2) The stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogengas were formed. If caffeine is made of carbon, hydrogen, nitrogen, and oxygen, determine theempirical and molecular formulas of caffeine, and write the balanced equation for thecombustion of caffeine.Balanced Combustion Equation

The molar mass of caffeine, CO2, H2O, and N2 is 194 g/mol, 44 g/mol, 18 g/mol, and 28 g/mol…

Answered: 2) The stimulant in coffee and tea is…

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter3: Mass Relations In Chemistry; Stoichiometry

Section: Chapter Questions

Problem 38QAP: Determine the simplest formulas of the following compounds: (a) tetraethyl lead, the banned gasoline...

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2) The stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376
g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen
gas were formed. If caffeine is made of carbon, hydrogen, nitrogen, and oxygen, determine the
empirical and molecular formulas of caffeine, and write the balanced equation for the
combustion of caffeine.
Balanced Combustion Equation

Expert Solution

Step 1

The molar mass of caffeine, CO₂, H₂O, and N₂ is 194 g/mol, 44 g/mol, 18 g/mol, and 28 g/mol respectively.

The number of mole of each is calculated as follows:

$n_{caffeine} = \frac{0.376 g}{194 g / mol} = 0.002 mol n_{{CO}_{2}} = \frac{0.682 g}{44 g / mol} = 0.0155 mol n_{H_{2} O} = \frac{0.174 g}{18 g / mol} = 0.01 mol n_{N_{2}} = \frac{0.110 g}{28 g / mol} = 0.004 mol$

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