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AP Chemistry Day 4 Unit 1 Atomic Structure Topic 1.3 Elemental Composition Cont

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AP Chemistry Day 4 Unit 1 Atomic Structure Topic 1.3 Elemental Composition Cont AP Chemistry Day 4 Unit 1 Atomic Structure Topic 1.3 Elemental Composition cont. Topic 1.4 (Briefly) mixtures 1.5 Atomic Structure intro Agenda - Sept 4, 2019 ● Topic 1.3 Elemental Composition of Pure Substances ○ Combustion Analysis Pogil ● Topic 1.4 Composition of Mixtures ● Topic 1.5 Atomic Structure and electron configuration ● Topic 1.6 Photoelectron spectroscopy ● Remember 1st ions quiz 10th Sept Topic 1.3 Elemental Composition of Pure Substances Combustion Analysis How can burning a substance help determine the substance’s chemical formula? Why? Scientists have many techniques to help them determine the chemical formula or structure of an unknown compound. One commonly used technique when working with carbon-containing compounds is combustion analysis. Any compound containing carbon and hydrogen will burn. With an ample oxygen supply, the products of the combustion will be carbon dioxide and water. Analyzing the mass of CO2 and H2O that are produced allows chemists to determine the ratios of elements in the compound. Model 1 - Combustion Reactions CH4 + O2 → CO2 + H2O C2H6 + O2 → CO2 + H2O Name the hydrocarbons C2H4 + O2 → CO2 + H2O C3H8 + O2 → CO2 + H2O Model 1 - Combustion Reactions CH4 + O2 → CO2 + H2O methane Draw C2H6 + O2 → CO2 + H2O molecular ethane structures C H + O → CO + H O 2 4 2 2 2 On ethene whiteboard C3H8 + O2 → CO2 + H2O propane Model 1 - Combustion Reactions CH4 + O2 → CO2 + H2O methane 1. What C2H6 + O2 → CO2 + H2O ethane reactant is always C2H4 + O2 → CO2 + H2O ethene required? C3H8 + O2 → CO2 + H2O propane Model 1 - Combustion Reactions CH4 + O2 → CO2 + H2O methane Oxygen is C2H6 + O2 → CO2 + H2O ethane always required for C2H4 + O2 → CO2 + H2O ethene combustion. C3H8 + O2 → CO2 + H2O propane Model 1 - Combustion Reactions CH4 + O2 → CO2 + H2O methane 2. Balance the equations- C2H6 + O2 → CO2 + H2O ethane but keeping coefficient for C2H4 + O2 → CO2 + H2O ethene hydrocarbons 1. C3H8 + O2 → CO2 + H2O propane Model 1 - Combustion Reactions CH4 + 2O2 → CO2 + 2 H2O methane 2. Balance the equations- C2H6 +7/2 O2 → 2 CO2 + 3 H2O ethane but keeping coefficient for C2H4 + 3O2 → 2 CO2 + 2 H2O ethene hydrocarbons 1. C3H8 + 5O2 → 3 CO2 + 4 H2O propane Model 1 - Combustion Reactions CH4 + 2O2 → CO2 + 2 H2O methane 3. How is coefficient of C2H6 +7/2 O2 → 2 CO2 + 3 H2O ethane CO2 related to formula of h/c C2H4 + 3O2 → 2 CO2 + 2 H2O ethene being analyzed? C3H8 + 5O2 → 3 CO2 + 4 H2O propane Model 1 - Combustion Reactions CH4 + 2O2 → CO2 + 2 H2O methane 3. Coefficient of CO is C2H6 +7/2 O2 → 2 CO2 + 3 H2O 2 ethane same as subscript for C2H4 + 3O2 → 2 CO2 + 2 H2O ethene C is h/c formula. C3H8 + 5O2 → 3 CO2 + 4 H2O propane Model 1 - Combustion Reactions CH4 + 2O2 → CO2 + 2 H2O methane 4. How is C2H6 +7/2 O2 → 2 CO2 + 3 H2O coefficient of ethane H2O related to C H + 3O → 2 CO + 2 H O 2 4 2 2 2 formula of ethene h/c? C3H8 + 5O2 → 3 CO2 + 4 H2O propane Model 1 - Combustion Reactions CH + 2O → CO + 2 H O 4 2 2 2 4. Coefficient methane of H2O is ½ C2H6 +7/2 O2 → 2 CO2 + 3 H2O ethane subscript of H in the h/c. C2H4 + 3O2 → 2 CO2 + 2 H2O ethene 4 = ½ (8) 2 = ½ (4) etc. C3H8 + 5O2 → 3 CO2 + 4 H2O propane Read This! Schematic diagram for a combustion analysis experiment. Don’t miss the particles in the 2nd and 3rd chamber! Read This! In a combustion analysis experiment, a hydrocarbon sample is heated in a stream of oxygen gas. As the sample burns, water and carbon dioxide is pushed through a series of chambers with materials that absorb each of the respective products. The chambers are each weighed before and after the combustion to determine the mass of each product. Model 2 - Combustion Analysis of CxHy Unknowns 10.0 Mass of Moles Moles Mass of Moles Moles Sample Total 0g CO2 of CO2 of C H2O of H2O of H s Mass of sam produced atoms produc atoms Empiric C and ple ed al H formula atoms 1 27.42 g 22.46 CH4 g 2 29.46g 17.97 CH3 g Showing work (best practice) 27.42gCO2 x 1moleCO2 = 0.6230 mol CO2 (12.01+2(16.00))g →0.6230mol Catoms Showing work (best practice) 22.46gH2O x 1moleH2O = 1.247 