Handout of practical Lectures in Animal Nutrition Directions ...

Animal Nutrition | Dr. Ali AL-Kuhla Page | 1

Ali Abdalwahab. M.AL-Kuhla, PhD

Lecturer, Department of Public Health

College of Veterinary Medicine, University of Mosul, Mosul, Iraq

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Animal Nutrition | Part I | 2st year 2019

Handout of practical Lectures in Animal Nutrition

Directions and instructions about working in nutrition laboratory

Nutrition Laboratory is similar to chemistry laboratory. It contains chemical and glass wares,

food analysis equipment's. Hence, such systems evoke certain instructions and directions which should

strictly be followed by both personnel, staff, beginners as well as researchers to ensure their safety and

security and to maintain and preserve the laboratory tools.

These Instructions can be emphasized in the following steps:

1. The working students must wear their special aprons during their work inside the lab. To keep their

clothes, suit’s, dresses and bodies away from contact and exposure to solutions and chemicals and

harmful materials.

2. Strict a war ness and caution at using concentrated chemicals and glass wares to avoid them from

breaking. Never change their taps or their indicating labels.

3. The glass wares must be thoroughly washed after their application and use. Never leave chemicals

inside glass wares unwashed.

4. Chemicals, distilled water and burning gas should be consumed in the least quantities. Glass wares

should be first washed by tap water and detergents followed by distilled water. These glass wares should

be kept reserved until being dried or being kept in drying oven before their re use.

5. Each group of student should work in certain place of lab. To keep their places clean as well as their

places and equipments. The student should read the procedures of each experiment carefully and should

follow strictly the instructions regarding the required addition and concentration and period of time

needed for each experiment. The students should bring their notebooks to record their measurements,

weights, and the other notes about each experiment in special notebook for calculations and results.

Notebook: Students should work their experiment under supervision of their in charge lectures directly

in the nutrition lab.

https://orcid.org/0000-%200001-%208320-%200558

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Tools, equipment's, and glass wares used in nutrition lab:-

1. Digital electrical scale:

Electrical scale is used for correct and proper device for accurate weights. Hence, the following steps

should be carried out:-

a. Put the scale on fixed , stable and unmovable table to prevent any agitation.

b. Clean all parts of the scale before measurement using special small brush to remove debris, waste and

dust specially those present on the scale.

c. To open the instrument, press on the button "power", to tare the instrument press on the button "tar".

d. Never handle the sample intended for weighing by hands because it negatively effects on their

weights. So, special handlers are used for this purpose.

e. Scaling process is carried out quickly with care to prevent the acquirement of the sample to climatic

moisture weight is immediately recorded.

f. Re-cleans the instrument after use from remains of the sample attached to the scale using special

brush.

g. The accuracy of the scale is (0.00) after the point.

2. The desiccator:

The relative humidity of food samples vary from time to other due to the climate. The moisture

differs from time to another which causes large mistakes so such food samples should be placed in the

desiccator to prevent these cases. The desiccator is a glass bowl having a strict closed- cover. The

desiccator contains CaCL2,Which acts as moisture absorption.

The desiccator is widely used in nutrition lab. For various purpose e.g. cooling of the samples,

keeping the tools before carrying out scaling. When food samples taken out from drying or burning

ovens should be put in the desiccator.

The cover of the desiccator should not strictly closed because these hot samples cause extension

of air inside the desiccator leading to push the cover with consequence of partial loss of the sample. The

cover may be strictly closed after partial cooling of the sample. The sample should be preserved until

being weighed.

Instruments used for volume measurement:


1. The Burette:

It is a thin glass cylinder with various capacities viz., 50 ml, 25 ml. the burette is closed from the

bottom by a valve. Each one ml. is further subdivided in to (10) portions. Measurement followed by

burette is up to 0.04- 0.05 ml and this amount is equal to one drop. (two drops in the burette equals one

line).

