4) Nicotine, the stimulant in tobacco, causes a very complex set of physiological effects in the body. It is composed of carbon, hydrogen, and nitrogen, and has a molar mass of 162 g/mol. When a 0.395 g sample of nicotine was burned, 1.072 g of carbon dioxide, 0.307 g of water, and 0.068 g of nitrogen gas were produced. What are the empirical and molecular formulas of nicotine? Write the balanced equation for the combustion of nicotine.
4) Nicotine, the stimulant in tobacco, causes a very complex set of physiological effects in the body. It is composed of carbon, hydrogen, and nitrogen, and has a molar mass of 162 g/mol. When a 0.395 g sample of nicotine was burned, 1.072 g of carbon dioxide, 0.307 g of water, and 0.068 g of nitrogen gas were produced. What are the empirical and molecular formulas of nicotine? Write the balanced equation for the combustion of nicotine.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:4) Nicotine, the stimulant in tobacco, causes a very complex set of physiological effects in the
body. It is composed of carbon, hydrogen, and nitrogen, and has a molar mass of 162 g/mol.
When a 0.395 g sample of nicotine was burned, 1.072 g of carbon dioxide, 0.307 g of water,
and 0.068 g of nitrogen gas were produced. What are the empirical and molecular formulas of
nicotine? Write the balanced equation for the combustion of nicotine.
Balanced Combustion Equation
Expert Solution

Step 1
From the amount of CO2, H2O and N2 formed, we find out the amount of C, N and H present in the given nicotine sample.
Empirical formula - C5H7N
Molecular formula - C10N2H14
Balanced combustion equation -
2 C10H14N2 + 17 O2 = 20 CO2
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