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Empirical Formula

The empirical formula or simplest formula gives the smallest whole number ratio of atoms present in a compound. This formula gives the relative number of atoms of each element in the compound.

Empirical formulas are the simplest form of notation. They provide the lowest whole-number ratio between the elements in a compound. Unlike molecular formulas, they do not provide information about the absolute number of atoms in a single molecule of a compound. The molecular formula for a compound is equal to, or a whole-number multiple of, its empirical formula.

To find the Empirical Formula of a compound we must find the whole number ratio of atoms of the elements in a sample of the compound. When the numbers in the ratio are reduced to their lowest terms that is when they have no common divisor they are the subscripts in the empirical formula.

Empirical Formula an be calculated by the following steps.
  • Assume a definite starting quantity of the compound, if not given and express the mass of each element in grams.
  • Convert the grams of each element into moles using each elements molar mass. This conversion gives the number of moles of atoms of each element in the quantity assumed in previous step.
  • Divide each value obtained in previous step by the smallest values. If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula.
  • Multiply the values obtained in the previous step by the smallest number that will convert them to whole numbers. Use these whole numbers as the subscripts in the empirical formula.

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Empirical Problems

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Solved problems based on empirical formula are given below.

Solved Examples

Question 1: Calculate the empirical formula of a compound containing 11.19% hydrogen (H) and 88.79% oxygen(O).
Solution:
 
1. Assume 100.0g of material. We know that the percent of each elements equals the grams of each element
11.19g H
88.79g O

2. Convert grams of each element to moles
H: (11.19g)[$\frac{1mol\ H\ atoms}{1.008g\ H}$] = 11.10 mol H atoms
O: (88.79g)[$\frac{1mol\ O\ atoms}{16.00g\ O}$] = 5.549 mol O atoms
The formula could be expressed as H11.10O5.549. However its customary to use the smallest whole number ratio of atoms.

3. Change the numbers to whole numbers by dividing by the smallest number.
H = $\frac{11.10\ mol}{5.549\ mol}$ = 2.000
O = $\frac{5.549\ mol}{5.549\ mol}$ = 1.000
The simplest ratio of H to O is 2:1
Empirical formula = H2O

 

Question 2: A sulfide of iron was formed by combining 2.233g of iron (Fe) with 1.926g of sulfur(S). What is the empirical formula for the compound?
Solution:
 
1. The mass of each element is known, so we use them directly

2. Convert grams of each element to moles
Fe: (2.233g Fe)($\frac{1mol\ Fe\ atoms}{55.85g\ Fe}$) = 0.03998 mol Fe atoms
S: (1.926g S)($\frac{1mol\ S\ atoms}{32.07g\ S}$) = 0.06006 mol S atoms

3. Change the numbers to whole numbers by dividing by the smallest number.
Fe = $\frac{0.03998\ mol}{0.03998\ mol}$ = 1.000
S = $\frac{0.06006\ mol}{0.03998\ mol}$ = 1.502

4. We still have not reached a ratio that gives whole numbers in the formula, so we multiply by a number that will give us whole numbers.
Fe: (1.000)2 = 2.000
S: (1.502)2 = 3.004

Empirical formula = Fe2S3


 

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