PRACTICAL 4 (PART B)

TITLE:

Particle Size and Shape Analysis Using Microscope

AIM:

To analyse and compare the shape and size of five different samples

INTRODUCTION:

The particle shape and size can be analysed by many methods. One of it is by using a microscope, which has been done in this experiment. This analysis can be used to determine the diameter, shape, and surface area of a particle. During the experiment, we were given different type of sand to be observed and analysed such as the size, shape and arrangement. We are using mixed size of sand, 150µ, 355µ, 500µ, and 850µ. We are also using lactose and MCC powder. In this experiment, we observe different sizes of particles under a microscope and sketch our observation.

EXPERIMENTAL METHOD:

MATERIAL AND APPARATUS:

Sands (150µ, 355µ, 500µ, 850µ, various size), Lactose powder, MCC powder, Microscope, Spatula, Glass slide, Cover slip

PROCEDURE:

1) 5 different types of sands with different sizes which are 150µ, 355µ, 500µ, 850µ and various sizes were analysed by using microscope, by observing the size and shape of given particle. The size and shape of 2 different types of powder, which are lactose powder and MCC powder were analysed too.
2) The samples were examined first by 4X10 magnification, followed by 10X10 magnification.
3) The particle shape is sketched and the overall particle shape of that material is stated.

RESULTS:

Sands with various size

Magnification: 4 x 100

Sands with 850µ

Magnification: 4 x 100

Sands with 500µ 

Magnification: 4 x 100

Sands with 355µ 

Magnification: 4 x 100

Sands with 150µ 

Magnification: 4 x 100

Lactose powder

Magnification: 4 x 100

MCC powder

Magnification: 4 x 100

QUESTIONS:

1. Explain in brief the various statistical methods that you can use to measure the diameter of a particle.

There are a few statistical methods to measure the diameter of a particle. Those methods are includes Martin’s diameter (M), Feret’s diameter (F), Projected area diameter (da or dp), longest dimension, perimeter dimension and maximum chord.

            Martin’s diameter (M) is the length of the line which bisects the particle image.  The lines may be drawn in any direction which must be maintained constant for all image measurements. Feret’s diameter (F) is the distance between two tangents on opposite sides of the particle, parallel to some fixed direction. Projected area diameter (da or dp)  is the diameter of a circle having the same area as the particle viewed normally to the plane surface on which the particle is at rest in a stable position. Longest dimension is a measured diameter equal to the maximum value of Feret's diameter. Perimeter diameter is the diameter of a circle having the same circumference as the perimeter of the particle. Maximum chord is a diameter equal to the maximum length of a line parallel to some fixed direction and limited by the contour of the particle.

2. State the best statistical method for each of the samples that you have analysed.

The best statistical method for each sample that has been analysed is Feret’s diameter and Martin’s diameter. Both methods will give the average diameter over many different orientations to produce a mean value for each particle diameter. This will give an average value of diameter in more orientation and giving an average diameter value which is more accurate.

DISCUSSION:

            Optimum production of efficacious medicine is significantly affected by the size, shape and dimension of particulates. Since solid particle is often considered to approximate to a sphere and can be characterised by the determination of its diameter. More than one dimension could be generated for a given irregular particle, such as projected perimeter diameter and projected area diameter. Feret’s and Martin’s diameters produce a mean value for each particle diameter.  Feret’s diameter refers to the mean distance between 2 parallel tangents to the projected particle perimeter. Martin’s diameter is the mean chord length of the projected particle perimeter which can be considered as the boundary separate equal particle areas.

There are various methods to determine particle size and shape. A right method should be chose to analyse the particles. Firstly, the nature of material that has to be analysed should be consider. On the other hand, the cost specific requirements and time restriction should be taken into consideration. Microscope is among the excellent technique as operator could able to examine each particle individually. This is a relative cheap method to be use. However, it is not suitable for quality control as operator need to elaborate sample preparation and slow. The operator would experience rapid fatigue and it may cause variability to analysis the size of particle in the same sample. Operator bias may present. There is also lack of information on 3D shape in this method.

The microscope used in this experiment is light microscope and the magnification used throughout the experiment is 4x for the 7 sample analysed. The samples are dispersed evenly to prevent the presence of agglomeration. From the experiment, the size of material used in ascending order is MCC, lactose, 150µm, 355µm, 500µm, 850µm and lastly various size. The general shape for sands are irregular shape and for MCC and lactose are in cylindrical shape.

