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Dilution Factor Calculations

Key Concepts

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Definition of Dilution Factor

You may come across something like, "prepare a 1:50 dilution of the solution".
What it means is, take a known volume of the stock solution (Vinitial) and add enough solvent to it so that the solution has a new volume, Vfinal, of 50 x Vinitial.
Vfinal = 50 × Vinitial

The "1:50" tells you the dilution factor, the ratio of volumes, to use to prepare the new solution.
V1 : V2
1 : 50

In this case it tells us that V1 = 1 and V2 = 50
so the dilution factor, DF, = V2 ÷ V1 = 50 ÷ 1 = 50
That is, the new, diluted solution will have a volume 50 times greater than the volume of the original, undiluted, solution:
Vfinal = DF × Vinitial

A dilution factor does not tell you what the initial volume is, neither does it tell you what the final volume is, it only tells you what the ratio of the initial to final volume is.
You can use the equation Vfinal = DF × Vinitial to find the final volume of solution after dilution if you know the initial volume of the solution.
The table below gives you a number of different options for preparing a 1:50 dilution:

Vintial (initial volume) 1 mL 1 L 0.1 mL 2 mL 25 μL
Vfinal (final volume) 50 mL 50 L 5 mL 100 mL 1250 μL
ratio of volumes used Vinitial : Vfinal
V1 : V2		 1 : 50
1 : 50		 1 : 50
1 : 50		 0.1 : 5
1 : 50		 2 : 100
1 : 50		 25 : 1250
1 : 50
dilution factor
DF = V2 ÷ V1		 50 ÷ 1
= 50		 50 ÷ 1
= 50		 50 ÷ 1
= 50		 50 ÷ 1
= 50		 50 ÷ 1
= 50

Note that in each example above, the final volume = 50 times the initial volume:

that is, Vfinal = 50 × Vinitial
or, Vfinal = dilution factor × Vinitial
or, Vfinal = DF × Vinitial

Now when the ratio of volumes is known, that is V1:V2 is known,

then dilution factor, DF = V2 ÷ V1
so Vfinal = DF × Vinitial
or Vfinal = V2/V1 × Vinitial

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Effect of Dilution Factor on Concentration

Remember, dilution factor tells us the ratio of the initial (undiluted) volume of solution to the final (diluted) volume of solution:

V(undiluted) : V(diluted)     or     V1 : V2
and, concentration in moles per litre (c)
c = amount of solute in solution (n) ÷ total volume of solution (V)
c = n ÷ V

Consider a 250 mL volumetric flask containing ONLY a 5 mL aliquot of a solution with a concentration of cinitial (this will look like a thin layer of solution in the bottom of the otherwise empty flask).

Vinitial = the initial volume of undiluted solution = 5 mL
cinitial = ? (we have not been told the concentration of this solution)

The amount of solute in the solution (n) = concentration (cinitial) × volume (Vinitial)
n(solute) = cinitial × 5 mL

When more solvent is added to the volumetric flask until the bottom of the meniscus just sits on the mark on the neck of the flask, the new volume of the diluted solution, Vfinal, is 250 mL

Vfinal = final volume of diluted solution = 250 mL

Note that adding more solvent to the flask does NOT add more solute, neither does it remove any solute.
The same amount of solute is still present in the volumetric flask, that is
amount of solute in diluted solution = amount of solute in solution before dilution
amount of solute in diluted solution = n(solute) = cinitial × 5 mL

The concentration of the final solution after dilution (cfinal) = amount of solute in solution ÷ total volume of solution

cfinal = n(solute)
    Vfinal		   from the definition of concentration
  = cinitial × 5 mL
      250 mL		   substituting for n and Vfinal used to make the dilute solution
  = cinitial × 5 mL
      250 mL		   simplifying the expression by dividing by 5, that is, 5/5 and 250/5
Notice that the units for volume also cancel out.
  = cinitial × 1
        50		   the simplified expression
  = cinitial × 1
50		   separating the concentration term from the ratio of volumes term
  = cinitial × Vinitial
Vfinal		   generalising the expression

This equation shows us the relationship between concentration and the ratio of the inital volume, Vinitial, and the final volume, Vfinal.
Now the ratio of the volumes actually used to make the dilute solution, Vinitial : Vfinal, will be the same as the ratio of volumes used to determine the dilution factor,
V1 : V2
So
Vinitial
Vfinal		 = V1
V2

So we can also write cfinal = cinitial × V1/V2

We can rearrange this new equation to find the relationship between dilution factor and concentration:

cfinal = cinitial x V1
V2		   the expression
cfinal × V2 = cinitial × V1 × V2
V2		   multiply both sides of the expression by V2
cfinal × V2
V1		 = cinitial × V1
V1		   divide both sides of the expression by V1
cfinal × V2
      V1		 = cinitial     rearranged expression