The burette should be thoroughly washed before it’s use by tap water and then by distilled water

and finally by the solution intended for use in the test. The burette is quietly filled from the top using

small glass funnel avoiding air bubbles formation inside the burette. The valve located at it’s bottom is

opened to ensure that it’s full of solution when transparent solution is used in titration process, the

reading is recorded when the solution is in lower surface (the concave surface of solution).

When coloured solution are used in titration process, the reading is recorded at the upper surface

of the solution (the convex surface of the solution).

It is better to use white paper behind the reading area of the burette to prevent the leakage of

light and to clarify the reading. Eye sight level should be horizontal at the area of reading.

2. pipette:

It is a standard slender graduate glass cylinder having narrow opening. It is used for measurement

of fluids in tests and experiments. The thumb is used for regulating the evacuation of the intended

volume.

There are two types of pipettes i.e standard pipette and graduate pipette. When the fluids are

holding using pipette, the end of the pipette should be placed in the inner surface of the flask intended

for conveying. It is advised not to blow the pipette in order to evacuate it. It is not necessary to evacuate

the last drop of the pipette because the pipette is designed to hold the last drop.

Some pipettes have a swelling calling safety bubble which prevent the arrival of the solution to

the mouth of the handler during the working.

3. Measuring flasks:-

It is a rounded glass flask having slender narrow neck. The volume of the flask is indicated by

special mark. Measuring flasks are used to prepare certain standard solution or for their dilution. These

flasks are not intended for heating or boiling solutions.


Also, hot solutions should not be put in these flasks because these operations may change their standard

volumes. These changes are due to the fact that these flask are not manufactured from pyrex glass which

is heat resistant and heat stable. There are many volumes of these flasks e.g. 100ml and 250 ml.

4. Conical flasks:

These are conical flask made from glass having broad bottom and slender opening. Usually,

these flasks are used for dilution of stock and standard solutions. Also, it can be used as a receiving

flask located bottom the burette or under the distiller.


5. Graduate cylinders:

It is a graduate glass cylinder and different volumes (10-1000ml). These cylinders are used in

lab usually for preparing stock and standard solutions or for their dilutions.

The repeat in Lab. Analysis

Feed and chemical samples are analyzed following two samples of each (Duplicate).

but it may be triplicate (three samples) of the species in the sometime. Results are calculated of

feed samples with no more than 3-5% of the experimental error is accepted. The percentage of the

experimental error can be estimated following these steps:

1. The difference between two results = maximum result – minimum result.

2. mean of the result = 𝒇𝒊𝒔𝒓𝒕 𝒓𝒆𝒔𝒖𝒍𝒕+𝒔𝒆𝒄𝒐𝒏𝒆𝒅 𝒓𝒆𝒔𝒖𝒍𝒕

𝟐

3. percentage of experimental error: = 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒘𝒐 𝒓𝒆𝒔𝒖𝒍𝒕𝒔

𝒎𝒆𝒂𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒔𝒖𝒍𝒕 ∗ 𝟏𝟎𝟎

Notebook: in case of triplicate, the result of any experiment which doesn’t conform with other

results should be excluded.


Example:

Determination of fat percentage of concentrates sample was: 5.6% fat percentage of the first

result, 6.3% of the second result and 5.8% of the third result.

Calculate the experimental error.

Solution:

Since the 2nd result does not conform with 1st and 3rd results, so it should be excluded.

The difference between two results = 5.8- 5.6=0.2

Mean of result= 5.6+5.8

2= 5.7

Percentage of error = 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑟𝑒𝑠𝑢𝑙𝑡𝑠

𝑚𝑒𝑎𝑛 𝑜𝑓 𝑟𝑒𝑠𝑢𝑙𝑡

= 0.2

5.7∗ 100

= 3.5%

Feed stuffs and methods of sampling:

Firstly, Feed stuffs: they are considered that any feed can be used for animal feeding. These materials

may include any material that farm animal can digest, absorbed, assimilated either for maintenance or

for production. Generally, animal feedstuffs can be divided into two main parts which are:-


1. Concentrates:

These types of feed contain high percentage of protein and soluble carbohydrates and low ratio

of crude fiber (less than 18%).

