The way you do the calculation for this circuit is to check the specs to see what current you want through the LED. Say you want 20mA. Since the LED will drop 3V, you will have a 2V drop over the resister. E=IR 2=0.020R R=100 Ohms.

Now, figure the resister may be +/- 10%. So if the resister is 90 Ohms will the current overdrive the LED and burn it out? I'll leave the math for this as an exercise for the reader.

First off, on the practical issue of how to set up an LED, let's not forget that this is usually a straightforward matter. The original question is why doesn't all the voltage drop over the resistor. And that is a deeper question than what resistance should be put in series.

So, as to the practical issue. Current running through an LED will induce a voltage across the device. LEDs have the behavior that the voltage developed across the device stays about the same regardless of current, roughly about 2 Volts. So they don't as like a resistor, where voltage is directly proportional to current. The rules of circuit analysis is that voltage drops across the circuit equal to the supply voltage. Another rule of thumb is run your LEDs at 10 mA. Thus, a resistor and LED in series with a 5V supply will have 2V across the LED and the Voltage across the resistor will have to be 3V. (3+2=5). Since there will be 3V across the resistor and you want 10 mA running through the circuit, E=IR; 3=0.010R; R=300. So with 5 volt circuits you see the experimenter with 330 1/8 Watt resistors, since 330 is a standard resistor value. If you go with more modern 3-3.2 V logic, you will want 100 Ohm resistors. The 10 mA drive current is selected because it gives LEDs are usually bright enough at that drive and 10 mA is way below the absolute maximum of 20mA, which is how a lot of LEDs are rated. This will work any of the colored LEDs you want.

For the deeper question of why doesn't all the voltage over the resistor, you may have heard someone tell you that the LED acts like it has no resistance, just develops a voltage. Then, knowing that zero resistance has no voltage drop I could see that you'd think all the voltage should drop over the resistor.

An example of a device that has a voltage without resistance is a capacitor. When there is a charge on the capacitor there will be a voltage. To limit the speed of the current into the capacitor a resistor can be placed in series. Of course, as the current flow there is a voltage across the resistor, but there is a voltage but no resistance acorss the capacitor. As the current charges the capacitor the voltage drop across the capacitor increases and the current drops until the capacitor reaches the voltage of the current supply. This is the basic RC circuit that you will learn about some time.

The nature of the LED is to maintain a roughly constant voltage regardless of current. This is true of any diode which is forward biased and with current flowing. This forward voltage drop is a relative constant regardless of current. The LED is special because it glows! But is just a forward biased diode in usual use. If you wanted you could build a rectifier out of one, but they aren't too useful as a rectifier due to the high voltage drop and limited current flow. That doesn't mean there is no resistance. After all, if an LED has 2V over it and it is passing 10mA, it is acting as if it has a resistance of 200 Ohms. t is interesting that if you drop the current to 5 mA the voltage is still about 2V, so the LED is acting like a 400 Ohm resistor. That's right, the LED acts like a variable resistor serving to maintain voltage over the device at about 2V.

There have been some simplifications in the above discussion. In truth, the forward voltage increases slightly as current increases. At a current of 0.1mA forward voltage is 1.7V, 1mA/1.8V, 10mA/1.9V in one datasheet I pulled up. Also, the forward voltage can vary quite a bit, from 2V to up to 3V in some LEDs even with the same part number. The rule of thumb of driving the LED at 1/2 to 3/4 maximum current generally gives satisfactory results and long life.