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Empirical formula of a compound helps to give the simplest ratio of the number of atoms. While, the molecular formula provides the actual number of every atom in a molecule.
- The Empirical Formula has the smallest whole-number ratio of every atom present in a compound.
- This formula provides the relative number of atoms of every element available in the compound.
- With the help of the mass of each element in the compound, as well as the percentage composition, we can determine the empirical formula of each compound.
- The Empirical Formula of Benzene is C6H6. The Molecular formula of Benzene is C6H6 as well.
The general statement which helps to relate the molecular formula and the empirical formula can be expressed as:
| Molecular Formula = n × Empirical Formula |
In case a formula is simplified, then it can be termed an “Empirical Formula”. The Molecular formula is typically mostly used, and it is a multiple of the empirical formula.
Read More: Classification of Organic Compounds
| Table of Content |
Key Terms: Empirical formula, Atoms, Molecular formula, Compound, Mass, Element, Molar Mass, Moles, Sulphur Monoxide, Acetylene, Benzene
What is Molecular Formula?
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Molecular formula can be derived from molecules and is the formula representing the total number of single atoms in a molecule of a compound.
- A molecular formula generally uses a subscript which states the actual number associated with each type of atom in a molecule of the compound.
- Molecular formulas associated with gram molecular masses are simplified whole number multiples of their corresponding empirical formula mass.
- Molecular Formula Example includes glucose, which is, C6H12O6.
In case the molar mass value is known, the Molecular Formula can be evaluated on the basis of the empirical formula. Thus:
| \(n = \frac{molar\ mass}{mass\ of\ the\ empirical\ formula}\) |
The molecular formula is typically equivalent to the empirical formula or is an exact multiple of it.
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What is an Empirical Formula?
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The Empirical Formula can be defined as:
| “The simplest whole-number ratio of any individual type of atom in any compound” |
- The empirical formula is known to be the simplest formula of a compound which is the ratio of the subscripts of the smallest whole number of the element in the given formula.
- Empirical formula is also known as the Simplest Formula.
- It helps to give information about the ratio of the number of atoms found in a compound.
- It's possible that it's the same as the compound's molecular formula. However, this is not always achievable.
- Thus, the empirical formula calculations can be done using details regarding the mass of each element in the compound, as well as the percentage composition.
- Simply, this formula specifies the relative number of atoms in each element in the compound.
Empirical Formula
Example of Empirical FormulaExample. Show an empirical formula example by using the compound “Sulphur Monoxide”. Ans. An example of the empirical formula is Sulphur Monoxide. The empirical formula of sulphur monoxide is simply SO, as is the empirical formula of Disulphuric dioxide, S2O2. So, Sulphur monoxide and Disulphuric dioxide, both sulphur and oxygen compounds, have a similar empirical formula. Example. Show an example of Empirical formula by considering Example (Glucose Molecular Formula Vs Glucose Empirical Formula). Ans. By considering the example of glucose, we can say: The molecular formula of glucose is C6H12O6, while the empirical formula of glucose is CH2O. Thus, a relation between the Molecular formula and the empirical formula of glucose can be derived. \(\therefore\) Empirical Formula & Molecular Formula of Butane & Octane ⇒ C6H12O6 = 6 × CH2O Derivation of a general expression: Molecular formula = n × empirical formula (here, n = whole number) Thus, the empirical formula and molecular formula can often be the same. |
Steps of Empirical Formula
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The steps to determine empirical formula are:
- First, we collect the information the problem gives us, especially the number of grams of each element.
- If percentages are given, we will assume that the total mass is 100 grams, so that the mass of each element equals the percentage provided.
- Then change the mass of each element into moles with the help of the periodic table’s molar mass.
- Now we divide each mole value with the smallest number of moles calculated.
- Always round to the nearest whole number. This is the mole ratio of the elements and is signified as subscripts in the empirical formula.
- If the number is too big to round off, multiply each response by the very same factor to have the lowest whole number multiple. (x.1 ~ x.9).
| Example: If one solution is 1.25, multiply each answer in the problem by 4 to obtain 5. Example: If one solution is 1.5, multiply each answer in the problem by 2 to obtain 3. |
- If the compound's molar mass is known when the empirical formula is determined, the molecular formula can be computed.
- The ratio between the molecular formula and the empirical formula must be established.
- The mass of the empirical formula is calculated by dividing the compound's molar mass by it.
- To find the molecular formula, multiply each atom (subscripts) by this ratio.
