Plato degree to specific gravity conversion question

Question

I have a question regarding the conversion from Plato degrees to specific gravity, and the widely used value of 46ppg for sugar dissolved in water.

According to wikipedia, plato degrees are defined by the relative amount of sugar by mass (and not volume).

Consequently, 1 pound of sugar (0.453 kg) in 1 gallon (3.785 liters) gives 0.453/(3.785*0.998+0.453)=10.71% sugar in mass, taking a density of water of 0.998 kg/liter at 20°C. So this would be 10.71 Plato degree.

However, formulas or tables for conversion between Plato degrees and specific gravity give 11.5% sugar for a specific gravity of 1.046.

It appears that there is something I haven't taken into account, but what?

Answer 1

I had to look up where you got the 1.046 value from. It seems it's common knowledge that "one pound of sugar dissolved in one gallon of water" yields a solution with a specific gravity of 1.046. However, that's not exactly correct. It is one gallon of a solution of one pound of sugar in water that has a SG of 1.046.

The distinction is that the mass of "one pound of sugar dissolved in one gallon of water" is the sum of the mass of one pound of sugar (0.453 kg) and the mass of one gallon of water (3.785 * 0.998) while the mass of "one gallon of a solution of one pound of sugar in water" is the sum of the mass of one pound of sugar (0.453 kg) plus the mass of however much water is required to reach a total volume of one gallon.

To put it another way, in the first case you're pouring one pound of sugar and one gallon of water into a container. In the second case, you're pouring one pound of sugar into a one-gallon container and then topping up with water.

Another problem is that since you're working with pure sugar and water, you should actually be referencing Brix values. Brix and Plato don't match up exactly, though I can't explain the difference between the two.

Wikipedia phrases it succinctly as, "One degree Brix is 1 gram of sucrose in 100 grams of solution." Assuming one milliliter of water has a mass of one gram, then a total mass of 100 grams would mean the sugar was added to only 99 ml of water. So, to use your formula above, 1 / (99 * 0.998 + 1) = 1.00% Brix.

Working backward with your formula from a 1.046 SG (11.4% Brix), we can calculate the amount of water you would need to add to one pound of sugar to get one gallon of solution.

0.453 / (V * 0.998 + 0.453) = 0.114
0.453 = 0.114 * (V * 0.998 + 0.453)
0.453 / 0.114 = V * 0.998 + 0.453
0.453 / 0.114 - 0.453 = V * 0.998
(0.453 / 0.114 - 0.453) / 0.998 = V
3.52773969 = V

So you'd need to add one pound of sugar to 3.53 liters (0.93 gal) of water to get one gallon of solution at 1.046 SG.

I'm sorry I can't link to a definitive reference. This is what I've cobbled together from various sources.

Answer 2

I would like to add,that this is an 'American' question with an 'American' answer. As the US gallon is 3.78541 Litres, and the UK gallon is 4.54609 Litres, we see that making up a solution to one gallon in each case will result in the UK measurement being lower than the US measurement. and so for the SG case in question,ie,- 1046, we get, 3.78541 divided by 4.54609, and then multiplied by 46. answer,- 38.1786182547, If you want to be exact. However, different manufacturers of ord' cane sucrose vary greatly. The scale on my UK suchrometer (hydrometer) gives me less than 6% ABV from one pound of sugar solution, with the mark on it at 1036.5. As ord' sugar ferments to near zero it can be seen that as the multiplying factor for sugar is .1632 then,- 36.5 x .1632 = 5.9568%, hope this helps someone.