
What are the empirical and molecular formulas of nicotine?
Answer
518.4k+ views
Hint :An empirical formula is the simplest ratio of the molecule while the molecular formula gives the actual number of atoms of different elements present in the molecule. If we know the molar amounts of elements we can determine the empirical formula easily.
Complete Step By Step Answer:
An empirical formula of a compound tells us about the relative ratios of different atoms in a compound. It gives the simplest formula in which the different atoms are present. In order to know the Empirical formula of a compound or molecule, we first need to know the mass percentages of the elements in the compound.
A molecular formula represents the total number of atoms of each element in the given molecule or compound.
The empirical formula of nicotine is $ {{C}_{5}}{{H}_{7}}N $ .It tells us that there are $ 5 $ atoms of carbon for each $ 7 $ atoms hydrogen and $ 1 $ atom of nitrogen in a nicotine molecule.
While the molecular formula of nicotine is $ {{C}_{10}}{{H}_{14}}{{N}_{2}} $ .It provides us the information that the ratio of atoms of each of the elements present are $ 5:7:1 $ .It also provides the actual number of atoms that are present in the molecule. It tells us that there are $ 10 $ carbon atoms for each $ 14 $ hydrogen and $ 2 $ nitrogen atoms in a molecule of nicotine.
Note :
Always remember the difference between molecular formula and empirical formula is that empirical formula gives the simplest ratio and molecular formula gives the actual no of atoms present in the molecule but both doesn’t say anything about the structure of the compound.
Complete Step By Step Answer:
An empirical formula of a compound tells us about the relative ratios of different atoms in a compound. It gives the simplest formula in which the different atoms are present. In order to know the Empirical formula of a compound or molecule, we first need to know the mass percentages of the elements in the compound.
A molecular formula represents the total number of atoms of each element in the given molecule or compound.
The empirical formula of nicotine is $ {{C}_{5}}{{H}_{7}}N $ .It tells us that there are $ 5 $ atoms of carbon for each $ 7 $ atoms hydrogen and $ 1 $ atom of nitrogen in a nicotine molecule.
While the molecular formula of nicotine is $ {{C}_{10}}{{H}_{14}}{{N}_{2}} $ .It provides us the information that the ratio of atoms of each of the elements present are $ 5:7:1 $ .It also provides the actual number of atoms that are present in the molecule. It tells us that there are $ 10 $ carbon atoms for each $ 14 $ hydrogen and $ 2 $ nitrogen atoms in a molecule of nicotine.
Note :
Always remember the difference between molecular formula and empirical formula is that empirical formula gives the simplest ratio and molecular formula gives the actual no of atoms present in the molecule but both doesn’t say anything about the structure of the compound.
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