Ratios and Proportions

HOME

Course Chapters

Calculator Fundamentals

Mathematics Review

Section Tests

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

Contact Webmaster

Problem 3 Solution


(a) 313 carbon atoms and 1252 hydrogen atoms
(b) Not if all of the atoms are involved in methane molecules.

Solution Steps for Part (a):

If a sample of methane contains 1565 atoms, how much carbon and hydrogen are present?

Answer:

This is different from the last problem because we are given the total number of atoms instead of the amount present of one of them. What do we know?

carbon + hydrogen = 1565 and

Let's cross multiply this proportion as usual to get:

4 carbon = hydrogen.

So, we can substitute 4 carbons in for the hydrogen in the first equation:

carbon + 4 carbon = 1565

Put the carbons together: (4 and 1 makes 5)

5 carbon = 1565 and divide both sides by 5 carbon = 313.

So what's left must be hydrogen: 1565 - 313 = 1252.

Solution Steps for Part (b):

Can a sample contain 1566 atoms?

Answer:

Look at the work from part (a): We know that the total number of atoms must be divisible by 5.

Think about what a ratio tells you when you can't have fractional parts of the items: The items must come in groups of 5 -- 1 carbon for every 4 hydrogens in this case.

Using this idea, there can't be 1566 atoms because 1566 is not divisible by 5 because 5 carbon = 1565(total). Try it on your calculator; do you get a whole number? No!

Try another problem like this one.

Next Try It Out Problem.

Developed by

The Shodor Education Foundation, Inc.
in cooperation with the Department of Chemistry,
Appalachian State University