LABORATORY CALCULATIONS
These exercises are designed to give you practice in the basic mathematical concepts
required for successful completion of the course. The maths is generally not difficult
but the application of the techniques to the chemical world may challenge even the
best mathematician . Lots of practice will ensure you are competent and therefore able
to successfully complete the assessments.
You are expected to be competent at the following and should tick off in the table
when you feel competent:
Concept
Week
Yes
No
Week
Yes
No
Week
Yes
No
1. Use of your calculator
2. Estimation of result
3. Rounding
4. Order of operations
5. Scientific notation
6. Directed numbers
7. Significant numbers
8. Substitution into formula
9. Rearranging formula
10. Unit conversions
11. Graphing
12. Formula weights
13. Mole calculations
14. Concentration calculations
15. Stoichiometry
16. Putting it all together
The first assessment covers concepts 1-11.
The second assessment covers concepts 12-16
1
Use of the calculator
All calculators operate slightly differently but all should give the same answer for
a particular calculation. If the answer is not correct then it is most likely that the
operator, you, has incorrectly entered data or not followed appropriate
mathematical logic.





You should know how the following functions operate on your calculator
square root
log 10 (normally called log or log10)
10x (often on the same button as log10)
memory (store, add, recall)
EXP for scientific notation
The best way to work our how your calculator operates is to read the manual…but
they are very thick and this will most likely not occur.
The next best way is to try some calculations that you know the expected answer.
Exercise
6 +2 x 3 =
(12)
5-4 x 3 =
(-7)
16 x 4 + 2 x 3 =
(70)
2
Estimation of result
The calculator result is only as good as the operator ie if you enter an incorrect number or do
not use correct mathematical logic the answer will be wrong. You should have an idea of the
expected answer, not just trust the calculator.
Basic process for estimation
1. Round all numbers to one non-zero digit
2. Calculate the values inside any brackets and round the results.
3. Cancel out any zeroes where possible.
4. Calculate estimate.
Example
Estimate the answer to the calculation from Example 1.2.
(63.8  1.94) x 0.0547
0.626 x 10
1.
Round all the numbers
(60  2) x 0.05
0.6 x 10
2.
Calculate inside the brackets and round result.
60 x 0.05
0.6 x 10
3.
Remove spare zeroes.
6 x 0 .5
6 x1
4.
Calculate estimate.
0.5 (the exact answer was 0.541).
Now try these
(i)
0.112 x 25.3
=
12.5
(ii)
24,300 x 16.8
=
8.314 x 298
iii)
(23.93  181.69) x 0.123
 77.7 =
87.6  1.45
3
Rounding
To round off to a stated accuracy the last figure to be retained is corrected depending on the
digit which follows. A typical rounding system states that if the following digit is 5, 6, 7, 8
or 9 then the last figure is taken up one.
Example
Round off correct to the nearest whole number
2463.62  2464
Round off correct to two decimals places
7.2048  7.20
Round off correct to the nearest thousand
849801  850000
Exercise
1. Round off to the nearest whole number

2344.62

16.403

7.9

0.735

829.99
2. Round off correct to the nearest figure as shown in brackets)

6.4137 (2 d.p)

36.45071 (3 d.p)

0.003682 (4 d.p)

5.20196 (2 d.p)

463.9 (tens)

8 420 (hundreds)

46 375.9 (hundreds)

68 420 (ten thousands)

4 724 361 (millions)

89 840 000
4
Order of Operations
The rules for order of operations are:
1. Brackets first
2.  and  from left to right second
3. + and – from left to right third
Exercise
1. 5 x 4 - 1
=
2. 6 + 10  2
=
3. 27 - 4 x 2
=
4. 2 x (3 + 6)
=
5. 6 x 4  2
=
6. 3 + 4 -2
=
7. (2 + 3 x 6) + 1 =
8. 4 x 2 + (3 + 1) =
9. 24  4  2
=
10. [(3 + 2) + 4 x 2] + 2 x 2 =
11.
52  4
8
=
12.
63
63
=
13.
3x 2  3
3
=
14.
8  2 1
5  2 x2
=
15.
4  24  6
64  8
=
16.
16  (4 x 2)
56  7
=
17.
4  4 x8
=
18. 122  52
=
19. 12 – (20 – 3 x 6)2 =
20.
13  31  2 x3
2
=
5
Scientific Notation
Scientific notation is a method for writing very large and very small numbers in the
form of a number between 1 and 10 and a power of 10.
Example…
Avagadro’s Number, the number of entities in a mole is a number that the mind has
difficulty understanding.
602000000000000000000000
Written in Scientifc notation the number is 6.02 x 10 23
If the index number is positive it indicates that the number is large and the decimal
point has been moved right to left
If the index number is negative it indicates the original number is small (less than 1)
and the decimal point has been moved left to right
Exercises
Write in scientific notation
1. 720 000
2. 15 000 000
3. 896.5
4. 8076
5. 0.005
6. 0.000 001 3
7. 0.000 009
8. 10 100 000
9. 0.000 246
10. 21 000
Write the original number for
11. 5.2 x 106
12. 2.015 x 10-3
13. 4.6 x 104
14. 1.11 x 1011
15.
3.33 x 10-4
6
Directed numbers
To multiply or divide

