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# 1.3 Elemental Composition of Pure Substances

jeremy kiggundu

dalia savy

dylan black

(editor)

⏱️ August 5, 2020

📅

## Pure Substances

Pure substances are generally defined by the three common states of matter:

### 🧊Solids

Solids are composed of densely packed atoms which have slight movements known as vibrations. Solids have a fixed volume and fixed shape.

### 💧Liquids

Atoms in liquids flow to assume the shape of the container that they are occupying. Liquids have a fixed volume, but no fixed shape.

### ♨️Gases

Atoms in gases flow freely and expand to fill their container. Gases have no fixed volume or shape.

### Formula Units

Here’s where things can get a little tough, but yay more calculations!

You can look at individual particles in a compound, but you could also look at formula units. A formula unit is the smallest collection of atoms from which the formula of the compound can be established. Understanding formula units is useful as it can be used as another means of conversion.

Example Problem

How many formula units of KCl would be needed if 1.76mol KCl were used in a chemical reaction?

## Law of Definite Proportions

The law of definite proportions states that the ratio of mass of the constituent elements in any pure sample of that compound is always the same. Constituent elements are the elements that the compound is made up of. For example, no matter what sample of water you obtain, you will always have the same ratio of hydrogen to oxygen. Whether it be a 100 g sample, a 200 g sample, or a 5000 g sample, the ratio will always reduce to be the same. The empirical formula uses this concept.

### Empirical Formula

The empirical formula of a compound is the simplest whole number combination of elements in a compound. In other words, it is a way to simplify the chemical formula of a compound.

Image courtesy of scienceabc

No matter how absurd the numbers in the molecular formula are, like C90H180O90, the ratio will always lead to the empirical formula of CH2O.

Empirical Formula - Practice Question

Sometimes, you could be asked to find the empirical formula from a percent composition. Let's try this:

(1) A carbohydrate, which contains C, H, and O, has a % composition of 33.3% C and 7.4% H. Find the empirical formula of this carbohydrate.

My teacher always had this poem for us that lined out the steps:

Percent to grams, Grams to Moles, Divide by Small, Times to whole

Step 1 and 4 are optional but in this problem, we are going to have to do both. Step 1 is only necessary if the composition is given in percents. If it is already given in grams, you don't have to do it.

Step 1) Drop the % sign. 33.3% C --> 33.3g C and 7.4% H --> 7.4 g H.

But what about oxygen? Let's use the 100% concept: all percents have to add up to 100. Therefore, we can simply subtract 100%-33.3%-7.4% to obtain the percent composition that is oxygen. This would get us 59.3% or 59.3 g O.

Step 2) Now we have 33.3g C, 7.4g H, and 59.3g O and we have to convert all of these to moles. This is done simply by dividing each of these gram amounts by the element's molar mass.

C: 33.3g / 12g = 2.775 moles C

H: 7.4g / 1g = 7.4 moles H

O: 59.3g / 16g = 3.70625 moles O

Step 3) Now, we have to divide each mole amount by the smallest mole amount that we have. 2.775 is the smallest amount so...

C: 2.775 / 2.775 = 1

H: 7.4 / 2.775 = 2.66

O: 3.70625 / 2.775 = 1.33

Step 4) Since the ratios for hydrogen and oxygen aren't full numbers, we have to "times the numbers to whole numbers." Whenever the numbers end in 0.5, 0.33/0.66, or 0.25/0.75, multiply all by 2, 3, and 4, respectively. Here, since the values end in .33 or .66, we have to multiply all of them by 3 to get whole numbers.

C: 1 x 3 = 3 C

H: 2.66 x 3 = 8 H

O: 1.33 x 3 = 4 O

Therefore, the empirical formula for this carbohydrate is C3H8O4. This question didn't ask for the molecular formula but an example would be C6H16O8. There are millions of possibilities for a molecular formula.