1Week 2

Week 2CHAPTER 1

Sample preparation-continued14 SEPT 2015

2Week 2

Preparing solutions

Concentration is the quantity of solute in a known amount of volume or mass of solution or solventConcentration = Concentration = Amount of soluteAmount of solute

Amount Amount of solutionof solution

Expression of concentration

Molarity (M) Formality (F) Normality (N) (not common) Molality (m) Part per thousand (ppt) Parts per million (ppm) Parts per billion (ppb) Percent concentration (%w/w, %w/v, %v/v)

Unit for concentrations

How to prepare a solution? How to prepare a solution with a certain concentration?

3Week 2

Molarity (M) = number of moles of soluteliters of solution

concentration of a particular chemical species

Formality (F) Formality (F) = number of moles of soluteliters of solution

substance's total concentration without regard to its specific chemical form

•• MolalityMolality (m)(m) = = number of number of moles of solutekilograms of solution

4Week 2

• For substances that DO NOT ionize in solution, such as glucose, molarity and formality are the SAME

• For substances that ionize in solution, such as NaCl, molarity and formality are DIFFERENT.

NaCl (aq) → Na+ + Cl–initial 0.1 mol - -final 0.0 0.1 mol 0.1 mol

The molarity of NaCl zero since there is no undissociated NaCl in solution. The solution, instead, is 0.1 M in Na+ and 0.1 M in Cl–. When we state that a solution is 0.1 M NaCl we understand it to consist of Na+ and Cl– ions. The formality of NaCl0.1 F because it represents the total amount of NaCl in solution.

The unit of formality is used only when it provides a clearer descriptionof solution chemistry.

4

5Week 2

• Normality concentration in terms of an equivalent, which is the amount of one chemical species reacting stoichiometrically with another chemical species

Normality (N) = number of equivalents of soluteliters of solution

6Week 2

The concentration of substance in commercial aqueous reagents, organic solvents and commercial household products are usually expressed in percent composition

• Weight percent = Weight of solute x 100%(%w/w) Weight of solution

• Volume percent = Volume of solute x 100%(%v/v) Volume of solution

• Weight-volume = Weight of solute x 100%(%w/v) Volume of solution

Example: 37% is labeled on a HCl reagent bottle. This means that it contains 37 g HCl per 100 g solution

Concentration in terms of percent composition

7Week 2

• Parts per thousand = g of solute 103 g solution

Or = mg/g

• Parts per million (ppm) = g solute106 g solution

Or = µg/g

When the sample is in liquid form, we can use w/w % in volume unit as well. How?

As the density of aqueous solution often very close to 1.00 g/L, we usuallycorrelates 1 g water with 1 mL water (approximation)

Therefore, 1 ppm = 1 µg/g can also be converted to 1 µg/mL = 1 mg/L

The w/w % unit may also be expressed as a fractione.g. 37% (w/w) can be expressed as 37 parts per hundredHow about smaller fractions??

Both of these units are for

SOLID SAMPLE

8Week 2

• Parts per billion (ppb) = g solute109 g solution

Therefore,1 ppb = 1 ng/g = 1 ng/mL = 1 µg/L

• Parts per trillion (ppt) = g solute1012 g solution

Therefore,1 ppt = 1 pg/g = 1 pg/mL = 1 ng/L

For trace analysis in liquid sample~Analysis of very very very little amount of analytes

9Week 29

Converting percent composition to molar concentration-converting ‘commercial term’ to ‘easier laboratory term’-

• Example: A commercial aqueous ammonia (NH3 ) bottle is labeled as 28.0% w/w,and has a density of 0.899 g/mL and MW of 17.03 g/mol. Let say you want to convert the concentration into molarity for easier calculation in your lab book, how can you do that?