Concentrates include cereals, grains of different origins such as wheat, barley, corn (maize) as

well as meals. Meals are the residues of oily crops after the extraction of oil from these crops such as

sesame meal, cotton seed meal, sun flower meal, soy bean meal, and safflower meal. Concentrates may

contain protein of animal origin as well as animal by products such as blood and bone meal, feather

meal. Of plants and vegetative sources it may include date palm residues, sugar beet residues, wheat

bran, molasses, rice husks and rice residues.

2. Roughages:

It refers to those feed which contain high percentage of crude fibers (more than 18%). It is

nutritional value is low and is given to the ruminants rations as bulky material. There are two types of

roughages:

A. Green roughages:

It includes grasses of various kinds and the clover, alfalfa as well as the silage which are

fermented grass preserved away from air in special siloes. Silage is considered as a palatable feed for

animals. Yellow corn silage is the most famous silage.

B. Dry roughages:

Straw and plant stems are examples of dry roughages which are produced after the harvest of

field crops. Also, hay is dry roughages, these materials are dry feed substance. The purpose of drying is

to prolong time of preservation, so it is stored at summer and being given to the animals at winter.

Examples of hay is alfalfa hay and oat hay.

Secondly: approximate analysis of feedstuffs:

It is regarded as a series of chemical analysis carried out on feedstuff to determine ratio and

percentages of essential feed compounds such as protein and fat and carbohydrates. The purpose of

feedstuff analysis as follows:

1. To determine the real nutritional value of the ration or feedstuff.


2. To detect adulteration (cheat) of the feedstuff.

3. To determine the suitable price of the feedstuff.


Procedure of feedstuff analysis can be explained within the following schedule:

Feed stuff

Crude fat

(Ether Extract)

moisture Dry matter

Ash Organic matter

Nitrogenous substances

(crude protein )

Non Nitrogenous

substances

Non – protein

nitrogenous

(NPN)

Non- soluble carbohydrates (crude fibers )

Carbohydrates

True protein

Soluble Carbohydrates

( N. F . E )

Lignin

Hemicellulose

s

Cellulose


1. Moisture:-

It is the free water present in the feedstuff. It can be estimated by drying feedstuff sample in

drying oven.

2. Dry Matter:-

It is the residual part of feedstuff sample after the total exclusion of moisture. The dry matter

contains all the portions of feedstuff sample except the water.

3. Ash:-

It is the non-organic part of feedstuff sample which is the residue after the burning of the sample

in muffle furnace. Ash contains salts , minerals and silica .

4. Organic matter:-

It is the non- miniralic portion of dry feedstuff involving crude protein, fat and carbohydrates.

5. Crude protein:-

It is all the nitrogenous substances present in feedstuff sample. It includes true protein and non-

true protein (non- protein nitrogen) such as urea.

6. Crude fat:-

It includes all compounds that can be dissolved in organic dissolvents (such as

ether, benzene, hexane, etc.). Fats, Oil, waxes, and plant dyes are examples of crude fat.

7. Carbohydrates:-

It includes all types of saccharides such as mono saccharides e.g. glucose and di- saccharides.

E.g. sucrose, lactose, and maltose.

Carbohydrates can be divided into two main parts:

A- Nitrogen- Free extract (N.F.E) (Soluble carbohydrates).

These are soluble carbohydrates which are capable of dissolving in diluted acids and bases.

Glucose, sucrose, and starch are examples of this type of carbohydrates.

B- Non- soluble carbohydrates (Crude fibers):


These carbohydrates do not dissolve in diluted acids and alkalines but it dissolves in concentrated

acids and alkalines Cellulose and hemicellulose and lignin are examples of this type of carbohydrates.

