QUANTITIES, MOLES & EQUATIONS CONVICTIONS ARE MORE DANGEROUS THAN LIES. NIETZSCHE Ch 4.1 J.C. Rowe Windsor University School of Medicine

LEARNING OBJECTIVES  to understand molecular and formula masses and be able to relate them to real quantities  to understand the mole and be able to use it quantitatively  to understand molar mass and be able to use it to convert between grams and moles of a substance  to understand how to calculate and use percent composition  to understand the difference between empirical and molecular formulas and be able to calculate both from chemical analyses

MOLECULAR AND FORMULA MASSES  the sum of the mass of all elements in a chemical formula is the mass of that formula or its formula mass (formula weights).  the sum of the mass of all elements in a molecule is the mass of that molecule or its molecular mass. (molecular weight)

 Molecular formula of vitamin C is C 6 H 8 O 6  Thus molecular mass is (6 x 12.0)+(8 x 1.0)+(6 x 16.0) = = 72.0 amu amu amu =176.0 amu  when we are in the lab, how do we weigh out amu? What is an amu relative to a mass that a lab balance can weigh? This is where the definition of the mole is handy!

Actual Atomic Mass (Ar)  C-12 = 2.00 x kg  Oxygen = 2.66 x kg  Bromine = 1.33 x kg  Silver = 1.8 x kg  Ar = (Element) x 12 Mass of one C-12

MOLE  We can measure masses in amu but how do we relate that to mass in grams?  Since the masses of atoms are too small to be able to weigh individually on a balance, we measure a "gaggle, gross, bunch, dozen" of atoms on the balance.  In other words, we define a quantity of atoms which have the same numerical mass in grams as the numerical mass in amu.

So how many atoms does it take to make, say, 1.00 g of H?  We know the mass of an H atom is about 1.7 x g  Using dimensional analysis: 1.0 g H x 1 atom H ~= 6.0 x atoms of H 1.7 x g H  12.0 g C x 1 atom C ~= 6.0 x atoms of C 2.0 x g C  Why did I choose 1.0 g of H and 12.0 g of C? Notice that these respective numerical values are the same numerical values of the masses of these atoms in amu: H = 1.0 amu and C = 12.0 amu from the periodic table

 Also notice the number of atoms required in both cases--6.0 x atoms--is the same!  This is showing us a pattern— if we want to equate the mass of any atom on the periodic table in amu to its respective mass in grams, we need 6.0 x of the atoms for it to work.

Pick any atom on the periodic table:  Br, atomic mass 80 amu from the periodic table: How many atoms of Br does it take to make 80 grams of Br? 6.0 x atoms = 80 grams of Br  Mn, atomic mass 55 amu from the periodic table: How many atoms of Mn does it take to make 55 grams of Mn? 6.0 x atoms = 55 grams of Mn

 K, atomic mass 39 amu from the periodic table: How many atoms of K does it take to make 39 grams of K? 6.0 x atoms = 39 grams of K If we did the calculations above more exactly the actual number would be x and ……

 we define that number as a mole -- therefore, 1 mole of any element = x atoms of that element. You've used equalities like this before, it is just another definition of an equality: 1 dozen of anything (donuts, for example) = 12 of anything (donuts) 1 mole of anything ( atoms, molecules or donuts, for example) = x atoms, molecules, or donuts (a lot of donuts!) 1 mole of cows = x cows 1 mole of Y atoms = x Y atoms

this number x is a very important number and is called AVOGADRO'S NUMBER- memorize it!  Let's review some of the relationships that we have discussed:  x atoms = one mole of atoms  1 atom C = 12.0 amu, and 1 mole C atoms = 12.0 g  1 molecule CO = 28.0 amu, and1 mole CO molecules = 28.0 g  1 atom O = 16.0 amu, and 1 mole O atoms = 16.0 g  1 molecule O 2 = 32.0 amu, and1 mole O 2 molecules = 32.0 g

Let's review some of the relationships  1 mole C = 12.0 g = x C atoms  1 mole CO = 28.0 g = x CO molecules  1 mole O = 16.0 g = x O atoms  1 mole O 2 = 32.0 g = x O 2 molecules  1 mole cows = x cows

MOLAR MASS  From this information we can define something called the molar mass (MM) of an atom:  from the equality, 1 mole C = 12.0 g C we get the molar mass of C or 12.0 g C = MM 1moleC  Likewise, if the formula mass of vitamin C, C 6 H 8 O 6, is amu, then the mass in grams of 1 mole of vitamin C molecules is g

 1 mole C 6 H 8 O 6 = g C 6 H 8 O 6 and from that we get the  The molar mass of C 6 H 8 O 6 is g C 6 H 8 O 6 1 mole C 6 H 8 O 6  or g C 6 H 8 O 6 =MM 1 mole

How many moles of vitamin C are contained in 5.00 g of vitamin C?  We use molar mass just like a conversion factor in solving problems using dimensional analysis:  Begin with the given amount of 5.00 g and use the conversion factor (MM) to cancel grams and obtain the required moles

 -remember, Our two conversion factors 1 mole = 176.0g C 6 H 8 O g C 6 H 8 O 6 1 mole  5.00 g vitamin C x 1mole = mole vitamin C g vitamin C

The difference between moles and molecules is one of scale, for example :  if we completely decompose one molecule of vitamin C, C 6 H 8 O 6, (take it apart into its constituent elements) what do we have?  6 carbon atoms  8 hydrogen atoms  6 oxygen atoms

if we completely decompose one mole of vitamin C what do we have?  6 moles carbon atoms (or 6 x x C atoms) 1 mole C atoms 8 moles hydrogen atoms (or 8 x x H atoms) 1 mole H atoms 6 moles oxygen atoms (or 6 x x O atoms) 1 mole O atoms

 THE MOLAR MASS IS ALWAYS THE RATIO OF THE MASS (gram mass from the periodic table) OF AN (atom, molecule or formula) TO ONE MOLE OF THE (atom, molecule or formula) AND IS USED TO CONVERT BETWEEN MASS AND MOLES

Hint for solving chemistry problems  Atom Molecules Mole Mass (in gram)  (1) formula  (2) Avogadro number ( 6.022x10 23 )  (3) molecular weight (M.W.) 1 2 3

It is the attitude of mind developed in the student as he proceeds with his studies, and not primarily the information he acquires, which determine the character and extent of his education. Luther Eisenhardt