= 28 g/100 g x 0.899 g/mL

= 0.2517 g/mL (part per thousand)

= 0.2517 g/mL / 17.031 g/mol

= 0.01478 mol/mL x 1000

= 14.8 mol/L or 14.8 M

density

MW

10Week 2

Example: How to prepare 250 mL of 0.100 M NaOH from NaOH solid? [MW: NaOH = 40]

Preparation of solution

1. Calculate the weight (g) of NaOH pellet using MW to produce 0.1 M solutionCalculations ???

2. Weigh ??? g of solid (generally 0.1 mg, ie up to 4 decimal places in grams)

3. Dissolve in water, transfer (quantitatively with rinsing) to a 250 mL volumetric flask, and dilute to the mark

Calculate Weigh Dissolve and transfer

11Week 2

Dilution of solution

Example: Describe the preparation of 50 mL of 0.100 M NaOH solution from a 0.5 M solution

Calculations… How to ???… Glassware requirement: ?? mL pipet and ?? mL volumetric flask

The moles solute in concentrated (1) solution equals the moles in dilute (2) solution

M1 V1 = M2 V2

M1 : Initial concentration of solutionV1 : Volume of concentrated solution transferredM2 : Concentration of final solution (diluted)V2 : Volume of concentrated solution transferred

12Week 2

13Week 2

Week 2+3CHAPTER 2

Data analysis and statistics

14Week 2

Normal phrases in describing results of an analysis“pretty sure”“very sure”“most likely”“probably”

Replaced by using statistical tests important to understand the significance of data and therefore to set limitations on each step of analysis.

Is there such a thing as “ERROR FREE ANALYSIS”? How reliable is our data??

Errors in chemical analysis

Impossible to eliminate errors.What is the maximum error that we can tolerate?Can only be approximated to an acceptable precision.

DATA ANALYSIS

15Week 2

Accuracy is how close you Accuracy is how close you get to the bullseye. Precision

is how close the repetitive shots are one to another. It is nearly impossible to have

accuracy without good precision

ACCURACY•degree of agreement between measured value and the true value (which may not be known!)•Therefore, it is the degree of agreement between measured value and the accepted true value

PRECISION•Degree of agreement between replicate measurements of the same quantity; repeatability of a result.•Expressed by standard deviation, the coefficient of variation, the range of the data or as confidence interval (e.g. 95%) about the mean value •How similar are values obtained in exactly the same way?•Useful for measuring deviation from the mean.

xxd ii

16Week 2

High PrecisionHigh accuracy

High precisionLow accuracy

Low precisionLow accuracy

Low precisionHigh accuracy

x

17Week 2

Accuracy is expressed in terms of errorA. Systematic (determinate) Error•Determinable and that presumably can be either avoided or corrected.•Can be constant (e.g uncalibrated weight used in all weighing) or variable (e.g.buret whose volume readings are in error by different amount at different volumes)•Readings all too high or too low that can affect accuracy. •How to detect?analysis of reference sample or determine recovery after adding known standard

Types of error in chemical analysis

C. Gross Errors•Serious but very seldom occur in analysis.•Usually obvious - give outlier readings.•Detectable by carrying out sufficient replicate measurements.•Experiments must be repeated.e.g. Instrument faulty, contaminated reagent

B. Random/accidental (indeterminate) Error•Represents experimental uncertainty occurred in any measurement•Random and cannot be avoided. Affects precision, can only be controlled•Revealed by small differences in successive measurements made by the same analyst under identical conditions can’t be predicted/estimated but can be dealt with statistics•Data scattered approximately symmetrically about a mean value.

18Week 2

Personal Error1. Sources: Physical handicap, prejudice, not competent2. Examples:

a) Insensitivity to colour changesb) Transfer of solutionsc) Incomplete drying of samplesd) Tendency to estimate scale readings to improve precisione) Math error in calculations

Instrumental Error1. All measuring apparatus contribute to systematic error; therefore need frequent calibration –

pipettes, volumetric flasks, burettes 2. Also for electronic devices such as spectrometers, pH meter3. Examples:

a) Temperature changes

SOURCES OF SYSTEMATIC ERROR

Method Error1. Due to inadequacies in physical or chemical behaviour of reagents or reactions (e.g. slow or

incomplete reactions, instability of reacting species, occurrence of side reaction/interference).

2. Difficult to detect and the most serious systematic error (can be changed if the conditions of determination are altered)

3. Correction by running a) reagent blank- analysis on the added reagent onlyb) 2 independent methods or SRM

19Week 2

SF in multiplication &Division•Use the same number of digits asthe number with the fewest numberof digits.