The important technique throughout the experiment includes operator should take a very least amount of particles to analysed under microscope to prevent agglomeration and ensure the shape and size to be seen clearly. While handling the microscope, correct techniques should applied to ensure microscope in good condition. Slides should be clean properly so that smear does not present and affect the observation.

CONCLUSION:

The size and shape of a particle can be analysed through microscopy analysis. Different types of particles have different sizes and shapes. In this experiment, it can be concluded that the size and shape of the particles are irregular and asymmetrical. 

REFERENCE:

1) Carlton, R. A.   2011.    Pharmaceutical Microscopy. Springer Science & Business Media.

2) Stanley-Wood, N. & Lines, R. W.   1992.    Particle Size Analysis. Royal society of chemistry.

3) Rolston Gordon Communications, 2001. Microscopy and Analysis.

PRACTICAL 4 (PART A)

TITLE:

Sieving

AIM:

To determine particle size distribution by sieve analysis

INTRODUCTION:

The oldest and best-known method of particle size determination is by sieve analysis. The particle size distribution is defined via the mass or volume. Sieve analysis is used to divide the particulate material into size fractions and then to determine the weight of these fractions. Sieves are commonly used to break down agglomerates and determine the size and size distribution of; a particular powder. Different particle sizes of powder have different flow and packing properties which alter the volume of powder during each tablet compression event. The particles that have small dimension will tend to increase the rate of dissolution. In this practical, we are required to use a sieve nest to determine the particle size and size distribution of two common excipients used in tablet formulations, namely lactose and microcrystalline cellulose (MCC).

EXPERIMENTAL METHOD:

MATERIALS AND APPARATUS:

Lactose, MCC, sieves, weighing boats and electronic balance.

PROCEDURE:

1) 100mg of lactose was weighted.

2) The sieve nest was prepared in descending order (the largest diameter to the smallest, from top to bottom)

3) The powder was placed at the uppermost sieve and the sieving process was allowed to proceed for 20 minutes.

4) After sieving was completed, the powder collected at the every sieve was weighted and the particle size distribution was plotted in form of histogram.

5) The above process was repeated using MCC.

RESULTS:

Size of     perture  (μm )	Particle size range, x ( μm )	Weight (g)
		MCC	Lactose
<53	0 < x <53	7.4930	3.7975
53	53 < x < 150	85.4171	93.7731
150	150 < x < 200	3.3120	1.4084
200	200 < x < 300	1.8168	0.8361
300	300 < x < 425	0.1288	0.0091
425	x >425	0.0820	0.0059
Total		98.2497	99.8301

QUESTIONS:

1. What are the average particle size for both lactose and MCC?

The average particle size for lactose and MCC is between 53 < x < 150 range.

2. What other methods can you use in determine the size of particle?

There are many other methods can be used to determine the size of particle. One of them is by microscopy which is able to examine each particle individually. Thus, make it being considered as an absolute measurement of particle size. However, this process is a time consuming process. .Perhaps the most obvious and accurate method for determining the particle size and shape characteristics of a small sample is microscopy. Unfortunately the operator time required to analyse a sufficient number of particles to be representative is prohibitive except in the highest value applications. This is an offline method of particle characterization with very limited throughput which may make it unsuitable for a number of applications. Next is sedimentation technique. This method depends on the fact that the terminal velocity of a particle in a fluid increases with the size.

DISCUSSION:

Based on the result of the experiment, the weight of the MCC obtained is the highest at the range of particle size 53 < x < 150 µm which is 85.4171 g. For the lactose, the highest weight is obtained at the finer particle size range of 53 < x < 150 µm with the value of 93.7731 g. It is known that if the particles cannot pass certain sieves, it is because the particles are bigger than the aperture of the sieve. By this, it can be deduced that most particles of MCC were finer than those of lactose as they are two different materials with different physical properties.

There are some possible causes of error while conducting this experiment as the total weight of lactose and MCC originally are not equivalent to the weight of powders obtained in individual sieves after being summed up. This could be due to the situation where there was had still amount of powder left in the sieves after the process was carried out.  Besides that, some of powders are spilled out from the container since the machine is not closed correctly which will affect the result obtained. Next, there were some powder left on newspaper when we choose an alternative to use newspaper in order to reduce the risk of the power spill out, but in the same time there was few of the powder still on the newspaper.

Therefore, before we start the experiment, we have to make sure that there was no other soil or other foreign substance. This precaution step is very important for us to get the most accurate and favorable reading when weighing the powder.

