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# Empirical Formula Calculator

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Empirical formula is a chemical formula which describes the whole number ratio of each element in the compound. If you plug in the composition of each elemnt into the empirical formula calculator will determine the empirical formula of the given compound. The composition of elements can be in percentage also.

## Steps

Step 1 : Check the given data from the problem. If the masses of elements is in grams then go to step 2. If the composition is given in percentage then the mass of element is equal to the percent.

Step 2 : Using the molar mass, calculate the moles of each element.

Step 3 : Divide each mole of elements by the smallest moles value calculated in step 2.

Step 4 : If the number is not in whole number, round off into the nearest whole number.

## Problems

Here are some solved problems of empirical formula below.

### Solved Examples

Question 1: What is the empirical formula of a compound that contains 16.7 g of Iridium and 10.3 g of Selenium?
Solution:

Given :
Mass of Iridium = 16.7 g
Mass of Selenium = 10.3 g

Moles of Iridium = $\frac{16.7 g }{192.217 g/mole}$ = 0.086881 mol

Moles of Selenium = $\frac{10.3 g }{78.96 g/mole}$ = 0.130446 mol

Dividing each mole by the smallest number of moles,

Iridium : $\frac{0.086881}{0.086881}$ = 1

Selenium : $\frac{0.130446}{0.086881}$ = 1.5

Multiplying 1 : 1.5 ratio by 2,

2 : 3 $\Rightarrow$ $Ir_{2}Se_{3}$

Empirical formula = $Ir_{2}Se_{3}$

Question 2: What is the empirical formula of a compound that contains 64.80% Carbon, 13.62% Hydrogen and 21.58% Oxygen?
Solution:

As mass is given in 100% , so
Mass of Carbon = 64.80 g
Mass of Hydrogen = 13.62 g
Mass of Oxygen = 21.58 g

Moles of Carbon = $\frac{64.80 g }{12 g/mole}$ = 5.4 mol

Moles of Hydrogen = $\frac{13.62 g }{1 g/mole}$ = 13.62 mol

Moles of Oxygen = $\frac{21.58 g }{16 g/mole}$ = 1.349 mol

Dividing each mole by the smallest number of moles,

Carbon : $\frac{5.4}{1.349}$ = 4

Hydrogen : $\frac{13.62}{1.349}$ = 10

Oxygen : $\frac{1.349}{1.349}$ = 1

Empirical formula = $C_{4}H_{10}O$