Units and Calculations used in the lab

This article should serve as an introduction to: units of volume, mass, amount of substance and concentration in the lab and various formulae linking these variables. There is also a table on the various models of Gilson pipette and the volumes of fluid that they disperse.


Units of Volume and the use of Gilson Pipettes

Volume is measured in litres (L). Volume can also be measured in decimetres cubed (dm3), where 1dm3 is equal to 1 litre.
The following table shows some other units of volume commonly used in the lab:
1mL
1 x 10-3 L (1/1000 L)
1µL
1 x 10-6 L (1/1,000,000 L)
1nL
1 x 10-9 L (1/1,000,000,000 L)

The Gilson’s pipettes we use in the lab each dispense different ranges of volumes. This table shows you the differences between the various models:
Model
Min. volume
Max. volume
On the dial...
P10
1µL
10µL
075 means 7.5µL
P20
2µL
20µL
125 means 12.5µL
P100
20µL
100µL
075 means 75µL
P200
50µL
200µL
125 means 125µL
P1000
200µL
1000µL (= 1 mL)
075 means 750µL (= 0.75mL)
P5000
1000µL (= 1 mL)
5000µL (= 5 mL)
125 means 1250µL (= 1.25mL)


Units of mass and amount of substance

Mass is measured in grams (g).

The following table shows some units of mass commonly used in the lab:
1 mg
1 x 10-3 g (1/1000 g)
1 µg
1 x 10-6 g (1/1,000,000g)
1 ng
1 x 10-9 g (1/1,000,000,000g)
Note: 1mg/mL is the same as 1ng/nL, or 1g/L

1 mole (mol) of a substance is the amount of that substance that is equal to the Avogadro’s constant (= 6.022 x 1023 particles of the substance).
Similarly to the above:
1 mmol
1 x 10-3 mol (1/1000 mol)
1 µmol
1 x 10-6 mol (1/1,000,000mol)
1 nmol
1 x 10-9 mol (1/1,000,000,000mol)

Molar mass is the mass (in grams) of one mole of a substance. Thus, molar mass is measured in g mol-1.

We can relate mass (g), molar mass (g mol-1), and the amount of substance (mol) using the following formulae:


Moles mol = mass g / molar mass g mol-1
Mass g = moles mol * molar mass g mol-1
Molar mass g mol-1 = mass g / moles mol



Some handy examples...

EXAMPLE 1:
I have 45mmol of Substance X. Substance X has a molar mass of 121.56 g mol-1. What mass, in grams, of substance X do I have?

Mass (g) = moles (mol) x molar mass (g mol-1)
Mass (g) = 0.045 mol x 121.56
Mass (g) = 5.4072g

I have 5.41g of Substance X, correct to 3 significant figures.


EXAMPLE 2:
I have 45mg of Substance Y. Substance Y has a molar mass of 68.54 g mol-1. How many millimoles (mmol) of substance Y do I have?

Moles (mol) = mass (g) / molar mass (g mol-1)
Moles (mol) = 0.045g / 68.54g
Moles (mol) = 6.57 x 10-4mol = 0.000657 mol or 0.657mmol

I have 0.657mmol of Substance Y.


Units of concentration and calculations involving concentration

Concentration involves both mass of solute and volume of solution, so we will be using units from both of the two previous sections now.

Concentration is measured by molarity (M). A 1 Molar (1M) solution has 1 mole of solute dissolved in 1 litre of solution. Thus 1M = 1 mol L-1

The following table shows some other units of concentration commonly used in the lab:
1 mM
1 x 10-3 M (1/1000 M)
1 mmol solute per litre of solution
1 µM
1 x 10-6 M (1/1,000,000M)
1 µmol solute per litre of solution
1 nM
1 x 10-9 M (1/1,000,000,000M)
1 nmol solute per litre of solution

If we want to make a solution of known volume and concentration, then we need to know what mass of solute needs to be weighed out. Molarity (M), moles (mol) and volume (L) are linked by the following set of formulae:

Molarity M = moles mol / concentration L
Moles mol = Molarity M * concentration L
Concentration L = moles mol / molarity M

The best way to demonstrate this is with some examples...

EXAMPLE 3:
If I wanted to make 50ml of a 1mM solution of glycine (molar mass: 75) how much glycine would I need to weigh out?

1mM solution is equal to 0.001M
50ml is equal to 0.05L

We can work out how many moles we need using the formula:
Moles (mol) = Concentration (M) x Volume (L)
Moles (mol) = 0.001M x 0.05L = 0.00005mol

Now, knowing the moles required, we can use the previous formula to calculate mass (g):
Mass (g) = moles (mol) x molar mass (g mol-1)
Mass (g) = 0.00005mol x 75g mol-1
Mass (g) = 0.00375g

So we need to weigh out 0.00375g (or 3.75mg) of glycine.

EXAMPLE 4:
You have weighed out 5g of galactose (molar mass = 180) and dissolved it in 100ml water. What is the molarity of the solution?

First we need to know how many moles (mol) of galactose we have used:
Moles (mol) = mass (g) / molar mass (g mol-1)
Moles (mol) = 5g / 180g mol-1
Moles (mol) = 0.0278mol

Now we can work out the concentration (M):
Concentration (M) = moles (mol) / volume (L)
Concentration (M) = 0.0278mol / 0.1L
Concentration (M) = 0.278M



Now we’ll look at concentrations involving w/w, v/v or w/v.

w/w and v/v must have the same units on the top and bottom.

w/v uses the same units for top and bottom too – since the volume is usually made up of water, we can say that 1L of water is equal to 1kg of water. Thus, 3g solute per 100mL water is the same as saying 3g solute per 100g water.

Keep in mind the total weight (or volume) of your compound or solution. The %v/v of 30mL ethanol in 100mL water is 30/130, not 30/100. However, the %v/v of 30mL ethanol used to make up 100mL final solution is 30/100.


EXAMPLE 5:
What is the %w/w of copper in an alloy when 25g of Cu is mixed with 125g of Zn?
25/150 = 16.7%


EXAMPLE 6:
What is the %w/v of 5g glycine used to make 100mL (that is, 100g) solution with water?
5/100 = 5%



Calculations involving dilution factors

In this final section we will look at dilution and dilution factors.

Another set of formulae to remember:

Dilution factor = initial concentration M / final concentration M
Initial concentration M = dilution factor * final concentration
Final concentration M = initial concentration M / dilution factor

Really not much understanding here, so we’ll dive straight into some examples:

EXAMPLE 7:
After a 40-fold dilution, a trehelose solution has a concentration of 0.05M. What was its original concentration?

Initial conc (M) = dilution factor x final conc (M)
Initial conc = 40 x 0.05
Initial conc = 2M



And, finally, an example that uses everything you’ve learned in this article....

EXAMPLE 8:
How much water do you need to add to 70mg of aspartic acid (molar mass = 133) to make a solution of 50mM?

50mM is 50mmol in 1 litre

In 70mg of aspartate the number of moles we have is:
Moles (mol) = mass (g) / molar mass (g mol-1)
Moles (mol) = 0.07g / 133g mol-1 = 5.26 x 10-4­­ mol or 0.526mmol

Now we need to work out the dilution factor from 50mM to 0.526mM:

Dilution factor = initial conc / final conc = 50 / 0.526 = approx. 95

The volume required is thus 1L/95 = 0.0105L = 10.5mL