In this chapter we will learn about the composition of chemicals, how to quantify those chemicals and their composition and how to work quantitatively with chemical formulas. We will learn the meaning of the "mole" and how to use it quantitatively.

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:

• this is actually a review, we know that the sum of the mass of all elements in a chemical formula is the mass of that formula or its formula mass and the sum of the mass of all elements in a molecule is the mass of that molecule or its molecular mass (your book calls these formula weights and molecular weights, but recognizing the difference between mass and weight, we will more correctly call them masses)


vitamin C         C₆H₈O₆            6 x 12.0  = 72.0 amu
                                                 8 x   1.0  =    8.0 amu
                                                 6 x 16.0 =   96.0 amu
                                                                  176.0 amu

• however, when we are in the lab, how do we weigh out 176.0 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!

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 (however many atoms is necessary) 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?

• Br, atomic mass 80 amu from the periodic table:


How many atoms or Br does it take to make 80 grams of Br? 
6 x 10²³ atoms = 80 grams of Br

• Mn, atomic mass 55 amu from the periodic table:


How many atoms or Mn does it take to make 55 grams of Mn? 
6 x 10²³ atoms = 55 grams of Mn

• K, atomic mass 39 amu from the periodic table:


How many atoms or K does it take to make 39 grams of K? 
6 x 10²³ atoms = 39 grams of K

• If we did the calculations above more exactly the actual number would be 6.022 x 10²³ and---


we define that number as a mole -- therefore, 1 mole of any element = 6.022 x 10²³ 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) = 
  6.022 x 10²³ atoms, molecules, or donuts (a lot of donuts!)
• 1 mole of cows = 6.022 x 10²³ cows
• 1 mole of Ar atoms = 6.022 x 10²³ Ar atoms

This number is a very important number and is called AVOGADRO'S NUMBER-- memorize it!

Let's review some of the relationships that we have discussed:

• 6.022 x 10²³ 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, and     1 mole CO molecules  = 28.0 g
• 1 atom O = 16.0 amu,   and             1 mole O atoms  = 16.0 g
• 1 molecule O₂ = 32.0 amu, and       1 mole O₂ molecules = 32.0 g
• 1 mole C = 12.0 g = 6.022 x 10²³ C atoms
• 1 mole CO = 28.0 g = 6.022 x 10²³  CO molecules
• 1 mole O = 16.0 g = 6.022 x 10²³  O atoms
• 1 mole O₂ = 32.0 g = 6.022 x 10²³  O₂ molecules
• 1 mole cows = 6.022 x 10²³  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
                                                                                                                   1 mole C

• Likewise, if the formula mass of vitamin C, C₆H₈O₆, is 176.0 amu, then the mass in grams of 1 mole of vitamin C molecules is 176.0 g or 1 mole C₆H₈O₆ = 176.0 g C₆H₈O₆ and from that we get the molar mass of C₆H₈O₆ is 176.0 g C₆H₈O₆             or 176.0 g C₆H₈O₆ = MM
   1 mole C₆H₈O₆                   1 mole

We use molar mass just like a conversion factor in solving problems using dimensional analysis:

• How many moles of vitamin C are contained in 5.00 g of vitamin C?


Begin with the given amount of 5.00 g and use the conversion factor (MM) to cancel grams and obtain the required moles--remember,

1 mole              =            176.0 g C₆H₈O₆       =       1              Our two conversion factors
176.0 g C₆H₈O₆             1 mole

5.00 g vitamin C    x      1 mole           =       0.0284 mole vitamin C
                                176.0 g vitamin C

The difference between moles and molecules is one of scale, for example:

• if we completely decompose one molecule of vitamin C, C₆H₈O₆, (take it apart into its constituent elements) what do we have?

• if we completely decompose one mole of vitamin C what do we have?
  
                         6 moles carbon atoms (or 6 x 6.022 x 10²³ C atoms)
                                                                         1 mole       C atoms
  
                        8 moles hydrogen atoms (or 8 x 6.022 x 10²³ H atoms)
                                                                           1 mole        H atoms                
                      
                        6 moles oxygen atoms (or 6 x 6.022 x 10²³ 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
 

Problems:

• In 25 g of C₁₂H₃₀O₂, THC, (tetrahydrocannibinol) how many moles are there? How many molecules are there?


we wish to convert between grams and moles of the single substance, we need a conversion factor, that conversion factor is the molar mass: 206 g/mol C₁₂H₃₀O₂

25 g C₁₂H₃₀O₂       x       1 mole C₁₂H₃₀O₂    =      0.12 mol C₁₂H₃₀O₂
                                        206 g C₁₂H₃₀O₂

0.12 mol C₁₂H₃₀O₂  x    6.022 x 10²³ molecules C₁₂H₃₀O₂    =        7.2 x 10²² molecules
                                       1 mol     C₁₂H₃₀O₂                                                        C₁₂H₃₀O₂

• 17.5 mg of cocaine (C₁₇H₂₁NO₄) per kg of body weight is a lethal dose. How many moles is that?  How many molecules?


again, the MM is the conversion factor: 303 g/mol C₁₇H₂₁NO₄ but we also need a conversion factor between grams and milligrams (which you should know!)

