Converting between concentration, moles and volume

Many chemical reactions, including the vast majority of those that happen in the human body, occur in solution. Concentration is used to describe the composition of a solution. The concentration is the amount of stuff that is present. A highly concentrated solution contains a great deal of stuff and a very dilute solution contains very little! Their properties will be very different.

The concentration is defined as the amount of solute dissolved in a particular volume of the solvent.

molarity. Concentration is indicated using square brackets around the substance to which it refers to. It has units of mol L^-1 or M. (Try not to confuse the molarity unit "M" with the symbol for molar mass "M" - italics are used to symbols and are not used to units.) When performing any calculations in science, an appreciation of the accuracy and significance of the measurements is vital - just ask an athlete under suspicion of taking performance enhancing drugs. If you are unfamilar with the proper way to handle signficant figures, read the iChem online tutorial (identical to the E9 week 4 pre-laboratory information)

Dealing with concentrations is very important and you will do so many times. There are three possible scenarios:

You know the amount of the substance and the volume of the solution and want to work out the concentration

You know the amount of the substance and the volume of the solution and want to work out the concentration:

concentration (in mol L^-1) = amount (in mol) or c = n

volume (in L) V

You know the volume of the solution and the concentration and want to work out the number of moles it contains

You know the volume of the solution and the concentration and want to work out the number of moles it contains:

amount (in mol) = concentration (in mol L^-1) × volume (in L) or n = c × V

You know the amount of the concentration and the number of moles and want to work out the volume required

You know the amount of the concentration and the number of moles and want to work out the volume required:

volume (in L) = amount (in mol) or V = n

concentration (in mol L^-1) c

Remembering these relationships

Remember these relationships using the triangle method:

Use a finger to cover up the quantity you want to calculate. The other two values show you how to do the calculation.

To calculate the concentration, cover up c and you are left with n / V. Hence, c = n / V

To calculate the number of moles, cover up n and you are left with c V. Hence, n = c × V

To calculate the volume, cover up V and you are left with n / c. Hence, V = n / c

Examples

Work through these examples carefully to see how the rules are applied and then test yourself with the self test quiz.

1. What is the concentration of a 1 L of glucose solution containing 2 moles of glucose?

1. A 1 L solution containing 2 moles of glucose has [glucose] = 2 M:

Using c = n ,   [glucose] = 2 mol = 2 mol L^-1 = 2 M

V 1 L

2. What is the concentration of the ions in a 200 mL sodium chloride solution containing 4 moles of NaCl?

2. A 200 mL solution of sodium chloride containing 4 moles of NaCl has [Na⁺] = 20 M and [Cl^-] = 20 M:
As 1000 mL = 1 L, 200 mL = 0.2 L. NaCl dissolves to give one Na⁺(aq) and one Cl^-(aq). Hence 4 mol of NaCl will dissolve to give 4 mol of both Na⁺ and Cl^-. Hence,

Using c = n ,   [Na⁺] = [Cl^-] = 4 mol = 20 mol L^-1 = 20 M

V 0.2 L

3. What is the concentration of H⁺ in a 20 mL sulfuric acid solution containing 2 moles of H₂SO₄?

3. A 20 mL solution of sulfuric containing 0.2 moles of H₂SO₄ has [H⁺] = 20 M:
As 1000 mL = 1 L, 20 mL = 0.02 L. H₂SO₄ dissolves to give two H⁺(aq) and one SO₄^2-(aq). Hence 0.2 mol of H₂SO₄ will dissolve to give 0.4 mol of H⁺ and 0.2 mol of SO₄^2-. Hence,

Using c = n ,   [H⁺] = 0.4 mol = 20 mol L^-1 = 20 M

V 0.02 L

4. What is the concentration of OH^- in a 550 mL calcium hydroxide solution containing 0.25 moles of Ca(OH)₂?

4. A 550 mL solution of calcium hydroxide containing 0.25 moles of Ca(OH)₂ has [OH^-] = 0.91 M:
As 1000 mL = 1 L, 550 mL = 0.55 L. Ca(OH)₂ dissolves to give one Ca²⁺ and two OH^-(aq). Hence 0.25 mol of Ca(OH)₂ will dissolve to give 0.25 mol of Ca²⁺ and 0.50 mol of OH^-. Hence,

Using c = n ,   [OH^-] = 0.50 mol = 0.91 mol L^-1 = 0.91 M

V 0.55 L

Note that the measurements (550 mL and 0.25 mol) are both known to 2 significant figures. The accuracy of the final answer is limited by the least accurately known measurement and so the answer has been given to 2 significant figures.

5. How many moles of glucose are required to make 500 mL of a 0.2 M solution?

5. A 500 mL solution with [glucose] = 0.2 M requires 0.1 mol of glucose:
As 1000 mL = 1 L, 500 mL = 0.5 L. With [glucose] - 0.2 M = 0.2 mol L^-1,

Using n = c × V,   amount of glucose = (0.2 mol L^-1) × (0.5 L) = 0.1 mol

6. How many moles of sodium chloride are required to make 100 mL of a 5 M solution of Na⁺?

6. A 100 mL solution with [Na⁺] = 5 M requires 0.5 mol of sodium chloride:
As 1000 mL = 1 L, 100 mL = 0.1 L. With [Na⁺] = 5 M = 5 mol L^-1,

Using n = c × V,   amount of Na⁺ = (5 mol L^-1) × (0.1 L) = 0.5 mol

NaCl dissolves to give one Na⁺(aq) and one Cl^-(aq). Hence 0.5 mol of Na⁺ requires 0.5 mol of NaCl.