Thirdly: Take the samples for feed analysis:

The sample is part of feedstuff substance which represent a proper part of the whole feedstuff

components samples taken should be free from adulteration to determine their actual values in a correct

way.

It is preferable to write the chemical composition of these samples on the packs or bags or sacks

containing these feedstuff.

Rations or feedstuffs stored in the form or displayed in local markets have different types, forms,

or shapes such as sacks, grains, cubicles, hay, and straw bales and green vegetative. plants cultivated in

the field and liquid dietary substances. Hence, each types of this feed should be sampled in its specific

method for ideal sample representative.

Basic conditions necessary for obtaining ideal sample:-

1. Samples should be taken from places having sufficient light to recognize it’s morphological characters

such as homogeneity, storing methods. One authorized worker should carry out sampling to ensure

proper samples.

2. Preparation of requirements and special equipment's used for sample receiving prior to the operation.

3. Determination of places in which samples will be taken from them.

Methods of sample taking:

There is an important notice that sample taken from the fields or sacks filled with animal feed

for food analysis should represent (10%) of the feed. This percentage should be taken or included from

different parts of the ration and should be random without any discrimination.

There are different methods to take samples owing to shape, form of animal feed as follows:-

1. Taking samples from grains, cereal, and seed sacks:

This operation is carried out by obtaining (10%) of number of available sacks. Samples should

be taken from different location of the sack with multiple dimensions and directions using a special pen

designed for sample picking. The primary sample should be to small less size sample called


"subsample". Subsample is a sample prepared from the total blending of the primary sample. The

primary sample is located on clean and smooth surface in a square form. The large square is then

subdivided into multiple small squares. Subsequently, the sample intended for analysis is taken from

each square as shown in the figure.

** ** ** **

** ** ** **

** ** ** **

The collection of this sample weighs about 1.kg. later, this sample is packed in three special jars

until the operation of analysis takes place.

2. Taking samples from feedstuffs having pellets, cubes or tablet forms:

Similar procedures are followed. Some tablets, pellets, are taken from different regions of the

sacks and from multiple directions. The primary sample is then reduce into (1 Kg.) applying the same

method.

3. Taking samples from hay bales:-

The previous method is followed as in case of grains. Ten percent of samples is taken i.e. from

ten bales, only one bale is taken. However, if the bales are less than ten bales, sample should be taken

from most bales. Primary sample should be taken from different places and directions. The primary

sample show later be reduced after blending and chopping followed the same mentioned method.

4. Taking samples from heaps of straw or other plant byproducts:-

Some secondary products which do not constitute the main product such as the remaining of

plant stems, plant leaves of tuber (underground Stem or rhizome) as well as straws gathered in forms

of heaps in the field. To obtain the primary sample, (10)points (spots) of different regions and directions

were determined. Later, samples were taken from the primary sample. Lastly, these samples are further

reduced into one sample following the previous method.

5. Taking samples from green roughage of the field:-

Thirty- fifty spots of green roughages are randomly determined within the farm. Three- five

plants were selected from each area. These plants are cut using sharp instrument at the ground


level. All parts of the plant should be preserve from falling and this represented the primary sample.

This sample will further be reduced after it’s chopping in the lab. to small pieces (2-3)cm. length using

sharp scissors. These operations should be done to secure that the heat should be distributed to all pieces

of the samples for drying. One- two hundred grams are taken for estimation of primary moisture.

6. Taking samples of liquid substances:

The quantity and volume of primary sample depends on the volume of liquid substance for the

purpose of analysis. If the purpose is to carry out total laboratory estimation of all the matters. Hence,

the weight of dry matter within the liquid substance should not be less than (150 gram). The sample is

taken after careful and total mixing and blending of the liquid by glass bar and then by special instrument

used for taking liquid sample.

