Q3.05 Carbon, hydrogen, and oxygen are not the only elements that can be charact
ID: 1081709 • Letter: Q
Question
Q3.05 Carbon, hydrogen, and oxygen are not the only elements that can be characterized by combustion analysis. If a compound also contains sulfur or nitrogen, then it will form CO2 (from the carbon), H2 0 (from the hydrogen), N2 (from any nitrogen), and SO2 (from any sulfur). The amount of oxygen in the original sample is determined from subtracting the masses of the other elements from the total, as in the combustion analysis described earlier. If a 0.500 g sample of a compound yields 0.814 gCO2,0.204 gH20,0.0288 g N2, subscript. and 0.132 g SO2 when burned, then what is its empirical formula? Give your answer in the form C8H#NaO#5#, where the number is the Word AnswerExplanation / Answer
Sample weight = 0.500 grams
weight of CO2 obtained due to combustion = 0.814 grams
44 grams CO2 contains 12 grams C
0.814 grams CO2 contains ---? grams C = (0.814/44)x12 = 0.222 grams
weight of C in the sample = 0.222 grams
weight of H2O obtained due to combustion = 0.204 grams
18 grams H2O contains 2 grams H
0.814 grams H2O contains ---? grams H = (0.204/18)x2 = 0.0226 grams
weight of H in the sample = 0.0227 grams
weight of N in the sample = 0.0288 grams
weight of SO2 obtained due to combustion = 0.132 grams
64 grams SO2 contains 32 grams S
0.132 grams SO2 contains ---? grams S = (0.132/64)x32 = 0.066 grams
weight of S in the sample = 0.066 grams
weight of O in the sample = 0.500 -0.222-0.0226-0.0288-0.066 = 0.1604
C H N S O
Weights in the sample: 0.222 0.0226 0.0288 0.066 0.1604
Weight/atomic weight : 0.222/12 0.0226/1 0.0288/14 0.066/32 0.1605/16
0.0185 0.0226 0.002 0.002 0.01
Simple ratio: 9 : 11 : 1 : 1 : 5
Empirical formula = C9H11NSO5
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.