Activity Combustion Analysis

Transcription

1 Activity Combustion Analysis Directions: This Guided Learning Activity (GLA) focuses on completing determining the chemical formula for organic compounds using combustion analysis. Part A reviews how to determine the empirical formula from the mass of each component, Part B discusses combustion analysis of compounds that contain only hydrogen and carbon, and Part C discusses using combustion analysis to find the empirical formula for a compound with carbon, hydrogen and oxygen. The worksheet is accompanied by an in-depth key. See for additional materials. Part A Finding the Molar Amounts of Carbon and Hydrogen To find the empirical formula for a compound, first the compound must be separated into its elements. One way to separate carbon from hydrogen in a hydrocarbon sample is by using combustion analysis. This procedure relies on the combustion of the sample: CC XX HH YY + OO 2 CCOO 2 + HH 2 OO (nnnnnn bbbbbbbbbbbbbbbb) In this reaction all of the carbon from the compound is converted into CO 2, and all of the hydrogen is converted into H 2O. If these two products are collected, we can measure the mass of CO 2 and H 2O to calculate the amount of carbon and hydrogen in the original sample. Figure 1. Diagram representation of a combustion analyzer, showing how the CO2 and H2O are collected. (Source: Chemistry, published by Openstax College, 2016) For example, consider the combustion of a hydrocarbon sample that yields 3.05 g of CO 2 and 0.84 g of H 2O. Based on these measurements, determine the molar amount of carbon and hydrogen in the original sample. In combustion analysis, the mass of CO 2 comes from the carbon of the sample and oxygen from the combustion. To separate these, we will use the relationship that for every 1 CO 2 molecule there is 1 carbon atom. Activity Page 1 of 8

2 gg CCOO 22 = CC We also note that in every one H 2O molecule, there are two hydrogen atoms gg HH 22 OO = mmmmmm HH Example # mg of ethylbenzene is burned to yield 663 mg of CO 2 and 167 mg of H 2O. How many moles of carbon and hydrogen were present in the original sample? 11 gg CCCC mmmm CCOO mmmmmm CCCC mmmmmm CC = CC mmmm CCOO gg HH 22 OO mmmm HH 22 OO = HH mmmm HH 22 OO Part B Determining the Empirical Formula for Compounds with only Hydrogen and Carbon We can use the moles of carbon and hydrogen calculated above to determine the empirical formula for the compound. Recall that the empirical formula for a compound represents the constituent elements in the compound, and the molar ratio between those elements. To find the molar ratio, we simply divide all molar amounts by the lowest molar amount we calculated. In part A, we found that a hydrocarbon sample that yielded 3.05 g of CO 2 and g of H 2O contained mole of carbon and mole of hydrogen in the original sample. To find the empirical formula: Divide by smaller number of moles. HH = ffffff HH mmmmmm CC Write formula and multiply to achieve whole numbers. Based on our calculations, the empirical formula for the compound is C 1.00H Because we cannot have 0.35 hydrogen atoms, we will multiply the entire formula by 3. (CC HH ) 33 = CC HH CC 33 HH 44 The empirical formula for the compound is C 3H 4. This maintains the ~1.35:1.0 ratio, while including only whole number of atoms. Activity Page 2 of 8

