# Amount of Substance Concentration (Molarity) Calculations Chemistry Tutorial

## Key Concepts

⚛ Concentration of a solution refers to the amount of solute dissolved in a given solvent to make a solution.
When water is the solvent, the formula of the solute is immediately followed by aq enclosed in round brackets, (aq).

term description example
solute substance that dissolves solid sodium chloride, NaCl(s)
solvent substance that enables solute to dissolve liquid water, H2O(l)
solution homogeneous mixture of solute dissolved in solvent sodium chloride dissolved in water, NaCl(aq)

⚛ Concentration of a solution can be given in moles of solute per litre of solution (mol L-1 or mol/L or M), or, in moles of solute per cubic decimetre of solution (mol dm-3 or mol/dm3)(1)

 solute solution moles (mol) moles per litre (mol L-1 or mol/L or M) moles (mol) moles per cubic decimetre (mol dm-3 or mol/dm3)

⚛ Molarity is the term used to describe a concentration given in moles per litre.

Alternative names for molarity are
· amount of substance concentration (IUPAC preferred term)
· amount concentration
· molar concentration

⚛ Amount of substance concentration, molarity, has the units mol L-1 (or mol/L or M) or the equivalent SI units of mol dm-3 (mol/dm3)

⚛ Amount of substance concentration, molarity, the concentration of a solution in mol/L, mol L-1, mol dm-3 or mol/dm3, is given the symbol c (sometimes M).

For a 0.01 mol L-1 HCl(aq) solution we can write :

(i) [HCl(aq)] = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M
(amount concentration implied by square brackets around solvated solute formula)

(ii) c(HCl(aq)) = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M
(c stands for amount concentration, solvated solute formula given in round brackets or parentheses)

⚛ Mathematical equation (formula or expression) to calculate the molarity of a solution (concentration in mol L-1) is

c = n ÷ V

c = concentration of solution in mol L-1 (mol/L or M)
n = amount of substance being dissolved (moles of solute)
V = volume of solution in litres (L) or cubic decimetres (dm3)

⚛ This equation (formula or expression) can be re-arranged to find:

(i) moles of solute given molarity and volume of solution:

n = c × V

(ii) volume of solution given moles of solute and molarity:

V = n ÷ c

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## Molarity, amount of substance concentration, Concepts

Amount of substance concentration, molarity, is the term given for concentrations of solutions that are given in units of moles per litre (mol L-1 or mol/L or M) or, in SI units, moles per cubic decimetre (mol dm-3 or mol/dm3).

Consider the diagram on the right.

The black box represents a 1 litre or 1 cubic decimetre container.

The white space in the box represents the volume occupied by the solvent.

Each blue X represents 1 mole of sugar molecules.

How many moles of sugar molecules are shown in this container? 1 mole
What is the volume of the container in litres? 1 L

The amount of substance concentration, or molarity, the concentration of the solution in units of mol L-1 or mol dm-3, tells us the amount of solute (moles of solute) present in 1 L or 1 dm3 of solution.

What is the concentration of sugar molecules in the solution?
There is 1 mole of sugar molecules per 1 litre of solution, or, 1 mole of sugar molecules per 1 cubic decimetre of solution
That is 1 mole per litre, or 1 mol L-1, or, 1 mole per cubic decimetre or 1 mol dm-3
Concentration of sugar molecules = 1 mol L-1 (1 mol/L or 1 M) = 1 mol dm-3

concentration of sugar solution = moles of sugar ÷ volume of solution (in L or dm3)

 X
Consider the diagram on the right.

The red box represents a 0.5 litre or 0.5 cubic decimetre container.

The white space in the box represents the volume occupied by the solvent.

Each blue X represents 1 mole of sugar molecules.

How many moles of sugar molecules are shown in this container? 1 mole
What is the volume of the container in litres? 0.5 L = 0.5 dm3

What is the concentration of sugar molecules in the solution?
There is 1 mole of sugar molecules per 0.5 litre of solution, or, 1 mole of sugar molecules per 0.5 cubic decimetres of solution.

The amount of substance concentration or molarity, the concentration of the solution in units of in mol L-1 or mol dm-3, tells us how many many moles of solute are present in 1 L of solution or in 1 dm3 of solution .
If we divide both the moles of sugar molecules and the volume by the volume, that is divide by 0.5:
There are (1 ÷ 0.5 = 2) moles of sugar molecules per (0.5 ÷ 0.5 = 1) litres of solution (or per 1 cubic decimetre of solution)
There are 2 moles of sugar molecules per 1 litre of solution (2 moles of sugar molecules per 1 cubic decimetre of solution)
That is 2 moles per litre, or 2 mol L-1 (2 moles per 1 cubic decimetre of solution, 2 mol dm-3)
Amount concentration or molarity of sugar molecules = 2 mol L-1 = 2 mol dm-3

amount concentration of sugar solution = moles of sugar ÷ volume of solution ( in L or dm3)