Two positive numbers give a positive number

Two negative humbers give a positive number

A positive and a negarive number gives a negative number
1. 6 + 2 (-2)
=
2. (-3) + 3
=
3. (-5) x 6
=
4. -15  5
=
5. -8 x -3
=
6. Given a = -1, b= 2, c = -4 d = -3
Evaluate

abc
=

d2
=

a3
=

a+b+c+d =

2b c
=

ab – cd
=

-(-d)
=
7
Significant Figures
Significant figures are defined as follows:
1. All non-zero digits are significant
2. Zeros at the end of a whole number or at the beginning of a decimal are not
significant
3. Zeroes between non-zero digits are significant
4. Zeroes at the end of a decimal are significant
Examples
1. 34 100 has 3 significant figures
2. 6010 has 3 significant figures
3. 0.0042 has 2 significant figures
4. 0.0380 has 3 significant figures
Exercise
1. Give the number of significant figures in each of these numbers

478 200

4 606

0.01003

3000

863.9462
2. Rewrite each of these numbers
i) correct to 2 sig fig
ii) correct to 1 sig fig

62 481.69

459

0.0003816

0.00204

3.004

86 200 000

0.00004937

2450
8
Substitution into a formula
Often the calculations you will have to do use provided formulae which are supposed
to make the process of getting the right answer easier. A formula is simply a
calculation where the numbers are replaced by symbols, such as the formula for
moisture content in Example 1.1.
EXAMPLE
You are analysing soil for its moisture (water) content. You weigh out soil into a basin
weighing 75.2826 g, until the balance records a weight of 85.9837 g. After 2 hours of
drying, the basin and dried soil weighs 82.5832 g.
 m  md 
 x 100
Moisture content (%)   w
 mw  mb 
where
mb is the mass of the empty basin = 75.2826
md the mass of the basin with dried soil = 82.5832
mw the mass of the basin with wet soil = 85.9837
 85.9837  82.5832 
 x 100  31.78%
Moisture content (%)  
 85.9837  75.2826 




Other formulae often used in the laboratory include:
C1V1 = C2V2
the dilution equation
C1 is the concentration of the first (more concentrated) solution
V1 is the volume of the first (more concentrated) solution
C2 is the concentration of the second (diluted) solution
V2 is the volume of the second (diluted) solution
C



m
V
a concentration equation
C is the concentration is g/L,
m the mass in grams
V the volume in L.
Exercise
Substitute the data into the concentration equation to determine the value for C
1. m = 15g, V = 2L
2. m = 0.5g, V = 0.05L
3. m = 525g, V = 200L
9
Rearranging formula to solve equations
If the unknown in an equation is not the subject of the formula than the formula must
be rearranged in order to obtain an answer. The process is simple as long as the rules
are followed:
Equations may be simplified and solved by:
1. adding the same number to both sides
2. subtracting the same number from both sides
3. multiplying both sides by the same number
4. dividing both sides by the same number
Example
x + 8 =4
x = 8–8 = 4–8
x = -4
6 x d = 42
6 x d = 42
6
6
(to remove the 8 we need to subtract 8, but from both sides)
(to remove the 6 we need to divide both sides by 6)
d =7
Exercise
Solve the following equations
1. c + 18 = 13
2. t – 8 = - 8
3. 4a = 2.4
4. 8 t = 4
5. y – 2.8 = 0.6
10
The same process can be used when substituting into formula
Example 1
C
m
V
rearrange to make m the subject of the formula
In order to remove the V we need to multiply each side by V
CxV = m xV
V
CxV=m
Example 2
C1V1 = C2V2 rearrange to make V1 the subject of the formula
In order to remove C1 from the left hand side we need to divide both sides by C1
C1V1 = C2V2
C1
C1
then
V1 = C2 V2
C1
Exercise
Rearranging the following formula to make the symbol in brackets the subject.
m
(V)
V
(a)
C
(b)
C1V1 = C2V2
(c)
Tc = 0.556Tf – 17.8
(d)
C
(C2)
(Tf)
m
(m)
V
11
General Revision of Concepts
Given the following data and formulae, calculate the values of the unknown, reporting
the result to an appropriate number of significant figures
Formula
Data
(a)
C1V1 = C2V2
C1 = 1.2 g/L, V1 = 5 mL, V2 = 250 mL
(b)
PV = nRT
P = 1.024 x 105, V = 0.234, R = 8.314, T = 298.2
(c)
Tc = 0.556 (Tf – 32)
Tf = 213.3
(d)
(e)
C
8000 (b  2a)
V
C1V1 = C2V2
b = 65.2, a = 21.3, V = 10.0
C1 = 1000 mg/L, C2 = 50 mg/L, V2 = 250 mL
12
Unit conversions
There are many different nits that could be used to state a measurement. For example,
if your height is 1.8 m, you could also say you are 180 cm, or 1800 mm, or 1.8 x 10 6
um or 0.0018 km. And then there are all the non-metric measures for length: 5.91 feet,
70.9 inches, 1.12 x 10-3 miles etc. Which is correct? They all are!
It is important to be able to convert units to provide numbers that are reasonable. The
process for linear , area and volume conversions is similar, You should practice the
method regularly to ensure that for even the most difficult conversions the answer is
correct.
The metric prefix system (not complete)
Prefix
Abbreviation
How Many
Base Units
How Many in
One Base Unit
nano
n
10-9
109
micro

