Maths handout

Question 1

Scientific notation

Answer 1

It is inconvenient to write out large and small numbers, and copying numbers with lots of zeros
can lead to mistakes.

E.g. 123000000 = 1.23 x 10 to the power of 8
0.00000123 = 1.23 x 10 to the power of -6

There is always one number before the decimal point in scientific notation.

Count the number of places that you need to move the decimal places to the left or right to determine n in 10 (to the power of) n and make sure that the answer is written as 10 (to the power of) -n if the original number is less than 1.

Question 2

Concentration units

Answer 2

The concentration of solutions can be expressed in many ways, and it is common to choose units that are most convenient: e.g. mg/ml or kg/m 3 might be used at small and large scale, respectively.

Question 3

Molar concentrations

Answer 3

1 mole - Relative Molecular Mass (RMM) of a compound in grams

1 Molar solution - 1 mole of the compound in 1 litre of solution (1 mole/litre)

Do not confuse mol (an amount) with M (a concentration)

Lower concentrations are expressed as mM (millimolar, or 10 -3 M) or µM (micromolar, or 10 -6 M).

Example:

1 mol of NaCl = 23 grams (Na) + 35.5 grams (Cl) = 58.5 grams

So 1M solution of NaCl = 58.5 grams of NaCl in 1 litres of H 2 O

Or 1 mM solution of NaCl = 0.059 grams of NaCl in 1 litres of H 2 O

Molar concentrations are useless for some purposes: you could not express the concentration of plant/animal tissue or microbial suspension in this way.

Question 4

Percentage Concentrations

Answer 4

Expressing concentrations in g l -1 or M is unambiguous.
When using percentage concentrations, you must specify

% weight/volume (w/v), % weight/weight (w/w) or % volume/volume (v/v)

because of changes in volume or density.

Example:

mixing 500 ml glycerol with 500 ml water does not produce a litre of solution. At low concentrations (<1%), the differences become insignificant and it is
acceptable to refer, for example, to 0.5% NaCl.

5%(w/v) glucose is equal to 50 g glucose per litre of solution

5%(w/w) glucose is equal to 50 g glucose per kg solution

In the case of solutions containing liquids, concentrations can also be expressed in terms of volume, so that 50%(v/v) ethanol is a mixture of equal volumes of ethanol and water.

Question 5

Working with concentrations

Answer 5

• Preparation of solutions
• Calculations involving dilution of solutions
• Expressing concentrations in alternative units

Question 6

Concentrations

Answer 6

aqueous solutions unless stated otherwise

Question 7

Mole

Answer 7

a mole of any compound contains the same number of molecules. This number is known as Avogrado’s number (6.023 x 10 (to the power of) 23).

Question 8

Relative molecular mass (RMM) -> formerly known as molecular weight (MWt)

Answer 8

This is calculated by adding the atomic weights of the elements present in a compound.

Example:

Glucose C6H12O6
-> (12 x 6) + (1 x 12) + (16 x 6) = RMM of 180

1 mol of glucose is 180 g of glucose