# Calculating the weight of a chemical within a liquid mixture

I want to know the weight of ethylene glycol in a mixture of Shellzone 50/50 Antifreeze. I have taken the mid-range of 40-60% from the safety data sheet(SDS) for the amount of ethylene glycol in the mixture and used the number 50 to estimate that 50% of the mixture is ethylene glycol. The specific gravity of the Shellzone 50/50 Antifreeze is 0.909, also taken from SDS, so I have calculated the solution to be 7.59 lbs/gal.

Would I be correct in saying that 1 gallon of Shellzone 50/50 Antifreeze contains 3.79 lbs of antifreeze?

EDIT:

Would I be correct in saying that 1 gallon of Shellzone 50/50 Antifreeze contains 3.79 lbs of ethylene glycol?

See below for calculations.

I believe that the amount of ethylene glycol in one gallon of Shellzone 50/50 Antifreeze would be 3.79 lbs because the concentration is a (% w/w) according to the SDS. Therefore, 40 - 60 percent of the mixture should contain 3.03 - 4.55 lbs of ethylene glycol or 3.79 lbs at its midpoint.

A coworker discussed with me that this may be incorrect because I did not use the specific gravity of ethylene glycol. I chose not to use the specific gravity of ethylene glycol because ethylene glycol and Shellzone Antifreeze 50/50 are not equivalent. Even though both solutions have a chemical in common, they are not the same solution.

I do not think it would be appropriate to use that number to calculate its mass since they are not equivalent. Similarly, if I were to apply the same logic to the other constituents, I would need to use the specific gravity for water and diethylene glycol to calculate the weight of each chemical. I think that this would be unnecessary since the relative density is already given and that the mixture is designed to be homogenous. It would seem fair to suggest that the weight of each constituent can be calculated by using the concentration and the relative density.

The only other potential problem with this calculation is the variation in the actual specific gravity due to temperature but that is not important for the purpose of this calculation.

• Provide detailed calculation and reasoning why it should be correct and why it may not be correct. Start with replacing the word antifreeze with ethylene glycol, as the antifreeze is the whole mixture. Jul 14 '21 at 16:06
• You procedure is correct. 50% w/w means 1/2 of total mass. You would need specific gravity of the antifreeze AND of ethylene glycol, if you worked with V/V%, similarly as for ethanol. As 50% V/V ethanol is 1 volume of ethanol + 1 volume of water, giving 1.96 volumes of the mixture(volumes are not additive). Jul 14 '21 at 17:45
• @Poutnik, I am glad that my edit was able to help you understand more clearly. I appreciate you taking the time to answer my question. Jul 14 '21 at 17:54
• Please give us values in usual units ! And not in lbs per gallon. Is lbs a pound ? And why is the word antifreeze crossed ? What is the meaning of your first sentence? I copy : "a mixture of Shellzone 50/50 Antifreeze" ? Try to be clear ! Jul 14 '21 at 20:40
• @Maurice I wanted to ask if he means (UK wine/US) gallon OR Imperial gallon, omitting corn/dry gallon as probably not applicable to liquids. OTOH lbs means pounds I guess, which are hopefully same everywhere. Jul 15 '21 at 7:58