mol H2O (2(1.008)+16)g Mol H atoms in original sample = 2 x mol H2O 2 x 1.247 mol →2.393 mol H atoms Model 2 - Combustion Analysis of CxHy Unknowns 10.0 Mass of Moles Moles Mass of Moles Moles Sample Total 0g CO2 of CO2 of C H2O of H2O of H s Mass of sam produced atoms produc atoms Empiric C and ple ed al H formula atoms 1 27.42 g 0.623 0.623 22.46 1.247 2.493 CH4 0 0 g 2 29.46g 0.665 0.665 17.97 0.997 1.994 CH3 g Model 2 - Combustion Analysis of CxHy Unknowns 30.0 Mass of Moles Moles Mass of Moles Moles Sample Total 0g CO2 of CO2 of C H2O of H2O of H s Mass of sam produced atoms produc atoms Empiric C and ple ed al H formula atoms 1 94.11 g 2.139 2.139 38.53 2.139 4.277 CH2 g 2 89.90g 2.041 2.041 49.03 2.721 5.442 C3H8 g 6. How the data from Model 2 could be used to find the lowest whole number ratio between carbon and hydrogen atoms. Which should give us the empirical formulas of the sample substances. 0.6230mol C : 2.493 mol H 0.6230 0.6230 Gives 1 mol C : 4 mol H CH4 7) Did we need the balanced chemical combustion equations? 8. Did we need to know the mass of the samples to find the empirical formulas of the unknowns? 9. What other information do we need to determine the molecular formula? 7) Did we need the balanced chemical combustion equations? Well we didn’t use the equations directly here, because we had noticed the relationships in the equations we looked at earlier. 8. Did we need to know the mass of the samples to find the empirical formulas of the unknowns? No, we didn’t use the mass of the samples of the unknowns in our calculations. 9. We need to know the molar mass of the compounds if we are to find the molecular formulas. 10. A 15.00g sample of an unknown h/c is analyzed by combustion analysis. 50.70g of carbon dioxide and 10.42 g of water are produced. Showing work (best practice) 50.70gCO2 x 1moleCO2 = 1.1152 mol CO2 44.01g →1.1152 mol Catoms Showing work (best practice) 10.42gH2O x 1moleH2O = 0.5783mol H2O 18.016g 2 x 0.5783 →1.157 mol H atoms 10. A 15.00g sample of an unknown h/c is analyzed by combustion analysis. 50.70g of carbon dioxide and 10.42 g of water are produced. 1.152mol C : 1.157mol H 1.152 1.152 Gives 1 mol C : 1 mol H Empirical Formula is CH 11. Calculate the total mass of carbon atoms and hydrogen atoms for each sample in Model 2. (This could be Good practice, but too much on one day for one person, I think. Use later when studying for tests and need a quick reminder). Showing work (best practice) For sample in Qu.11 1.152 molC x 12.01gC = 13.84 g C 1 mol C Showing work (best practice) 1.157mol H 1.008 g H= 1.166 g H 1molH 1.166g + 13.84g = 15.006g = 15.01g w/i experimental limits same as original sample. Law of conservation of mass. Showing work (best practice) Would the sum of the 1.157mol H 1.008 g H= 1.166 gmass H of the C atoms and H atoms be the 1molH same as the mass of the original sample if the sample included 1.166g + 13.84g = 15.006g = 15.01gother atoms? (Like N, w/i experimental limits same asO original or S, for example).sample. Law of conservation of mass. Showing work (best practice)If other elements were present in the sample, then the total mass of 1.157mol H 1.008 g H= 1.166the g carbonH and 1molH hydrogen atoms would be less than the mass of the sample (since 1.166g + 13.84g = 15.006g =combustion 15.01g products for the other elements w/i experimental limits same aswere original missing.) sample. Law of conservation of mass. See google classroom for 12-15. Extension Questions 16. It is critical in combustion analysis procedures that the sample is dry. What errors in data would occur if the sample itself contained moisture (H2O)? How might this affect the final empirical formula? 16. If the sample contains moisture (water impurities), then that water would evaporated from the sample during the combustion and be absorbed by the water absorbent material. The mass of water found by the analysis would be higher than it would be if the sample had been dry. This would lead to a higher value for moles of H than were really present in the compound and so the Empirical formula would also have too many H atoms in it. 17. What errors in data would occur if the sample contained a carbon-based impurity? How might this affect the final empirical formula? 17.
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