## Wednesday, December 16, 2020

### Finding Empirical Formulas

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to perform calculations to determine the composition of a compound or mixture when given the necessary information and to apply lab data to determine the empirical and molecular formula of a compound.

Finding Empirical Formulas
This does NOT sound fun…

We probably should start off with some definitions and examples, eh?

Empirical Formula: The integer ratio of elements in a compound reduced to lowest multiples.

Examples:

H2but not H4O2

CO but not C2O4 https://en.wikipedia.org/wiki/Empirical_formula

It is important to note that more than one compound can have the same empirical formula. What? This is because it represents the ratio of parts. It does not specify exactly how the parts are combined, as a molecular formula does.

"Glucose (C6H12O6), ribose (C5H10O5), acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O. This is the actual molecular formula for formaldehyde, but acetic acid has double the number of atoms, ribose has five times the number of atoms, and glucose has six times the number of atoms" (Source).

Given any percent composition, it's not terribly hard to find the empirical formula. Really.

How does it work?

In addition to the examples on the linked page above, Google will provide many, many examples. We should add one more! Yeah, let's do that!

EXAMLE:

A compound is made up of 74.20% sodium (Na) and 25.80% oxygen (O). What is its empirical formula. I'm taking a stab at it and guessing this is sodium oxide…

Step 1: Assume there is 100 grams of the substance AND (This means that 74.20% of 100 grams is the mass of the Na… etc.)  convert the percents into masses.

Na = 74.20 grams

O = 25.80 grams.

Step 2: Convert grams to moles of atoms. Moles to grams you multiply, grams to moles divide.

Moles of Na = 74.20 grams • (1 mol / 22.99 g/mol) = 3.2275 moles

Moles of O = 25.80 grams • (1 / 15.99 g/mol) = 1.6135 moles

Now, you have the ratio of moles, albeit rendered in very unwieldy decimals. Not to fear!

Step 3: Divide the very unwieldy decimals by the smallest value to get something… better:

1.6135 moles of O / 1.6135 = 1        <--- THIS was the plan. Get this guy to 1

3.2275 moles Na / 1.6135 = 2      <-- Rounding this off from 2.00030, because… rounding.

Step 4: SOMETIMES you have to multiply the results by 2 in order to get integers.

For instance in C3H6O2, you will end up with:

1.5 moles C

3 moles H

1 mole O

Multiply by 2 to get rid of the 1.5:

3 moles C

6 moles H

2 moles O

The empirical formula is the atomic symbols and the calculated number of moles. For the two examples above in Step 3 and Step 4:

Na2O

C3H6O2

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If you are given the composition in terms of the masses, the work is even easier.

The process is to find out how many moles you have of each thing, then use the smallest of those to convert to an integer ratio.

Let's look at the math in isolation…

Say you have something that is made up of .2 moles of X and .4 moles of Y.

You can probably see that you have twice as many moles of Y as you do X, so the answer would be

XY2

Probably…

So, what's the math on that?

You have, based on the masses, a ratio of

.2 : .4

To make that into an integer ratio, divide everything by the smallest number…

.2  : .4  ÷ .2 = 1 : 2

Now, back to the chemistry…

Find the empirical formula given a 7.315 gram sample containing 0.815 grams of hydrogen and 6.500 grams of oxygen.

Use these atomic masses:

O = 16.0

H = 1.0

Step 1: Find the mole ratio present in the sample.

Step 1a: How many moles of each thing do you have?
Moles to grams you multiply, grams to moles divide!

Divide grams given by atomic masses;

Moles O = 6.50 grams ÷ 16.0 g/mole

Moles O = 0.40625

Moles H = 0.815 grams ÷ 1.0 g/mole

Moles H = 0.815

So, knowing how many moles you have, you now… have the mole ratio!

The ratio of O to H is

0.40625 : 0.815

Step 2: Divide that ratio by the smallest number to get an Integer ratio

0.40625 : 0.815 ÷ 0.40625 = 1 : 2.006

Step 3: The integer ratio (rounded off) becomes the subscripts for the empirical formula:

H2O   <--- the subscript on the O would be 1, but we don't write that.

Last thought… Sometimes, you will have to multiply by 2 (just like before, above) to get a whole number, but that's not a big deal.

___________________________

That's it! Wasn't so bad, afterall!

### Finding Percent Composition From Molecular Formulas

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to perform calculations to determine the composition of a compound or mixture when given the necessary information and to apply lab data to determine the empirical and molecular formula of a compound.

Finding Percent Composition From Molecular Formulas
Well, now… That's a whole lot of words!

We probably should start off with some definitions and examples, eh?

Molecular Formula: The integer ratio of elements in a compound that represents the numbers of each type of atom in the compound. https://www.google.com/search?q=chemistry+molecular+formula

This is not new!

So, what about that percent composition, then? What's that?

Percent Composition: The percent of the mass of a compound made up by the mass of each of the constituent elements. https://www.google.com/search?q=chemistry+percent+composition

Another way to look at it is
to think of it this way… What percent of the mass of XXXX is made up by YYYYY? What percent of water is made up by oxygen? What percent of sucrose is made up by carbon?

We can make it look more like chemistry this way…

Find the percent composition of each element in CuBr2.

So, percent… in this case, you'll be doing

Part ÷ Whole X 100%          <--------  Look at the bottom of this page for more explaining!

Using the masses from a periodic table:

Cu = 63.5

Br = 79.9

we find the molecular mass (the whole) of CuBr2

CuBr2 = 63.6 + (2 X 79.9)

CuBr2 = 223.3

Now, we have everything we need! Recap? Sure?

Cu = 63.5

Br = 79.9

CuBr= 223.3

What percent is the Cu? Let's do that Part ÷ Whole X 100% thing.

Percent Cu = 63.5 / 223.3 X 100%
Percent Cu = 28.44%

What percent is the Br? Let's do that Part ÷ Whole X 100% thing. BUT!!

There are TWO Br in the molecule. So, the mass of the Br is not 79.9, but rather 159.8.

Percent Br = 159.8 / 223.3 X 100%
Percent Br = 71.56%

Let's try another, but follow some steps… Steps are good, right?

Step 1: Find all the masses of the elements.

Step 2: Find the mass of the compound.

Step 3: Find the "part" masses of the compound.

Step 4: Do that Part ÷ Whole X 100% thing

Find the percent composition of each element in C2H4O2.

Step 1:

C = 12.01

H = 1.01

O = 15.99

Step 2:

C2H4O2 = 60.04

Step 3

C2 = 24.02

H4 = 4.04

O2 = 31.98

Step 4:

Percent C = 24.02 / 60.04 X 100%
Percent C = 40.01 %

Percent H = 4.04 / 60.04 X 100%
Percent H = 6.73%

Percent O = 31.98/60.04 X 100%
Percent O = 53.26%

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What is this Part ÷ Whole X 100% thing?

Math… If you want to know what percent of something you have, you… do that.

A classroom has 10 red desks and 30 blue desks. What percent of the desks are red?

The "part" you are looking at is the red desks. The "whole" is ALL of the desks.

Percent red desks =  Part ÷ Whole X 100%

Percent red desks = 10 ÷ 40 X 100%

Percent red desks = .25 X 100%

Percent red desks = 25%

It works like this for all "what percent of" questions. Math…