
What is the simplest formula for copper $ (II) $ nitride?
Answer
519.3k+ views
Hint :We know that to answer this question, we must firstly recall the concept of empirical formula. It is the simplest whole number ratio of the number of atoms present in the molecule. Nitride is an Anion of nitrogen. It gains electrons in order to attain the configuration of a noble gas, which means it gains three electrons.
Complete Step By Step Answer:
The elemental constituents of a compound define its chemical identity, and chemical formulas are the most appropriate way of representing this elemental makeup. When a compound’s formula is not known to us, measuring the mass of each of its constituent elements is obviously the first step in the process of determining the formula experimentally.
Atoms, by default, have the same number of protons as electrons, resulting in a balance between positive and negative charged particles. Because it gains three electrons, there are more electrons than protons, so it has a charge of $ 3-. $
The easiest way of thinking about the combination of ions is by using the method I recall as being called, the swap and drop. So we have two ions; $ C{{u}^{2+}} $ and $ {{N}^{3-}} $
Now we swap the charge of each and drop, Note that the two ions have to be of opposite charge. Thus we have $ C{{u}_{3}} $ and $ {{N}_{2}} $
Now just by combining both the compound we have; $ C{{u}_{3}}{{N}_{2}} $
Therefore, the simplest formula for copper $ (II) $ and nitride $ C{{u}_{3}}{{N}_{2}}. $
Note :
We should remember that with regard to deriving empirical formulas; we must consider examples in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. This will make our calculations much easier.
Complete Step By Step Answer:
The elemental constituents of a compound define its chemical identity, and chemical formulas are the most appropriate way of representing this elemental makeup. When a compound’s formula is not known to us, measuring the mass of each of its constituent elements is obviously the first step in the process of determining the formula experimentally.
Atoms, by default, have the same number of protons as electrons, resulting in a balance between positive and negative charged particles. Because it gains three electrons, there are more electrons than protons, so it has a charge of $ 3-. $
The easiest way of thinking about the combination of ions is by using the method I recall as being called, the swap and drop. So we have two ions; $ C{{u}^{2+}} $ and $ {{N}^{3-}} $
Now we swap the charge of each and drop, Note that the two ions have to be of opposite charge. Thus we have $ C{{u}_{3}} $ and $ {{N}_{2}} $
Now just by combining both the compound we have; $ C{{u}_{3}}{{N}_{2}} $
Therefore, the simplest formula for copper $ (II) $ and nitride $ C{{u}_{3}}{{N}_{2}}. $
Note :
We should remember that with regard to deriving empirical formulas; we must consider examples in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. This will make our calculations much easier.
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