10-6
106
milli
m
0.001
1000
centi
c
0.01
100
deci
d
0.1
10
kilo
k
1000
0.001
mega
M
6
10
10-6
The process will work for all conversions…..you just need to practice
Example
1. How many metres are in 24.7 km
24.7 km = 24.7 km x 1000 m
km
thus
24.7 km = 24.7 x 1000 m
(km will cancel out)
2. How many grams in 356 mg
356 mg = 356 mg x
1g .
1000 mg
(mg will cancel out)
356 mg = 0.356 g
The same system will work for imperial conversions as well
13
Perform the following conversions
(a) 57 metres to centimetres
(b) 20 metres to millimetres
(c) 45000 ug to g
(d) 1.24 hectares to m2
(e) 72 cm3 to m3
(f) 23 m to feet, given 1 m = 3.28 ft.
(g) 2682 acres to hectares, given 1 acre = 0.405 ha
(h) 5983 Pa to psi, given 1 psi = 6891 Pa
(i) 1.24 pounds to grams, given that 1 pound = 454 g
Some more to try
(a) 457 nm to cm.
(b) 0.4982 kHz to MHz.
(c) 25 mg to kg
(d) 2.54 x 104 inches to miles, given 36 inches = 1 yard and 1760 yards = 1
mile.
And some more difficult ones to try
Convert:
(a)
6932 ug/L to g/mL.
(b)
0.0487 mg/g to mg/kg.
(c)
181 ug/mL to g/L.
14
And these also
Questions
1 foot (ft) = 0.305 metre (m)
1 pound (1b) = 0.454 kilogram (kg)
1 gallon (gal) = 4.55 litres (L)
1 acre (ac) = 4047 square metres (m2)
1 foot (ft) = 12 inches (in)
1 inch = 2.54 cm
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
205.6 nanometres to metres.
306,500 grams to kilograms.
3.45 MJ to kJ
856 mm2 to m2.
5075 dm3 to cm3.
500 m3 to ML.
4.56 x 103 L to kL
2350 mm3 to cm3.
50 g/L to mg/mL.
1.822 g/mL to kg/m3.
27,740 kJ/kg to J/g.
106 feet to metres.
78 lb/ft3 to kg/m3.
2500 m2 to acres.
20 millilitres per minute cubic centimetres per hour.
560 square inches to square centimetres.
15 litres per second to gallons per minute.
15
Graphing
Graphs are pictorial versions of numerical data. They are intended to tell the “story”
in the data in a visual way. The can also allow certain types of calculations to be done
more easily.
Types of graphs
There are many different types of graphs, each with a particular purpose. Some can
be used for all types of data, others will only suit particular types of data sets. When
you study spreadsheets in computing, you will learn how to construct graphs using a
computer program. The main graph types are:

column

bar

pie

line

X-Y scatter
Examples of each are shown in Figure 4.1. In this chapter, we will concentrate on XY scatter graphs.
Column
Pie
Bar
Line
X-Y scatter
Graph types
You may think that line and X-Y scatter graphs are the same - they look similar, but
there is one important difference. In column, bar and line graphs, each data point is
equally spread from the next one, whereas in X-Y scatter, they are separated by
differences in the X-value. Figure 4.2 shows this more clearly.
16
points evenly
spread across
horizontal axis
Data
X
Y
0.2
0.4
0.8
1.6
1
2
4
8
points spread across
horizontal axis
according to X-values
X-Y Scatter
Line
FIGURE 4.2 The difference between line and X-Y scatter graphs
X-Y scatter graphs are used where the X-value is as important as the Y-value. This is
particular the case in so-called calibration graphs, where an instrument response is
checked against solutions of known concentrations. In line graphs, the X-value is
only important as an identifier, eg months of the year.
4.2 How to draw graphs
The are a number of points to consider when drawing graphs, so that they are correct,
informative and not misleading:

which axis is which - the y- (usually vertical) axis plots the measurement while
the x- (usually horizontal) axis plots the set value (the solution concentration) this does not apply to pie charts; for bar charts, the y-axis is the horizontal axis

labelling - axes should be labelled the name and units of the data variable

labelled and sensible axis scale - where an axis scale is important (the y-axis in
all cases and the ax-axis in X-Y graphs), the scale should be clearly labelled and
sensible - make sure you fill up the sheet of graph paper as bets as possible, but
don’t choose a scale where 1 cm = 3.45 units just to fill the page!

consistent scale - the scale (what each cm, for example, is worth) should be the
same across the whole in most circumstances (see Figure 4.3)
1.2
1.2
1
1
0.8
0.8
Reading
Reading
Calibration Graph
(b)
Calibration Graph
(a)
0.6
0.4
0.2
0.6
0.4
0.2
0
0
0
5
10
15
20
Concentration
25
0
1
2.5
5
10
Concentration
FIGURE 4.3 (a) Correct scale (b) incorrect scale