17.5 mg       x     1 g      x       1 mol C₁₇H₂₁NO₄    =    5.78 x 10 ^-5 mol C₁₇H₂₁NO₄
                     1000 mg            303 g C₁₇H₂₁NO₄

5.78 x 10 ^-5 mol C₁₇H₂₁NO₄     x     6.022 x 10²³ molecules C₁₇H₂₁NO₄
                                                          1 mol C₁₇H₂₁NO₄

= 3.48 x 10¹⁹ molecules C₁₇H₂₁NO₄

• How many grams of nitrogen (N) are contained in 70.0 g of C₆H₅NO₃?


use the MM = 139 g/mol C₆H₅NO₃ to get moles of C₆H₅NO₃

70.0 g C₆H₅NO₃    x         1 mol C₆H₅NO₃       =       0.504 mol C₆H₅NO₃
                                         139 g C₆H₅NO₃

take apart the molecule to determine the ratio of moles of N to moles of C₆H₅NO₃ the conversion factor is the ratio 1 mole N/1 mole C₆H₅NO₃

0.504 mol C₆H₅NO₃       x          1 mol N          =          0.504 mol N
                                                 1 mol C₆H₅NO₃

use the MM of N as a conversion factor to convert to grams of N:

0.504 mol N           x         14.0 g N        =         7.05 g N
                                          1 mol N

• How many moles of O are contained in 1.50 moles of C₆H₅NO₃?


coversion factor:   ratio of moles of O atoms/moles of C₆H₅NO₃

1.50 mol C₆H₅NO₃        x       3 mol O atom       =          4.50 mol O atom
                                              1 mol C₆H₅NO₃

PERCENT CALCULATIONS:

A percent is defined as a fraction based on 100:         part        x       100      =      %
                                                                                whole

for example:        50%         =           50 (part)      =        0.50      =       1/2
                                                       100 (whole)

We use percents to calculate the percent (or part) of the mass of a compound or formula that is due to the mass of a particular atom, for example:

• What is the % N in N₂O₅?


Molar Mass of N₂O₅ is 108.0 amu or 108.0 g
                                                             mole

the part of the mass of the molecule due to N is the %N:

%N      =    (2 x 14.0 amu)     =      28.0 amu        x      100      =       25.9% N
                     108.0 amu                 108.0 amu

since the unit of mass cancels out in the ratio, we can use either amu or g/mol as the mass unit. All we need is the ratio of the mass of the part (total mass of all N atoms in the formula) to the mass of the whole (total mass of the formula).

%O     =   (5 x 16.0 amu)      =          80.0 amu O     x      100       =      74.1% O
                108.0 amu N₂O₅              108.0 amu N₂O₅
notice that the sum of the percentages of each element in a formula must equal 100!

• Two iron ores that have been used as sources of iron are magnetite (Fe₃O₄) and hematite (Fe₂O₃). Which one contains the higher percentage of iron?


first we need the MM of both comounds:      231 amu Fe₃O₄          160 amu Fe₂O₃

next we need the ratio of mol Fe / mol compound for each:

Fe₃O₄     3 mol Fe                         Fe₂O₃      2 mol Fe
             1 mol Fe₃O₄                                   1 mol Fe₂O₃

now compare the masses:

Fe₃O₄     167 amu Fe     x       100      =         72.5% Fe
               231 amu Fe₃O₄

Fe₂O₃     112 amu Fe      x       100     =         70.0% Fe
             160 amu Fe₂O₃

Fe₃O₄ has the greater percentage of Fe

You cannot always simply count atoms, as shown in the following example:



• Which has the greatest percentage of N, N₂O or N₂O₅?


N₂O        28.0 amu N       x         100       =        63.6%
              44.0 amu N₂O

N₂O₅      28.0 amu N        x          100       =        25.9%
           108 amu N₂O₅

although both compounds have the same number of N atoms, N₂O has, by far, the greater percentage of N by mass.

