How to rationalize load cell readings
You are going to have to amplify the signal in any case, so you can scale the voltage there. 

In your case you have a full scale voltage of 5.0986*1.7576 = 8.9613mV for 5kg. 

If you have a 16 bit ADC (65536 counts) and a 2.5V reference and you want 20000 counts for 5kg (full scale of the load cell) you need to amplify that voltage to 0.76295V, so you need a gain of 85.138. Generally you want to use as much of the ADC range as possible (allowing for possible overrange) so you might want to double that gain for 40,000 counts so you are using more than 60% of the ADC range. 

On the other hand, for a display ADC with 20,000 counts (0-19.999) you would probably prefer to display 5.000 for 5kg so you will amplify it to 0.50000 of the reference voltage (assuming the ADC displays 10000 for input = reference). 



Since you are using ratiometric measurement, the excitation voltage cancels out (though, of course it matters for the noise floor, and hopefully it is low-noise enough not to add excess noise in itself). 

So you look up the relationship of the data to input voltage of your ADC. Full scale (assuming bipolar mode) is typically something like 

\$\pm\frac{Vref}{2*PGA}\$, so with a gain of 128 full scale is +/-Vref/256
Full scale of your load cell is 1.7576E-3*Vref, so you will get a reading in fraction of full scale of 1.7576E-3 * 256 = 0.44994, which is 3774370, or 0x3997A2 in hex (for 5kg). You can easily scale that number to be in engineering units or whatever else you need by a multiplication. 

Of course there will likely be offset, gain trims and perhaps tare weight that have to be applied in the calculation in a real application.