Week 3/Th: Lecture Units 6 & 7 Unit 5: The Mole -- definition for counting, molar mass -- stoichiometry and nomenclature -- elemental composition Unit 6: Stoichiometry for Cpds. -- mass, moles and composition -- Balancing reactions Unit 7: Calculations for Chemical Reactions -- molar mass in reactions Unit 8: Limiting reagents Issues:? Homework Set 2 due on Saturday @ 08:00AM http://www2.chemistry.msu.edu/faculty/morrissey/
Week 3/Th: Elemental Composition -5-5) 1.256 g of sulfur reacts with F 2 to produce 5.722 g of a substance SF x. What is X? Let s write a word equation for now. Sulfur + F 2 à SF X Mass 1.256 g? 5.722 g Molar Mass 32.07 g/mol 38.00 g/mol? g/mol = 1.256 g / 32.07g/mol = 4.466 g / 38.00 g/mol = 0.03916 mol = 0.1175 mol Mole Ratio =0.03916/0.03916 =1 = 0.1175/0.03916 = 3.001 of F 2 X is 6 (5.722-1.256) = 4.466 g Conservation of Mass
Week 3/Th: Elemental Composition -6-6) A compound was analyzed and found to contain 53.30% carbon, 11.19% hydrogen, and 35.52% oxygen by mass. The molar mass was determined (in a separate expt.) to be about 90 g/mol. What is the empirical formula and what is the molecular formula of this compound? C H O Amount 53.30 11.19 35.52 g Molar Mass 12.01 1.008 16.00 g/mol 4.438 11.10 2.220 moles Mole Ratio 1.999 5.00 1 Empirical Formula: C 2 H 5 O, Formula Mass = 45.06 g/mol but Molar Mass ~ 90 g/mol was given so Molecular Formula is: C 4 H 10 O 2
Week 3/Th: Composition -7-7) Suppose a hydrocarbon cpd. (only C & H) is burned in air to produce CO 2 and H 2 O. The number of moles of CO 2 is 1.5 times the number of moles of water. What is the empirical formula of this compound? If the molar mass = 120 g/mol what is its molecular formula? C x H y + excess O 2 à x CO 2 Amount + (y/2) H 2 O Molar Mass 120. 32.0 44.01 18.02 g/mol Mole Ratio 1.5 1 of carbon = 1.5 * 1 carbon in CO 2 = 1.5 of hydrogen = 1 * 2 hydrogens in H 2 O = 2 Emp.Formula: C 3 H 4, Formula Mass = 40.04 g/mol but Molar Mass ~ 120 g/mol was given so Molecular Formula is: C 9 H 12
Week 3/Th: Composition -8-8) A 15.67 g sample of a hydrate of magnesium carbonate was carefully heated in an oven to drive off the water. The mass reduced to 7.58 g. What is the formula of the hydrate? [MgCO. 3? H 2 0] à MgCO 3 +? H 2 O Amount 15.67 7.58 8.09 g Molar Mass?? 84.31 18.02 g/mol 0.08991 0.4489 Mole Ratio 1 0.4489/0.08991=4.99 Formula: [ MgCO 3. 5 H 2 0 ]
Week 3/Th: Composition -9-9) A cpd of (only) carbon, hydrogen, and nitrogen was analyzed and found to contain 45.42% C, 12.20% H, and 42.38% N. What is the empirical formula of the compound? C H N Amount 45.42 12.20 42.38 g Molar Mass 12.01 1.008 14.01 g/mol 3.782 12.1 3.025 mol Mole Ratio 1.25 4.00 1 Interger ratio 5 16 4 Emp. Formula: C 5 H 16 N 4, Formula Mass = 132.22 g/mol
Week 3/Th: Stoichiometry, Balancing Eqs. A Chemical Equation is a collection of symbols that represent a complete and balanced chemical process. Atoms and Mass MUST be conserved both MUST balance. How to balance a chemical equation: 1) Start with the correct formulas for all of the participants. (duh!) 2) Select the most complicated component, balance those atoms first. 3) Work with atoms the only appear once on each side of the reaction. (not always possible) 4) Leave the simple (elemental, H or O containing) molecules until last. Al 4 C 3 + 12 H 2 O à 4 Al(OH) 3 + 3 CH 4 4 NH 3 + 5 O 2 à 4 NO + 6 H 2 O C 3 H 8 + 5 O 2 à 3 CO 2 + 4 H 2 O DJMorrissey, 2oo9
Week 3/Th: Stoichiometry, Balancing Eqs. A Chemical Equation is a collection of symbols that represent a complete and balanced chemical process. Atoms and Mass MUST be conserved both MUST balance. DEMO: the barking dog reaction gas phase reaction of carbon disulfide with nitrogen dioxide 4 CS 2 + 4 NO 2 à 4 CO 2 + 2 N 2 + S 8 DJMorrissey, 2oo9
Week 3/Th: Stoichiometry-1- Combustion of propane, C 3 H 8, produces CO 2 and water What mass of CO 2 can be produced from 110 g of propane? balance rxn. first! C 3 H 8 + 5 O 2 à 3 CO 2 + 4 H 2 O Mass 110 g Molar Mass 44.09 32.00 44.01 18.02 g/mol Present 2.50 mol Rxn 2.50 3*2.50 mol Mass Rxn 3*2.50*44.01? g = 329 g
Week 3/Th: Stoichiometry-2- Combustion of propane, C 3 H 8, produces CO 2 and water How many moles of H 2 O are produced from 110 g of C 3 H 8? C 3 H 8 + 5 O 2 à 3 CO 2 + 4 H 2 O Mass 110 g Molar Mass 44.09 32.00 44.01 18.02 g/mol Present 2.50 mol Rxn 2.50 4*2.50=10.0? mol Mass Rxn g
Week 3/Th: Stoichiometry-3- Combustion of propane, C 3 H 8, produces CO 2 and water What is the mass of oxygen required for 110 g of C 3 H 8? C 3 H 8 + 5 O 2 à 3 CO 2 + 4 H 2 O Mass 110 g Molar Mass 44.09 32.00 44.01 18.02 g/mol Present 2.50 mol Rxn 2.50 5*2.50 mol Mass? g Rxn 5*2.50*32.00=?. g
Week 3/Th: Limiting Reagents The NO 2 needed for the Demonstration was produced by the reaction of nitric acid with copper. The reaction ended when the nitric acid was used up, there is copper left in the bottom of the flask. Nitric Acid was the limiting reagent, copper was in excess Cu + 4 HNO 3 à Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O The nitric acid was not pure but in aqueous solution, something that we will have to address next week. DJMorrissey, 2oo9