Fourthly: preparation of samples for analysis:

Typical, sample is prepared for lab. analysis operations as follows:

1. Exclusions of all foreign bodies such as fine stones, gravels, dust, sand, debris and straws. If the

sample is of grain origin.

2. The sample is grinding using cereal mill for small amount of grain. Willy mills or ball mills are used

for coarse grain grinding. The later mills are manufactured from smooth stainless metals. Temperature

of the sample should not be elevated during the grinding. The sample is then sieved and is regrind for

those of rough and coarse parts again in order to be homogenous.

3.The green forages and plant (grasses) of should be dried first to get rid of humidity. The process of

drying is done by putting it in oven (60-70C) for (16) hours (overnight). Then, the sample is ground,

kept in special glass packages until food analysis of the sample will be carried out.

4. Meat and animal byproducts samples. The samples are exposed to mincing and crushing

"pulverization" in special mortar or using special blenders after separation meat from bone.

Fively: Preservation of samples until carrying out the food, analysis:

Some feedstuffs are characterized by rapid deterioration and spoilage due to rapid autolysis leading

to changes in the ratio and amounts of its feeding components. Hence, many methods are suggested to

prevent changes in moisture, enzymatic activity and microbial development.

- To prevent changes occurring in moisture:-


Such as green. Feeds (grasses): can be avoided by drying, followed by grinding and then packed in

special glass jars.

- To prevent enzymatic activity occurrence:

It can be preserved by grinding the sample and mincing and crushing operations. Then, the samples

should be stored and kept after the addition of (95%) hot alcohol. Calcium carbonate are added to

neutralize the acidity by precipitation. The sample will later warm in water bath (60C) for half an hour.

Lastly, the sample is cooled and is kept in low temperature.

- To prevent Microbial development occurrence:

It is done by addition of preservatives such as acetic acid or sodium benzoate by sufficient amounts to

prevent microorganism growth.

- To preserve the milk:

It is carried out by addition of (2ml) of mercuric chloride to each liter of milk.


Determination of Moisture in feedstuffs

Moisture is defined as the free amount of water present in any feedstuff. Moisture can be avoided

in any feedstuffs sample by putting such sampling in oven. The remaining quantity which remains after

this process is called "Dry Matter".

The most important causes for determination of moisture:

1. For preparation of feed sample for different chemical analysis in case of green or plant feeds.

2. To record the result according or on basis of total dry matter.

3. To preserve the feedstuff because high moisture lead to growth of fungi and rapid deterioration of the

feed, also, high amount of moisture causes rapid chemical reaction and autolysis of feedstuffs.

Procedure of the experiment:

1. Empty, labeled, clean moisture can is placed in oven (100C) for an hour. The aim of this process is

for cleaning and sterilization.

2. the can is transferred from drying oven into desiccator for cooling.

3. the can is weighed by digital electrical scale with recording the weight.

4. one- two gm. Of feed sample intended for moisture estimation is weighed and is put in the moisture

can of previously weighed and recorded.

Notebook: there is Two types followed with weighing of sample with the can in the digital electrical

scale.

a. At the beginning, the empty can weighed (e.g. 12gm.). If a sample weight is about (1.5gm.) wanted,

then the sample weight will be added into can weight with the sample to be (13.5gm.).

b. Empty can is weighed and recorded (12gm). Then, the scale is tarred and the sample will be added

(1.5gm).

the second type of weighing is preferable in the laboratory in sample weighing.

5. moisture can containing the sample is put in the drying oven on (130- 150C) for half an hour.

6. At the end of drying period, the can is transferred into desiccator for cooling.

Then, the can is weighed by digital scale with recording the can weight.

Calculations:-


1. Moisture can is weighed when it’s empty.

2. Weight of moisture can containing the sample (the weight before drying ).

3. Weight of the sample =

(weight before drying – weight of empty moisture can).

