Since there is a voltage drop occur across a resistor in series, what happens when there is also a capacitor in series?
Consider the situation like this: the capacitor is an open circuit. That means that through the 1 ohm resistor, there is no current (which is also why we don't factor it in when calculating the voltage over the 9 ohm resistor). Since there is no current, there is also no voltage drop - V = I*R=0*1=0 V.

This is why the voltage over the capacitor is Vcap = 15-Vr = 15-0 = 15 V.

EDIT: As has been pointed out, the capacitor becomes an open circuit for DC only after some time, during which it charges. After that no current passes through the resistor as the capacitor is now fully charged.