Combustion analysis is a method used to determine the empirical formula of a compound, especially hydrocarbons. By measuring the masses of carbon dioxide and water produced during combustion, the amounts of carbon and hydrogen in the original sample can be calculated.
No, determining the molecular formula generally requires knowing the empirical formula first, as it provides the simplest ratio of the elements. Once you have the empirical formula and molar mass, you can calculate the molecular formula.
An empirical formula represents the simplest whole-number ratio of atoms in a compound, while a molecular formula shows the exact number of each type of atom in a molecule. The molecular formula is often a multiple of the empirical formula.
To determine the molecular formula, you need the empirical formula and the compound’s molar mass: Calculate the empirical formula mass. Divide the molar mass by the empirical formula mass to find the multiplier. Multiply the subscripts in the empirical formula by this number to get the molecular formula.
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Empirical Formula and Molecular Formula
Empirical Formula shows the simplest whole-number ratio of the elements in a compound. Molecular Formula indicates the exact number of each type of atom in a molecule. It may be the same as the empirical formula or a multiple of it.
1.0Molecular Formula
Represents the actual number of each type of atom present in a molecule.
Uses subscripts to indicate the exact count of each atom in a molecule.
Reflects the true composition of a compound, not simplified.
The molecular mass (or molar mass) of a compound corresponds to a whole-number multiple of the empirical formula mass.
Example: The molecular formula for glucose is C6H12O6, showing that each molecule contains 6 carbon, 12 hydrogen, and 6 oxygen atoms.
2.0Empirical Formula
The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present in the compound. It gives the relative number of atoms rather than the actual number in a molecule.
Represents the simplest whole-number ratio of the elements in a compound.
Provides a reduced or simplified version of the formula, focusing on the ratio, not the exact count.
Known as the "simplest formula" because it only shows the basic proportional relationship between elements.
The empirical formula is derived directly from the per cent composition of the compound.
Example: The empirical formula for glucose is CH2O, indicating the simplest ratio of carbon, hydrogen, and oxygen is 1:2:1.
Related Video:
3.0Steps to Calculate the Empirical Formula
Determine the Mass of Each Element: Use the given data or assume a 100 g sample if percentages are provided.
Convert Mass to Moles: Use the formula: Moles of an element=Mass of the element (g)Atomic mass of the element (g/mol)Moles of an element=Atomic mass of the element (g/mol)Mass of the element (g)Moles of an element=Atomic mass of the element (g/mol)Mass of the element (g)
Find the Simplest Whole-Number Ratio: Divide all mole values by the smallest number of moles calculated.
Write the Empirical Formula: Use the ratio of moles as subscripts in the formula. If the ratios are not whole numbers, multiply by the smallest integer that converts them to whole number
Example
Glucose:
Percent composition: C: 40%, H: 6.7%, O: 53.3%
Empirical formula: CH2O
Hydrogen Peroxide:
Simplest ratio: HO
Solved Example
Calculation of the Empirical Formula of Phosphoric Acid
Given Composition:
Hydrogen (H): 3.06%
Phosphorus (P): 31.63%
Oxygen (O): 65.31%
Solution:
Assume 100 g of the compound:
This simplifies the percentages directly into masses:
Mass of H = 3.06 g
Mass of P = 31.63
Mass of O = 65.31 g
Calculate the number of moles for each element:
Using the formula: Moles=AtomicMassMass
For H, Moles of H = (3.06)/1 =3.06
For P. Moles of P = (31.63)/31 = 1.02
For O, Moles of O = (65.31)/16 =4.08
Find the simplest ratio of moles:
Divide all the mole values by the smallest number of moles, 1.02
Ratio of H = 1.023.06=3
Ratio of P = 1.0231.63=1
Ratio of O = 1.0265.31=4
Write the empirical formula:
Use these ratios as subscripts for the elements:
Empirical formula = H3PO4
Solved Example
Glucose:
Empirical formula: CH2O
Empirical formula mass: 12+2(1)+16=30 g/mol
Molar mass: 180 g
n = 180/30 = 6
Molecular formula: C6H12O6
Hydrogen Peroxide:
Empirical formula: HO
Empirical formula mass: 1+16=17 g/mol
Molar mass: 34 g/mol
n = 34/17 = 2
Molecular formula: H2O2
4.0Process to Determine the Molecular Formula from the Empirical Formula
Determine the Empirical Formula:
Start with the per cent composition or the masses of each element in the compound.
Convert these masses to moles.
Find the simplest whole-number ratio of the elements, which gives you the empirical formula.
Calculate the Empirical Formula Mass:
Add up the atomic masses of all the elements in the empirical formula. This gives you the empirical formula mass.
Determine the Molecular Weight (Molar Mass)of the Compound:
The molar mass (usually given in the problem) is the actual mass of one mole of the compound.
This value is needed to relate the empirical formula to the molecular formula.
Find the Integer Multiple (n):
Use the formula: n=EmpiricalFormulaMassMolecularWeight
This integer n tells you how many times the empirical formula must be multiplied to obtain the molecular formula.
Calculate the Molecular Formula:
Multiply each subscript in the empirical formula by the integer n.
The result gives the molecular formula.
Solved Example
Empirical formula: CH2
The molar mass of the compound: is 84 g/mol
Steps:
Empirical Formula Mass:
Carbon (C): 12 g/mol
Hydrogen (H): 2×1 g/mol = 2 g/mol
Empirical formula mass = 12+2 = 14 g/mol
Find the Integer Multiple n: n=EmpiricalFormulaMassMolecularWeight
n = 8414 = 6
Determine the Molecular Formula:
Multiply each subscript in CH2 by 6.
Molecular formula: C6H12.
5.0Differences Between Empirical and Molecular Formulas
Aspect
Empirical Formula
Molecular Formula
Definition
The simplest whole-number ratio of atoms in a compound.
The exact number of each type of atom in a molecule.
Information Provided
The basic proportion of elements in the compound.
Complete and detailed composition of the molecule.
Relation
A simplified form of the molecular formula.
Often a whole-number multiple of the empirical formula.
Example
CH2 for ethene.
C2H4 for ethene.
Table of Contents
1.0Molecular Formula
2.0Empirical Formula
3.0Steps to Calculate the Empirical Formula
4.0Process to Determine the Molecular Formula from the Empirical Formula
5.0Differences Between Empirical and Molecular Formulas