CONCLUSION:

Based on experiment that had been carried out, sieving is one of the method to determine particle size. The particle size distribution can be analyzed after finished the experiment. This process is important because it will affect the bioavailability and activity of drugs. In addition it is very important in formulation of drugs.

REFERENCE:

1) Chang, R. & Goldsby, K. A. 2014. Chemistry. Edisi ke-11. New York: McGraw- Hill Education.

2) https://www.sympatec.com/EN/Science/Characterisation/22_SievingMethods.html

3) Martin, A. N. & Sinko, P. J. 2011. Martin's Physical Pharmacy and Pharmaceutical Sciences, Kluwer/Lippincott Williams & Wilkins.

PRACTICAL 3: PHASE DIAGRAM (PART A)

TITLE:

Determination of Phase Diagram for Ethanol/Toluene/Water System Theory Three-Component Systems

AIM:

1) To determine the solubility limit in ternary system of water and two other liquids (ethanol & toluene), one of which is completely miscible (ethanol) and the other is partly miscible with water (toluene).

2) To construct the solubility curve of the system being studied on triangular diagram.

INTRODUCTION:

In pharmaceutical formulation, multiple substances need to be mixed together to form homogenous solution. Homogenous solution can be formed by knowing the exact ration of each component to be mixed with concern of other conditions like temperature and pressure. In this particular experiment, three components involved are ethanol, toluene and water. Water and toluene mixture is insoluble. When the mixture is added with ethanol, all these three components can become a homogenous solution at equilibrium if proper proportions are used.

Ternary phase diagrams are 3 component systems. To construct a ternary diagram it is necessary to know the three binary systems for the three components. Ternary diagrams have a vertical temperature axis. The composition of the points lie inside the area of triangle can determined by using triangle grid method. Triangle grid is constructed on the diagram. This grid is most commonly set up representing a 10 % incremental increase in the components. To determine the composition of a point within the triangular area of the diagram a series of three lines are drawn through the point of intersection with each line parallel to a side of the triangle. With these lines in place the percentage of each component in the composition of the point can readily be determined.

EXPERIMENTAL METHOD:

MATERIAL AND APPARATUS:

Retort stand with clamp, conical flasks, burette, measuring cylinder, toluene, ethanol, distilled water

PROCEDURES:

1) Mixture of ethanol and toluene was prepared in conical flasks containging the following   percentage of ethanol( 10,25,35,50,65,77,90,95).

2) 20 ml of each mixture was prepared accurately by filling a certain volume using a burette.

3) Each mixture was titrated using distilled water until cloudiness appears due to the existence of a second phase.

4) Conical flasks were shaken well after addition of a little water.

5) The volume of water added into the mixture was recorded.

6) Few measurements were taken to get accurate readings.

7)  The room temperature was measured.

8) The percentage based on the volume of each component when the second phase starts to appear was calculated.

9) Points were plotted in the triangular paper to obtain the triple phase diagram at the recorded temperature.

RESULTS: 

Conical Flask	Component of analyte				Volume of water used, mL
	Ethanol		Toluene
	%	mL	%	mL
A	10	2.00	90	18.00	0.40
B	25	5.00	75	15.00	0.80
C	35	7.00	65	13.00	1.10
D	45	9.00	55	11.00	1.20
E	65	13.00	35	7.00	2.80
F	75	15.00	25	5.00	4.60
G	90	18.00	10	2.00	11.00
H	95	19.00	5	1.00	17.10

Percentage of the components after titration:

Conical Flask	Percentage of each Component of titration product
	Ethanol(%)	Toluene(%)	Water(%)
A	9.80	88.24	1.96
B	24.04	72.16	3.85
C	33.18	61.61	5.21
D	42.45	51.89	5.66
E	57.02	30.70	12.28
F	60.98	20.33	18.70
G	58.06	6.45	35.48
H	51.21	2.70	46.09

The percentage of the components was calculated using the formula:

Percentage = Volume of the component ÷ total volume x 100%

                          

 

PRACTICE:

1) Does the mixture containing 70% ethanol, 20% water and 10% toluene (volume) appear clear or does it form two layers?

The mixture remains clear and no cloudy solution appeared forming one homogenous liquid phase.

2) What will happen if you dilute 1 part of the mixture with 4 parts of (a)water (b)toluene (c)ethanol?