We can check this equation for our example:
amount sugar (moles sugar) = 1 mol
volume = 0.5 L = 0.5 dm3
amount concentration = moles ÷ volume = 1 ÷ 0.5 = 2 mol L-1 ( 2 mol/L or 2 M) = 2 mol dm-3 (2 mol/dm3)

 X

## Molarity Equation (amount of substance concentration equation)

The amount of substance concentration, or the molarity of a solution, is given by the equation :

 c = n V

c = amount concentration (molarity) in mol L-1 (or mol dm-3)
n = amount of solute in mol
V = volume of solution in L (or dm3)

Note that we often use square brackets around the formula of solvated species to indicate the molarity of a solution (the concentration of the solution in mol L-1):

for example an aqueous solution of sodium chloride, NaCl(aq), with a molarity of 0.154 mol L-1 (or 0.154 mol dm-3) could be represented as

c(NaCl(aq)) = 0.154 mol L-1 = 0.154 mol dm-3

[NaCl(aq)] = 0.154 mol L-1 = 0.154 mol dm-3

If you know the moles of solute in a solution, and, you know the volume of the solution, you can calculate the concentration of the solution in mol L-1 (or mol dm-3) using the mathematical equation c = n ÷ V

But what if you know the concentration of the solution in mol L-1 (or mol dm-3) and the volume of the solution in L (or dm3), can you calculate how many moles of solute are present in the solution?
Yes! You just need to rearrange the equation by multiplying both sides of the equation by V (the volume of the solution):

 c × V = n × VV c × V = n

You can even calculate the volume of the solution if you know how many moles of solute are present and its concentration in mol L-1 (or mol dm-3).
Just rearrange the equation above by dividing both sides by c (amount concentration or molarity of solution in mol L-1 or in mol dm-3)

 c × V c = n c V = n c

In summary, to calculate

(a) amount of substance concentration (molarity) in mol L-1 (or mol dm-3):

c = n ÷ V

(b) amount of solute in solution in mol :

n = c × V

(c) volume of solution in L (or dm3):

V = n ÷ c

## Examples of Molarity Calculations with Worked Solutions

Apply the following 5 steps to solve the molarity problems below:

1. Step 1: What is the question asking you to calculate? (note the units of measurement!)
2. Step 2: What information has been given in the question? (extract the data from the question including units of measurement and convert if necessary)
3. Step 3: What is the relationship between what you know and what you need to find? (write the mathematical equation)
4. Step 4: Substitute the values into the equation and solve (check the number of significant figures)
5. Step 5: Write the answer (include units of measurement!)

### (1) Calculating Amount of Substance Concentration, Molarity (c = n ÷ V)

Question: Calculate the amount of substance concentration in mol L-1 (molarity) of an aqueous sodium chloride solution containing 0.125 moles sodium chloride in 0.50 litres of solution.

Solution:

Step 1: What is the question asking you to calculate?

c(NaCl(aq)) = molarity (concentration in mol L-1) of solution = ? mol L-1

Step 2: What information has been given in the question?

Extract the data from the question
n(NaCl) = moles of solute = 0.125 mol
V(NaCl(aq)) = volume of solution = 0.50 L

Step 3: What is the relationship between what you know and what you need to find?

Write the equation:
c(NaCl(aq)) = n(NaCl) ÷ V(NaCl(aq))

Step 4: Substitute the values into the equation for molarity and solve:

[NaCl(aq)] = c(NaCl(aq)) = 0.125 mol ÷ 0.50 L = 0.25 mol L-1 (or 0.25 mol/L or 0.25 M)

(Note: only 2 significant figures are justified)

[NaCl(aq)] = 0.25 mol L-1

### (2) Calculating Amount of Solute (n = c × V)

Question: Calculate the moles of copper sulfate in 250.00 mL of 0.020 mol L-1 copper sulfate solution, CuSO4(aq).

Solution:

Step 1: What is the question asking you to calculate?

n(CuSO4) = moles of solute = ? mol

Step 2: What information has been given in the question?

Extract the data from the question:
c(CuSO4(aq)) = amount concentration (molarity) of solution = 0.020 mol L-1
V(CuSO4(aq)) = volume of solution = 250.00 mL
Convert volume in mL to volume in L (there are 1000 mL in 1 L)
V(CuSO4(aq)) = 250.00 mL ÷ 1000 mL/L = 250.00 × 10-3 L = 0.25000 L

Step 3: What is the relationship between what you know and what you need to find?

Write the equation:
n(CuSO4) = c(CuSO4(aq)) × V(CuSO4(aq))

Step 4: Substitute the values into the molarity equation and solve:

n(CuSO4) = 0.020 mol L-1 × 0.25000 L = 0.005000 mol = 0.0050 mol

(Note: only 2 significant figures are justified)

n(CuSO4(aq)) = 0.0050 mol

### (3) Calculating Volume of Solution (V = n ÷ c)

Question: Calculate the volume in litres of a 0.80 mol L-1 aqueous solution of potassium bromide containing 1.60 moles of potassium bromide.

Solution:

Step 1: What is the question asking you to calculate?

V(KBr(aq)) = volume of solution in litres = ? L

Step 2: What information has been given in the question?