Since dilution factor = V2 ÷ V1 = DF
we see that:
cinitial = dilution factor × cfinal
cinitial = DF × cfinal

and by rearranging this equation,
cfinal = cinitial ÷ dilution factor
cfinal = cinitial ÷ DF

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Calculations

Let
cinitial = initial concentration of undiluted solution cfinal = final concentration of solution after dilution
Vinitial = initial volume of undiluted solution Vfinal = final volume of solution after dilution

and
dilution factor as a ratio of volumes is V1 : V2
dilution factor as an equation is DF = V2 ÷ V1

then
Vinitial : Vfinal
≡ V1 : V2
Vinitial
Vfinal		 = V1
V2
Vfinal = Vintial × V2
          V1
Vfinal = Vintial × DF

and
cfinal = cinitial × V1
V2
cinitial = cfinal × V2
V1
cfinal = cinitial
  DF
cinitial = cfinal × DF

A word of warning, the units used in the calculations must be consistent!

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Examples

Calculating Volumes

Question 1. What volume of stock solution must be added to a 100 mL flask to produce a 1:25 dilution once the flask is filled to the mark with solvent?

  1. What is the question asking you to do?
    Calculate the initial volume of solution, Vinitial.

  2. What information (data) has been given in the question?
    Vfinal = final volume of solution after dilution = 100 mL
    dilution factor as a ratio of volumes, V1:V2 is 1:25
    therefore V1 = 1 and V2 = 25

  3. What is the relationship between what you know and what you need to find out?
    (i) DF = dilution factor = V2 ÷ V1
    (ii) Vfinal = DF × Vinitial
        So, Vinitial = Vfinal ÷ DF

  4. Calculate the dilution factor, DF
    DF = V2 ÷ V1 = 25 ÷ 1 = 25

  5. Calculate the initial volume of the stock solution used
    Vinitial = Vfinal ÷ DF = 100 mL ÷ 25 = 4 mL

Question 2. A 5 mL pipette is used to transfer an aliquot of stock solution to a volumetric flask.
What is the volume of the volumetric flask used to produce a 1:200 dilution of the solution?

  1. What is the question asking you to do?
    Calculate the final volume of the solution, Vfinal.

  2. What information (data) has been given in the question?
    Vinitial = intial volume of solution = 5 mL
    dilution factor as a ratio of volumes, V1:V2 = 1 : 200
    therefore V1 = 1 and V2 = 200

  3. What is the relationship between what you know and what you need to find out?
    (i) DF = dilution factor = V2 ÷ V1
    (ii) Vfinal = DF × Vinitial

  4. Calculate the dilution factor, DF
    DF = V2 ÷ V1 = 200 ÷ 1 = 200

  5. Calculate the final volume of solution
    Vfinal = DF × Vinitial = 200 × 5 mL = 1000 mL

Calculating Concentrations

Question 1. What is the concentration of the diluted solution after 0.020 mol L-1 NaCl(aq) is diluted 1:150 ?

  1. What is the question asking you to do?
    Calculate the final concentration of the solution, cfinal.

  2. What information (data) has been given in the question?
    cinitial = initial concentration = 0.020 mol L-1
    dilution factor as a ratio of volumes, V1:V2 is 1:150
    So, V1 = 1 and V2 = 150

  3. What is the relationship between what you know and what you need to find out?
    (i) DF = dilution factor = V2 ÷ V1
    (ii) cfinal = cinitial ÷ DF

  4. Calculate dilution factor, DF
    DF = V2 ÷ V1 = 150 ÷ 1 = 150

  5. Calculate final concentration
    cfinal = cinitial ÷ DF = 0.020 ÷ 150 = 0.00013 = 1.3 × 10-4 mol L-1

Question 2. What was the concentration of the stock solution if, after 1:75 dilution, the final solution has a concentration of 4.85 × 10-3 mol L-1 ?

  1. What is the question asking you to do?
    Calculate the initial concentration, cinitial.

  2. What information (data) has been given in the question?
    dilution factor as a ratio of volumes, V1:V2 is 1:75
    so V1 = 1 and V2 = 75
    cfinal = final concentration of solution after dilution = 4.85 × 10-3 mol L-1

  3. What is the relationship between what you know and what you need to find out?
    (i) DF = dilution factor = V2 ÷ V1
    (ii) cinitial = DF × cfinal

  4. Calculate dilution factor
    DF = V2 ÷ V1 = 75 ÷ 1 = 75

  5. Calculate initial concentration
    cinitial = DF × cfinal = 75 × 4.85 × 10-3 = 0.364 = 3.64 × 10-1 mol L-1

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Problem Solving : Dilution Factor

The Problem:

The site of a meteorite which fell to Earth millions of years ago is suspected of having enough iron in it to make it worthwhile trying to mine it.
Jo the Geologist has taken a sample of the meteorite to Chris the Chemist to determine how much iron is present.
Chris has dissolved 10.00 grams of the meteorite sample in 5.00 L of acid to produce a stock solution.
Chris ran a trial of the analytical procedure on the stock solution, but found the solution was too concentrated to be used.
So Chris prepared a 1:250 dilution of the sample solution, and, after running the analytical procedure successfully, found the solution contained 1.62 x 10-5 mol L-1 Fe2+(aq).
What is the mass of iron in the meteorite sample?

