CH 111, Week 4 Notes
CH 111, Week 4 Notes CH111
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This 2 page Class Notes was uploaded by Jordyn Meekma on Friday September 30, 2016. The Class Notes belongs to CH111 at Northern Michigan University taught by Dr. Tom Getman in Fall 2016. Since its upload, it has received 2 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.
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Date Created: 09/30/16
CH 111 Notes: Week 4 Combustion Reactions: an organic substance combines with oxygen and is burned, giving off heat energy and creating carbon dioxide and water o Propane (C H 3 8 C 3 +85O → 32CO + 4H O2 2 Empirical Formula: the formula of a molecule with the smallest whole number ratio of atoms o C 5 4 vs. C10 8 → related by a whole number multiple of 2 In a combustion reaction, 135.0 mg of a hydrocarbon reacts with excess oxygen to produce 440.0 mg of carbon dioxide and 135.0 mg water. The hydrocarbon MM = 270 g/mol. Determine the empirical and molecular formulas. o C x +yO → C2 + H O 2 2 -3 0.4400 g CO (1 2ol CO /44.012g)(1 mol C/1 mol CO ) = 9.9977x12 mol C -2 0.1350 g H O 2 1 mol H O/18.22 g)(2 mol H/1 mol H O) = 1.4983210 mol H X = 9.9977x10 mol C/9.9977x10 mol C = 1 (x2) = 2 -2 -3 *Divide C and H by the lower number of Y = 1.4983x10 mol H /9.9977x10 mol C = 1.499 (x2) = 3 Empirical Formula: C H 2 3 moles and multiply to get to the nearest whole number MM C H 2 37.05 g/mol (270 g/mol)/(27.05 g/mol) = 10 C 2 3x10) = … Molecular Formula: C H 20 30 Lithium metal reacts with nitrogen gas to form a compound. If 0.437 g Li are completely reacted to form 0.725 g of the compound, what is the empirical formula? o Li + N →2Li N x y 0.725 g compound – 0.437 g Li = 0.288 g N X = 0.437 g Li (1 mol Li/6.941 g) = 6.296x10 mol Li/2.056x10 mol N -2 = 3.06 Y = 0.288 g N (1 mol N/14.01 g) = 2.056x10 mol N/2.056x10 mol N -2 = 1 Empirical Formula: Li N 3 A compound containing C, H, and O is burned. If 125.0 mg of the compound produces 177.5 mg of Co2 and 108.8 mg of H2O, what are the empirical and molecular formulas? MM of the compound = 62.08 g/mol. C ? O?+?O → C2 + H O 2 2 -3 o C: 0.1775 g CO (1 m2l CO /44.01 2)(1 mol C/1 mol CO ) = 4.033x10 2 mol C -3 -2 4.033x10 mol C (12.01 g/1 mol C) = 4.844x10 g C 4.033x10 mol C/4.03x10 mol O = 1.002 -2 o H: 0.1087 g H O (2 mol H O/18.02 g)(2 mol H/1 mol H O) = 1.206x12 mol H 1.206x10 mol H (1.01 g/1 mol H) = 1.216x10 g H -2 -2 -3 1.206x10 mol H/4.03x10 mol O = 3.00 o O: 0.1250 g compound - 4.844x10 g C - 1.216x10 g H = 6.44x10 g -2 O 6.44x10 g O (1 mol O/16.00 g) = 4.03x10 mol O -3 -3 -3 4.03x10 mol O/4.03x10 mol O = 1 Empirical Formula: CH O M3 = 31.04 g/mol Molecular Formula: (62.08 g/mol)/(31.04 g/mol) = 2.000 CH 3 (x2) = C 2 O6 2 Chapter 4: Solutions Solvent: the component in a solution with the greatest quantity (of moles) Solute: the component in a solution that is dissolved in the solvent Molarity: (M) moles of a component divided by the volume in liters (mol/L) o How many grams of sucrose (C H O ) are 12qu22e11to make 250.0 mL of a 1.25 M sucrose solution? MM C H O 12 22 11 = 342.34 g/mol 1.25 mol C H12 22 L110.2500 L)(342.34 g/1 mol C H O ) = 112 22 11 C 12O22 11 o Using this 1.25 M solution, how could we prepare 1.00 L of a 0.20 M sucrose solution? *M 1 =1M V *2 2 dilution equation (1.25 M)(V )1= (0.20 M)(1.00 L) V1= 0.260 L Spectrophotometry: using light (absorbance) to determine the concentration of a solute in solution o Beer’s Law *You will be using this concept in lab* Absorbance (no unit) Concentration of solute
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