)sf 3( 10 21.3 228 . 204

1633.0 1.40

2

Significant figures (SF)•The SF of a number are those digits that carry meaning contributing to its precision.• Zero is significant only when:A. It occurs in the middle of a number. Examples:a) 401 - 3 SFb) 6.0015 - 5 SFB. It is the last number to the right of the decimal point. Examples:a) 3.00 - 3 SFb) 6.00 102 - 3 SFc) 0.0500 - 3 SF

= 13.6 (1 dp)

Rounding off in the step of addition & subtraction•Use the same number of decimal places (dp) as the number with the fewest decimal places. Example: 12.2 + 0.365 + 1.04 = 13.605

(1 dp) (3 dp) (2 dp)

20Week 2

•Defining the confidence interval (RANGE) of values around a set mean (x) within which the population mean () can be expected with a given probability

•Determining the number of replicates required to assure (at a desired probability) that an experimental mean (x) falls within a predicted interval of values around the population mean ().

•Estimating the probability that the experimental mean (x) and true value () are different or two experimental mean are different (t test).

Application of statistics in data analysis

•Estimating the probability that data from two experiments are different in precision (F-test).

•Deciding when to accept/reject outliers among replicates (Q-test/ Dixon’s).

•Treating calibration data.

21Week 2

Sample Mean/Average

N

x x

N

1 i

i

Xi = individual values of x N = number of replicate measurements

MedianData in the middle if the number is odd, arranged in ascending order.For even numbers, the mean of the median pair is used.used when a set of data contains an outlier

Errors•Absolute error = measured value – true valueE.g. if a 2.72 g sample is analysed to be 2.62g, the absolute error is -0.10g

•Relative error is the absolute error expressed as % of the true valueE.g. (-0.10g/2.72 g) x 100% = -3.7%

RangeThe difference between the highest and lowest result.

Statistical treatment of random error

If for large data set, it is called population

mean (µ)

22Week 2

Each set of analytical results should be accompanied by an indication of the precision of the analysis standard deviation which is measuring of the precision of a population of data.

Smaller SD means more precise the analysis is

Standard Deviation (SD) – a very important precision indicator

Can be calculated using Excel spreadsheet

1N

x ix s

2

Small sample size (N=20)

Xi = individual values of x = mean

N = number of replicate measurementsx

N

ix

2

Population (N= >20)

Xi = individual values of x = mean

N = number of replicate measurements

23Week 2

Varian, V

The square of standard deviation.

For sample, V = s2

1

) x x()( 1

2i

2

NVs

N

i

Relative Standard Deviation (RSD)– also a very important precision

indicator

%100 xs RSD

Acceptable RSD % depends on thetype of analysis. I.e.•Trace analysis, RSD % should be <5%•Environmental analysis <15%•Pharmaceutical analysis <10 %

24Week 2

25Week 2

Examples:HPLC chromatogram of toxins

Replicate 1

Replicate 2

Replicate 3

1

1

1

2

2

2

Can you guess which result (peak) with higher precision?

example sd rsd.xlsx

26Week 2

Confidence limits (CL) and confidence interval (CI)– how sure are you?

•Calculation of SD for a set of data provides indication of the precision inherent in particular procedure.

•If for large data set, it doesn't by itself give any info about how close the experimentally determined mean ( ) to the true mean value (µ).

•Statistic allows to estimate range within which true value might fall, within a given probability, defined by the experimental mean and sd.

•This range confidence interval and the limits of this range confidence limit

x

The likelihood that the true value falls within the range is called probability or confidence level, usually expressed as %

27Week 2

Confidence interval for small data set (N = 20)

x )( IN

tsC • Values of t depend on degree of freedom, v (N - 1) and confidence level (from Table).

• t also known as ‘student’s t’ and will be used in hypothesis test.

