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CH 111, Week 3 Notes

by: Jordyn Meekma

CH 111, Week 3 Notes CH111

Jordyn Meekma

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About this Document

Covers the basics of stoichiometry
General Chemistry 1
Dr. Tom Getman
Class Notes
25 ?




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This 3 page Class Notes was uploaded by Jordyn Meekma on Thursday September 29, 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 3 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.


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Date Created: 09/29/16
CH 111 Notes: Week 3 Polyatomic Ions: charged ions composed of two or more atoms, covalently bonded, that act as a single unit o Hydroxide OH - o Carbonate CO 2- 3 - o Bicarbonate HCO 3 o Nitrate NO 3- o Nitrite NO - 22- o Sulfate SO 4 o Sulfite SO 32- o Acetate CH CO - 33- 2 o Phosphate PO 4 o Ammonium NH 4+ - o Thiocyanate SCN- o Cyanide CN o Permanganate MnO 4 - o Hypochlorite - ClO o Chlorite ClO 2 o Chlorate ClO 3- - o Perchlorate ClO2- o Peroxide O 2 *memorize these polyatomic ions and their charges* Molecular Mass: the mass of one (covalently bonded) molecule o CO 2 1C: 1(12.01) values are the elements’ masses from the periodic table + 2O: 2(16.00) 44.01 amu Formula Mass: the mass of one formula unit of an ionic compound o Fe(SCN) 2 1 Fe: 1(55.85) 2 S: 2(32.07) 2 C: 2(12.01) + 2 N: 2(14.01) 172.03 amu Mass Percentage: the percent composition of a compound o (Mass of one type of atom/total mass of compound) x100%  %Fe = (55.85 amu/172.03 amu) x100% = 32.47% Fe *This concludes all the material that will be on the first exam* Chapter 3: Stoichiometry  SO 3(g)+ H 2 (g)→ H S2 4(l) States of matter abbreviations Reactants Products (g) – gas (l) - liquid 23 (s) – solid (aq) -  1 mole = 6.022x10 molecules aqueous  Molar Mass: (MM) molecular weight in g/mol o Molar mass of PCl = 3 1P: 1(30.97 g/mol) 3Cl: 3(35.45 g/mol) 137.32 g/mol  The is an alloy containing Iron and Nickel, with 75% Iron and 25% Nickel. What is the mass percentage of Nickel? Molar Mass 75.0 g Fe (1 mol Fe/55.85 g) = 1.34 mol Fe 25.0 g Ni (1 mol Ni/58.69 g) = 0.426 mol Ni 0.426 mol Ni/(1.34 mol + 0.426 mol)x100% = 24.1% Nickel  How many moles of CaCO are in an an3acid tablet that contains 0.600 g CaCO ? 3 MM = 100.09 g/mol -3 0.600 g CaCO (1 m3l CaCO /100.09 g3 = 5.99x10 mol CaCO 3 o What is the number of Carbonate ions in this tablet? -3 - 23 5.99x10 mol CaCO (1 mol 3O /1 mol CaCO3)(6.022x10 3 ions/1 mol) = 3.61x10 ions  CaCO 3(s) 2HNO 3(aq)→ Ca(NO ) 3 2(aq)+ CO 2(g)+ H 2 (l) (MM HNO = 63.02 g/mol) (MM CO = 44.01 g/mo2) If 5.00 g of HNO react3 with excess CaCO , how many grams3of CO 2 are produced? Limiting reactant: The reactant in a chemical reaction that limits the amount of product that can be formed 5.00 g HNO (1 3ol HNO /63.02 g3(1 mol CO /2 mol HNO )(24.01 g/1 3 mol CO ) 2 1.75 g If 5.00 g HNO reac3s with 5.00 g of CaCO , which is the l3miting reactant? 5.00 g CaCO (1 3ol CaCO /100.09 3)(1 mol CO /1 mol CaCO ) 2 2 -2 4.996x10 mol CO 2 5.00 g HNO (1 3ol HNO /63.02 g3(1 mol CO /2 mol HNO ) 2 3.967x10 3 -2 mol CO 2 *Limiting o How many grams of CO are produce2? reactant* 3.967x10 mol CO (44.012g/1 mol CO ) = 1.75 g C2 2  TiO (s)+ 2C (s)2Cl 2(g)  TiC4(g)+ 2CO (g) If there are 10.0 g of each reactant, how many grams of TiCl  are4produced? 10.0 g TiO 2(1 mol TiO 2 79.87 g)(1 mol TiCl 41 mol TiO2) = 0.1252 mol TiCl 4 10.0 g Cl2 (1 mol Cl2/70.90 g)(1 mol TiCl4/2 mol Cl2) = 0.0705 mol TiCl4 * 10.0 g C (1 mol C/12.01 g)(1 mol TiCl 42 mol C) = 0.4163 mol TiCl 4 0.0705 mol TiCl 4(189.67 g/ 1 mol TiCl 4 = 13.4 g TiCl4 Percent yield: (Actual yield/Theoretical yield)x100% (10.3 g TiCl4/13.4 g TiCl4)x100% = 76.9% yield


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