CHEM 499 Lecture Notes - Molar Concentration, Triethylene Glycol, Analytical Chemistry
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Relating the Numbers of Moles of Reactant and Products
Problem: How many moles of CO2 are produced in the combustion of 2.72 mol of triethylene glycol,
C6H14O4, in an excess of O2?
An excess of oxygen gas means that triethylene glycol is the limiting reactant.
Balance the equation: 2 C6H14O4 + 15O2 12CO2 +14H2O
? mol CO2 = [2.72 mol C6H14O4 x (2 mol CO2 / 2 mol C6H14O4)] = 16.3 mol CO2
Chemical Reactions in Solutions
Most chemical reactions take place in solution partly because mixing the reactants helps to achieve the
close contact between atoms, ions, or molecules necessary for the reaction to occur.
Solvent determines whether the solution exists as a solid, liquid, or a gas. However, for learning
purposes, the aqueous (aq) solution will be used as a solvent.
Solute is the material being dissolved by the solvent.
Aqueous reactions can be grouped into three general categories: Precipitation reactions, acid-base
neutralization reactions and oxidation-reduction (red ox) reactions.
Molarity is a solution property defined as the number of moles solute per liter of solution
Molarity (M) = (amount of solute, in moles / volume of solution, in liters)
The term M stands for the term molar or mol/L.
Problem: A solution is prepared by dissolving 25.0 mL ethanol, C2H5OH (d= 0.789 g/mL), in enough water
to produce 250.0mL solution. What is the molarity of ethanol in the solution?
Solution: To determine the molarity, we must first find how many moles of ethanol exists in a 25.0 mL
? mol C2H5OH = [25.0 mL C2H5OH x ( 0.789g C2H5OH/ 1 mL C2H5OH) x ( 1 mol C2H5OH/ 46.07 g C2H5OH)]
= 0.428 mol C2H5OH
Now apply the definition of molarity to this answer. Thus divide the moles of C2H5OH by the solution
Molarity = (0.428 mol C2H5OH / 0.2500 L soln) = 1.71M C2H5OH
Preparation of a Solution
1. Weigh the solid sample.
2. Dissolve it in a volumetric flask partially filled with solvent.
3. Carefully fill to the mark.