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In this article, we will explore the chemical basis of how antifreeze (ethylene glycol) lowers the freezing point of water.  We will also discuss the amount of antifreeze required to protect your vehicle to -40˚C by using some fundamental chemical calculations.  If you are not interested in the calculations, simply go to the very end of this article to see the proper percentage of antifreeze as compared to water (it is not simply 50/50 !!).

ΔT(freeze) = K(freeze) x m(solute)

Great, so how does this help me?

Well, if we consider that we know the temperature at which we want the antifreeze to protect water to and we know the cryoscopic constant of water, we can calculate the molality (concentration) of ethylene glycol / water solution required.  With the molality, we can calculate the exact volume of antifreeze required by:  volume, mass, or percentage.  Convenient huh?

Rearranging the above equation to isolate the molality of antifreeze required:

m(solute) = ΔT(freeze) / K(freeze)
m(solute) = | -40˚C |  / (1.86˚C kg / mol)
m(solute) = 21.5 mol / kg solvent

Imagine now that we want to protect 4 liters of water (sorry imperial folks, science is metric 8-)

Knowing that the molality of ethylene glycol must be 21.5 mol / kg solvent, we can calculate the number of moles of ethylene glycol required:

4.0 L H2O = 4.0 kg H2O  (at 4˚C because the density of water is 1.0 g / mL at 4˚C)
# moles ethylene glycol = m(solute) x (mass H2O)
# moles ethylene glycol = (21.5 mol / kg solvent) x (4kg H2O)
# moles ethylene glycol = 86 moles

Ok, now that we have the number of moles of antifreeze, we can calculate the mass of antifreeze required.

mass antifreeze = (# moles ethylene glycol) x (molecular mass ethylene glycol)
mass antifreeze = (86 mol) x (61.2 g / mol)
mass antifreeze = 5263.2g = 5.26 kg of antifreeze

Now that we have the mass of ethylene glycol required, we can calculate the proper % mixture by mass:

% antifreeze by mass = (mass antifreeze) % (mass antifreeze + mass water) x 100%
% antifreeze by mass = [5.26kg / (5.26kg + 4.0kg)] x 100%
% antifreeze by mass = (5.26 / 9.26) x 100%
% antifreeze by mass = 57%

Assuming the density of ethylene glycol to be 1.114 g / mL we can calculate the volume of antifreeze required to protect water to -40˚C

volume antifreeze = (mass antifreeze) x (density antifreeze)
volume antifreeze = (5263 g) / (1.114g / mL)
volume antifreeze = 4724 mL = 4.72 L of antifreeze

Knowing the volume of ethylene glycol required, we can calculate the proper % mixture by volume:

% antifreeze by volume = (volume antifreeze) % (volume antifreeze + volume water) x 100%
% antifreeze by volume = (4.72 L) / (4.72 L + 4.0 L) x 100%
% antifreeze by volume = (4.72  / 8.72) x 100%
% antifreeze by volume = 54%

So, in summation .. to protect 4 L of water to -40˚C a volume of 4.72 L antifreeze (ethylene glycol) is required.  Put another way, if you mix your antifreeze by percentage, you should use roughly 50% water, 50% antifreeze!

Isn't chemistry great!?


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