Week 3/Th: Lecture Units 6 & 7

Transcription

1 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 08:00AM

2 Week 3/Th: Elemental Composition -5-5) g of sulfur reacts with F 2 to produce g of a substance SF x. What is X? Let s write a word equation for now. Sulfur + F 2 à SF X Mass g? g Molar Mass g/mol g/mol? g/mol = g / 32.07g/mol = g / g/mol = mol = mol Mole Ratio = / =1 = / = of F 2 X is 6 ( ) = g Conservation of Mass

3 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 g Molar Mass g/mol moles Mole Ratio Empirical Formula: C 2 H 5 O, Formula Mass = g/mol but Molar Mass ~ 90 g/mol was given so Molecular Formula is: C 4 H 10 O 2

4 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 g/mol Mole Ratio 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 = g/mol but Molar Mass ~ 120 g/mol was given so Molecular Formula is: C 9 H 12

5 Week 3/Th: Composition -8-8) A 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 g Molar Mass?? g/mol Mole Ratio / =4.99 Formula: [ MgCO 3. 5 H 2 0 ]

6 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 g Molar Mass g/mol mol Mole Ratio Interger ratio Emp. Formula: C 5 H 16 N 4, Formula Mass = g/mol

7 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 H 2 O à 4 Al(OH) CH 4 4 NH O 2 à 4 NO + 6 H 2 O C 3 H O 2 à 3 CO H 2 O DJMorrissey, 2oo9

8 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 NO 2 à 4 CO N 2 + S 8 DJMorrissey, 2oo9

9 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 O 2 à 3 CO H 2 O Mass 110 g Molar Mass g/mol Present 2.50 mol Rxn *2.50 mol Mass Rxn 3*2.50*44.01? g = 329 g

10 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 O 2 à 3 CO H 2 O Mass 110 g Molar Mass g/mol Present 2.50 mol Rxn *2.50=10.0? mol Mass Rxn g

11 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 O 2 à 3 CO H 2 O Mass 110 g Molar Mass g/mol Present 2.50 mol Rxn *2.50 mol Mass? g Rxn 5*2.50*32.00=?. g

12 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