# Moles, Molarity and Molality

Chemistry usually involves tiny molecules and huge numbers. Using the humble mole unit makes otherwise complex calculations easy! Learn what moles represent, and their usefulness in molarity and molality calculations.

## Chemical Moles

A typical chemical reaction involves more atoms and molecules than is feasible to count, hence chemists use moles to represent them. 1 mole = 6.022 x 1023 units of whatever it is you’re measuring. That’s 602 200 000 000 000 000 000 000, which isn’t exactly a pleasant number to work with as is. But don’t be intimidated! The mole is simply a unit of measurement, just like ‘dozen’ or ‘grand’.

Note: 6.022 x 1023 is also known as Avogadro’s number.

Because they come in different sizes and masses, 1 mole of a certain element will not weigh the same as 1 mole of another element. For example, 1 mole of carbon weighs 12 grams, while 1 mole of uranium weighs 238 grams. These weights are known as the atomic mass, which is a number you can usually find below each element in the periodic table. In the case of molecules, their mass per mole is known as their molar mass.

Remember that molar mass has units of grams, so remember to convert all masses to grams before performing mole calculations!

No. of \: moles = \frac{Mass}{Atomic\:or\:Molar\:Mass}

### Worked Example 1

How many moles are there in 80g of nitrogen gas?

Nitrogen gas has a chemical formula of N2, with a molar mass of 28.

No. of \: moles = \frac{80}{28}= 2.857\: moles \:of\:N_{2}

### Worked Example 2

What is the mass of 14 moles of iron?

Mass=Moles \:\ast \:Atomic \:Mass\\[0.1in]
Mass=14 \:\ast\:55.85=781.9g


## Molarity

While we can work with moles alone, most chemical reactions occur in some sort of solvent. When we swallow a pill, for example, the drug is distributed throughout our bloodstream. Therefore, the concentration of a substance – the solute – can be more relevant than the number of moles. Usually expressed in the form of moles per liter (uppercase M), the molarity of a solute is also known as its molar concentration.

Remember that units for molarity are moles per liter! So make sure that volumes are converted into L during calculations.

Molarity=\frac{No. \:of\:moles\:of\:solute }{Volume\:of\:solution}

### Worked Example 1

What is the molarity of 400g of NaCl dissolved in 2.5 L of water?

No. of \: moles\: NaCl = \frac{Mass}{Molar\:Mass}= \frac{400}{58.44}=6.85\:moles\\[0.1in]
Molarity\: NaCl=\frac{No. \:of\:moles\:of\:solute }{Volume\:of\:solution}=\frac{6.85}{2.5}=2.74\:M


### Worked Example 2

You need to make 300 ml of a 0.15 M solution of sucrose (MW: 342.2). How many grams of sucrose do you need?

First, find the number of moles of sucrose required:

Molarity=\frac{No. \:of\:moles\:of\:solute }{Volume\:of\:solution} \\[0.1in]
0.15\:M=\frac{No. \:of\:moles\:of\:sucrose }{0.3\:L}\\[0.1in]
No. \:of\:moles\:of\:sucrose=0.15*0.3=0.045\:moles\\


Next, use the molar mass equation to determine the mass of sucrose needed:

No. of \: moles = \frac{Mass}{Atomic\:or\:Molar\:Mass}\\[0.1in]
0.045=\frac{Mass\:of\:sucrose\:required}{342.2}\\[0.1in]
Mass\:of\:sucrose\:required=0.045*342.2=15.4g

## Molality

Similar to molarity (and perhaps confusingly), molality is also a measure of solute concentration. However, in this case, the number of moles of solute is divided by the mass of the solvent. Note that the mass of the solvent is used in the denominator, not the mass of the entire solution! This sometimes involves additional steps during calculation, as you’ll see in the worked example 2.

With units of moles/kg, molality is useful in calculations because both the solute and the solvent are measured by units of mass, hence are unaffected by changes in temperature and pressure.

Molality=\frac{No. \:of\:moles\:of\:solute }{Mass\:of\:solvent}\\

### Worked Example 1

What is the molality of 82g benzoic acid dissolved in 1.5 kg of ethanol?

Calculate the number of moles of benzoic acid we have, then use it in the molality equation to determine mol/kg.

No. of \: moles\:benzoic \:acid = \frac{Mass}{Molar\:Mass}= \frac{82}{122.12}=0.67\:moles\\[0.1in]
Molality\:benzoic \:acid=\frac{No. of \: moles\:benzoic \:acid}{Mass\:of\:ethanol}=\frac{0.67}{1.5}=0.45\:mol/kg


### Worked Example 2

You are required to concoct 240g of a 1.22 mol/kg solution of potassium permanganate (KMnO4, MW: 158) in water. How many grams of KMnO4 do you need?

Remember that molality is the number of moles of solute divided by the mass of the solvent (not the solution!) Therefore, we need to first find the total mass of 1.22 moles of KMnO4 in 1kg of water.

1.22\:mol/kg\:KMnO_{4}\:solution=1.22\: moles\:KMnO_{4}+1\:kg \:H_{2}O\\[0.1in]
Mass\:of \:1.22\:moles\:KMnO_{4}\:in\:1\:kg\:H_{2}O=(1.22\:*158)+1000g=1192.8g


We want 240g of the final solution, so we need to determine the mass fraction to calculate the mass of KMnO4 required:

Fraction\:of\:total\:mass\:required=\frac{240}{1192.8}=0.2\\[0.1in]
Mass\:of \:KMnO_{4}\:required=1.22\:*\:158\:*0.2=38.55g