plain paper vs graph paper – if you have a computer, then you can produce
professional-looking graphs; however, the grid on graph paper is useful where
you need to make a measurement from the graph
computer vs hand-drawn – draw the graph yourself on graph paper if you have
to make a measurement from the graph, eg a calibration graph
17
20
4.3 Calibration graphs
As mentioned above, these are X-Y scatter graphs, used to calibrate an instrument –
measuring its response to standards (solutions of known concentration) for the
purpose of using the graph to determine the concentration of samples. Normally, a
calibration graph will be a straight line, but the data you have from the standards will
probably not be perfect.
A line of best fit is a straight line drawn through the data points being as close
to each as possible. A computer program, such as Excel, will not only graph your data,
but also work out the correct line of best fit. A calibration graph does not:
 join the dots,
 plot samples – they have an unknown concentration
EXAMPLE 4.1
Plot a calibration graph from the data below.
Conc.
0
2
4
6
8
Respons
e
5
75
141
235
311
Graph on next page.
CLASS EXERCISE 4.2
Draw a calibration graph from the data below on the paper provided.
Concentration
(mg/L)
0
5
10
20
Response
0
0.25
0.51
1.05
350
Graph for Example 4.1
300
Response
250
200
150
100
50
0
-50
0
2
4
6
8
10
Conc.
18
How to find the concentration of a sample from the calibration graph
If you are using graph paper, then you draw a horizontal line across from the vertical
axis at the response value for the sample. When this hits the calibration graph, drop a
vertical line to the concentration axis. Where the line hits the axis is the sample
concentration.
EXAMPLE 4.2
Using the calibration graph from Example 4.1, determine the concentration of
a sample with a response of 100.
350
300
Response
250
200
150
100
50
0
-50
0
2
4
conc. = 2.5
6
8
10
Conc.
CLASS EXERCISE 4.3
Determine the concentration of the sample with an instrument response of
0.44, using your graph from Exercise 4.2.
19
You can see in Example 4.2 that the plain paper and computer combination is not
ideal for measuring the sample concentration. If you are using a computer then it is
better to use an equation, which describes the straight line (Equation 4.1).
response  y int ercept
Eqn 4.1
slope
where the slope is the rise ÷ run of the straight line, and the y-intercept is where the
line hits the vertical axis (when the concentration is zero).
The computer program will calculate the slope and y-intercept for you, and
you will b shown how to do this in the computing subject where you cover
spreadsheets. However, we will use the equation here, given the necessary data.
concentration 
EXAMPLE 4.3
For the graph in Example 4.1, the slope is 38.6 and the y-intercept is –1.0.
What is the concentration for the sample with a response of 100.
conc. 
100  ( 1)
 2.61
38.6
CLASS EXERCISE 4.4
Calculate the sample concentration for the following:
Slope
y-intercept
Exercise 4 page 47
0.0433
0.0005
Exercise 5 page 47
0.0137
-0.0002
Concentration
Calculating the slope of a line from graph paper
The slope is equal to the vertical rise divided by the horizontal run. This is not the rise
and run as measured by your ruler, but from the measured values from the scale.
EXAMPLE 4.4
Calculate the slope of the graph in Example 4.1 using rise/run.
350
300
Response
250
rise = 160 - 80 = 80
200
150
100
run = 4 - 2 = 2
50
0
-50
0
2
4
6
8
10
Conc.
Slope 
rise 80

 40
run
2
CLASS EXERCISE 4.5
20
Calculate the slope of your graph in Exercise 4.2.
Questions
1.
Determine the sample concentrations for the following.
(a)
Standard Conc.
Response
(mg/L)
0
0.00
5
0.24
10
0.48
15
0.72
Sample
0.36
(b)
Standard Conc. %
0
20
40
60
80
Sample
Response
0
0.2567
0.4099
0.5621
0.7436
0.4488
(c)
Std Conc. (mole/L)
0
1 x 10-4
2 x 10-4
3 x 10-4
4 x 10-4
Sample
Response
0
0.17
0.36
0.53
0.71
0.40
(d)
Standard Conc.
mg/L)
0
4
8
12
Sample
Response
0.000
0.171
0.353
0.516
0.264
21
1.
2.
(e)
Standard Conc. (%)
0
5
10
20
Sample
Calculate the sample concentration, given the following line-of-best fit data.
(a)
(b)
(c)
3.
Response
0
157
324
462
333
Slope
0.00567
168
0.0594
y-intercept
0.001
-12
-0.005
Sample Reading
0.167
1810
0.333
Calculate the slope of the your graphs in Q1.
22
Formula weights
The formula weight of a pure chemical substance is the sum of the atomic weights of
the elements that make up the substance. The Table gives you the atomic weights of
common elements: if you need the weight of a less common element, use a periodic
table.
Atomic weights of selected elements
Element
At. Weight
Element
At. Weight
Element
At. Weight
Hydrogen
1.008
Aluminium
26.98
Iron
55.85
Carbon
12.01
Phosphorous
30.97
Copper
63.54
Nitrogen
14.01
Sulfur
32.06
Bromine
79. 91
Oxygen
16.00
Chlorine
35.45