3 Example #2. In example #1 we calculated the moles of carbon and hydrogen in a mg sample of ethylbenzene. Now, calculate the empirical formula for this compound. HH = ffffff HH CC Based on our calculations, the empirical formula for the compound is C 1.00H Because we cannot have 1.23 hydrogen atoms, we will multiply the entire formula by 4. (CC HH ) 44 = CC 44 HH 55 Part C Combustion Analysis of Compounds with Hydrogen, Carbon and Oxygen Many organic compounds consist of carbon, hydrogen and oxygen. When these samples are burned the only two products are CO 2 and H 2O, so combustion analysis can also be used to find the empirical formula for C xh yo z compounds. Aspirin is an example of a substance that consists of carbon, hydrogen and oxygen. When a 250. mg sample of aspirin is combusted, 550. mg of CO 2 and 100. g of H 2O are produced. As before, we can use the mass of CO 2 to calculate the moles of C in the original sample, and the mass of H 2O to calculate the moles of H in the original sample. Step 1. Convert the mass of each element to its molar amount. 11 gg CCCC mmmm CCOO mmmmmm CCCC mmmmmm CC = CC mmmm CCOO gg HH 22 OO mmmm HH 22 OO = HH mmmm HH 22 OO Because the oxygen in the products comes from both the sample and the additional O 2 gas, we need to use another method to find the moles of oxygen. For this, we rely on the conservation of mass. We know: mm ssssssssssss = mm cccccccccccc + mm hhhhhhhhhhhhhhhh + mm oooooooooooo The mass of carbon and the mass of hydrogen in the original sample can be found from the molar amount. To convert each element into its molar amount, simply multiply by the atomic mass gg CC CC = gg CC 11 mmmmmm CC gg HH HH = gg HH 11 mmmmmm HH To find the mass of oxygen in the original sample, we subtract the mass of carbon and hydrogen from the total mass. mm oooooooooooo = gg gg gg = gg OO Activity Page 3 of 8

4 To find the empirical formula, we need to know the moles of oxygen. The original sample contained g of oxygen, which can be converted to moles using the molar mass gg OO = OO gg OO Since we now know the molar amounts of carbon, hydrogen and oxygen, we can use these to find the mole ratios between components and the empirical formula. The sample of aspirin contains mol C, mol H and mol O. What is the empirical formula for aspirin? Divide by smaller number of moles. CC = ffffff CC HH = ffffff HH OO Write formula and multiply to achieve whole numbers. Based on our calculations, the empirical formula for the compound is C 2.23H 1.98O Because we cannot have 0.23 carbon atoms, we will multiply the entire formula by 4. (CC HH OO ) 44 = CC HH OO CC 99 HH 88 OO 44 The empirical formula for aspirin is C 9H 8O 4. Example #3. Vitamin C is an organic acid containing only carbon, hydrogen and oxygen. A 1.00 g sample of vitamin C is analyzed by combustion. The compound produced 1.50 g of carbon dioxide and 0.41 g of water. What is the empirical formula for vitamin C? gg CCOO gg CC = gg CC gg CCOO gg HH 22 OO gg HH 11. = gg HH 11 mmmmmm HH We need to add the gram of C and H and subtract that mass from the total 1.00 g sample to be able to determine the mass of oxygen g g = g C & H 1.00 g g = g of O We can then convert the mass of oxygen to moles of oxygen and find the empirical formula: gg oooo OO = OO gg OO HH mmmmmm = ffffff HH Activity Page 4 of 8

5 mmmmmm CC mmmmmm OO mmmmmm Based on our calculations, the empirical formula for the compound is C 1.00H 1.33O Because we cannot have 1.33 Hydrogen atoms, we will multiply the entire formula by 3. (CC HH OO ) 33 = CC 33 HH 44 OO 33 Part D Extra Practice Calculate the empirical formula for each of the following compounds. C XH Y Compounds 1. A compound that contains only carbon and hydrogen is collected from a chemical feedstock. A g sample of the compound is subjected to combustion analysis, and produces 1.63 g of carbon dioxide and g of water. What is the empirical formula for the compound? gg CCOO 22 = CC gg HH 22 OO = HH HH = ffffff HH CC Based on our calculations, the empirical formula for the compound is C 1.00H Because we cannot have 1.50 hydrogen atoms, we will multiply the entire formula by 2. (CC HH ) 22 = CC 22 HH A hydrocarbon sample is analyzed by combustion. Combusting 2.00 g of the sample produced 6.28 g of CO 2 and 2.56 g of H 2O. What is the empirical formula for the sample? gg CCOO 22 = mmmmmm CC gg HH 22 OO = mmmmmm HH mmmmmm HH mmoooo = ffffff HH mmmmmm CC mmmmmm Based on our calculations, the empirical formula for the compound is CH 2. Activity Page 5 of 8