7. How many moles of sulfuric acid are required to make a 250 mL of a 0.10 M solution of H⁺?

7. A 250 mL solution with [H⁺] = 0.10 M requires 0.13 mol of sulfuric acid
As 1000 mL = 1 L, 250 mL = 0.25 L. With [H⁺] = 0.1 M = 0.1 mol L^-1,

Using n = c × V,   amount of H⁺ = (0.1 mol L^-1) × (0.25 L) = 0.025 mol

One H₂SO₄ ionizes to give two H⁺(aq) ions. Hence 0.025 mol of H⁺ requires 0.0125 mol of H₂SO₄.

The measurements (250 mL and 0.10 M) are both known to 2 significant figures. The accuracy of the final answer is limited by the least accurately known measurement. Hence to 2 significant figures, 0.013 mol of sulfuric acid is required.

8. How many moles of calcium hydroxide are required to 12 mL of a 0.2 M solution of OH^-?

8. A 12 mL solution with [OH^-] requires 0.01 mol of calcium hydroxide:
As 1000 mL = 1 L, 12 mL = 0.012 L. With [OH^-] = 0.2 M = 0.2 mol L^-1,

Using n = c × V,   amount of OH^- = (0.2 mol L^-1) × (0.012 L) = 0.024 mol

Ca(OH)₂ dissolves to give two OH^-(aq) ions. Hence 0.024 mol of OH^- requires 0.012 mol of Ca(OH)₂.

The measurements, 12 mL and 0.2 M, are known to 2 and 1 significant figures respectively. The accuracy of the final answer is limited by the least accurately known measurement. Hence to 1 significant figure, 0.01 mol of calcium hydroxide is required.

9. What volume of a 0.5 M glucose solution contains 0.4 mol of glucose?

9. 0.8 L of a solution with [glucose] = 0.5 M contains 0.4 mol:
With [glucose] = 0.5 M = 0.5 mol L^-1,

Using V = n ,   volume of solution = 0.4 mol = 0.8 L

c 0.5 mol L^-1

10. What volume of a 0.25 M Na⁺ solution contains 0.50 mol of Na⁺?

10. 2.0 L of a 0.25 M Na⁺ solution contains 0.50 mol of Na⁺:
With [Na⁺] = 0.25 M = 0.25 mol L^-,

Using V = n ,   volume of solution = 0.50 mol = 2.0 L

c 0.25 mol L^-1

11. What volume of a 2.0 M H⁺ solution contains 1.5 mol of H₂SO₄?

11. 0.75 L of a 2.0 M H⁺ solution contains 1.5 mol of H₂SO₄:
As H₂SO₄ ionizes to give two H⁺, 1.5 mol of H₂SO₄ will produce 3.0 mol of H⁺. As the solution has [H⁺] = 2.0 M = 2.0 mol L^-1, the volume that contains 3.0 mol of H⁺ can be found

Using V = n ,   volume of solution = 3.0 mol = 1.5 L

c 2.0 mol L^-1

12. What volume of a 0.20 M OH^- solution contains 0.40 mol of Ca(OH)₂?

12. 4.0 L of a 0.20 M OH^- solution contains 0.40 mol of Ca(OH)₂:
As Ca(OH)₂ ionizes to give two OH^-, 0.40 mol of Ca(OH)₂ will produce 0.80 mol of OH^-. As the solution has [OH^-] = 0.20 M = 0.20 mol L^-1, the volume that contains 0.80 mol of OH^- can be found

Using V = n ,   volume of solution = 0.80 mol = 4.0 L

c 0.20 mol L^-1

The measurements (0.20 M and 0.40 mol) are both known to 2 significant figures. The accuracy of the final answer is limited by the least accurately known measurement. Hence to 2 significant figures, the volume is 4.0 L.

Self test quizzes

Test your understanding using the quizzes below.

Each quiz is made up of 10 questions randomly drawn from a large set so you can repeat the quiz many times over. The results are not stored!
Practice multiple choice questions (not an assessment)
Practice fill in the blank (not an assessment)

Solutions

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Converting between concentration, moles and volume

concentration (in mol L^-1) =	amount (in mol)	or	c =	n
	volume (in L)			V

volume (in L) =	amount (in mol)	or	V =	n
	concentration (in mol L^-1)			c

Using c =	n	, [glucose] =	2 mol	= 2 mol L^-1 = 2 M
	V		1 L

Using c =	n	, [Na⁺] = [Cl^-] =	4 mol	= 20 mol L^-1 = 20 M
	V		0.2 L

Using c =	n	, [H⁺] =	0.4 mol	= 20 mol L^-1 = 20 M
	V		0.02 L

Using c =	n	, [OH^-] =	0.50 mol	= 0.91 mol L^-1 = 0.91 M
	V		0.55 L

Using V =	n	, volume of solution =	0.4 mol	= 0.8 L
	c		0.5 mol L^-1

Using V =	n	, volume of solution =	3.0 mol	= 1.5 L
	c		2.0 mol L^-1

Using V =	n	, volume of solution =	0.80 mol	= 4.0 L
	c		0.20 mol L^-1