Question #e42f1

Feb 14, 2018

$\text{CH"_2"O}$

Explanation:

Here is a little rhyme to help you remember the steps:

Percent to mass
Mass to mole,
Divide by small,
Multiply 'til whole

The source of this quote is:
A Simple Rhyme for a Simple Formula
by Joel S. Thompson
Journal of Chemical Education
Vol. 65, No. 8; August 1988, p. 704

Now, what does it mean?

Percent to mass: Take the percents you are given and convert them to grams. If you assume you have 100-gram samples, all you need to do for this step is change the percent sign to g (grams.)

Mass to mole: Write a conversion factor to convert from grams to moles. It will have 1 mol on the top and the molar mass of the element on the bottom. Do the math. (The number of grams you have times the 1 mol divided by the molar mass of the element. Grams cancel out and you have mol.)

Divide by small: Look at the answers you got from that calculation. take the smallest number of moles and divide ALL of the answers by that one.

Multiply til whole: If the answers are all whole numbers, you are done. If they are not, multiply ALL of the answers by the SAME number until they all are whole numbers. If the decimals on the numbers are around .25 or .75, try multiplying everything by 4. If the decimals are around .33 or.66, try multiplying everything by 3. If the decimals are around .50, multiply by 2. The resulting whole numbers are the number of atoms of each respective element in the empirical formula.

Example:

$\text{%" -> "mass"," mass" -> "mole"," divide by small"," multiply 'til whole}$

$\text{C: " 40% = "40 g" xx ("1 mol") /("12.01 g") = ("3.33 mol C")/("3.33 mol") = "1 C}$

$\text{H: " 6.7% = "6.7 g" xx ("1 mol") /("1.01 g") = ("6.63 mol H")/("3.33 mol")= "2 H}$

$\text{O " :53.3% = "53.3 g" xx ("1 mol") /("16.00 g") = ("3.33 mol O")/("3.33 mol") = "1 O}$

Since you now have all whole numbers, no multiplying is needed. the empirical formula has one carbon, two hydrogens, and one oxygen.

Here is a great resource for more information: https://www.chemteam.info/Mole/EmpiricalFormula.html