized Hydrocarbons Functional group derives from the functionality or chemical character that a specific atom or group of atoms imparts to an organic compound Group of organic compounds with the same functional group are a family Chemical Equations Short hand for chemical reactions 2Mg(s) + O 2g) 2MgO(s) Reactants are on the left side and the product is on the right side Physical state- (s)- solid (l)-liquid (g)-gas (aq)- aqueous Reading 2Mg(s) + O 2g) 2MgO(s) Two atoms of Mg plus one molecule of O 2akes 2 formula units of MgO Balancing Number of atoms on each side of arrow must be equal You can only add coefficients Steps Step 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O Step 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2 C2H6 NOT C4H12 Step 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O C2H6 + O2 2CO2 + H2O C2H6 + O2 2CO2 + 3H2O Step 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2CO2 + 3H2O C2H6 + 7/2O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O Step 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O Combustion Analysis Common technique used for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made By knowing the mass of the product and composition of constituent element in the product, the original amount of constituent can be found Once masses of all constituent elements in the original compound is determined the Empirical formula can be found Finding Empirical Formula Step 1 Write down as given the masses of each combustion product and the mass of the sample (if given). Given: 1.83 g CO2 , 0.901 g H2O Find: empirical formula Step 2 Convert the masses of CO2 and H2O from step 1 to moles by using the appropriate molar mass for each compound as a conversion factor. Step 3 Convert the moles of CO2 and moles of H2O from step 2 to moles of C and moles of H using the conversion factors inherent in the chemical formulas of CO2 and H2O. Step 4 If the compound contains an element other than C and H, find the mass of the other element by subtracting the sum of the masses of C and H (obtained in step 3) from the mass of the sample. Finally, convert the mass of the other element to moles. The sample contains no elements other than C and H, so proceed to the next step. Step 5 Write down a pseudoformula for the compound using the number of moles of each element (from steps 3 and 4) as subscripts. C0.0416H0.100 Step 6 Divide all the subscripts in the formula by the smallest subscript. (Round all subscripts that are within 0.1 of a whole number.) Step 7 If the subscripts are not whole numbers, multiply all the subscripts by a small whole number to get whole-number subscripts. The correct empirical formula is C5H 12 Law of Definite Proportions Two samples of carbon dioxide are decomposed into their constituent elements. One sample produces 25.6 g of oxygen and 9.60 g of carbon, and the other produces 21.6 g of oxygen and 8.10 g of carbon. Show that these results are consistent with the law of definite proportions. Solution To show this, calculate the mass ratio of one element to the other for both samples by dividing the mass of one element by the mass of the other. For convenience, divide the larger mass by the smaller one. For the first sample: 25.6 9.60 = 2.66 For the second sample: 21.6 = 2.66 8.10 The ratios are the same for the two samples, so these results are consistent with the law of definite proportions Law of Multiple Proportions Nitrogen forms several compounds with oxygen, including nitrogen dioxide and dinitrogen monoxide. Nitrogen dioxide contains 2.28 g oxygen to every 1.00 g nitrogen, while dinitrogen monoxide contains 0.570 g oxygen to every 1.00 g nitrogen. Show that these results are consistent with the law of multiple proportions. Solution To show this, calculate the ratio of the mass of oxygen from one compound to the mass of oxygen in the other. Always divide the larger of the two masses by the smaller one. 2.28 = 4.00 .570 The ratio is a small whole number (4); these results are consistent with the law of multiple proportions
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