EMPIRICAL AND MOLECULAR FORMULAS:

We know what a molecular formula is: it gives the type and exact number of atoms in a molecule. An empirical formula is a little different: it gives the type and simplest ratio of atoms in a molecule. The difference is exemplified below:

molecular formula                                empirical formula

     H₂O₂ (2:2)                                                   HO (1:1)

      H₂O (2:1)                                                   H₂O (2:1)

     C₆H₆ (6:6)                                                    CH (1:1)

     C₂H₆ (2:6)                                                    CH₃ (1:3)

     N₂O₅ (2:5)                                                   N₂O₅ (2:5)

notice that ratios for the molecular and empirical formulas are the same except that in most cases, the empirical formula exhibits the simplest ratio. In some cases the ratios are exactly the same and the molecular and empirical formulas are exactly the same (the molecular formula already is the simplest ratio).

If we are given the percent masses of each element in a compound, we can use that information to calculate the empirical formula.

If we are given the percent masses of each element in a compound and the molar mass of the compound, we can calculate both the empirical and the molecular formula.

Empirical Formula Problems:

• A new compound is synthesized and sent to the lab for analysis. The lab report says it contains


52.2% C + 13.1% H + 34.7% O = 100.0%. Calculate the empirical formula of the compound.

assume we have 100 g of the new compound (it doesn’t matter how much we have of it, it will still be 52.2% C, 13.1% H, 34.7% O, by mass)
if we have 100 g of compound then 52.2 g of that will be due to the mass of C, 13.1 g will be due to the mass of H and 34.7 g will be due to the mass of O (why did I pick 100 g of compound to begin with?):

52.2 g C     x      1 mole/12.0 g C      =       4.35 mole C

13.1 g H     x      1 mole/1.01 g H     =       13.0 mole H

34.7 g O     x      1 mole/16.0 g O     =         2.17 mole O

The ratio of atoms in the molecule are 4.35 mol C : 13.0 mol H : 2.17 mol O

or C_4.35H_13.0O_2.17

but we never use decimal or fractional numbers as atom ratios in molecular formulas, so
divide all the subscripts by the smallest number to get the simplest ratio C₂H₆O or the empirical formula                                                       4.35 : 13.0 : 2.17         =        2 : 6 : 1
                                             2.17   2.17   2.17



• Calculate the empirical formula of malonic acid which has 34.6% C, 3.9% H, 61.5% O


assume you have 100 g of malonic acid, therefore, you have

34.6 g C      x      1 mol C       =      2.88 mol C
                           12.0 g C

3.9 g H       x       1 mol H      =       3.86 mol H
                           1.01 g H

61.5 g O    x        1 mol O       =      3.84 mol O
                           16.0 g O

divide by the smallest number (2.88) to get the simplest mole ratio of atoms in malonic acid

1 mol C : 1.34 mol H : 1.33 mol O

as you can see, we do not have a whole number ratio and we cannot get a whole number ratio by further dividing by 1. The ratio, itself, is correct, now we have to get it into the simplest set of whole numbers. Notice that our fractions (in the form of decimals) occur in approximately thirds. If we multiply all the numbers by 3, we do not change the basic ratio, but we get whole numbers: 3 mol C : 4 mol H : 4 mol O

The empirical formula for malonic acid is: C₃H₄O₄

Molecular Formula Problems:

We calculated the empirical formula C₂H₆O for the first example above, but what is the molecular formula? All we know is that the simplest ratio of atoms is 2:6:1. There are many possible molecular formulas:

(C₂H₆O)_n where n = 2,3,4….etc. C₄H₁₂O₂, C₆H₁₈O₃ etc.

n         =          actual molar mass
               empirical formula molar mass

We can only determine the correct, actual or exact atom ratio if we know the actual molar mass of the compound:
                Proposed Formula                      Mass

• Calculate the molecular formula of adipic acid which consists of 49.3% C, 6.9% H, 43.8% O and has a molar mass of 146 amu.


assume 100 g of adipic acid, therefore: 49.3 g C; 6.9 g H; 43.8 g O

49.3 g C      x      1 mol C/12.0 g C         =        4.11 mol C

6.9 g H        x      1 mol H/1.01 g H        =        6.83 mol H

43.8 g O      x      1 mol O/16.0 g O        =        2.74 mol O

simplest ratio: 4.11/2.74 : 6.83/2.74 : 2.74/2.74 = 1.5 : 2.5 : 1

multiply by 2 to get whole numbers: 3 : 5 : 2

empirical formula for adipic acid is: C₃H₅O₂

compare the actual molar mass to the empirical formula molar mass:

actual MM 146 amu          =       2       =        n
  E.F. MM 73 amu

(C₃H₅O₂)₂ the real molecular formula for adipic acid:     C₆H₁₀O₄

Click here to go to Chapter 8 Quantities in Chemical Reactions