Iodine
126.9
Sodium
22.99
Potassium
39.10
Barium
137.3
Magnesium
24.31
Calcium
40.08
Lead
207.2
What is the formula weight of copper nitrate – Cu(NO3)2?
You have to include all atoms, not just one of each element, so in this substance there
are:
 1 copper,
 2 nitrogen, and
 6 oxygen atoms
FW
= (1 x 63.54) + (2 x 14.01) + (6 x 16.00)
= 187.56 au
Exercise
Calculate the formula weights of the following:
(a)
sodium phosphate, Na3PO4
(b)
glucose, C6H12O6
(c)
magnesium nitrate, Mg(NO3)2
(d)
ammonium iron (II) sulfate, (NH4)2Fe(SO4)2
23
The Mole
The mole is an important concept in Chemistry. It is the Unit used by chemists and
engineers to determine the quantities involved in reactions. There are a number of
relationships that would have been explained in your Chemistry subject. These
include:
 Avagadro’s Number
 Mass
 Formula weight (or formula mass or molar mass or molecular mass)
 Concentration
 Volume of gas
Complete the mole map below :
Mass
(g)
No of Entities
Moles
Vol of Gas
(L)
Concentration
(mol/L)
24
THE MASS OF ONE MOLE OF A SUBSTANCE IS EQUAL TO ITS
FORMULA WEIGHT.
moles 
mass (g)
FW
Exercise
Calculate the number of moles in the following (the FWs were calculated
previously
(a)
0.159 g of sodium phosphate
(b)
15.74 g of glucose
(c)
0.369 g of magnesium nitrate
(d)
1.81 g of ammonium iron (II) sulfate
Calculate the mass of the following:
(e)
0.0125 moles of sodium phosphate
(f)
5.09 x 10-4 moles of glucose
(g)
2.5 moles of magnesium nitrate
(h)
7.84 x 10-5 moles of ammonium iron (II) sulfate
25
Some more to try
1.
2.
What is the formula weight of:
(a) NaNO3
(b) Al2(SO4)3
(c) CuSO4. 5H2O
How many moles are in the following?
(a) 0.25 g of NaNO3
(b) 34.2 mg of CaCO3
(c) 21.2 mL of 0.103 M HCl
(d) 250 mL of 2.5 M H2SO4
26
Concentration calculations
The concentration of something is how much of that something there is in a given
amount of material. To be slightly more technical, we call the something we are
interested in the analyte, and the material containing it the sample.
Concentration is always a ratio of the analyte to the sample, as in Equation
3.2:
concentration 
amount of analyte
amount of sample
Eqn 3.2
The units for the amounts will vary depending on the type of sample (solid, liquid or
gas) and the situation of the analysis. In general,
conc. in solid sample 
mass of analyte
mass of sample
conc. in liquid sample (solution ) 
Eqn 3.3
moles OR mass of analyte
volume of sample
Eqn 3.4
Solutions
A solution is a mixture of a solute (the dissolved substance) in a liquid (normally, but
not always), known as the solvent. The concentration is the amount of the solute per
unit of volume of the solution, or the ratio of amount to volume.
Common solution concentration units
Concentration unit
Amount of solute
Volume of solution
molarity (mole/L)
moles
litres
g/L
mass – grams
litres
g/100 mL (%w/v)
mass – grams
100 mL
mg/L (ppm)
mass – milligrams
litres
ug/L (ppb)
mass – micrograms
litres
ml/100 mL (%v/v)
volume - mL
100 mL
EXAMPLE
1.
Calculate the concentration of a solution (in g/100 mL) if 1.45 g is dissolved in
250 mL of solution.
Amount unit is g: 1.45 g
Volume unit is 100 mL: 250 ÷ 100 = 2.5 (100 mL)
Concentration = amount ÷ volume = 1.45 ÷ 2.5 = 0.58 g/100 mL
2.
What mass of solute is required to prepare 300 mL of 250 mg/L?
Volume unit is L: 300 mL = 0.3 L
Amount unit will be mg.
Amount = concentration x volume = 250 x 0.3 = 75 mg
27
3.
What volume of solution is required to make a 15 mL/100 mL (%v/v) solution
from 25 mL of solute?
Amount unit is mL: 25
Volume unit will be 100 mL lots.
Volume = amount ÷ concentration = 25 ÷ 15 = 1.67 x 100 mL = 167 mL.
CLASS EXERCISE 3.5
1.
Calculate the solute concentration in the following solutions in the specified
unit:
2.
Amount
Volume
Unit
(a)
3.25 g
75 mL
g/L
(b)
27.3 mg
500 mL
mg/L
(c)
0.5732 g
21.4 mL
g/100 mL
(d)
6.9 mL
250 mL
ml/100 mL
Concentration
What amount of solute is required to prepare the following solutions:
(a) 1.25 L of 0.46 g/L
(b) 50 mL of 4.56 g/100 mL
(c) 250 mL of 0.1 mole/L
(d) 1500 mL of 1 mL/100 mL
3.
What volume of solution (to the nearest multiple of 10) is required to
make the following solutions from the given solute amount?
Amount
Concentration
(a)
350 mg
200 mg/L
(b)
45 g
20 g/100 mL
(c)
0.55 g
1.0 g/L
(d)
25 mL
2 mL/100 mL
Volume
28
Solids
The concentration of a chemical in a solid is just as important as that in a liquid. However,
you won’t normally be making up a solid mixture of a known concentration, but rather trying
to work out the concentration in a solid sample from an analysis, eg the moisture content in
soil, the lead content in fallout.
The basic format for concentration in a solid is the same, but this time the amount of
sample is a mass, not a volume.
mass of analyte
Concentrat ion 
mass of sample
The two common concentration units for solids are:

