# Seeking formula for adding sugar dissolved in water to produce specific gravity

I have seen several tables, notably:

But they all differ slightly.

I'd like to test my hydrometer and refractometer with known gravities by adding sugar to a known volume of water (say 1 gallon for example) and taking measurements (like with no sugar, then the correct amount for 1.010/2.5Brix, etc), then increase the sugar and continue taking measurements, just to verify the readings.

However, I can't seem to find where they are getting the numbers from for oz/gallon of sugar. Notably, Jack's table includes sugar added to water, which I believe would be the most useful, since I'll measure the water first, then add the sugar.

This post from adjelange has some more substantial detail, along with a "tl;dr:" that says basically: 1.046 for 1 lb in 1 gallon, and "The density does depend on the type of sugar but is not something you would be able to detect with a hydrometer."

(I found this by doing a google search for "sucrose gravity"; it was the first result.)

• Heh, nice. I will read through that and see if I can glean the formula. Sep 4, 2013 at 15:45

OK, it appears the answer can be derived from the simple fact that "The [Plato] scale expresses the density as the percentage of sucrose by weight". By that token, the sugar content in oz is simply

8.3*16 * degrees plato or brix

To account for adding to water, you could say

(8.3*16)P%+(P%(8.3*16)*P%)

Obviously, the formula you use determine brix would play a large factor as well as the fact that the relationship between degrees Plato and specific gravity (SG) is not linear. I guess that's why Ball, Ack and Plato used tables...

I'm having the same question, and I guess they were trying to say this:

"3.8 lt of solution with 454 gr of sugar is = 1.045~1.046 SG(Relative Density)", which means: Density of solution/Density of water

Bc the solution being 3.8 liters is the only way it makes sense (not diluting 454 gr of sugar in 3.8 lt of water), look:

Using the metric system would be 119.9 gr of sugar per liter (of solution, not water), the sugar has 1.59 gr/cm3 and water is .99821 gr/cm3 at 20ºC so to get the density would be

Weights 119.9 gr sugar and 922.9362 gr water cm3 (ml) 75.4088 cm3 sugar and 924.5912 cm3 (ml)water = 1000 ml of solution

Remember we are trying to get the solution measure, so we are counting that the solution is made by putting sugar and then adding water until it reaches 1000 ml, so we calculate the space sugar occupies and substract it from the 1000 cm3(ml), so we get less water than 1 liter and end with less weight in gr bc water weights less than 1gr (.99821) at 20ºC

So 1042.8362gr/1000cm3(total ml of the solution)= 1.0428362/.9982(density of solution7density of water) = 1.0447~1.045

I've checked this formula with two different sources and its very close, although i would love to be super close

https://www.engineeringtoolbox.com/density-aqueous-solution-organic-sugar-alcohol-concentration-d_1954.html

https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780471790990.oth1

I hope we get close to what they mean