6 3. Biphenyl is a stable hydrocarbon that can be used to preserve citrus fruits. A 50.0 mg sample of biphenyl is analyzed by combustion analysis and produces 171 mg of carbon dioxide and 29.2 mg of water. What is the empirical formula for biphenyl? Challenge: The molar mass of biphenyl is g. Use this to calculate the molecular formula for biphenyl. 11 gg mmmm CCOO 22 = CC mmmm 11 gg mmmm HH 22 OO = HH mmmm HH = ffffff HH CC = ffffff CC Based on our calculations, the empirical formula for the compound is C 1.19H Because we cannot have 1.19 carbon atoms, we will multiply the entire formula by 5. (CC HH ) 55 = CC 66 HH 55 Challenge: Our empirical formula mass is g, and the molecular mass formula is g. therefore the ratio of molecular to empirical formula is = To obtain the molecular formula we need to multiply the empirical formula by 2 which gives us CC 1111 HH 1111 the formula of Biphenyl. C XH YO Z Compounds 4. Glycerol is a common food and cosmetic additive that consists of carbon, hydrogen and oxygen. A g sample of glycerol produces g CO 2 and g H 2O upon combustion. What is the empirical formula for glycerol? gg CCOO 22 = CC gg HH 22 OO = HH gg CC CC = gg CC 11 mmmmmm CC gg HH HH = gg HH 11 mmmmmm HH Activity Page 6 of 8

7 We need to add the gram of C and H and subtract that total value from the mass of the original sample (0.100 g) to be able to determine the mass of oxygen g g = g C & H g g = g of O gg OO = mmmmmm OO gg OO HH = ffffff HH CC OO Based on our calculations, the empirical formula for the compound is C 1.00H 2.67O Because we cannot have 2.67 O atoms, we will multiply the entire formula by 3. (CC HH OO ) 33 = CC 33 HH 88 OO A waste gas is collected and analyzed. Combustion of a 1450 μg sample resulted in 2120 μg CO 2 and 867 μg H 2O. What is the empirical formula for the compound? 11 gg CCOO μμμμ CCOO = mmmmmm CC μμμμ CCOO gg HH 22 OO μμμμ HH 22 OO μμμμ HH 22 OO = mmmmmm HH gg CC mmmmmm CC 11 mmmmmm CC = gg CC gg HH mmmmmm HH 11 mmmmmm HH = gg HH We need to add the gram of C and H to be able to determine the amount of oxygen from the 1450 µg sample. Mass of O 11 gg SSSSSSSSSSSS μμμμ oooo OO = gg SSSSSSSSSSSS gg gg CC gg HH = gg OO gg OO gg OO = mmmmmm OO Activity Page 7 of 8

8 mmmmmm HH = ffffff HH mmmmmm CC mmmmmm CC mmmmmm CC mmmmmm OO mmmmmm CC Based on our calculations, the empirical formula for the compound is C 1.00H 2.00O CCHH 22 OO 6. A 15.0 g sample of an organic solvent is analyzed by combustion analysis and produces g of carbon dioxide and 14.0 g of water. What is the empirical formula? Is the solvent acetone or ethanol? gg CCOO 22 = mmmmmm CC gg HH 22 OO = mmmmmm HH gg CC mmmmmm CC = gg CC 11 mmmmmm CC gg HH mmmmmm HH = gg HH 11 mmmmmm HH We need to add the gram of C and H to be able to determine the amount of oxygen from the 15.0 g sample g g = g C & H 15.0 g g = g of O gg oooo OO = mmmmmm OO gg OO mmmmmm HH mmmmmm = ffffff HH mmmmmm CC mmmmmm = ffffff CC mmmmmm OO mmmmmm Based on our calculations, the empirical formula for the compound is C 3.01H 6.05O CC 33 HH 66 OO The solvent is acetone Activity Page 8 of 8