4. Weight of moisture can with the sample (the weight after drying).

5. Weight of moisture =

(the weight before drying – weight after drying) .

6. The percentage of moisture in the sample=

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑜𝑖𝑠𝑡𝑢𝑟𝑒

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 ∗ 100

7. The percentage of dry matter of the sample =

= 100- moisture percentage.


Determination of moisture in green feedstuffs samples

Determination of moisture in green feedstuffs samples should be dried in two stages. In the first

stage, 100- 200 gm of feedstuff is dried after it’s chopping. Drying of feedstuffs should be at (60-70C)

for 16 hours (overnight). Then, the sample should be cooled, weighed and determined the primary

moisture. In the second stage, the dried sample is ground finely and (1-2) gm. of the sample is taken.

The secondary moisture is estimated following the fore mentioned previous procedures. The purpose of

drying of green feedstuffs following two stages is to prevent the occurrence (the case of hardness).

Case of hardness by formation of outer solid layer to prevent oozing and leakage of moisture from the

lower layers of the sample. Hence, little fraction of the moisture may remain in the sample leading to

false and improper results. Total moisture of the green feedstuffs is estimated following the equation:-

Total percentage% =

primary moisture (%)+ 𝒔𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒎𝒐𝒊𝒔𝒕𝒖𝒓𝒆 (%)∗(𝟏𝟎𝟎−𝒑𝒓𝒊𝒎𝒂𝒓𝒚 𝒎𝒐𝒊𝒔𝒕𝒖𝒓𝒆%)

𝟏𝟎𝟎

Dry matter (%)= 100- total moisture.


Example No. ( 1 )

Two hundred grams of green alfalfa feed sample was dried primarily on (60-70C) for 16 hours.

After drying the weight becomes (40gm). Two gm. of the primarily dried sample was taken and was

further dried on (150C) for half an hour to be (1.6gm).

Estimate primary, secondary and total percentage of moisture in sample.

Solution:-

Primary moisture %= 𝑤𝑒𝑖ℎ𝑡 𝑜𝑓 𝑚𝑜𝑖𝑠𝑡𝑢𝑟𝑒

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒∗ 100

= 𝑤,𝑡.𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑟𝑦𝑖𝑛𝑔−𝑤.𝑡.𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑎𝑓𝑡𝑒𝑟 𝑑𝑟𝑦𝑖𝑛𝑔

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑏𝑒𝑓𝑜𝑟𝑒 𝑑𝑟𝑦𝑖𝑛𝑔 ∗ 100

= 200−40

200∗ 100

= 80%

Secondary moisture%=

𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆 𝒃𝒆𝒇𝒐𝒓𝒆 𝒅𝒓𝒚𝒊𝒏𝒈−𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆 𝒂𝒇𝒕𝒆𝒓 𝒅𝒓𝒚𝒊𝒏𝒈

𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆 𝒃𝒆𝒇𝒐𝒓𝒆 𝒅𝒓𝒚𝒊𝒏𝒈 ∗ 𝟏00

= 2−1.6

2∗ 100

= 20%

Total moisture percentage(%)=

primary moisture (%)+ 𝒔𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒎𝒐𝒊𝒔𝒕𝒖𝒓𝒆 (%)∗(𝟏𝟎𝟎−𝒑𝒓𝒊𝒎𝒂𝒓𝒚 𝒎𝒐𝒊𝒔𝒕𝒖𝒓𝒆(%))

𝟏𝟎𝟎

= 80 + 20∗(100−80)

100

= 80 + 400

100

= 84%

Example No. ( 2 )

Estimate the percentage of dry matter in a feed sample containing (70%) primary moisture and 5%

secondary moisture.

Example No.( 3 )


The weight of empty moisture can is (13.5gm). fifteen gram is the weight the can and the

concentrate sample added to the first sample. The sample is put in oven (130C) for half an hour. The

sample was transferred from the oven and was weight. The weight is became (14.9 gm.). Estimate dry

matter percentage of the sample.