N=Number of measurements/replicates

t=Student’s t values (constant and given)

s= SD

28Week 2

Values of t at various confidence level

Degree of

Freedom, v

Confidence Level

90 % 95 % 99 %

1 6.31 12.70 63.66

2 2.92 4.30 9.92

3 2.35 3.18 5.84

4 2.13 2.78 4.60

5 2.02 2.57 4.03

6 1.94 2.45 3.71

7 1.90 2.26 3.50

8 1.86 2.31 3.36

9 1.83 2.26 3.25

10 1.81 2.23 3.17

11 1.80 2.20 3.11

12 1.78 2.18 3.06

13 1.77 2.16 3.01

14 1.76 2.14 2.98

15 1.75 2.13 2.95

16 1.75 2.12 2.92

17 1.74 2.11 2.90

18 1.73 2.10 2.88

19 1.73 2.09 2.86

20 1.72 2.09 2.85

infinity 1.64 1.96 2.58

29Week 2

Example:

Data for the analysis of calcium in rock are given by 14.35%, 14.41%, 14.40%, 14.32% and 14.37%. Within what range are you 95% confident that the true value lies??

Nts x ( CI )

0.05% 14.37% 50.037 2.78 x

Solution:Mean, = 14.37SD, s = 0.037From the table, at 95 % confidence level, N - 1 = 4, t = 2.78.

Therefore, CI is :

x

So you are 95% confidence that (in the absence of determinate error), the true value falls within 14.32% to 14.42%

30Week 2

At different confidence level,

Confidence Level Confidence Interval

90% = 14.37% 0.04

95% = 14.37% 0.05

99% = 14.37% 0.08

Summary:If the confidence level increased, the confidence interval (CI) also increased. The probability of the true mean value () appeared in the interval will increase

31Week 2

•since the exact value of population mean () cannot be determined, one must use statistical theory to set limits around the measured mean, that probably contain .

•CI only having meaning with the measured standard deviation, s, is a good approximation of the population standard deviation, , and there is no bias in the measurement.

•CI when is known (population),

N z

x

N = Number of measurements/replicates

z= the value from the standard normal distribution for the selected confidence level

Confidence interval for large data set (N >20)

For the standard normal distribution, P(-1.96 < Z < 1.96) = 0.95, i.e., there is a 95% probability that a standard normal variable, Z, will fall between -1.96 and 1.96

32Week 2

Table 1 Values for z at various confidence levels

Confidence Level, % z50 0.67

68 1.00

80 1.29

90 1.64

95 1.96

96 2.00

99 2.58

99.7 3.00

99.9 3.29

33Week 2

Examples:

Nσ2.58 x μ

,2.58 z level, confidence 99% At

Nσ1.96 x μ

,1.96 z level, confidence 95% At

Nσ1.64 x μ

1.64, z level, confidence 90% At

or,Nσ 2.58 x μ

Nσ 2.58 x

34Week 2

•To determine number of replicates needed for the mean to be within the confidence interval.•To determine systematic error.

A. To determine number of replicates

If is known (s ),

2

x

z N

N x μ

z

If is unknown

2

x s t N

Nts x

Other usage of confidence interval (CI)

35Week 2

ExampleA standard solution gave an absorption reading of 0.470 at a particular wavelength. Ten measurements were done on a sample and the mean gave a value of 0.461, with standard deviation (s) of 0.003. Show whether systematic error exists in the measurements at 95% confidence level.

SolutionAt 95% confidence level, N = 10, t = 2.26,

002.0 461.0 10003.0 26.2 461.0

N

ts )(

xCI

The calculation gives confidence limit of,0.459 < < 0.463

•Does the true mean 0.470 belong to the interval?

•Does systematic error present?

B. To determine systematic error1) Calculation method 1

36Week 2

Values of t at various confidence level

Degree of

Freedom, v

Confidence Level

90 % 95 % 99 %

1 6.31 12.70 63.66

2 2.92 4.30 9.92

3 2.35 3.18 5.84

4 2.13 2.78 4.60

5 2.02 2.57 4.03

6 1.94 2.45 3.71

7 1.90 2.26 3.50

8 1.86 2.31 3.36

9 1.83 2.26 3.25

10 1.81 2.23 3.17

11 1.80 2.20 3.11

12 1.78 2.18 3.06

13 1.77 2.16 3.01

14 1.76 2.14 2.98

15 1.75 2.13 2.95

16 1.75 2.12 2.92

17 1.74 2.11 2.90

18 1.73 2.10 2.88

19 1.73 2.09 2.86

20 1.72 2.09 2.85

infinity 1.64 1.96 2.58

37Week 2

ExampleA standard solution gave an absorption reading of 0.470 at a particular wavelength. Ten measurements were done on a sample and the mean gave a value of 0.461, with standard deviation (s) of 0.003. Show whether systematic error exists in the measurements at 95% confidence level.

