Empirical Formula Calculator

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Empirical formula gives the simplest ratio of the elements present in the given compound.**Empirical formula Calculator** Calculates the empirical formula for the given compound if the percentage of composition of each element in that compound is given. **Input should be given in the form of these following examples :**

40% carbon, 6.67% hydrogen and 53.3% oxygen

92.3% Carbon, 7.69% Hydrogen

45% Carbon, 55% Oxygen.

Below is a given default composition of a compound, click "Calculate Empirical Formula" to find the respective empirical formula with its mass composition.

40% carbon, 6.67% hydrogen and 53.3% oxygen

92.3% Carbon, 7.69% Hydrogen

45% Carbon, 55% Oxygen.

Below is a given default composition of a compound, click "Calculate Empirical Formula" to find the respective empirical formula with its mass composition.

Read the problem and note down the Composition of each element and check whether the sum of the composition of all the elements are 100%. If yes go ahead with the problem.

Assume 100% percent as 100 grams and Convert all the given percentage composition into grams. Then Convert the grams into moles by dividing it by atomic mass.

Check out which element is having least value in moles. Then divide every element by that element to get the empirical value. Then with the help of empirical values we get the empirical formula.

A Compund was found to contain the following percentage composition 40% carbon, 6.67% hydrogen and 53.3% oxygen.

Calculate its empirical formula (C=12, H=1, O=16).

**Step 1 :**Given: The Compound is Hydrocarbon and its composition is

C = 40 %

H = 6.67 %

O = 53.3 %

**Step 2 :**The total sum is 100%. Let us assume it as 100 grams.

$\therefore$ Carbon content is 40 grams, Hydrogen is 6.67 grams, Oxygen is 53.3 grams.

To Convert the grams into Moles, divide the element with its atomic mass.

C = $\frac{40}{12.0107}$ = 3.33

H = $\frac{6.67}{1.00794}$ = 6.623

O = $\frac{53.3}{16}$ = 3.33.

**Step 3 :**Here Oxygen and Carbon has least value. So lets divide every element by their value.

The Empirical value for

Carbon = $\frac{3.33}{3.33}$ = 1

Hydrogen = $\frac{6.623}{3.33}$ = 1.989

Oxygen = $\frac{3.33}{3.33}$ = 1.

**Answer :**$\therefore$ The Empirical formula is CH

_{2}O and ratio is Carbon = 1, Hydrogen = 1.989 and Oxygen = 1.The Carbon dioxide compound is found to contain the following percentage composition.

C = 45 %, O = 55 %.Calculate its empirical formula (C=12, O=16).

**Step 1 :**Given: The Compound is Carbon dioxide and its composition is

C = 45 %

O = 55%.

**Step 2 :**The total sum is 100%. Let us assume it as 100 grams.

$\therefore$ Carbon content is 45 grams and Oxygen is 55 grams.

To Convert the grams into Moles, divide the element with its atomic mass.

C = $\frac{45}{12.0107}$ = 3.7468

O = $\frac{55}{16}$ = 3.4375.

**Step 3 :**Here Oxygen has least value. So lets divide every element by the value of oxygen.

The Empirical value for

Carbon = $\frac{3.7468}{3.4375}$ = 1.09

Oxygen = $\frac{3.4375}{3.4375}$ = 1.

**Answer :**$\therefore$ The Empirical ratio value for Carbon is 1.09 and oxygen is 1 and Empirical formula is CO.