1 part mixture x 70% ethanol= 1 x 70/100= 0.7 part of ethanol

1 part mixture x 20% water= 1x 20/100= 0.2 part of water

1 part mixture x 10% toluene = 1x 10/100= 0.1 part of toluene

So, in the mixture, there are 0.7 part of ethanol, 0.2 part of water and 0.1 part of toluene.

(a)         Water: 1 part of mixture + 4 parts of water

Ethanol= 0.7 ÷ 5 x 100% = 14%  

Water= (0.2+4) ÷ 5 x 100% = 84%

Toluene= 0.1 ÷ 5 x 100% = 2%   

Hence, based on the phase diagram, this mixture is under the area of the binomial curve, so 2 phase is formed.

(b)        Toluene: 1 part of mixture + 4 parts of toluene

Ethanol= 0.7 ÷ 5 x 100% = 14%   

Water= 0.2 ÷ 5 x 100% = 4%  

Toluene= (0.1+4) ÷ 5 x 100% = 82%      

Hence, based on the phase diagram, this mixture is outside the area of the binomial curve, so 1 phase is formed.

(c)         Ethanol : 1 part of mixture + 4 parts of ethanol

Ethanol= (0.7+4) ÷ 5 x 100% = 94%  

Water= 0.2 ÷ 5 x 100% = 4%  

Toluene= 0.1 ÷ 5 x 100% = 2%

Hence, based on the phase diagram, this mixture is outside the area of the binomial curve, so 1 phase is formed.

DISCUSSION:

A phase diagram shows the phases existing in equilibrium at any given condition. According to the Phase Rule, a maximum of four intensive variables (intensive properties) must be specified to completely define the state of a three-component system. The intensive variables that are usually chosen are pressure, temperature and concentration.  According to the phase rule, a single phase in a three-component system may possess four degrees of freedom. The calculation involved is shown below:

F = C – P + 2 = 3 – 1 + 2 = 4

in which F is degree of freedom; C is component; P is phase.

The addition of a third component to a pair of miscible liquids can change their mutual solubility. If this third component is more soluble in one of the two different components, the mutual solubility of the liquid pairs is decreased. However, if it is soluble in both of the liquids, the mutual solubility is increased. Thus, when ethanol is added to a mixture of toluene and water, the mutual solubility of the liquid pair increased until it reached a point whereby mixture become homogenous.

Regions where one or two phases appear have been indicated in phase diagram for three-component system. When the three components are mixed to give an overall system composition that falls in the two phase region, the system will separate into two phases: a phase rich in water and another rich in toluene. The curve of the plotted graph is termed as binomial curve. The region bounded by the curve shows the presence of two phases. The mixture within this region is cloudy in appearance due to phase separation as the amount of ethanol is not sufficient for a homogenous mixture to be produced. The region above the curve boundary shows one phase of homogenous solution. Addition of ethanol which acts as surfactant allows the two phase solution to be in one phase. 

Based on the results obtained, when there is a higher percentage of ethanol compared to the percentage of toluene in the mixture, the volume of water needed to titrate the mixture until cloudiness is observed is higher. This proves that the ethanol has increased the miscibility of the other two components and more water is needed to break the homogeneity. A second phase will only be separated out when the proportion of water added exceed the theoretical percentage where three components are partially miscible.

There is no fixed indication. The appearance of cloudiness in the solution exists in variety of its intensity. So, it is hard to determine the most accurate volume of titrant that is correlated with the appearance of cloudiness that comes with different intensity.

The apparatus should be rinsed with the distilled water to avoid any external reaction. One of the precautions is the same person should observe all the cloudiness of mixture every time throughout the experiment. The observer should make sure the eyes are perpendicular to the lower meniscus of the liquids to avoid parallax error. The volatile liquids should be poured immediately from the container to the conical flasks to minimize loss of volume. The room temperature must be consistent. The apparatus must be clean and dry to avoid contamination. The mixture should be well agitated then allowed to attain equilibrium.

CONCLUSION:

The reaction of water, toluene and ethanol appears as two phase system due to the decreasing in solubility of the mixture.

REFERENCE:

1) http://www.chm.davidson.edu/vce/phasechanges/PhaseDiagram.html

2) http://www.csun.edu/~jeloranta/CHEM355L/experiment5.pdf

3) Martin's Physical Pharmacy and Pharmaceutical Science, Sixth Edition, Patrick J. Sinko,   Wolters Kluwer, Lippincott Williams & Wilkins.