In the fermentation process, for every molecule of glucose consumed, 2 molecules of CO2 are produced.
Five US gallons @1.085 = 4.276 kg sucrose. Sucrose is 342.30 g/mol, so you have 12.49 mol of sucrose. This gets converted into 2 molecules of glucose, so 24.98 mol glucose. There would thus be twice that amount - 49.97 mol - of CO2 produced @44.01 g/mol, so 2,199 kg CO2.
Initially, assuming you had a solution of 5 US gallon water plus 4.276 kg sucrose, the mass was (18.93L*1kg/L) + 4.276kg = 23.21 kg. The mass drop should thus be 9%.
Interestingly, it doesn't really match reality: We would also expect 2mol of ethanol per mol of glucose, so again, 49.97 mol ethanol @46.07 g/mol = 2302g @0.789g/ml = 2,917 liters. So %ABV should be roughly 2.917/18.93 = 15.4% ABV. Experimentally, it's rather 11.4%.
[taking this too seriously:] So if looking instead at the experimental data, 11.4%ABV is 2.16L ethanol/18.93L solution. 2.16L ethanol @0.789g/ml = 1704g @46.07g/mol = 36.99 mol. We should have about the same amount of CO2 produced (as said before, the fermentation process is: 1 mol glucose -> 2 mol CO2 + 2 mol ethanol). So 36.99 mol CO2 @44.01g/mol,
1.628 kg CO2 (again, ~9.3% mass drop) and that's what I'd bet on
Please weight it before/after and report your results