# Moles and molar concentration

### Moles and molar concentration

#### Lessons

In this lesson, we will learn:
• To relate the mole concept to amount of substance in aqueous solutions.
• To calculate the molarity of aqueous substances.
• To calculate amounts of substance from titration problems.

Notes:

• We saw in Moles, mass and gas calculations that for practical reasons, gas is usually measured in volume as opposed to mass which is used for solids. When we fix (keep constant) the temperature and pressure, at STP or RTP, the volume that one mole of an ideal gas occupies is also fixed, or constant.
• At standard temperature and pressure (STP) it is 22.4 litres per mole (L mol-1).
• At room temperature and pressure (RTP) it is 24 litres per mole (L mol-1).

Like gases, when it comes to aqueous substances (something dissolved in water), mass and volume are not used. Instead, concentration is used.

• Concentration effectively means “how much stuff in how much space?” which is why in chemistry concentration is measured in moles per litre, or moles per cubic decimetre (the value is equal).
Sometimes the term molarity is used. Molarity means concentration, for chemists – molarity is the number of moles of a chemical per amount of volume.
• Units of concentration are abbreviated “M”. It means moles per litre, written mol / L or mol L-1 or moles per cubic decimeters, written mol/dm3 or mol dm-3.
• Square brackets, e.g. [HCl] are used to show the concentration of a chemical. For example, you could write [HCl] = 0.1 mol dm-3 which means there is 0.1 mole of HCl per 1 litre of this solution.

You can use $c =$ $\large \frac{n}{V}$ to find concentration, where n = number of moles and $V$ is volume (in liters, L, or cubic decimeters, written dm3). You can then re-arrange for $n = c * v$.

• Because concentration is measured in one unit per another unit (moles per litre), using the conversion factor method will not cancel your units like the examples with mass or volume. It is first easiest to separate the two unit conversions, combining them at the end.
• For example, find the concentration of a solution of 54g of solid NaOH pellets being dissolved in 1500 mL of water.
Our target units are moles per litre: we will find moles using the mass and molar mass first, then find litres using the mL value given and then we will divide the moles by litres:

54 $g \, NaOH \,* \,$ $\large \frac{1 \, mol \, NaOH} {40 \, g \, NaOH}$ $=$ 1.35 $mol \, NaOH$

1500 $mL \, NaOH \,* \,$ $\large \frac{1 \, L \, NaOH} {1000 \, mL \, NaOH}$ $=$ 1.57 $L \, NaOH$

Using $C = n/v$ we can now find the concentration by dividing moles by volume:

$\large \frac{1.35 \, mol \, NaOH} {1.5 \, L \, NaOH}$ $=$ 0.9 $M \, NaOH$

• Using concentration to find the number of moles is very useful for knowing the amounts of substance in titration experiments.
A titration is an experiment used to find out the unknown concentration of an acid by reacting it with a base of known concentration, or vice versa (unknown base with known acid). We call the solution of known concentration a standard solution. The concentration of this is known precisely.
For an acid of known concentration reacting with a base of unknown concentration:
• A titration experiment runs until the number of moles of acid equals the number of moles of base, which is found by a colour change using an indicator.
• The known acid concentration and volume of acid is used to find the number of moles of acid, which is used to find the number of moles of the base of unknown concentration.
Look for the molar ratio from the chemical equation, for example if acid A and base B react in the equation 2A + B $\,$$\,$ C + H2O to make salt C, there is a 2:1 molar ratio. The moles of A will be twice the moles of B.
• You use the number of moles found and the volume of the unknown concentration to find the concentration of the unknown substance.

In the exercises of this lesson we will practice worked calculations using data from titration experiments, using the equation for concentration and unit conversions shown above. The detailed process of titration experiments is looked at in Acid-base titration

• Be careful with units of concentration, converting units if you need to. Volume is often given in mL but concentration is measured in moles per litre or moles per cubic decimeter, which has the same value. Dividing by 1000 converts from mL to L.
• Introduction
Expanding our moles calculations
a)
Recap - moles, molar volume and concentration

b)
Calculating concentration of a solution

c)
Example - calculating concentration

d)
Titration, an introduction

• 1.
Find the number of moles and concentration of substances used in chemical reactions.
a)
Calculate the number of moles in 100 mL of 0.2M HCl $_{(aq)}$

b)
Calculate the concentration of a solution of 1.7 litres of water with 0.75 moles of HCl dissolved in it.

c)
Calculate the new concentration of this solution when 3.4 extra litres of water are added to the solution.

• 2.
Find the number of moles and use it to find the volume of substances used in chemical reactions.
a)
Calculate the number of moles in 120mL of 0.05M H$_2$SO$_{4\; (aq)}$ solution.

b)
When H$_2$SO$_4$ and CaCO$_3$ are reacted, CO$_2$ gas is produced. What is the volume of CO$_{2\; (g)}$ produced when 120 mL of 0.1M H$_2$SO$_{4\; (aq)}$ is reacted with an excess amount of CaCO$_3$?

c)
What volume of 0.04M HCl $_{(aq)}$ is required to react completely to neutralise 6 g of Mg(OH)$_2$?

• 3.
Find the number of moles and use it to find the molarity of substances used in chemical reactions.
a)
In a titration reaction, 32 mL HCl $_{(aq)}$ of unknown concentration reacts with 25 mL of 0.5M NaOH $_{(aq)}$. Calculate the molarity of the hydrochloric acid being analysed.

b)
In another titration reaction, it took 14mL of NaOH $_{(aq)}$ of unknown concentration to react with 39 mL of 0.1M H$_3$PO$_{4\; (aq)}$. Calculate the molarity of the NaOH solution used.

• 4.
Find the number of moles and use it to find the quantities of substances used in chemical reactions.
Consider the reaction:
2 Al$_{\;(s)} +$2 NaON$_{\;(aq)} +$2 H$_2$O$_{\;(l)}$→2 NaAlO$_{2\;(aq)} +$3 H$_{2\;(g)}$
a)
A student needs to produce 20L of H$_2$ at STP using this reaction. Excess water and aluminium is available to react with 1.5M sodium hydroxide solution. How much sodium hydroxide solution will be required?

b)
When repeating the experiment, only 500 mL of 2.2M sodium hydroxide solution is available. How many litres of hydrogen gas can be produced given this limit?