Ok, so I did a bit of math for you, given an estimate of steel density found on the web. Apparently, steel- depending upon alloy- has a typical density between 4.48 and 4.65 ounces per cubic inch. A 1" long section of 1" diameter cylindrical steel with a 308 cal bore weights ~ 3.2 oz. A 1" long section of 0.75" cylindrical steel with a 308 cal bore weighs ~ 1.7 oz
Also online, I found that...
a medium palma barrel tapers from 0.92" to 0.82" over a length of 24.5". 30 inch barrel
a heavy palma barrel tapers from 1" to 0.9" over a length of 24.5". 30 inch barrel
an M24 barrel tapers from 1.2" to 0.9" over a length of 23". 26 inch barrel
http://precisionrifleblog.com/2014/11/18/rifle-barrels-what-the-pros-use/
To get to a 16" barrel you would need to cut 14" from the palma barrels and 10" from the M24 barrel.
This would remove
~ 31.5 oz from the med palma barrel
~ 39 oz from the heavy palma barrel
~ 30oz from the M24 barrel
The weights given in the article for the barrels are 5.35, 6.5, and 7 lbs respectively so you could expect (given no reprofiling of the barrel after cutting)...
a med palma barrel- cut to 16"- to weight 3.4 lbs
a heavy palma barrel- cut to 16"- to weight 4 lbs
an M24 contour barrel- cut to 16"- to weight 5.1 lbs
Given the specific example of a 26" med palma barrel cut down to 16"- you can expect a weight reduction of ~ 1.4 lbs.
If you can give me the diameter of YOUR barrel at the muzzle and at the proposed cut site, I can give you an estimate of the weight reduction within a couple of oz.
Just for reference...
volume of a cylinder = PI * (r^2) * h
volume of a tapered cylinder = (1/3) * PI * ((r1^2) + (r1*r2) + (r2^2)) * h
Further, I assumed that cutting the grooves removes (1/2) of the available material such that the void volume of the bore =
(PI * (0.300^2) * h) + (((PI * (.308^2) * h) - (PI * (0.300^2) * h)) / 2) where h = the length of the barrel section removed.