We'll do some empirical formula calculations and problems in just a second, but first let's get some definitions out of the way.

**Empirical Formula**

__Empirical Formula__ = the simplest whole-number ratio of atoms in a compound.

**Molecular Formula**

__Molecular Formula__ = the exact formula of a compound.

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Here's how the two formulas are related:

* Molecular Formula = n ( Empirical Formula)* -- where n = whole-number integer

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**Empirical Formula Calculations**

Let's take a look at a few sample problems...

As always, we learn best by watching example calculations and using them to help us with __our own__ problems and calculations.

*ex:* Suppose a compound is 71.65% chlorine, 24.27% carbon, and 4.07% hydrogen. It's molar mass is 98.96 g/mol. Find the empirical and molecular formulas.

_________

**answer:**➞ always find the empirical formula first. Always!

➞ even if you're only asked to find the molecular formula.

**Step 1.** Assume you have 100g of material and convert to moles.

**Step 2.** Divide by the lowest number of moles:

(71.65g Cl / 1) _{*} (1 mol Cl / 35.45g Cl) = 2.021 mol Cl / 2.021 = **1 mol Cl**

(24.27g C / 1) _{*} (1 mol C / 12.01g C) = 2.021 mol C / 2.021 = **1 mol C**

(4.07 g H / 1) _{*} (1mol H / 1.01g H) = 4.04 mol H / 2.021 = ** 2 mol H**

Here that is again, in a more relaxing easier-to-follow handwritten format:

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**Step 3.** Combine the moles of each atom into an empirical formula:

**CH _{2}Cl** =

__empirical formula__. Carbon (C) is always first if all elements are nonmetals.

Since we were asked to also find the __molecular formula__, two additional steps are needed...

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**Step 4.** Calculate the __empirical formula mass__.

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**Step 5.** Plug the empirical mass and the molar mass into the boxed-up formula from earlier in this post:

* Molecular Formula = n ( Empirical Formula)* -- where n = whole-number integer

98.96 g/mol = n ( 49.48g/mol)

n = 2 ➞ 2 [ CH_{2}Cl ] = **C _{2}H_{4}Cl_{2}** ➞ this is the

__molecular formula__.

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*ex:* A compound containing only nitrogen and oxygen is 30.4% nitrogen by mass. If the molar mass of the compound is 92 g/mol, what's the molecular formula?

_________

**answer:** ➞ always find the empirical formula first. Always!

➞ even if you're only asked to find the molecular formula.

**Step 1.** Assume 100g, so we have 30.4g N and 69.6g O. Convert to moles.

**Step 2.** Divide by the lowest number of moles.

**Step 3.** Combine the moles of each atom into an empirical formula:

(30.4g N / 1) _{*} (1 mol N / 14.01g N) = 2.17 mol N / 2.17 = **1 mol N**

(69.6g O / 1) _{*} (1 mol O / 16.00g O) = 4.35 mol O / 2.17 = **2 mol O**

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**Step 4.** Calculate the __empirical formula mass__.

**Step 5.** Plug the empirical mass and the molar mass into the boxed-up formula from earlier in this post:

92 g/mol = n ( 46.01g/mol)

n = 2 ➞ 2 [ NO_{2} ] = **N _{2}O_{4}** ➞ this is the

__molecular formula__.

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**QUESTION: ** What would you do if after *STEP 2* above, you had * 1.33 mol N* and

*?*

**2 mol O**_________

**answer:**Well... 1.33mol N = 1 1/3 mol N = 4/3 mol N

So we have:

4/3 mol N (x 3) = **4 mol N**

2 mol O (x 3) = **6 mol O**

➞ ** N _{4}O_{6} ** ➞ our "new"

__empirical formula__.

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Now, besides the "1.33 mol N" issue that can occur, also watch out for "1.67 moles" and "1.5 moles" situations. See here:

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That's it for my 3rd post covering **SECTION 3 - Chemical Quantities and Stoichiometry**.

The next topics in this Section ("Chapter") include: