Example
Let's say the fuel has energy density 44,558,176 Joules per kilogram. (I've taken that figure from a simple enthalpy calculation on dodecane.)
Just take the reciprocal and fuel use would be 1/44,558,176 kilograms per Joule for a perfect engine. Measuring energy in kilowatt hours instead of Joules - a simple change of units - and using grams instead of kilograms, gives 80.79 grams per kilowatt hour. That's still just the same energy density of the fuel, restated in other units.
If we assume 40% efficiency ....
80.79 / 0.40 = 201.98 g/kW-hr
So, you need to know
1. your fuel energy density,
2. your engine power (being generated), and
3. its efficiency.