SolutionAt 95% confidence level, N = 10, t = 2.26,

The tcalc >ttable

Does systematic error present?

49.9 003.010 470.0461.0

t

sNxt

B. To determine systematic error2) Calculation method 2

38Week 2

Observations

Hypothesis Model

Valid? Reject

Basis for further experiments

YES

NO

Testing a hypothesis

39Week 2

• Approach tests whether the difference between the two results is significant (due to systematic error) or not significant (merely due to random error).

Significant tests

Hypothesis Data Theory

•Normally, the measured data is not always the same and seldom agree.

•We use statistics to test the disagreement.

40Week 2

The values of two measured quantities do not differ (significantly) UNLESS we can prove it that the two values are significantly different.

“Innocent until proven guilty”

•The calculated value of a parameter from the equation is compared to the parameter value from the table.

•If the calculated value is smaller than the table value, the hypothesis is accepted and vice-versa.

Null hypothesis, ho

•Can be used to compare:

41Week 2

Values of t at various confidence level

Degree of

Freedom, v

Confidence Level

90 % 95 % 99 %

1 6.31 12.70 63.66

2 2.92 4.30 9.92

3 2.35 3.18 5.84

4 2.13 2.78 4.60

5 2.02 2.57 4.03

6 1.94 2.45 3.71

7 1.90 2.26 3.50

8 1.86 2.31 3.36

9 1.83 2.26 3.25

10 1.81 2.23 3.17

11 1.80 2.20 3.11

12 1.78 2.18 3.06

13 1.77 2.16 3.01

14 1.76 2.14 2.98

15 1.75 2.13 2.95

16 1.75 2.12 2.92

17 1.74 2.11 2.90

18 1.73 2.10 2.88

19 1.73 2.09 2.86

20 1.72 2.09 2.85

infinity 1.64 1.96 2.58

42Week 2

• Comparison between experimental mean and true mean ( and )

• To check the presence of systematic error• Can be used to compare replicate measurement• Can be used to compare individual difference

x

t-test

43Week 2

If is known,

σNμ x z

1) Comparing two mean values and

i) If is not known,

sN - xt

Nst x

x

x

ii) Calculate t or z (tcalc) from the data.

iii) Compare tcalc and ttable

iv) If tcalc > ttable

Reject Null Hypothesis (Ho) i.e.

• The difference is due to systematic error.

) ( x

v) If tcalc < ttable

Accept Null Hypothesis (Ho) i.e.

• The different is due to random error.

Steps in t-test

44Week 2

Example:

The sulphur content in a sample of kerosene was found to be 0.123%. A new method was used on the same sample and the following data is obtained:

%Sulphur : 0.112; 0.118; 0.113; 0.119

Show whether systematic error is present in the new method.

Since tcalc> ttable, Ho is rejected and the two means are significantly different and thus systematic error is present.

xx

)(t 38.4 0032.0

4 123.0116.0

sN x t

calc

Null Hypothesis, Ho : =

= 0.116%, = 0.123%, s = 0.0032

t table = 3.18 (95 %, N-1 = 3)

45Week 2

data) tal(experimen 007.0

0.123 - 0.116 x

Other Solution:

)value Table( 0051.0

40.0032 3.18

N

ts x

x

tablecalculated - x x

N

ts x

Since,

(i.e. 0.007 > 0.0051)

xHo is rejected, the difference ( ) is significant and there is systematic error in the measurement.

46Week 2

2x1x

2.Comparing two mean values and

•Normally used to determine whether the two samples are identical or not.

•The difference in the mean of two sets of the same analysis will provide information on the similarity of the sample or the existence of random error.

Data: , and s1, s2

Ho : =

We want to test whether - = 0

1x 2x

2

22

2

1

111

N

ts x

N

ts x

1x 2x

2x1x

47Week 2

2121 and Assume,

Calculate the value of t;

Compare tcalc with ttable; if tcalc< ttable, Ho is accepted.

