To boil 2 cups of water (68 F to 208 F) using 4.9 grams of fuel means that the efficiency is ~70%. That is a highly efficient system.
(sorry, cut and pasted from Excel)
Isobutane 45.2088 J/g
Fuel consummed 4.9 g
Total energy available 221523.12 Joules
M*Cp*DeltaT
2 cups of water 473 g
water 4.186 J/gC 4.186 J/g°C
Start 20.0 C 68 F
end 98.3 C 209 F
Energy Required 155098 Joules
Efficeincy 70%
If you believe in math (which I do)
Boiling the same 2 cups of water using 2.0 grams of fuel means that your efficiency is 172%.
Math is fairly predictable.  My 2 cents.
P.S. there is a simple test for this as well. At 100% efficiency, the exhaust air temperature should be the same as the ambient air temperature.