Solving the Problem

Using the StoPGoPS model for problem solving:

STOP!
State the question. What is the question asking you to do?
Determine the mass of iron in sample

What chemical principle will you need to apply?
Apply stoichoimetry (chemical calculations)

What information (data) have you been given?

  • m(sample) = mass of sample = 10.00 g

  • stock solution prepared by dissolving all the sample in acid

  • V(stock solution) = volume of stock solution = 5.00 L

  • diluted solution, dilution factor as a ratio of volumes, 1:250
    So, V1 = 1 and V2 = 250

  • cfinal = concentration of Fe2+ in diluted solution = 1.62 x 10-5 mol L-1

PAUSE!
Plan. Step 1. Calculate the dilution factor
DF = dilution factor = V2 ÷ V1 =

Step 2. Calculate the concentration of Fe2+(aq) in stock solution.
cinitial = DF × cfinal

Step 3. Calculate moles of Fe in sample
    c(Fe2+(aq)) = concentration of Fe2+(aq) in mol L-1
    V(Fe2+(aq)) = volume of Fe2+(aq) solution in L
    n(Fe2+(aq)) = amount of Fe2+(aq) in moles
n(Fe2+(aq)) = c(Fe2+(aq)) × V(Fe2+(aq))

Step 4. Calculate mass of Fe in sample
Assume the only source of Fe in the solutions was the meteorite sample.
Assume all the Fe in the meteorite sample was converted to Fe2+
    m(Fe) = mass of Fe in g
    M(Fe) = molar mass of Fe in g mol-1 from Periodic Table =
m(Fe) = n(Fe2+(aq)) × M(Fe)

GO!
Go. Step 1. Calculate the dilution factor
DF = dilution factor = V2 ÷ V1 = 250 ÷ 1 = 250

Step 2. Calculate the concentration of Fe2+(aq) in stock solution.
cinitial = DF × cfinal = 250 × 1.62 × 10-5 = 4.05 × 10-3 mol L-1

Step 3. Calculate moles of Fe in sample
    c(Fe2+(aq)) = concentration of Fe2+(aq) in mol L-1 = 4.05 × 10-3 mol L-1
    V(Fe2+(aq)) = volume of Fe2+(aq) solution in L = 5.00 L
    n(Fe2+(aq)) = amount of Fe2+(aq) in moles
n(Fe2+(aq)) = c(Fe2+(aq)) × V(Fe2+(aq)) = 4.05 × 10-3 × 5.00 = 2.03 × 10-2 mol

Step 4. Calculate mass of Fe in sample
Assume the only source of Fe in the solutions was the meteorite sample.
Assume all the Fe in the meteorite sample was converted to Fe2+
    m(Fe) = mass of Fe in g
    M(Fe) = molar mass of Fe in g mol-1 from Periodic Table = 55.85 g mol-1
m(Fe) = n(Fe2+(aq)) × M(Fe) = 2.03 × 10-2 × 55.85 = 1.13 g

PAUSE!
Ponder plausability. Does this solution answer the question that was asked?
Yes, we have determined the mass of iron, Fe, in the sample.

Is the solution reasonable?
The mass of iron calculated to be in the sample (1.13 g) is less than the mass of the sample (10.00 g), so the answer is reasonable.
We can work backwards to check the solution:
roughly calculate concentration of Fe2+ in stock solution using the mass of iron we calculated:
    c = n ÷ V = (m/M) ÷ V ≈ (1/50) ÷ 5 = 0.004 mol L-1
dilution factor was 1 :250, so concentration of dilute solution = concentration of stock solution ÷ 250 = 0.004 ÷ 250 = 1.6 × 10-5 mol L-1 which is approximately the same as that given in the question so our solution looks good.

STOP!
State the solution. The mass of iron in the meteorite sample was 1.13 g

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1. While it is possible to add a known volume of solvent to a known volume of solution, and then add these two volumes together to give you the final volume of solution, in practice this is usually not a good idea because it assumes additivity of volumes.
It is best to place a known volume of solution in a volumetric flask and then add solvent up to the mark so that you know the final volume of the solution.
This overcomes any difficulties there may be in assuming additivity of volumes.