Question

In 2006, Professor Galen Suppes, from the University of Missouri-Columbia, was awarded a Presidential Green Challenge Award for his system of converting glycerin, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), a by-product of biodiesel production, to propylene glycol, \(\mathrm{C}_{3} \mathrm{H}_{6}(\mathrm{OH})_{2}\). Propylene glycol produced in this way will be cheap enough to replace the more toxic ethylene glycol that is the primary ingredient in automobile antifreeze. (a) If \(50.0\) mL of propylene glycol has a mass of \(51.80 \mathrm{~g}\), what is its density? (b) To obtain the same antifreeze protection requires \(76 \mathrm{~g}\) of propylene glycol to replace each \(62 \mathrm{~g}\) of ethylene glycol. Calculate the mass of propylene glycol required to replace \(1.00\) gal of ethylene glycol. The density of ethylene glycol is \(1.12 \mathrm{~g} / \mathrm{mL}\). (c) Calculate the volume of propylene glycol, in gallons, needed to produce the same antifreeze protection as \(1.00\) gallon of ethylene glycol.

Short answer

The density of propylene glycol is \(1.036 \mathrm{~g/mL}\). To replace 1 gallon of ethylene glycol, \(5208.95 \mathrm{~g}\) of propylene glycol is required, with a volume of \(1.33 \mathrm{~gallons}\).

Step-by-step solution

Calculating the density of propylene glycol

To find the density of propylene glycol, we can use the formula: Density = mass / volume Given: mass = 51.80 g volume = 50.0 mL D = 51.80 g / 50.0 mL = 1.036 g/mL The density of propylene glycol is 1.036 g/mL. (b) Finding the mass of propylene glycol needed to replace 1.00 gal of ethylene glycol

Calculate the mass of 1.00 gal of ethylene glycol

First, we need to convert gallons to milliliters: 1.00 gal = 3,785.41 mL (1 gal = 3,785.41 mL) We know that the density of ethylene glycol is 1.12 g/mL, so: mass_ethylene_glycol = volume x density mass_ethylene_glycol = 3,785.41 mL x 1.12 g/mL = 4,239.66 g

Calculate the mass of propylene glycol needed

We are given that 76 g of propylene glycol is needed to replace 62 g of ethylene glycol. So we can use this ratio to find out the mass of propylene glycol required: propylene_glycol_needed = (76 g / 62 g) x mass_ethylene_glycol propylene_glycol_needed = (76 g / 62 g) x 4,239.66 g = 5,208.95 g The mass of propylene glycol needed to replace 1.00 gal of ethylene glycol is 5,208.95 g. (c) Finding the volume of propylene glycol needed to produce the same antifreeze protection as 1.00 gallon of ethylene glycol

Calculate the volume of propylene glycol needed

We know the mass and density of the propylene glycol needed, which we can use to find the volume: volume_propylene_glycol = mass / density volume_propylene_glycol = 5,208.95 g / 1.036 g/mL = 5,025.91 mL

Convert the volume from milliliters to gallons

Now, we will convert the volume from milliliters to gallons: volume_propylene_glycol_gal = 5,025.91 mL / 3,785.41 mL/gal = 1.33 gal The volume of propylene glycol needed to produce the same antifreeze protection as 1.00 gallon of ethylene glycol is 1.33 gallons.

Key concepts

Molar Conversion

Understanding molar conversion is essential in chemistry, especially when it comes to complex chemical processes like converting by-products into useful materials. In the case of transforming glycerin into propylene glycol, knowledge of molar conversion helps quantify the reactants and products in the reaction.

Molar conversion relates to the use of Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mol, to switch between the mass of a substance and the number of molecules or atoms it contains. Effectively, this conversion factor allows chemists to use the mole as a bridge between the microscopic world of atoms and the macroscopic world we can measure in the lab.

For example, if we know the molar mass of a substance and the amount of the substance in grams, we can determine how many moles we have. Vice versa, if we are given the number of moles, we can calculate the mass in grams. This step is essential in almost all stoichiometric calculations including the calculation of yields of chemical reactions and the amount of a by-product that can be converted into a useful compound.

Chemical By-product Utilization

The utilization of chemical by-products is a key concept in green chemistry and sustainable industrial practices. By transforming by-products, such as glycerin from biodiesel production, into valuable chemicals like propylene glycol, we not only reduce waste but also create economic value. This concept promotes the efficient use of resources and the creation of a circular economy.

In the given problem, glycerin is converted to propylene glycol, which can then be used as an antifreeze. This conversion process involves a series of chemical reactions, and its efficiency depends on many factors including reaction conditions and the purity of the by-product. Chemical engineers and chemists work together to optimize these processes to achieve the highest yield and purity of the target product while keeping costs low.

Furthermore, utilizing by-products effectively reduces the environmental impact of chemical manufacturing by decreasing the amount of waste that needs to be managed and can even mitigate the use of toxic or less sustainable substances, as in the case of replacing ethylene glycol with propylene glycol in antifreeze solutions.

Antifreeze Properties

Antifreeze agents play a crucial role in maintaining the function of engines and other systems under cold conditions. The properties of antifreeze are determined by its chemical composition, which needs to lower the freezing point of water to prevent the liquid from freezing and causing damage.

Propylene glycol, the focus of the textbook exercise, is valued for its lower toxicity and its ability to serve as a safer alternative to ethylene glycol in antifreeze formulas. To ensure the same level of antifreeze protection, chemists must understand the relationship between the mass of ethylene glycol currently used and the mass of propylene glycol required to achieve the same effect.

The efficacy of an antifreeze is related to its thermal properties, such as its freezing point depression, boiling point elevation, and heat capacity. When evaluating a new antifreeze agent, it is also necessary to consider its compatibility with the materials it will come into contact with, its toxicity, and how it disperses heat to protect the mechanism it serves.