g/100 g (also known as %w/w)

mg/kg (also known as ppm, used for low concentrations)
CLASS EXERCISE 3.6
Complete the following table.
Concentration unit
Mass unit of analyte
Mass unit of sample
g/100 g
mg/kg
EXAMPLE 3.4
1.
Calculate the concentration if there is 0.573 g of analyte in 2.982 g of solid sample in
g/100 g.
Mass of analyte should be in g: 0.573
Mass of sample should be in 100 g lots: 2.982 ÷ 100 = 0.02982.
Concentrat ion 
2.
0.573
 19.2
0.02982
What is the concentration of lead in fallout if a 1.0482 g sample was found to have 1.45
ug of lead?
Mass of analyte should be in mg: 1.45 ug = 0.00145 mg
Mass of sample should be in kg lots: 1.0482 ÷ 1000 = 0.0010482.
Concentrat ion 
3.
0.00145
 1.38
0.0010482
How much water is in a 50 g sample of soil which has a moisture concentration of 8.75
g/100 g?
This time, the unknown is the mass of analyte, not the concentration. Same equation,
but a bit re-arranging.
Mass of sample should be in 100 g lots: 50 ÷ 100 = 0.5
8.75 
mass of analyte
0.5
Mass of analyte = 8.75 x 0.5 = 4.375 g
29
4.
How much soil should be weighed out if it has an approximate lead concentration of 1
mg/kg, and the analysis requires 0.1 mg lead?
This time, the unknown is the mass of sample.
Mass of analyte should be in mg: 0.1.
5 
0.1
mass of sample
Mass of sample = 0.1 ÷ 5 = 0.02 kg = 20 g.
CLASS EXERCISE 3.7
1.
Calculate the analyte concentration in the following solid samples in the specified
unit:
Analyte mass
Sample mass
Unit
Concentration
2.
(a)
0.0545 g
10.444 g
g/100 g
(b)
14.6 mg
201.5 g
mg/kg
(c)
456 mg
0.9785 g
g/100 g
(d)
181 ug
2.045 g
mg/kg
What mass of analyte is contained in the following:
(a) 4.8452 g of 123 mg/kg
(b) 500 g of 1.05 g/100 g
(c) 22.22 g of 222 mg/kg
(d) 75 g of 1%w/w
3.
What mass of sample is required to contain:
(a) 10 mg from 50 mg/kg sample
(b) 1 g from 12.5 g/100 g
3.
What mass of sample is required to contain:
(c) 500 ug from 200 mg/kg
(d) 0.1 g from 5% w/w
30
3.4 Dilution of solutions
When a solution has more solvent added to it, it is said to be diluted. This means that
the concentration has decreased, because the amount of solute – the dissolved
substance – does not change, but the volume increases, as shown in Figures 3.1 and
3.2.
add 100 mL
extra solvent
1
g
1
g
1 g/100 mL
1 g/200 mL (or 0.5 g/100
mL)
FIGURE 3.1 Dilution of a solution
concentratio
nn
volume
amount
FIGURE 3.2 Summary of the effect of dilution on solution measures
If the dilution is done accurately – ie the volumes of the first (concentrated) solution
and the volume of the second (diluted) solution are volumetric (pipette, burette, vol.
flask) – then the relationship between the concentrations of the two solutions is simple,
as shown in Equation 3.1.
C1V1 = C2V2
Eqn 3.5
where C1 and C2 are the concentrations of the two solutions, and V1 and V2 the
volumes. The units for the concentrations must be the same, and likewise for the
volumes.
EXAMPLE 3.5
(a)
What is the concentration of a solution made by diluting 5 mL of 0.1 mole/L to
250 mL?
C1 = 0.1 M
V1 = 5 mL
C2 = ? V2 = 250 mL
0.1 x 5 = C2 x 250
C2 = 0.002 mole/L (we know the unit for C2 is mole/L because that is the unit
for C1).
31
(b)
What volume of a 1000 mg/L solution is needed to make up 1 L of 50 mg/L.
C1 = 1000 mg/L
V1 = ?
C2 = 50 mg/L
V2 = 1 L
1000 x V1 = 50 x 1
V1 = 0.05 L (or 50 mL).
CLASS EXERCISE 3.8
1.
What is the concentration of the solution if:
(a) 25 mL of 2.5 g/100 mL is diluted to 100 mL
(b) 5 mL of 400 mg/L is diluted to 200 mL
(c) 15 mL of 2.5 mole/L is diluted to 2.5 L
(d) 50 mL of 0.123 g/L is evaporated down to 10 mL
2
(a) How much 0.05 M solution is required to make 200 mL of 0.01 M?
(b) How much 2000 mg/L solution is required to make 500 mL of 100 mg/L?
(c) How much 25 %v/v solution is required to make 250 mL of 5 %v/v?
(d) How much 50 g/L solution is required to make 20 L of 100 mg/L?
32
Dilution factors
The dilution factor is a common term, which describes the ratio of the volumes of the
diluted solution to the concentrated solution, as shown in Equation 3.6. It becomes an
easy way of working the concentration of the concentrated solution if that of the
diluted solution is analysed (Eqn 3.7).
Dilution factor 
Volume of diluted solution
Volume of concentrated solution
Eqn 3.6
Conc. of original solution = dilution factor x conc. of diluted solution
Eqn 3.7
EXAMPLE 3.6
(a)
What is the dilution factor is 25 mL of solution is diluted to 500 mL?
Dilution factor = 500 ÷ 25 = 20.
(b)
What is the concentration of the original solution if it was diluted by a factor of
20 and the diluted solution had a concentration of 23.5 mg/L?
Conc. of original solution = 20 x 23.5 = 470 mg/L
CLASS EXERCISE 3.9
1.
What is the dilution factor in the following cases?
(a) 10 mL is diluted to 250 mL
(b) 25 mL is diluted to 1 L
(c) 5 mL is diluted to 250 mL
(d) 2.5 mL is diluted to 250 mL
2.
Using the dilution factors in Q1, find the concentration of the original solutions,
if the diluted solutions are analysed and found to have the following
