Missing known sinusoid in FFT of sensor data
If your sampling frequency is 5Hz (Which I don't think is!) and the signal contains a known frequency of 5Hz. So, the sampled signal will have aliasing for sure. 

You are correctly mapping your positive FFT spectrum from 0 to 2.5Hz, because digital frequency of sampled signal can only map from $[-f_s/2, f_s/2]$ which is $[-2.5hz, 2.5hz]$. 

The question is where do you want to see your 5 Hz in this. You can only see a high DC which is actually the negative folded image of the 5Hz tone at $f_s$. And, you can actually see a peak very close to DC.


  Based on the Plots you have shown :


The plot of time domain data and assuming you have injected a 5Hz tone, I think the sampling frequency is close to 500Hz and not 5Hz. Simply, because you have ~2.5 cycles of 5Hz sinusoidal in 250 samples. 5Hz sinusoidal wave will take 1 second for 5 cycles, hence ~2.5 cycles of such wave will take 0.5 second. 250 samples/0.5 second means 500 samples/second. Hence, your sampling frequency is 500 Hz.

And, if you are taking 250 length FFT, frequency resolution will be of :
$$\frac{2\pi}{250} = \frac{2}{250}.\frac{f_s}{2} = 2Hz$$
Hence, your 5Hz tone will be seen as a peak between $2^{nd}$ and $3^{rd}$ coefficient of the 250-Point DFT, which you are able to see.

MATLAB Simulation showing your situation:

fd = 5;             % data tone of 0.25Hz   
fs = 500;           % sampling frequency of 5Hz
Ts = 1/fs;
n_samps = 250;      % Number of samples
time = n_samps/fs;  % duration of data collection (0.5 secs)
n = linspace(0,time-Ts,n_samps); % sample points in the duration [0sec, 0.5sec)

% Carrier Frequency fc = 5Hz + a slow linear ramp as it seems from the
% time-domain plot you have attached.
signal = (n/250 + sin(2*pi*fd*n));

% AWGN Noise added 
signal = awgn(signal, 15);
figure(1)
plot(signal)

Y = fft(signal);
figure(2)
plot(5/n_samps * (0:1:n_samps/2-1), (abs(Y(1:length(Y)/2))));





  The other small peaks are most probably being generated due the
  harmonics which are present as a consequence of clipping in sampled
  data due to fixed bit-width ADC. If you lower your amplitude a little
  to accommodate your sinusoidal wave within the dynamic range of ADC,
  these small peaks might go away.


Adding analysis of the time-domain samples you have pasted :

Assumption : Sampling Frequency = 500Hz

I have plotted FFT of the samples centered around 0 Hz(DC). And, based on number of samples (256) and assumed $f_s = 500Hz$, frequency spacing between 2 FFT samples is $\frac{f_s}{n_samps} = \frac{500}{256} = 1.9531$. Since 5Hz is not an integral multiple of this spacing, therefore, you will see the tone energy spread around into 1.9531*2Hz and 1.9531*3Hz i.e. 2nd and 3rd FFT coefficient.

You are also seeing a strong peak at DC (0Hz) because your sinusoid of 5Hz is having an amplitude shift of a non-zero value. 

You can see that your tone is not missing at all. It is present right where is should be.

MATLAB Code to plot the signal and it's DFT with correct $f_spacing$ is as follows:

plot(500/256*(-128:127), abs(fftshift(fft(x))))