**How to find Empirical Formula from a
Molecular Formula**

The empirical formula is the simplest __whole numbers__
non-reducible ratio formula for a molecular formula or compound.
The empirical formula of the molecular formula C_{4}H_{6} is
**C _{2}H_{3} **(divided by 2) because it
can't be divided by other numbers than

**Example** - **How to find Empirical Formula
From Element Mass**

We can determine the empirical formula (simplest formula) by using mass of each element in the compound data. A compound is composed of 5.045g of carbon, 0.847g of hydrogen, and 3.36g of oxygen. Find the empirical formula for this compound knowing that H = 1 g/mole , O = 16 g/mole and C = 12 g/mole

First, convert the grams of each element to moles:

5.045/12 = 0.42 mole C , 0.847/1 = 0.847 mole H , 3.36/16 = 0.21 mole O

Divide each of the three mole figures by the lowest of the three in order to simplify the mole ratio.

0.21/0.21 = 1 mole O , 0.42/0.21 = 2 mole C , 0.847/0.21 = 4 mole H

Well, that's **1 **O , **2 **C and **4**
H

The resulting SIMPLEST WHOLE NUMBER empirical formula : **C _{2}H_{4}O**

**NOTE**: In this case all mole figures represent whole
numbers, so we can say that this is the SIMPLEST WHOLE NUMBER
formula. If one of the numbers was (say) 1.5, we have to multiply
all the mole figures by 2.

**Finding the Empirical Formula** **using
CHEMIX** **School**

A compound is composed of 5.045g of carbon, 0.847g of hydrogen, and 3.36g of oxygen. Find the empirical formula

Step 1) Determine moles: Insert **5.045gC** (0.42
moles) , **0.847gH** (0.84 moles) ,
**3.36gO ** (0,21 moles)

Step 2) Divide all moles figures by **0.21**: 0.42/0.21
= **2**C 0.84/0.21 = **4**H
0.21/0.21 = **1**O

The resulting SIMPLEST WHOLE NUMBER empirical formula : **C _{2}H_{4}O**