Input current in an ic
An IC is a load, it's just more complex than a simple resisitve or capacitive load.  There are a few things to note about op-amps, though.  They will consume some current to keep their circuitry biased properly, but they will also source current from the power supply pins to whatever is connected to the output.  This makes it a little difficult to characterize an op-amp's current requirements in isolation as it really depends on what ciruit is built up around the op-amp.  Looks like the power consumption of that amplifier is max 85 mW, so you can work out the current draw from that.  

However, you probably don't want to use a 741 for your project if your only power supply is 9.6 volts.  Here's why.  The output voltage swing with +/- 15 volt rails is only guaranteed to +/- 12 volts into a 10k load and +/- 10 volts into a 2k load.  We'll call 9.6 volts 10 volts here, so a split rail supply would be +/- 5 volts.  If you assume the margin is independent of voltage, powering the amp on +/- 5 volts will provide minimum 0 volts of output swing.  In other words, nada.  You either need to use a higher voltage power supply, or what is called a 'rail to rail' op amp.  A rail to rail op amp can drive its output to within less than a volt of the power supply rails.  A rail to rail op amp powered on +/- 5 volts should easily provide a +/- 4 volt swing into a reasonable load.  

For input offset current and other op-amp parameters, take a look at chaper 11 of http://www.ti.com/lit/an/slod006b/slod006b.pdf .  This guide discusses many of these parameters.  Also, take a look at the rest of the guide - it's a good reference for anyone using op-amps.