What are the most common questions about moles that often comes out in the exams?

1 Answer
Jan 2, 2015

I'll try and provide some concept questions that often come up when dealing with moles. The answer will be quite long, so bear with me...

$\text{What is a mole?}$

A mole is simply a way of counting atoms and molecules. Think of it like this:. let's say cents are the equivalent of atoms, and dollars are the equivalent of moles.

In the same way one dollar has 100 cents, 1 mole has $6.022 \cdot {10}^{23}$ atoms (or molecules). This will also give you a good analogy for just how many atoms are there in a mole:

1 dollar has ${10}^{2}$ cents
1 million dollars has ${10}^{8}$ cents
1000 trillion dollars has ${10}^{17}$ cents...and so on

$\text{What's the difference between one mole of oxygen}$ $\text{and one mole of diatomic oxygen?}$

One mole of $O$ contains $6.022 \cdot {10}^{23}$ oxygen atoms, while one mole of ${O}_{2}$ contains $6.022 \cdot {10}^{23}$ ${O}_{2}$ molecules.

$\text{How many moles of N contain}$ $4.35 \cdot {10}^{25}$ $\text{atoms of N?}$

Since one mole has $6.022 \cdot {10}^{23}$ atoms, more atoms must mean more than one mole.

$4.35 \cdot {10}^{25}$ "atoms of N" * (1 "mole")/(6.022 * 10^23 "atoms of N") = 72.2 moles

$\text{How many atoms of hydrogen are in}$ $0.0722$ $\text{moles of water?}$

First, determine how many molecules of water are in 0.0722 moles of water

$0.0722$ "moles of water" * (6.022*10^23 "molecules of water")/("1 mole of water")

yields $4.35 \cdot {10}^{22}$ molecules of water. Since water's molecular formula is ${H}_{2} O$, we have 2 atoms of hydrogen for every 1 molecule of water

$4.35 \cdot {10}^{22}$ "molecules of water" *("2 H atoms")/("1 water molecule") =

$8.70 \cdot {10}^{22}$ atoms of $H$.

$\text{How many moles of oxygen are in 54.3 g of oxygen?}$

Here is where molar mass comes into play. We know that one mole of oxygen weighs approximately 16.0 grams, so

$54.3$ "g of oxygen" * ("1 mole")/("16.0 g") = 3.39 $\text{moles}$

$\text{How many moles of oxygen are needed for the combustion}$ $\text{of 12.0 g of methane?}$

Questions like this make use of mole ratios.

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$

So, 1 mole of methane needs 2 moles of oxygen, so we could set it up like this

$12.0$ "g of methane" * ("1 mole")/("16.0 g") * ("2 moles oxygen")/("1 mole methane") =1.50

moles of oxygen needed.

"How many moles of hydrogen are in 17.2 L at STP"?

Always think molar volume when you see STP; 1 mole of any ideal gas occupies 22.4 L at STP, so

$n = \frac{V}{V} _ \left(\text{molar") = ("17.2 L")/("22.4 L}\right) = 0.768$ $\text{moles}$

For reaction stoichiometry, always remember that moles allow you to go from grams of one substance to grams of another by using mole ratios. Use molar masses to determine the number of moles from grams.

Here's a link to some more example problems:

http://antoine.frostburg.edu/chem/senese/101/moles/faq.shtml