Understanding Moles, Molar Mass, and Chemical Formulas, Assignments of Dimensional Analysis

Download Understanding Moles, Molar Mass, and Chemical Formulas and more Assignments Dimensional Analysis in PDF only on Docsity! The Mole and Chemical Formulas A mole (also known as Avogadro's Number) is the number that is used in making calculations involving atoms and molecules. Given the relatively microscopic size of the particles, it would be rather difficult (read: utterly pointless) to calculate amounts in terms of numbers of atoms. So, we use the mole. A mole, (mol for short) is equal to (all together now) 6.022 x 1023 atoms or molecules. The atomic mass given for an element on the periodic table measured out in grams is equal to one mole of atoms of that element. Thus, the Molar Mass (MM) of elements and compounds is the mass, in grams, equal to the atomic and formula masses of those elements and compounds. The unit of Molar Mass is grams/mole. Empirical and Molecular Formulas Believe it or not, there is more than one kind of chemical formula (insert a sharp gasp of surprise here). Judging from the heading of this section, they would be empirical and molecular formulas. Empirical formulas give the lowest whole number ratio of the atoms in a compound, whereas the molecular formula gives the exact composition of one molecule. Empirical Molecular HO H2O2 CH C2H2 CH C6H6 NO NO NO2 N2O4 Empirical formulas can be calculated using experimental data: • Given that a certain compound is 69.58% Ba, 6.090% C and 24.32% O, calculate the empirical formula of this compound. • Assume that you have 100.00 g of the compound: • 69.58% Ba = 69.58 g Ba • 6.090% C = 6.090 g C • 24.32% O = 24.32 g O • Convert the mass of each element to moles of that element: • Ba = (69.58 g Ba)(1 mol Ba/137.3 g Ba) = 0.5068 mol Ba • C = (6.090 g C)(1 mol C/12.01 g C) = 0.5071 mol C • O = (24.32 g O)(1 mol O/16.00 g O) = 1.520 mol O • Divide through each value by the smallest number of moles to get a 1 : 1.001 : 2.999 ratio, which rounds of nicely to give the formula BaCO3 Molecular formulas can be calculated using empirical formulas: • Given that the empirical formula of a compound is CH and the molar mass is 104 g/mol, calculate the molecular formula. mass of C = 12.0 g/mol mas of H = 1.01 g/mol empirical formula mass = 13.0 g/mol • to find the number of CH units in the compound: • # CH units = (104 g/mol)(1 mol/13.0 g) = 8.00 • Molecular formula = 8(CH) or C8H8 Calculating Molarity The properties and behavior of many solutions depend not only on the nature of the solute and solvent but also on the concentration of the solute in the solution. Chemists use many different units when expressing concentration; however, one of the most common units is molarity. Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution: Molarity (M) = moles solute liters solution For example, a 0.25 M NaOH solution (this is read as 0.25 molar) contains 0.25 moles of sodium hydroxide in every liter of solution. Anytime you see the abbreviation M you should immediately think of it as mol/L. In order to calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution. To calculate molarity: 1. Calculate the number of moles of solute present. 2. Calculate the number of liters of solution present. 3. Divide the number of moles of solute by the number of liters of solution. Instead of calculating the moles of solute and liters of solution present individually, you can also string all the calculations together in one problem: 1. Divide the amount of solute by the volume of solution (regardless of the initial units given). 2. Use dimensional analysis to convert the amount of solute to moles of solute. 3. Use dimensional analysis to convert the volume of solution to liters of solution (if necessary). The following example will illustrate both methods. Example: What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in enough water to make a total of 225 mL of solution? Method 1: • Calculate the number of moles of solute present. mol NaOH = 15.0g NaOH x 1 mol NaOH 40.0 g NaOH mol NaOH = 0.375 mol NaOH • • Calculate the number of liters of solution present. L soln = 225 mL x 1 L = 0.225 L soln 1000 mL • Divide the number of moles of solute by the number of liters of solution. M = 0.375 mol NaOH = 1.67 M NaOH 0.225 L soln Method 2: • Divide the amount of solute by the volume of solution (regardless of the initial units given). M = 15.0 g NaOH 225 mL soln • Use dimensional analysis to convert the amount of solute to moles of solute. M = 15.0 g NaOH x 1 mol NaOH 225 mL soln 40.0 g NaOH • Use dimensional analysis to convert the volume of solution to liters of solution (if necessary). M = 15.0 g NaOH x 1 mol NaOH x 1000 mL soln 225 mL soln 40.0 g NaOH 1 L soln M = 1.67 mol NaOH = 1.67 M NaOH L soln As you can see, both methods give exactly the same result. Choose whichever method is most comfortable for you. Just remember that ultimately your units must be mol/L (= M). Practice Problems 1. What is the molarity of a solution that contains 1.724 moles of H2SO4 in 2.50 L of solution? 2. What is the molarity of a solution prepared by dissolving 25.0 g of HCl (g) in enough water to make 150.0 mL of solution?