The pooled standard deviation, sp is calculated using:

2 1-1-

21

222

211

N+N)s+(N)s(N=s p

where,N1, N2 are numbers of data in sets 1 and 2

21

21

p

21calc NN

NNs

x-xt

48Week 2

Example;

Since tcalc < ttable, Ho is accepted and the source of the urine statistically is the same.

49Week 2

Values of t at various confidence level

Degree of

Freedom, v

Confidence Level

90 % 95 % 99 %

1 6.31 12.70 63.66

2 2.92 4.30 9.92

3 2.35 3.18 5.84

4 2.13 2.78 4.60

5 2.02 2.57 4.03

6 1.94 2.45 3.71

7 1.90 2.26 3.50

8 1.86 2.31 3.36

9 1.83 2.26 3.25

10 1.81 2.23 3.17

11 1.80 2.20 3.11

12 1.78 2.18 3.06

13 1.77 2.16 3.01

14 1.76 2.14 2.98

15 1.75 2.13 2.95

16 1.75 2.12 2.92

17 1.74 2.11 2.90

18 1.73 2.10 2.88

19 1.73 2.09 2.86

20 1.72 2.09 2.85

infinity 1.64 1.96 2.58

50Week 2

F-testComparing the precision of two measurements

•Is Method A more precise than Method B?

•Is there any significant difference between both methods?

22

21

2

1

s

s

vvF

•With the degree of freedom = N – 1, Ho : the precision is identical; s1 = s2

•Then, if Fcalc < Ftable , Ho is accepted.

Do you remember?V= s2

Since the values of F (from table) are always greater than 1, the smaller variance (the more precise) always become the denominator.V1 > V2, so 1F

nominator

denominator

51Week 2

F values

22

21

2

1

s

s

vvF

52Week 2

Example:The determination of CO in a mixture of gases using the standard procedure gave an s value of 0.21 ppm. The method was modified twice giving s1 of 0.15 (10 degrees of freedom (N-1)) and s2 of 0.12 (10 degrees of freedom (N-1))

Are the two modified methods more precise than the standard?

In the standard method, s and the degrees of freedom becomes infinity.Refer the F table:Numerator = , and denominator = 10; giving the critical: Ftable= 2.54

Ho : s1 = sstd and Ho : s2 = sstd

96.115.021.0

ss

F 2

2

21

2std

1 06.312.021.0

ss

F 2

2

22

2std

2

Solution:

Since F1 < Ftable, Ho is accepted, while F2>Ftable, Ho is rejected.

53Week 2

Ho : s1 = sstd.

and

Ho : s2 = sstd.

Conclusions

54Week 2

The Dixon test (Q test)A way of detecting outlier a data that does not belong to the set.

Example:Data: 10.05, 10.10, 10.15, 10.05, 10.45, 10.10

By inspection,10.45 seem to be out of the data normal range.

It is easier to see it when the numbers are arranged in a decreasing or increasing order.

10.05, 10.05, 10.10, 10.10, 10.15, 10.45

At 95% confidence limit, should this data be eliminated??

If this data (10.45) is eliminated, the mean will change from the original value!

55Week 2

wx-x

Q nqtexp

where,xq = the questionable dataxn = its nearest neighbourw = the difference between the highest and the lowest value (range).

The Qexpt or Qcalc will be compared with the Qcritical or Qtable, and the Ho is checked.

56Week 2

Values of Q

57Week 2

10.05 - 10.4510.15 -45.10

Q texp = 0.75

Qcritical (95%, n = 6) = 0.625

Solution:

Qexpt > QcriticalData (10.45) can be rejected.

wx-x

Q nqtexp

58Week 2

An analysis on calcite gave the following percentage of CaO: 55.45, 56.04, 56.23, 56.00, 55.08

19.055.45 - 56.2356.08-23.56Qcalc

Qcalc<Qtable. Data cannot be rejected.

71.055.45 - 56.2356.00-45.55Qcalc

Qcalc = Qtable. Data cannot be rejected.

Q: Is there any that data should be rejected at 95% confidence level?

Example:

•Arrange data55.08, 55.45, 56.00, 56.04, 56.08, 56.23•Suspected data:Qtable from 5 determinations, 95% = 0.710

Solution:

55.45 OR 56.23