Discussion on health consequences of air particulates

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Postby dilerani » Mon Jun 04, 2012 12:52 am

How to calculate the empirical formula of nicotine? I need to know how to calculate this, I already know the answer but I can't figure out the process by which to work it out. A sample of nicotine which contains only carbon, nitrogen, and hydrogen is burned in excess oxygen. A 0.2340 g sample yielded 0.1286 g of H20 and .06329 g of CO2. What is the empirical formula of Nicotine?
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Nicotine Empirical Formula

Postby Wilberforce » Tue Jun 05, 2012 12:36 pm

This is off-topic, and I don't have time to do free online tutoring, but I'll do it just this once, since I need the practice.

Oops...your numbers are off. The H2O has transposed 2 and 8. The CO2 decimal point is in the wrong place. The correct numbers are given in my example, obtained by reverse-engineering the original molecular formula of nicotine, C10-H14-N2

A) Calculate the moles and grams of hydrogen and carbon, from the product masses

B) Add the carbon and hydrogen masses (from step A) plus unknown Nitrogen mass (X),
then set equal to nicotine reactant mass. This gives nitrogen grams and molar mass

C) Divide all three elements H,C,N, by the lowest molar mass (nitrogen)
These are the molar ratios and thus the empirical formula.

• The Surgeon General has determined that there is no safe level of exposure to ambient smoke!

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