# How should I study about moles and how they relate to things like stoichiometry?

##### 2 Answers
Nov 6, 2016

You should remember that the $\text{mole}$ is simply a $\text{number}$....

#### Explanation:

Instead of moles, we could use $\text{dozens}$, or $\text{100's}$ or $\text{1000's}$ or $\text{1000000's}$. The mole, however, has a special property. If I have a mole, $\text{Avogadro's number, } 6.022 \times {10}^{23}$ of ""^1H atoms, I have a mass of $1 \cdot g$ of ""^1H atoms. Likewise, if I have a mole, $\text{Avogadro's number, } 6.022 \times {10}^{23}$ of ""^12C atoms, I have a mass of $12 \cdot g$ of ""^12C atoms.

You won't ever be asked to remember these masses, because you should be provided with a copy of the Periodic Table, in which molar or equivalent masses are explicitly given, in every exam in chemistry or physics you ever sit. You still have to be able to use the table competently and correctly when you determine equivalent mass.

The $\text{mole}$ is thus the link between the micro world of $\text{atoms}$ and $\text{molecules}$, which we can't see but whose existence we can infer, to the macro world of $\text{grams}$, and $\text{kilograms}$, and $\text{litres}$, which we can measure by some means on a laboratory bench.

So the next time you hear the term $\text{mole}$ in a chemistry class, just think a large number, i.e. a few million, billion etc. dozen. That's all there is to it.

Nov 7, 2016

Warning! Long answer.
You should remember that the mole is central to all stoichiometry calculations.

#### Explanation:

Stoichiometry is the term applied to problems that ask, "How much of $\boldsymbol{\text{A}}$ is formed from or reacts with this much of $\boldsymbol{\text{B}}$?"

One thing we always need is a balanced equation for the reaction, because the equation tells us the ratios in which the molecules react with each other.

If we multiply the numbers of molecules by Avogadro's number, ${N}_{\text{A}}$, we get the same number of moles and therefore the molar ratio in which the molecules react.

All we have to do is to get the moles of $\boldsymbol{\text{A}}$ and $\boldsymbol{\text{B}}$ in the right ratio.

Usually, we have to do the calculation in three steps:

1. Use the molar mass of $\boldsymbol{\text{A}}$ to convert grams of $\boldsymbol{\text{A}}$ to moles of $\boldsymbol{\text{A}}$.
2. Use the molar ratio of $\boldsymbol{\text{B}}$ to $\boldsymbol{\text{A}}$ from the balanced equation to convert moles of $\boldsymbol{\text{A}}$ to moles of $\boldsymbol{\text{B}}$.
3. Use the molar mass of $\boldsymbol{\text{B}}$ to convert moles of $\boldsymbol{\text{B}}$ to grams of $\boldsymbol{\text{B}}$

Here's a chart to help you navigate a stoichiometry problem.

People call it the Mole Bridge, because you use moles to convert the known grams, particles, or litres of a substance $\boldsymbol{\text{A}}$ to grams, particles, or litres of a substance $\boldsymbol{\text{B}}$.

You always work from known values to the unknown (i.e. from left to right).

Here's how it works.

Sample Problem

What mass of hydrogen is needed to prepare 100.0 g of ammonia from nitrogen and hydrogen?

Solution

This is a stoichiometry problem.

I know that I will need the balanced equation and the molar masses, so I assemble all the numbers for later use.

${M}_{\text{r}} : \textcolor{w h i t e}{m m m m l l} 2.016 \textcolor{w h i t e}{m l l} 17.03$
$\textcolor{w h i t e}{m m m m l l} {\text{N"_2 + "3H"_2 → 2"NH}}_{3}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m m} x \textcolor{w h i t e}{m m l l} 100.0$

Here, the known compound is "NH"_3 (bb"A") and the unknown compound is "H"_2 (bb"B").

We start at the left of the mole bridge and work to the right.

Step 1: ${\text{Moles of NH"_3 = 100.0 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "5.872 mol NH}}_{3}$

Step 2: ${\text{Moles of H"_2 = 5.872 color(red)(cancel(color(black)("mol NH"_3))) × ("3 mol H"_2)/(2 color(red)(cancel(color(black)("mol NH"_3))) )= "8.808 mol H}}_{2}$

Step 3: ${\text{Mass of H"_2 = 8.808 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "17.76 g H}}_{2}$

You need ${\text{17.76 g H}}_{2}$ to prepare ${\text{100.0 g of NH}}_{3}$.