Determination of moisture in milks, meats, and eggs samples

The handling of these sample differs from those of ordinary and concentrate feedstuff because

they adhere with the can at drying process. So, some substances are added with these samples such as

filter paper with the milk, sand with the meat, asbestos with the eggs. The purpose of using these

substances are due to:-

1. Increase the surface area for more evaporation.

2. To avoid the adhesion of the sample with the can.

Procedure of the experiment:-

1. Weigh the moisture can with the substances

(filter paper, sand, asbestos).

2. Weigh the can with the substances + meat sample

(weight before drying).

3. weight of sample = weight before drying – weight of can with substances.

It is preferable that the weight of the original sample of meat, eggs and milk is (5 gm.).

4. The can containing the substances and the sample should be put in water bath for half an hour.

5. The can containing the substances and the sample should be inserted in drying oven on (150C) for

half an hour. Then, the sample is taken and is weighed which is considered as

" weight after drying".

6. Weight of moisture = weight before drying – weight after drying.

7. percentage of moisture = 𝑤𝑒𝑖ℎ𝑡 𝑜𝑓 𝑚𝑜𝑖𝑠𝑡𝑢𝑟𝑒

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 * 100

Example :- The weight of empty moisture can is (11.6gm) .(5gm.)of dry sand add to this can, then the

weight became (16.6gm.).add meat sample ,then the weight became (21.45gm.).the can inserted with

substances in oven ,then put out and weight becomes (18.76gm.).Estimate the moisture percentage in

this sample.

solution:- percentage of moisture = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑜𝑖𝑠𝑡𝑢𝑟𝑒


w.t of moisture =w.t before drying(can+meat+sand)_w.t after drying

= 21.45_18.76


= 2.69gm.

w.t of sample(meat)=w.t before drying (can+meat+sand)- w.t of can with sand

= 21.45-16.60 = 4.85

moisture(%)= 2.69

4.85 * 100

= 55.46 %

Determination of ash in feedstuffs

Ash is defined as residues of different elements remaining after the burning of feedstuff sample

at (600C) in muffle furnace.

The planning of feedstuff in the muffle furnace will lead to burning of all organic matter of the

feedstuff sample. The ash may contain some foreign elements of non- feed origin such as, stone, gravels,

glass pieces, dust, debris, sand (Silica).

In determination process of ash, crucible are used instead of ordinary metal cans. This is due to

the tolerance and durability of crucibles to high temperature of muffle furnace and for long time. In

concentrates feed sample, ash percentage should not exceed 10% , otherwise the sample is considered

as cheat.

Procedure of the experiment:

1. clean and labeled crucible is put in muffle furnace on (500-600C) for an hour, Then, it is transferred

into desiccator for cooling. The aim of this operation is to sterilize the crucible and to exclude any debris

mixed with the sample.

2. The cooled crucible is weighed using electrical digital scale and record the weight. This is the empty

crucible weight. Half gram (0.5gm) of the feedstuff sample intended for ash determination is weigh

by the digital scale and this weigh is considered as

( weight before burning).

3. The crucible containing the sample is placed inside muffle furnace on (500- 600C) for three hours.

The temperature of muffle furnace should be gradually elevated because such operation avoid the

occurrence of ( spattering of sample) .


spattering of the sample is a process characterized by the ejection and expelling of the sample

from the crucible. This case occurs when the temperature of the muffle furnace is abruptly elevated or

when the sample is immediately inserted inside the hot muffle furnace.

4. After the termination of burning process, the crucible is taken and should be lift for a while to be

cooled partially. The remaining ash is colored usually with white. Later, the sample is put on desiccator

until being exactly cooled. It is weighed and recorded as

" weight after burning". The crucible containing the ash is kept for determination of silica.

Note book: when ash percentage is elevated or exceed 15%, this sample is regarded as cheat

sample.