Extract the data from the question:
n(KBr) = moles of solute = 1.60 mol
c(KBr(aq)) = molarity (concentration in mol L-1) of solution = 0.80 mol L-1

Step 3: What is the relationship between what you know and what you need to find?

Write the equation:
V(KBr(aq)) = n(KBr) ÷ c(KBr(aq))

Step 4: Substitute the values into the equation and solve:

V(KBr(aq)) = 1.60 mol ÷ 0.80 mol L-1 = 2.00 L = 2.0 L

(Note: only 2 significant figures are justified)

V(KBr(aq)) = 2.0 L

## Problem Solving: Amount of Substance Concentration (molarity)

The Problem: Chris the Chemist has been given a 250.00 mL volumetric flask and asked to use it to make an aqueous solution of sodium chloride, NaCl, with a concentration of 0.100 mol L-1 for a corrosion experiment. The sodium chloride, NaCl, is available as an analytical reagent composed of white crystals. Determine the mass in grams of sodium chloride that Chris the Chemist will need to weigh out.

Solving the Problem using the StoPGoPS model for problem solving:

 STOP! State the question. What is the question asking you to do? Determine (calculate) the mass of sodium chloride in grams. m(NaCl(s)) = mass of sodium chloride = ? g PAUSE! Pause to Plan. What information (data) have you been given? solute is NaCl(s) solvent is water (we know this because we are told Chris makes an aqueous solution) V(NaCl(aq)) = volume of NaCl(aq) = 250.00 mL c(NaCl(aq)) = amount concentation (molarity) of NaCl(aq) = 0.100 mol L-1 What is your plan for solving this problem? Step 1: Calculate moles of NaCl in solution n(NaCl(s)) = c(NaCl(aq)) × V(NaCl(aq)) c(NaCl(aq)) = 0.100 mol L-1 V(NaCl(aq)) = 250.00 L Convert volume in mL to volume in L: V(L) = V(mL) ÷ 1000 (mL/L) Assume the temperature of the laboratory is the same as the temperature required by the volumetric flask (eg 25°C).(2) Assume the water used to make up the solution does not contain any NaCl.(3) Step 2: Calculate mass of NaCl(s) to be used moles(NaCl(s)) = mass(NaCl(s)) ÷ molar mass(NaCl(s)) Use the Periodic Table to find the relative atomic masses (atomic weights) for (i) Mr(Na) (ii) Mr(Cl) Find the molar mass, M, of NaCl: M(NaCl) = Mr(Na) + Mr(Cl) Assume the NaCl(s) that is to be used is 100% pure (no impurities) (4) GO! Go with the Plan. Step 1: Calculate moles of NaCl in solution c(NaCl(aq)) = 0.100 mol L-1 V(NaCl(aq)) = 250.00 L Convert volume in mL to volume in L: V(L) = V(mL) ÷ 1000(mL/L) = 250.00 mL ÷ 1000 mL/L = 0.25000 L Assume the temperature of the laboratory is the same as the temperature required by the volumetric flask (eg 25oC). Assume the water used to make up the solution does not contain any NaCl. n(NaCl(s)) = c(NaCl(aq)) × V(NaCl(aq)) = 0.100 mol L-1 × 0.25000 L = 0.0250 mol (Note: 3 significant figures are justified) Step 2: Calculate mass of NaCl(s) to be used Use the Periodic Table to find the relative atomic masses for Mr(Na) = 22.99 Mr(Cl) = 35.45 Find the molar mass, M, of NaCl: M(NaCl) = Mr(Na) + Mr(Cl) = 22.99 + 35.45 = 58.44 g mol-1 Assume the NaCl(s) that is to be used is 100% pure (no impurities) m(NaCl) = n(NaCl) × M(NaCl) = 0.0250 mol × 58.44 g mol-1 = 1.46 g (Note: 3 significant figures are justified) PAUSE! Ponder Plausability. Have you answered the question that was asked? Yes, we have calculated the mass of NaCl(s) that needs to be used to make the solution. Is your solution to the question reasonable? Let's work backwards to see if the mass of NaCl we calculated will give the right concentration: m(NaCl) = 1.46 g M(NaCl) = 58.44 g mol-1 n(NaCl) = m(NaCl) ÷ M(NaCl) = 1.46 g ÷ 58.44 g mol-1 = 0.0250 mol c(NaCl) = n(NaCl) ÷ V(NaCl(aq)) = 0.0250 mol ÷ (250.00 mL/1000 mL/L) = 0.100 mol L-1 1.46 g of NaCl in a 250.00 mL volume does make a 0.100 mol L-1 so we are confident our solution is correct. STOP! State the solution. What is the mass of NaCl(s) Chris the Chemist needs to weigh out? m(NaCl(s)) = 1.46 g

## Sample Question: Mass Concentration

A 25.00 mL aliquot of 0.0264 mol L-1 AgNO3(aq) is added to a 150.00 mL volumetric flask and distilled water is added until the meniscus sits on the mark when viewed at eye-level.
Determine the mass in grams of AgNO3 in the resultant solution.

m(AgNO3) = g