See
http://www.batteryshowdown.com - they tested a bunch of batteries at 200mA and 1A :
200mA : results and chart with all batteries:
http://www.batteryshowdown.com/results-lo.html![](http://www.batteryshowdown.com/static/images/curve-200ma-135.png)
1A : results and chart with all batteries :
http://www.batteryshowdown.com/results-hi.html![](http://www.batteryshowdown.com/static/images/curve-1000ma-135.png)
So it depends on how much current the battery needs to provide and the battery type and the quality of the battery.
I don't know the capacity of C or D type batteries but I suspect it would be around 3-4000mAh but then you're probably better off with some lithium 18650 or regular flat batteries.
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Now I don't get exactly what has to be powered from 9v, because you talk something about an amp and other things.
Measure the current consumption of the device that needs 9 amps then go from there. If that thing only needs 10-100mA, you could do something like I did for my multimeter, a dc-dc boost regulator:
https://www.youtube.com/user/HelpediaVideos Basically, the multimeter I have takes a 9v battery but it shows low voltage indicator at around 5.5v so I used a LT1307 to boost 1-3v to about 5.7v (just above the low voltage battery). The boost regulator can output 5v from 1v, from about 1.15-1.2v it can probably do 7-8v, well within a 9v battery range. So you could replace the 9v adapter with as little as one 1.5v alkaline battery (but it would be more efficient to use 2 or 3 batteries) and this boost dc-dc converter, as long as the device doesn't draw more than 100mA (the maximum limit this LT1307 can do)
If you need more current, you'd be better off with putting 8 batteries in series and just running it directly, or maybe through a switching regulator in SEPIC configuration.