HOME

Course Chapters

Calculator Fundamentals

Mathematics Review

Basic Concepts

Advanced Concepts


Section Tests

Pre-test

Post-test


Useful Materials

Glossary


Online Calculators

Redox Calculator

Kinetics Arrhenius Calculator

Thermodynamics Calculator

Nuclear Decay Calculator

Linear Least Squares Regression

Newton's Method Equation Solver

Compressibility Calculator

Units Conversion Calculator

Nomenclature Calculator


Related Information Links

Texas Instruments Calculators

Casio Calculators

Sharp Calculators

Hewlett Packard Calculators


Credits

Credits

Contact Webmaster


Problem 4 Solution

A sample thought  to be caffeine is tested and the resulting composition
was 53% carbon, 4% hydroge, 30% nitrogen and 13% oxygen.  We know from
theory that one mole of caffeine contains 96.08 g carbon, 10.08 g hydrogen,
56.04 g nitrogen and 32.00 g oxygen.  Was the sample really caffeine?
No, it was not caffeine.

Solution Steps:

We need to compare the percent composition of the unknown sample to the known amounts in caffeine. We can't compare amounts with percents so we need to change the amounts into percents.

First we'll find the total number of grams in one mole of caffeine:

96.08 + 10.08 + 56.04 + 32.00 = 194.20

Now we can figure the percent of each element in caffeine:

carbon: 96.08 / 194.20 = .494747683 = 49.47%

hydrogen: 10.08 / 194.20 = .051905252 = 5.19%

nitrogen: 56.04 / 194.20 = .288568486 = 28.86%

oxygen: 32.00 / 194.20 = .164778579 = 16.48%

Are the percents from the sample close to the percents from caffeine? Not really. How far off are they? Here's a good place to check % error! Let's look at the % error in the oxygen figure:

That's off by more than 20%! Not likely!


Another Problem Like This One.


Developed by
Shodor
in cooperation with the Department of Chemistry,
The University of North Carolina at Chapel Hill

Copyright © 1996-2008 Shodor
Please direct questions and comments about this page to
WebMaster@shodor.org