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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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Steps to Calculate:
- 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.
Tips and Tricks:
- 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.
Example -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
Related Questions
- A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical
- An organic compound containing C, H and O gave the following analysis
- The weight of one molecule of a compound C₆₀H₁₂₂ is
Subject: Chemistry (2512)
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