Stoichiometry Tutorial (not complete)

Question

Stoichiometry Tutorial  (not complete)

aaronq · Answer

$\sf \large \underline{What~ is~ a~ mole?}$

A $\sf \color{red}{mole~ is~ a~ number}$, it is equal to $\sf \large 6.022*10^{23}$ and it could be anything; atoms, molecules, cats - you name it! 

[By the way, this number is called $\sf \large Avogadro’s$ number ($\sf N_A$) (not avocado, the fruit!)]. 

The purpose of using $\sf moles$ is to simplify the numbers used in calculations, so instead of using really $\Large large$ numbers (atoms), we can use $\small smaller$ numbers (moles) that represent the same number of objects.

$\sf \large \underline{What~ is~Molar ~Mass?}$  
(also called Atomic Mass, Molecular Mass, Molecular Weight)

Molar mass ($\sf M_{m}$) is the average mass (in grams) of 1 mole of an $\sf element$ - the mass of $6.022*10^{23}$ atoms. The term $element$ represents all the possible $\sf \color{green}{isotopes}$ of an element, so $\sf M_{m}$ is the weighted average of all the isotopes that exist of that element. 

$\sf \color{green}{The~ term~ isotope~ indicates~ that~ two~ atoms ~of~ the~ same~ element ~(that ~is}$ $\sf \color{green}{with ~same ~number~ of~ protons) ~have ~different~ masses~ because~ they~ have}$$\sf \color{green}{different~amounts~ of~ neutrons ~in ~their ~nuclei.}$

A weighted average takes into account the percent an object represents from a whole. It’s the same concept that you use to calculate your marks in school. Suppose that in a course you take, you have 2 Term tests (each worth 25 marks) and 1 Final exam (marks 50) which add up to 100 marks. If you got 85% on Test 1, 80% on Test 2 and 75% on the Exam, then your mark is:

$\sf   Average~mark=0.85*(25~marks)+0.80*(25~marks)+0.75*(50~marks)$  

$\sf  Average~mark= 78.75~marks \approx 79~marks ~out~of~100$  (Good job!)

The molar masses of the elements are calculated in the same way.
For example, Oxygen has 3 isotopes $\sf ^{16}O$, $\sf^{17}C$ and $\sf ^{18}C$ with masses of 15.995 g/mol, 16.999 g/mol and 17.999 g/mol$^1$, and with varying natural abundances 99.757%, 0.038% and 0.205%$^2$, respectively. 

The Molar mass is the weighted average of these:

$\sf \small Average~ M_m=0.99757(15.995 ~g/mol)+ 0.00038*(16.999~ g/mol)+ 0.00205*(17.999~ g/mol)$

$\sf Average~M_m=15.999 ~g/mol$ 

This is the value that one would find on a $\sf periodic~ table~of~elements$.

The units are in grams per mole, so this is how many grams 1 mole of oxygen atoms is (the oxygen you breathe is $O_2$, so there 2 Oxygen atoms not only 1).

Note: You don't need to do this for every problem, the Molar masses of all the elements are listed on the periodic table, this is just an example to show where these values come from. 

The $\sf \color{red}{purpose}$ of using $\sf M_m$ is to convert the mass (grams) of a substance into a number of moles -or viceversa, convert a number of moles into mass (grams). These numbers allow for easier calculations.
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     To interconvert between mass and moles, use the relationship:

                                        $\Large \sf n=\dfrac{m}{M_m}$

        where, $M_m$=molar mass, $m$=mass, and $n$= moles.

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Example: How many moles of gold are in a 20 gram 24-karat (pure) gold bracelet?

Molar mass ($M_m$) of gold (Au)= 196.966 g/mol   (from Periodic Table)

          $\sf n_{Au}=\dfrac{20~g}{196.966~g/mol}=0.101540 \approx 0.102~ moles$


We can take this further and find the number of gold atoms present in the bracelet, using:

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                                           $\sf \Large n=\dfrac{N}{N_A}$

       where $n$=moles, $N$=particles, $N_A$=Avogadro’s number

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We already know the moles, and we know avogadro’s number.

                                $0.102~moles=\dfrac{N}{(6.022*10^{23})}$

                     $N=(0.102~moles)*(6.022*10^{23}~particles/mole)$

                                         $N=6.142*10^{22}~particles$

In this case particles are atoms (this will differ if you use molecules).

There are $6.142*10^{22}~atoms$ (61,424,400,000,000,000,000,000) of gold, or 61 billion trillions that make up the bracelet. (This is a HUGE number!! we don't want to be work with atoms… unless you wanna be here all day doing math! Let’s use moles instead.)

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Citations for isotope data:

1. G. Audi, A. H. Wapstra Nucl. Phys A. 1993, 565, 1-65 and G. Audi, A. H. Wapstra Nucl. Phys A. 1995, 595, 409-480. 

2. IUPAC Subcommittee for Isotopic Abundance Measurements by K.J.R. Rosman, P.D.P. Taylor Pure Appl. Chem. 1999, 71, 1593-1607.

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