Ethylene glycol is the common name for the liquid used to...

Ethylene glycol is the common name for the liquid used to keep the coolant in automobile cooling systems from freezing. It is $38.7 \%$ carbon, $9.7 \%$ hydrogen, and $51.6 \%$ oxygen by mass. Its molar mass is $62.07 \mathrm{g} / \mathrm{mol}$ and its density is $1.106 \mathrm{g} / \mathrm{mL}$ at $20^{\circ} \mathrm{C}$ a. What is the empirical formula of ethylene glycol? b. What is the molecular formula of ethylene glycol? c. In a solution prepared by mixing equal volumes of water and ethylene glycol, which ingredient is the solute and which is the solvent?

Ethylene glycol is the common name for the liquid used to keep the coolant in automobile cooling systems from freezing. It is $38.7 \%$ carbon, $9.7 \%$ hydrogen, and $51.6 \%$ oxygen by mass. Its molar mass is $62.07 \mathrm{g} / \mathrm{mol}$ and its density is $1.106 \mathrm{g} / \mathrm{mL}$ at $20^{\circ} \mathrm{C}$
a. What is the empirical formula of ethylene glycol?
b. What is the molecular formula of ethylene glycol?
c. In a solution prepared by mixing equal volumes of water and ethylene glycol, which ingredient is the solute and which is the solvent?

Key Concepts

Solute and Solvent in Solutions

Understanding which component serves as the solute (the substance present in a lesser amount) and which is the solvent (the substance present in a greater amount or capable of dissolving the solute) is critical. In a solution, the physical properties and concentration of each component determine their roles, particularly when mixing liquids of comparable properties by volume.

Molecular Formula Determination

This concept builds on the empirical formula by using the given molar mass to find the actual number of atoms in a molecule. Once the empirical formula mass is calculated, the molar mass of the compound can be divided by the empirical formula mass to determine the multiplication factor that gives the molecular formula.

Empirical Formula Determination

This concept involves using the percent composition of an element or compound to calculate the simplest whole-number ratio of atoms present in the substance. By converting each element's mass percentage to moles (using their atomic masses), one finds the ratio of the atoms, which is then simplified to yield the empirical formula.

Transcript

00:01 Problem 112 is a review of the calculation of an empirical formula from percent mass composition.

00:09 Then knowing the empirical formula and the actual molar mass of the compound, which in this case is ethylene glycol, you need to know how to convert the empirical formula into the molecular formula.

00:22 And then last of all, it asks you to recall the definitions of solute and solvent.

00:31 Ute is the one that is in least amount.

00:34 Amount can be expressed in moles most appropriately.

00:39 And then the solvent is the one in largest amount.

00:44 So to determine the empirical formula from mass percent, if it's 38 .7 percent carbon, then for every hundred grams of the compound, we have 38 .7 grams of carbon.

00:55 So we'll assume 100 grams of the compound, which then allows us to convert the 9 .7 percent hydrogen into 9 .7 grams and the 51 .6 % oxygen into 51 .6 grams.

01:10 We convert each of these from grams into moles by dividing by the corresponding molar mass.

01:16 Molar mass of carbon, molar mass of hydrogen, molar mass of oxygen.

01:22 And these are the mole values we get.

01:24 As you recall, once you know the moles that are associated with 100 grams of the compound, the moles of each element associated with the grams of the compound, you then divide by the smallest mole number in an attempt to get integers.

01:40 And in this case, we get one mole carbon for one mole oxygen for three moles hydrogen...

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