An alkane, any alkane, has a formula of #C_nH_(2n+2)#; try it with the regular alkanes, methane, ethane, propane........etc.

When given a formula, this is the first thing to check. As regards heteroatoms, oxygen in the formula may be ignored; if there is nitrogen, we subtract #NH# from the formula; #"halogens"# count for #"1 hydrogen"#.

Given these simple rules of thumb, we can assess the makeup of many organic formulae.

#"Methane, n=1"#, is saturated, so is #"ethane"#, and #"propane"#, and #"butane"#. What are the formulae here?

But now we go to ethylene, #H_2C=CH_2#, which clearly does not fit the alkane formula. Because ethylene has a #C_2H_4# rather than the #"saturated"# #C_2H_6# formula, ethylene is said to have #"1 degree of unsaturation"#, that is 2 hydrogens LESS than the saturated formula. #"Propylene"# has a formula of #C_3H_6#; again #1""^@# of unsaturation.

So what do we do with the definition? Every #"degree of unsaturation"# specifies a double bond (i.e. #C=O#, #C=N#, #C=C#), OR a ring junction: compare hexane, #C_6H_14#, no degrees of unsaturation, to cyclohexane, #C_6H_12#. The ring junction, the #C-C# bond, has reduced the hydrogen count by 2, so that cyclohexane has the 1 degree of unsaturation.

#"Benzene"# has 4 degrees of unsaturation given its cyclic #C_6H_6# formula. Given your question, a formula of #C_nH_(2n+2)O# is saturated, an alcohol or an ether, a formula of #C_nH_(2n)O# is unsaturated, and possesses a #C=C#, OR a #C=O# bond, OR a ring.

So when you are presented with a formula, it is always useful to assess it degree of unsaturation, in order to see what possible functionality it may possess.

And finally, I'll give some examples. How many degrees of unsaturation are there in #"acetylene"#, #H-C-=C-H#, in #"acetone, "C_3H_6O#, in #"cyclohexanone, "C_6H_10O#, in #"toluene, "C_7H_8#?

If this hasn't addressed your question, I apologize. I (and others) will certainly entertain further questions.