concentrations:
(a) 0.034 mole/L
(b) 3.67 mg/L
(c) 69 mg/L
(d) 0.103 %w/w
Remember that these dilution calculations only work for concentrations, not
masses!
Aliquots of solutions
If an accurate volume (aliquot) is taken from a known volume of solution (volumetric
flask), and then analysed for the amount (mass or moles), we can work how much was
in the total solution, by simple fractions. Remember that the concentration of the
aliquot is the same as the total solution.
Amount in total solution  Amount in analysed solution x
Volume of total solution
Volume of analysed solution
33
EXAMPLE 3.7
What mass of lead is contained in 250 mL of solution, if a 25 mL aliquot is found to
contain 15.3 mg?
The 25 mL aliquot is one-tenth of the original solution, so the mass will be onetenth, ie 153 mg in the main solution.
CLASS EXERCISE 3.10
What amount (moles or mass) is contained in the original solution if:
(a)
a 10 mL sample from a 500 mL solution contains 2.34 x10-3 moles
(b)
a 50 mL aliquot from a 200 mL solution contains 0.105 g
(c)
a 25 mL aliquot from a 500 mL solution contains 99.3 mg
(d)
a 1 mL aliquot from a 20 mL solution contains 4.3 ug
3.6 Titration calculations
When chemical substances react together, they do so by number of molecules, not
mass of compounds. Therefore, you have to work in moles. Remind yourself of the
two formulae for calculating the number of moles.
=
from mass
Number of
moles
=
from solution
A titration is a method of analysis which you will use practically in your first
laboratory subject. Titrations rely on knowing the number of moles of one of the
reacting substances (the standard), and using that to work out the moles of the other
substance (the analyte).
If the standard is a pure solid, then you need to know the mass AND the formula
weight.
If the standard is a solution, then you need to know the molarity (mole/L) and the
volume.
34
EXAMPLE 3.8
Identify the analyte and standard in the following.
(a)
25 mL of apple juice is analysed for acidity by titration with 26.7 mL of 0.111
M NaOH.
Both substances are in solution, so we need to know the molarity and volume
for the standard: the NaOH is therefore the standard, and the apple juice acid is
the analyte, because the molarity is not known.
(b)
12.3 mL of AgNO3 titrates 0.123 g of pure NaCl.
The molarity of the silver nitrate is not known, so it must be the analyte. The
sodium chloride is pure, and its mass is known, so it is the standard.
CLASS EXERCISE 3.11
Identify the standard and the analyte in the following.
Standard
(a)
(b)
(c)
(d)
(e)
(f)
Analyte
25 mL of HCl solution is titrated with 18.2 mL of 0.102 M
NaOH.
The calcium in 50 mL of tap water is titrated with 12.3 mL
of EDTA
0.157 g of pure sodium carbonate is titrated with 17.5 mL
of HCl.
The acidity in 25 mL of vinegar is analysed by titration
with 25.3 mL of 0.0987 M NaOH.
The KOH content in 0.105 g of washing powder is titrated
with 11.2 mL of 0.0534 M H2SO4.
The chloride content in 50 mL of blood serum is analysed
by titration with 18.1 mL of 0.0969 M AgNO3.
The titration calculation procedure
The following method shows you how to calculate the number of moles of analyte in
the titrated solution. What you do with it from there depends on what you are after:
concentration, mass, dilutions etc.
1.
Calculate moles
of standard
2.
Determine
reaction ratio
3.
Calculate moles
of analyte from
1&2
If that seems easy, that’s because it is! However, you need to be careful to make sure
you are using the right bit of information in the right place. For example, if you were
calculating the number of moles of NaOH standard delivered form the burette, then
make sure it is the volume and molarity of the NaOH that you use.
35
Step 1: use the appropriate formula (mass÷FW for solids, CV for solutions)
Step 2: this is critical; either write the balanced equation, or find out what the
reaction ratio for analyte to standard is
Step 3: look at the numbers in the reaction ratio: what do you have to multiply the
standard coefficient by to get to the analyte number; use that multiplication
for the actual moles of standard to get the moles of analyte
EXAMPLE 3.9
Referring back to Example 3.8
(a) The reaction ratio is 1 acid: 2 NaOH (you weren’t expected to know this).
moles of NaOH (std)
= 0.111 x 26.7 x 10-3 (volume must be in litres)
= 2.964 x 10-3
The multiplication factor is 0.5 (2  1).
moles acid (analyte)
(b)
= 0.5 x 2.964 x 10-3
= 1.482 x 10-3
The reaction ratio is 1:1.
moles of NaCl (std) = 0.123 ÷ 58.44
= 2.105 x 10-3
The multiplication factor is 1 (1  1).
moles AgNO3 (analyte)
= 1 x 2.105 x 10-3
= 2.105 x 10-3
CLASS EXERCISE 3.12
Calculate the moles of analyte for the titrations in Exercise 3.11, given the
following reaction ratios:
(a)
1 HCl : 1 NaOH
(b)
1 Ca : 1 EDTA
(c)
2 HCl : 1 Na2CO3
(d)
1 acid : 1 NaOH
(e)
1 H2SO4 + 2 KOH
(f)
1 AgNO3 : 1 Cl
36
1.
How much NaCl (FW 58.44) would be needed to make the following
solutions?
(a) 500 mL of 5 g/100 mL
