A useful idea to develop in this respect is the #"degree of unsaturation"#, which I will try to outline.
The general formula of an alkane is #C_nH_(2n+2)#. This works for the simplest alkane, #"methane"#, #CH_4#, and for ethane, #C_2H_6#, propane, #C_3H_8#, ......pentane, , #C_5H_12#, etc. We say that carbon is saturated with respect to hydrogen substitution.
If we substitute in a halogen, #X#, this stands in for #1xxH#. And thus for #CH_2Cl_2#, we still has a saturated molecule. If there is oxygen in the formula we assess the degree of saturation directly, i.e. #CH_3OH#, is still saturated. If there is nitrogen in the formula, we SUBTRACT #NH# from the formula, before we assess saturation, And thus for #"methylamine"#, #H_3CNH_2#, we work on #CH_4#, i.e. #"saturated"#.
That is the preliminary, now we work on the question you posed. An aldehyde or a ketone, or an olefin, contains an unsaturated bond, and whether #C=C# or #C=O#, this counts for #"ONE DEGREE of UNSATURATION"#; 2 hydrogens LESS than the saturated formula. Compare saturated #"propane"#, #C_3H_8#, with #"acetone"#, #C_3H_6O#, or saturated #"ethane"#, #C_2H_6#, with #"ethylene"#, #H_2C=CH_2#, or #"acetaldehyde"#, #H_3C-C(=O)H# both with the ONE DEGREE of unsaturation.
And so the moral: a #"degree of unsaturation"# CORRESPONDS to a double bond, #C=C#, or #C=O#, or #C=N#, or a RING JUNCTION. Every organic formula that you are given should be assessed on this basis. #"Cyclohexane"#, #C_6H_12# has ONE DEGREE of unsaturation. Where? How many degrees of unsaturation does #"benzene"# have? What about #"acetylene"#?
