[MK23]

Here is the long and short of the problem.  I was given H NMR , C NMR and IR for a compound with the formula C₅H₁₀O I was able to deduce it was pentanal/valeraldehyde due to the sharp c=o absorption at ~1725 and 2 peaks at ~2720.  Furthermore C NMR showed a peak at ~ 205 and H NMR showed a carbonyl hydrogen at ~9.7.  I am having a bit of difficulty assigning the peaks in the H NMR and C NMR spectras. 

What I have

H NMR  Integration 1:2:2:2:3
~9.7ppm   Singlet 1H
~2.4ppm   Triplet 2H
~1.6ppm   Mutiplet 2H
~1.3ppm   Multiplet 2H
~ 0.9ppm  Triplet 3H

Of course the 1H at 9.7 is the Hydrogen that makes the aldehyde an aldehyde, and the Triplet at 0.9 is shielded the most and is at the far end of the molecule. I am thinking that the triplet at 2.4 is the closest carbon to the c=o, the multiplet at 1.6 is next to it and the 1.3 is closest to the methyl group. 
IF I am incorrect please let me know.

C NMR.. this is where I am having a fit
I am showing 5 distinct carbons
~205ppm
~43ppm
~24ppm
~22ppm
~14ppm

The c=o is a no brainer at 205ppm, and the saturated methyl should be at the opposite end at 14ppm, and I am guessing that they will proceed downfield in the same order that the hydrogens did
H-C(=O)-   CH₂-   CH₂-   CH₂ -   CH₃
205           43    24      22      14

Thanks in advance.

[Yggdrasil]

Looks perfect to me.  Good job.