Read Also: Benzene Reaction
Difference between Empirical Formula and Molecular Formula
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The Empirical Formula Vs Molecular Formula differences are tabulated below:
| Empirical Formula | Molecular Formula |
|---|---|
| An empirical formula denotes the simplest whole-number ratio of different atoms in a compound. | The molecular formula indicates the exact number of various types of atoms in a molecule of a compound. |
| Empirical Formula Example: The empirical formula of Acetylene is C6H6. | Molecular Formula Example: The molecular formula of Acetylene is C2H2. |
Empirical Formula Statistics
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The Empirical formula statistics claim that:
- Roughly 95% of the observations in a normal distribution comprise three Standard Deviations from the Mean.
- Hence, the empirical formula rule helps in forecasting.
- The formula exhibits the predicted observation percentage which can be found in the Standard Deviation from the Mean.
As per the rule:
- 68% of the observations can be found to lie within 土 1 Standard Deviation (SD) from the mean.
- 95% of the observations can be found to lie within 土 2 SD from the mean.
- Around 7% of observations can be found to lie within 土 3 SDs from the mean.
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Things to Remember
- The simplest whole-number ratio of any individual type of atom in any compound is its empirical formula.
- Empirical formula specifies the relative number of atoms in each element in the compound.
- The empirical formula can be calculated using details regarding the mass of each element in the compound, as well as the percentage composition.
- Empirical formula states to you the simplest ratio of elements in a compound, whereas molecular formula helps you to determine how many atoms of each element are present in a compound.
- If somehow, the molecular formula of a compound cannot be reduced any further, then the empirical formula is similar to the molecular formula.
Also Read: Nucleus of an Atom and Atomic Mass
Previous Year Questions
- Describe the temperature vs time graph. [JEE MAIN 2019]
- Cp – Cv relation for diatomic gases. [AMUEEE1999]
- Find the value of ϒ for diatomic gases. [NEET 2019]
- Two Identical Bodies Are Made of A Material [NEET 2016]
- Steam at 100∘C is passed into 20g of Water [KCET 2014]
- What fraction of heat energy is converted to work? [KCET 2020]
- Heat evolved in neutralisation of HF. [UPSEE 2018]
- Hammer Of Mass 200 Kg Strikes A Steel Block Of M [DUET 2019]
- Hydrogen, Helium and Other Ideal Diatomic Gas [NEET 2019]
- What is calorie? [BHU UET 2010]
- Determine ratio of specific heats at constant volume. [AMUEEE 1998]
Sample Questions
Ques: What is the use of Empirical Formula? (1 mark)
Ans: Empirical formula is typically used to show what elements can be found in a molecule. It is generally very useful when an individual wants to know what elements can be found in a given compound.
Ques: What is the empirical and molecular formula of Acetylene? (2 marks)
Ans: The molecular formula of acetylene is C2H2.
As per the definition, the empirical formula can be defined as the simplest ratio of the number of each atom in a compound.
Thus, Molecular Formula = n × Empirical Formula
C2H2 = 2 × CH
Hence, the Empirical formula of acetylene is CH.
Ques: Show the empirical of a compound that has a molecular formula of C6H12O6. (2 marks)
Ans: It can be said: C6H12O6 = 6 × CH2O
As we are aware, Molecular Formula = n × Empirical Formula
Herein, n = 6
Thus, the empirical formula for the compound is CH2O.
Ques: A compound consists of 63.52 % iron and 36.48% sulfur. Determine its empirical formula. (2 marks)
Ans: 63.5F2g Fe × 1 mol / 55.85 g = 1.37 mol Fe
36.48g S × 1 mol / 32.06g = 1.138 mol S
1.137 mol Fe / 1.137 : 1.138 mol S / 1.137
=) 1 mol Fe : 1.001 mol S
Hence, Empirical formula = FeS
Ques: A compound consists of 88.79% oxygen (O) and 11.19% hydrogen (H) then calculate the empirical formula of the compound. (3 marks)
Ans: Let’s assume a substance of 100.0g . It’s easily visible that the percentage of each element is similar to the grams of each element
11.19g H
88.79g O
Now, Change the grams of each element into moles
H: (11.19/1.008) = 11.10 mol H atoms [molar mass of H=1.008g/mol]
O: (88.79/16.00) = 5.549 mol O atoms [molar mass of O= 16.00g/mol]
The formula can be specified as H11.10O5.549. Therefore, it’s general to use the smallest whole-number ratio of atoms.
Now divide the lowest number to alter the numbers into whole numbers.