(b) 500 mL of 5 g/L
(c) 50 mL of 200 mg/L
(d) 250 mL of 0.1 mole/L
2.
What is the concentration of 0.75 g of NaCl in 500 mL?
(a) in g/100 mL
(b) in g/L
(c) in mole/L
3.
What is the molarity (mole/L) of the following solutions?
(a) 5 g of NaOH in 250 mL
(b) 5 g of CuSO4. 5H2O in 250 mL
(c) 2.345 x 10-3 moles of NaCl in 15 mL
4.
What is the concentration of analyte in the following solid samples?
(a) 0.202 g of water in 1.568 g of soil (as %w/w)
(b) 6.78 mg of DDT in 50.292 g of soil (as mg/kg)
(c) 9.876 x 10-3 moles of NaCl in 1.0246 g of soil (as %w/w)
5.
(a) How much lead is in 250 kg of waste, which has a concentration of 23.4
mg/kg?
(b) How much nitrate is in 2 g of soil which has a concentration of
0.0437 %w/w?
6.
What is the concentration of the following diluted solutions:
(a) 25 mL of 100 mg/L is diluted to 250 mL
(b) 5 mL of 0.013 M is diluted to 100 mL
(c) 10 mL of 0.65 g/100 mL diluted to 250 mL
7.
What volume of concentrated solution is required in the following:
(a) 1 L of 10 mg/L from 1000 mg/L
(b) 200 mL of 0.05 M from 0.1 M
(c) 100 mL of 0.1 g/L from 20 g/L
8.
(a) 20 mL of sample was diluted to 500 mL and analysed. The diluted solution
was found to have a concentration of 2.45 mg/L. What was the concentration of
the original solution?
(b) A sample is diluted by a factor of 25. The diluted solution was found to
have a concentration of 0.0875 g/100 mL. What was the concentration of the
original solution?
9.
(a) 0.125 g of pure Na2CO3 was used to standardise a solution of HCl. 22.1 mL
of the HCl was required to reach endpoint. Calculate the molarity of the HCl
(b) A sample of soil was titrated with 14.2 mL of 0.0103 M iron (II) solution to
determine its chromium concentration. The reaction ratio is 3 moles of iron
reacts with 1 mole of chromium. Calculate the mass of Cr in the soil.
(c) The chloride content of IV saline solution was analysed by titration with
silver nitrate solution. The reaction is 1 Cl:1 Ag. Calculate the concentration of
NaCl in the sample as g/100 mL, if a 20 mL aliquot of sample was titrated by
30.75 mL of 0.102 M silver nitrate.
(d) 10 mL of sulfuric acid is titrated with 15.8 mL of 0.111 M NaOH. Given
that 2 moles of NaOH react with 1 mole of sulfuric acid, determine the
concentration of the acid as g/L.
37
Answers
Page 3
(i)
(ii)
(iii)
Estimation
0.17
(0.227 … 3 S.F)
167
(165 3 … S.F.)
80
(78.0 ….3 S.F
Page 4 Rounding
1.
2345
2.
6.41
8400
16
36.4561
46400
Page 5 Order of Operations
1. 19
2. 11
7. 21
8. 12
13. 3
14. 1
19. 8
20 4
8
0.0037
70000
3. 19
9. 3
15. 0
1
5.20
5000000
4. 18
10. 17
16. 3
Page 6 Scientific Notation
1.
7.2 x 105
4.
8.076 102
7.
9 x 10-6
10.
2.1 x 104
13.
46000
2.
5.
8.
11.
14.
Page 7 Directed Numbers
1. 2
2. 0
6. 8;
9;
-1;
-6;
3. -30
-16;
-14;
1.5 x 107
5 x 10-3
1.01 x 107
5200000
111000000000
Page 8 Significant Figures
1.
4;
4;
4;
1;
7
2.
62000 (60000);
460 (500);
0.0020 (0.002);
3.0 (3);
0.000049 (0.00005); 2500 (2000);
Page 9 Substitution into a formula
1. 7.5 g/L
2. 10 g/L
4. -3
-3
5. 12
11. 7
17. 6
3.
6.
9.
12.
15.
830
460
90000000
6. 5
12. 7
18. 13
8.965 x 102
1.3 x 10-6
2.46 x 10-4
0.002015
0.000333
5. 24
0.00038 (0.0004)
86000000 (90000000)
3. 2.625 g/L
Page 10 Rearranging formula to solve equation
1. 13
2. 0
3. 0.6
4. 0.5
5. 3.4
Page 11
a)
V = m/C
b) C2 = C1V1 /V2
c) Tf = (Tc + 17.8) /0.556
d)
m = CV
Page 12 General Revision of Concepts
a)
0.024 g/L
b) 9.66
c) 100.8
d) 18080
Page 14
a) 5700 cm
b) 20000mm c) 0.045 g (45mg)
d) 12400 m2
f) 75.44 ft
g) 1086 ha
h) 0.8682 psi
i) 563g
Some more to try
a) 4.57 x 10-5 cm
b) 4.98s x 10-4 MHz c) 2.5 x 10-5 kg
Some more difficult ones
a) 6.932 x 10-6 g/Ml
b) 48.7 mg/kg
c) 0.181 g/L
Page 15
1) 2.06 x 10-7 m
2) 306.5 kg
3) 3450 kJ
5) 5075000 cm3
6) 500 ML
7) 4.56 kL
9) 50 mg/ mL
10) 1822 kgm3
11) 27740 J/g
13) 1248kg/m3
14)0.6177acres
15) 1200cm3/hr
17) 197.8 gal/min
e) 12.5 mL
e) 7.2 x 10-5 m3
d) 0.40 mile
4) 8.56 x 10-4 m2
8) 2.35 cm3
12) 32.33 m
16) 3612.9 cm2
38
Page 19
Page 23
a) 163.94 au
b) 180.156au c) 148.33 au
d) 284.054au
Page 25
a) 9.699 x 10-4 b) 0.0874
c) 0.0024877 d) 0.00637
Page 26
1 a) 85 au
b) 342.1au
c) 249.7 au
-3
-4
2 a) 2.94 x 10 b) 3.42 x 10
c) 2.18 x 10-3 mol
d) 0.625 mol
Page 28
1 a) 43.3. g/L b) 54.6 mg/L c) 2.6785 mg/100mL d) 2.76 mL/100 mL
2 a) 0.575g
b) 2.28 g
c) 0.025 mol d) 15 mL
3 a) 1.75 L
b) 225 mL
c) 0.55 L
d) 1250 mL
Page 30
1 a) 0.5218 g/100g
b) 72.46 mg/kg
c) 46.60 g/100g
d) 88.51 mg/kg
2 a) 0.5960 mg
b) 5.25 g
c) 4.933 mg
d) 0.75 g
3 a) 200 g
b) 8 g
c) 2.5 g
d) 2.0g
Page 32
1 a) 0.625g/100mL
b)10 mg/L
c) 0.015 mol/L
d) 0.615 g/L
2 a) 40 mL
b) 25 mL
c) 50 mL
d) 40 mL
Page 33
1 a) 25
b) 40
c) 50
d) 100
2 a) 0.85 mol/L
b) 146.8 mg/L
c) 3450 mg/L
d) 10.3% w/w
Page 34
a) 0.117 mole
b) 0.42 g
c)1986 mg
d)86 ug
Page 35
a) Std NaOH Analyte HCl
b) Std EDTA Analyte Ca
c) Std Na2CO3 Analyte HCl
d) Std NaOH Analyte H+
e) Std H2SO4 Analyte KOH
f) Std AgNO3 Analyte ClPage 36
a) 1.856 x 10-3 mol
b) 0.123 x 10-3
c) 2.96 x 10-3
d) 2.50 x 10-3
-3
-3
e) 1.20 x 10
f) 1.75x 10
39