# What is CnH2n 2

Not only Lilly in GZSZ desperately in her Abitur preparations at the number combinations that describe the general sum formulas of the homologous series of hydrocarbons. It is now called C_{n}H_{2n} or C_{n}H_{2n + 2}, and where does the 2 come from anyway? It seems even more complicated with them *Alkenes* and only then with the *Alkynes* to become. And then what about them *Isomers?*

You can practice that. We want to show that here. First a note: Here you can also work nicely with molecular models - e.g. B. with the fabulous stick-ball model kit from CVK.

*1. Alkanes*

First we write down the structural formula (= constitutional formula) of butane:

We see that all four carbon atoms each have two hydrogen atoms. Be *n* the number of carbon atoms, then the number of hydrogen atoms is first *2n*. The two terminal carbon atoms, however, each have one more hydrogen atom. (So that's where the 2 comes from - which Lilly couldn't explain ...) So:

**(1c) General empirical formula of the alkanes = C _{n}H_{2n + 2}**

For butane is *n* = 4. This means that its molecular formula is C_{4}H_{2 x 4 + 2} = C_{4}H_{10}.

We then practice that, starting with methane (n = 1).

*2. Alkenes*

We choose butene-1 as the test substance.

We see that almost all C atoms have a different number of H atoms (exactly: 1, 2, 2, 3). Counting shows that in alkenes the number of C atoms is exactly half as large as that of H atoms. Hence:

**(2b) General empirical formula of the alkenes = C _{n}H_{2n}**

For butene is *n* = 4. This means that its molecular formula is C_{4}H_{8}.

To practice we have to go with you *n* = 2 or begin with ethene.

(The general formula for alkenes also applies to ** Cycloalkanes**how to z. B. can easily make clear on the basis of the cyclohexane.)

*3. Alkynes*

Here is the constitutional formula of butyne-1:

We see that all C atoms here really have a different number of H atoms (exactly: 0, 1, 2, 3). Again, we count down for the sake of simplicity. Is *n* the number of carbon atoms is the number of hydrogen atoms *2n*-2.

**(3b) General empirical formula of the alkynes = C _{n}H_{2n-2}**

For butyne is with *n* = 4 the empirical formula C_{4}H_{6}.

For practice we start with *n* = 2 or with ethyne.

*4. What about the isomers?*

We first draw the structural formulas of the two positional or constitutional isomers of **butane**.

Counting results in for *both* Molecular formulas C_{4}H_{10}.

The same applies to the three isomers of **Butene**:

The molecular formula *all* Butene isomer is C._{4}H_{8}.

Of **Butyne** there are two isomers:

The molecular formula of *both* Isomer of butyne is C_{4}H_{6}.

We learn from this:

**The general empirical formula is the same for positional or constitutional isomers.**

*What if you still can't remember the expressions for the general structural formulas?*

Then the following helps: You are certainly familiar with the hydrocarbons *n* = 2: *Ethane, ethene, ethine*. With these you can easily derive the general sum formulas.

Hydrocarbon group | Fabric withn = 2 | Semi-structuralformula | GeneralMolecular formula |

Alkanes | Ethane | CH_{3}-CH_{3} | C._{n}H_{2n + 2} |

Alkenes | Ethene | CH_{2}= CH_{2} | C._{n}H_{2n} |

Alkynes | Ethine | C._{n}H_{2n-2} |

*R diger flower*

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