H =11.10/ 5.549 = 2.000
O = 5.549/ 5.549= 1.000
The simplest ratio of H to O is 2:1
Empirical formula = H2O
Ques: A sulphide of iron was created by mixing 1.926g of sulphur(S) with 2.233g of iron (Fe). Determine the compound’s empirical formula. (4 marks)
Ans: As the mass of every element is already given, we can make use of them directly
Change grams of each element into moles
Fe: (2.233g /55.85g) = 0.03998 mol Fe atoms [molar mass of Fe =1.008g/mol]
S: (1.926 /32.07) = 0.06006 mol S atoms [molar mass of S =32.07g/mol]
Now by dividing with the smallest number, convert the numbers to whole numbers.
Fe =0.03998/0.03998 = 1.000
S = 0.06006/0.03998mol = 1.502
We multiply by a number that will provide us whole numbers because we still haven’t found a ratio that offers us entire numbers in the formula.
Fe: (1.000)2 = 2.000
S: (1.502)2 = 3.004
Empirical formula = Fe2S3
Ques. The 0.290g Compound of organic compounds having C, H, and O gave combustion of 0.270g of water and 0.66g of Co2. Calculate the empirical formula of the compound. (4 marks)
Ans: First determine the percentage of C, H and O
% of C = (12/44) ×100 = 62.07 %
% of H = (2/8) ×100 = 10.34 %
% of O = 100 - (10.34% +62.07%) = 27.59
Number of Moles of Carbon.
C = 5.1725 , H = 10.34 , and O = 1.72
Divide the number of moles by the least number 1.72. We will get a simple whole ratio.
C = 3.0, H = 6.0 and O = 1
Empirical Formula: C3H6O
Ques: An oxide of Aluminium is created with the reaction of 4.151 grams of Aluminium with 3.692 grams of Oxygen. Now, calculate the empirical formula of the compound. (4 marks)
Ans: Step 1: Calculate the masses
4.151 gram of Al and 3.692 gram of O
Step 2: Now find out the number of moles
4.151 g Al ×1molAl/26.98gAl = 0.1539 mol Al atoms
3.692 g O ×1molO1/6.00gO = 0.2398 mol O atoms
Step 3: Here, we divide the number of moles
0.1539molAl / 0.1539= 1.000 mol Al atoms
0.2398molO / 0.1539 = 1.500 mol O atoms
Step 4: Change the numbers into whole numbers
The compound consists of 2 Al and 3 O atoms
Hence, the Empirical Formula is Al2O3.
Ques. A Compound consists of 32.65% Sulphur, 65.3% Oxygen and 2.04% Hydrogen. State the empirical formula of the compound. (5 marks)
Ans: First we have to convert the percentage into grams.
32.65%→32.65g of S
65.3%→65.3g of O
2.04%→2.04g of H
Now divide all the given masses with their molar mass.
32.65g of S/ 32gm-1 = 1.0203 moles of S
65.3g of O/ 16gm-1 = 4.08 moles of O
2.04g of H/ 1.008gm-1 = 2.024 moles of H
Then, take the smallest answer in moles from the previous step and divide all the answers with that. Remember to round to the next whole number if you compute a number that is x0.9.
1.0203 moles of S/ 1.0203 = 1
4.08 moles of O/1.0203 = 3.998 ≈ 4
2.024 moles of H/1.0203 = 1.984 ≈ 2
Lastly, the coefficients determined in the previous step will act as subscripts in the chemical formula.
- S = 1
- O = 4
- H = 2
Thus, H2SO4
Ques: Given the mass of a reactant before a chemical reaction and the mass of a product after a reaction.
When 0.273g of Mg is heated in a Nitrogen (N2) environment a chemical reaction happens. The product of the reaction is 0.378g . Now Determine the empirical formula of compound. (5 marks)
Ans: In any empirical formula question you should first find the mass % of the elements of the compound. As the total mass of the final product was 0.378 we find that:
0.378g total-0.273g magnesium = 0.105g nitrogen
0.105g nitrogen/0.378g total (100) = 27.77%
0.273g magnesium/0.378g total (100) =72.23%
Now convert the percentage into grams
27.77%→27.77g of N
72.23%→72.23g of Mg
Here divide all the masses with the help of their respective molar masses.
27.77g/14gm-1 = 1.98 moles N
72.23g/24.31gm-1 = 2.97 moles Mg
Take the smallest answer of moles and divide all figures with that.
1.98 moles N/ 1.98 = 1
2.97 moles Mg/ 1.98 = 1.5
Because our answer for magnesium is not close enough just to round to the nearest whole number, we have to multiply all of the values by a factor that will produce whole numbers.
1*2= 2 N
1.5*2= 3 Mg
Lastly, the coefficients determined in the previous step will act as subscripts in the chemical formula.
1.98 moles N/ 1.98 = 1
1.98 moles N/ 1.98 = 1
N = 2
1.98 moles N/ 1.98 = 1
Mg = 3
Thus, Mg3N2.
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