Prepared by Chris Weller
Laboratory Glassware:

Mole Mass Conversion
 Grams/L x (1/MW) = Moles/L
Example:
Need to know how many grams of NaOH to make
1Litre of 2M.
Rearrange equation: to show that the grams is the
unknown we what to find.
Grams= Moles / (1/MW)
Grams =2M / (1/40.01)
Grams= 80.02 of NaOH /Litre of distilled water.
 Caution this is exothermic reaction.
Important to remember check the label you
need to know the starting strength (%)
Grams = Density x 1000ml x %
Example: HCl comes in two different strengths
32% and 36%.

Grams = 1.179g/ml x1000ml x 36%
Grams = 1.179g x1000 x0.36
Grams = 424.44
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Next step to work out the number of moles.
Grams x (1/MW) = Moles
Rearrange to find the number Moles.
M= (g x 1000ml) / (MW x ml)
M= (424.44g x 1000) /(36.46 /1000)
Moles = 11.64M
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This is the strength of your HCl
Now work out how many Moles is 32% HCl.

Next Step is dilution.
Example : 32% HCl = 10.17 M and you need to
make 500mls of 2M.
M1x V1 = M2 x V2
M1=The original number of moles of HCl 10.17
V1= How much do we need?
M2 = Moles needed 2M
V2 = Volume needed 500mls
10.17M x V1ml= 2Mx 500ml
Rearrange to find V1
V1= (2x500)/10.17
V1= 98.33ml
Sulphuric acid. 96-98%
1. Grams = Density x 1000ml x %
2. M= (g x 1000ml) / (MW x ml)
Step1.
Grams= 1.84g/mlx 1000mlx 96%
Grams=1.84g x 1000ml x0.98
Grams=1803.2
Step 2.
M= (1803.2 x1000ml) / (96.07x/1000ml)
M= 18.85
Then dilute to suit the moles you require.
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When mixing solutes what is the final Mole?
Example: 50ml of 0.5M NaOH with 250ml of
1M NaOH What is the final molar strength.
M1V1+M2V2= M3V3
M1= 0.5M ; V1 = 50ml; M2 = 1.0; V2 =250ml
M3 = unknown; V3 = 300ml
Rearrange to find M3
M3= (0.5Mx50ml+1.0M x 250ml) /300
M3=0.92M
Does the Chemical have water added?
This needs to be taken into consideration when
considering the Molecular Weight (MW).
 Read the information on the Chemical container label.
Example. Copper Sulphate comes in Anhydrous
CuSO4(pale green to white powder) and Pentahydrate.
CuSO4.5H2O (bright blue powder)
CuSO4 =159.62g/mol
CuSO4. 5H2O = 249.70g/mol
This will affect your accuracy of your chemical solution.
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
Liquids chemicals always check the % and density


NaOH example: 80.02g/L to make 2M
You only need 250mls.
 1000/250 = 4
 80.02/4 = 20.005g
Therefore you need 20.005g of NaOH to make
up 250mls.
Composition of concentrated reagent grade acids, ammonium hydroxide, and sodium and potassium hydroxide solutions (with dilution directions to prepare 1N solution)
Chemical
MF
Approx. Strength of
Molarity of
Milliliters of Concd. Reagent Necessary
Concd. Reagent
to Prepare 1 Liter of 1 Normal Soln.
99.8
17.4
57.5
Concd. Reagent
Acetic Acid, Glacial
Formic Acid
CH3COOH
a
c
HCOOH
90
23.6
42.5
Hydrochloric Acid
HCl
37.2
12.1
82.5
Hydrofluoric Acid
HF
49
28.9
34.5
Nitric Acid
HNO3
70.4
15.9
63
Perchloric Acid
HClO4
70.5
11.7
85.5
Perchloric Acid
HClO4
61.3
9.5
105.5
Phosphoric Acid
H3PO4
85.5
14.8
22.5
Sulfuric Acid
H2SO4
96
18
28
Ammonium Hydroxide
NH4OH
56.6b
14.5
69
NaOH
50.5
19.4
51.5
KOH
45
11.7
85.5
Sodium Hydroxide
Potassium Hydroxide
a - Representative value, w/w%.
b - Equivalent to 28.0% w/w NH3.
c - Rounded to nearest 0.5 ml.

Example Nitrate testing of water samples.
Students need a more accurate result to compare
samples as all the sample appear to be the same
reading. So you make up the standards with in the
range they ask for. E.g. 0-10 ppm then the Student
can use Beer-Lambert Law to make a calibration
curve.
0-10ppm = 0-10mg/L
 Can use Potassium or Calcium nitrate. To
make up the solutions. As K and Ca are found
in the water normal so the metals won’t affect
the results.
 Consider : how accurate are you scales? How
many digital placing? This will affect how you
approach making the standards.
For this exercise we will assume the scales read
two decimal places.
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Accuracy is important. Need to calculate the
standards require. A range between 0 and 10
mg/L. The more samples you have the better
the accuracy. The norm is between 5-10
samples for this range including your blank
sample.
0, 1,2,3,4,5,6,7,8,9,10 mg/L
0,1,2.5,5,10mg/L
If you don’t want to make up a Litre you can
scale it back to 100mls.
10mg/L equals 0.01grams in 1000mls.
Our problem is that our scale only reads 2 decimal
places. What are our options.
1. Go and buy better scales $$
2. We could do a series of dilutions.
Series Dilutions:
Step 1.
Make up 0.01g of CaNO3 in 1000mls of water.
This is our stock solution. You may have to may up
the fresh standards daily as this is a colorimetric
method. The colour can either fade or intensify,
depends on what it is.

Step 2.
Once you have thoroughly mixed the stock
solution. Decant 100mls in to a volumetric flask.
Mark it as 10mg/L. of CaNO3
 Step 3.
Into another 100ml volumetric flask pipette 50 mls
of stock solution and then fill with distilled water
to the mark. This is 5mg/L CaNO3
 Step 4.
Pipette 50mls of the 5mg CaNO3 into a 100mls
volumetric flask then fill with distilled water to the
mark. This is 2.5mg/L CaNO3.

Step 5.
Pipette 10mls of the 10mg standard into a 100ml
volumetric flask and fill to the mark. This is 1mg/L
of CaNO3.
 Step 6.
Pipette 5 ml of your 1st standard 10mg then add
the drops of indicators as per the instructions in
the kit. Wait the required time and them place in
cuvette and then read the sample in your
colorimeter.
 Repeat for all other standards.
 Then using excel you can plot your Standard
Curve Chart. Then test the unknown.
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This method can be done for all colorimetric
and UV Vis analysis.
For use in EEI the students can either make
the standards up themselves and compare
their standard curve results against yours.
Or the students can use your standard curve
results and then insert their own data to give
them a more accurate reading in ppm.

Oriakhi, Christopher O. Chemistry in
Quantiative Lanugage: Fundamentals of
General Chemistry.