Find Empirical Formula of compound when percentage composition is given

What Will Be Given: Percentage Composition of Each Element in The Compound

What To Calculate: Empirical Formula

Explanation :

The empirical formula of a compound is the minimum number of atoms required for each of its elements to form the bonding.

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Get the mass of each element by assuming a certain overall mass for the sample (use 100 g as mass as it is easy to work with percentages).
Find the moles for each element by using the atomic mass. Convert mass to moles.
Find the element which has the smallest number of moles(T)
Find the ratio or the moles of each element by dividing the number of moles of each by T(smallest number of moles identified in step 3).
Write the empirical formula based on the ratios of each element.

If the composition is missing for one element in the compound. Use 100 – (sum of other element’s composition) to find the missing element composition.
When mole ratio obtained in step 4 contains decimal numbers. Do not round it off to nearest whole number, as it will lead to wrong answers.
When mole ratio obtained in step 4 contains decimal numbers instead of whole numbers, a multiplication factor must be applied to get whole numbers for each element ratio. Refer example-1.
The key to finding out the multiplication factors lies in considering making the decimal digits into the whole number. For example, in 1.334,334 is one-third of 1. Similarly, .5 is 1/2 of 1 and .2 are 1/5 of 1.

A compound contains 48.38% carbon, 8.12% hydrogen, and 53.5% oxygen by mass. Find its empirical formula.

As per Step-1,

Mass of Carbon = 0.4838 x 100 g = 48.38 grams

Mass of Hydrogen = 0.0812 x 100 g = 8.12 grams

Mass of Oxygen = 0.535 x 100g = 53.5 grams

As per Step-2,

One mole of Carbon weighs 12.10 g(Atomic mass of Carbon)

X mole of Carbon weighs 48.38 gms

Moles of Carbon (X) = (48.38 g C) x 1 mol / 12.10 g C = 4.028 mol

One mole of Hydrogen weighs 1.008 g(Atomic mass of Hydrogen)

X mole of Hydrogen weighs 8 grams

Moles of Hydrogen (Y) = (8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol

One mole of Oxygen weighs 16.00 g(Atomic mass of Oxygen)

X mole of Oxygen weighs 53.38 grams

Moles of Oxygen (Y) = (53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol

As per Step-3,

the lowest number of moles is present in the oxygen

As per Step-4,

Ratio of Carbon = 4.028/3.336 = 1.2

Ratio of Hydrogen = 8.056/3.336 = 2.4

Ratio of Oxygen = 3.336/3.336 = 1

As per Tips and Tricks-2,

The least factor to be multiplied is 5 to get all the above ratios to the whole number.

(1 mol O) (5) = 5 mol O

(1.2 mol C) (5) = 6 mol C

(2.4 mol H) (5) = 12 mol H

As per Step-6,

Empirical Formula = C₆H₁₂O₅