If this is my circuit, how do I calculate voltage drop in the output ?
That's a voltage divider
You can only calculate it if you have no load drawing current from that. Otherwise, you need to know the load resistance, too.
To expand upon the two previous answers.
If your 'load' at that point has a high impedance (resistance), you get a good approximation of the voltage as Vout = (R2 / (R1+R2) * Vin. Or using your values Vout = (16000 / (25000) * 5 = 3.2V
However, if your 'load' at that point has a low impedance (resistance), say 1000 ohms the numbers change quite a bit since the load is in parallel with R2 (16k).
Two resistors in parallel have a resistence is Rnew = 1 / (1/R1 + 1/R2) = 1 / (0.001 + 0.0000625) = 941 ohms
So your out voltage would be 5V * (941 / 9941) or about 0.5V.. Think I did that math correctly...
'Corrected my error with R1/R1 in the voltage divider formula!
If your 'load' at that point has a high impedance (resistance), you get a good approximation of the voltage as Vout = (R1 / (R1+R2) * Vin. Or using your values Vout = (9000 / (25000) * 5 = 1.8V
Should have the 16 on top, not the 9.
But good that you both explained the difference between my assuming it was a "current free" voltage divider and a real load on the output.
Ease of calculation depends heavily on the schematic presentation.
There are internationnal standards:
Wherever possible :
-the ground rail is horizontal and placed to the bottom of the graph.
-If there is a power rail it is horizontal and placed to the top the graph.
-The inputs are on the left of the schematic.
-The outputs are on the right of the schematic.
Applying this we obtain the diagram attached.
If the internal resistor of IC connected to the OUT is very higher than 16 kohms there is no correction to do. Otherwise, if for example the connected following IC internal resistor is 50k ohms you must calculate the equivalent resistor:
Re = (50k 16k )/ (16k +50 k) = 11.5 k.
Now we consider that IC internal resistor is very high.
For calculation we will use Ohms Law U = RI.
We know U = 5V, we know R = (9k + 16k) = 25k
The current is calculated as follows: I = U / R = 5/25k I = 0.2 mA
It is now very simple for OUT
OUT = 16k * I
OUT = 0.2 mA*16k = 3.2 V
Ok, thank you, one last thing. If I'm using arduinos 5V to power this circuit (does it have a high impedence, so that Vout = (R2 / (R1+R2) * Vin ?) and if I'm reading the output with arduinos analog pin and than print it (analogRead, Serial.printIn), what values should I get ? I mean... what's the meaning of those values ?
It is the impedance/load of what you connect to Out, not the supply voltage from the Arduino.
As a rule, with resistances around 10K (like your example) you are fine with the theorectical value (no need to include load) when connecting the voltage divider to an analog in. If there is any concern, you can connect an op amp buffer to your out, which presents the out with that ideal high impedance, and provides an ideal 'low impedance' out to whatever device you need to measure the out.
Here is a schematic for a buffer; 
Out goes to an analog pin on arduino
And if it's a 30K resistance ?
arcadefire:
And if it's a 30K resistance ?
The analog inputs on the Arduino are reasonably high impedance, so your straight voltage divider isn't going to be much of an issue. Of course, that is unless you are doing all of the other hoop jumping required to get a accurate voltage reading from the ADC. Most folks who use the ADC are happy with 'close enough', which is what you see implemented in nearly all Arduino examples.
What would be the 30k resistance you are asking about?
My suggestion is to build your circuit and compare the values you are getting from the Arduino with what you read from a voltmeter, and what you 'think' you should be getting--for instance measuring the voltage being divided and doing the math for the voltage divider. If your measurements (either Arduino or Voltmeter) don't match come back with specific vallues if you have any questions.