Calculations:

1. weight of empty crucible.

2. weight of crucible with the sample (the weight before burning).

3. weight of sample =

weight before burning – weight of empty crucible.

4. weight of the crucible with sample (the weight after burning).

5. Weight of ash = weight after burning – weight of empty crucible.

6. the percentage of ash = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑠ℎ


7. the percentage of organic matter =

Percentage of dry matter – percentage of ash.

Or ) wt. before burning – wt. after burning


Example: -

Two grams of ground corn sample was put in muffle furnace at (500C)for three hours. The

weight became (0.1gm) after burning. Estimate the percentage of organic matter of the sample,

percentage of dry matter of the sample is 95%

Solution:

percentage of ash = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑠ℎ


= 0.1

2∗ 100

= 5%

Percentage of organic matter =

Percentage of dry matter – percentage of ash.

= 95-5

= 90%


Determination of Silica from Ash of feedstuffs

Silica is defined as a collection of foreign mineral substances of non- feed origin. These

substances are added deliberately for commercial cheat. Silica may involve sand, soft stones and dust.

Present of silica within feedstuff sample may be due to bad storage. Determination of silica ratio may

be carried out because of high percentage of Ash. Also, such determination may be an indicator for

cheat of feedstuffs, hence, such indication may determine the possibility and feasibility these feedstuff

for feeding farm animals. The height percentage of silica within feedstuffs may cause damage and injury

to animal health because of irritation of mucous membranes lining the digestive system. Silica ratio can

be determined in the ash of burning feedstuffs can be achieved by addition of hydrochloric acid (Hcl)

(concentrated and diluted Hcl). The addition of diluted and concentrated (Hcl) is to separate minerals of

feed origin from those present as foreign materials of the ash converting into chlorides of the elements

as follows:

Na Nacl (sodium chloride )

K + Hcl Kcl (potassium chloride)

Mg MgCl 2 (magnesium chloride

)


Notebook: when the percentage of silica exceeds 10%, the sample is considered as cheat.

Procedures of the experiment:

1. Empty crucible weight is recorded (as mentioned in previous laboratory).

2. Original sample weight is recorded (as mentioned in previous laboratory).

3. Careful addition of (1ml) of conc. Hcl is done to the crucible containing ash. It is better to put a

metallic can down of the crucible. The crucible with it is contents should be put in oven for twenty

minutes until being dried from the acid.

4. Diluted Hcl (previously prepared by addition of three quarters of water to one quarter of acid) to ash

sample in small amounts. The mixture should continuously be mixing using steering rod to homogenize

the mixture and to break down the formed masses.

5. Washing and filtration processes are carried out using Buchner funnel in which the remaining matter

is washed by distil water (preferably lukewarm distil water) to facilitate the operation process. This

operation cause a change in pressure leading to draw the salts (chlorides) down wards.

6. The crucible should be put in muffle furnace on (500-600C) for twenty minutes. The crucible is

drawn from the oven and weighed.

Calculations:-

1. Weight of empty crucible (as mentioned from previous laboratory).

2.Weight of original sample (as mentioned from previous laboratory).

3. The crucible put out from the muffle furnace and weighed (it is considered as weight of crucible after

treatment and burning).

4. Weight of Silica = weight of crucible after treatment and burning- weight of empty crucible.

5. Percentage of Silica = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑖𝑙𝑖𝑐𝑎


Example:-

Weight of feedstuff sample is (1.6gm). after burning of the sample in muffle furnace the weight

is (0.4gm). the remaining ash is treated with hydrochloric acid and is burning. The weight after burning

is (0.2gm). estimate the percentage of ash and silica in the sample. Does the sample is cheat ?


Solution:

Percentage of ash = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑠ℎ


= 0.4

1.6∗ 100 = 25%

percentage of Silica = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑖𝑙𝑖𝑐𝑎


= 0.2

1.6∗ 100 = 12.5%

Because silica percentage exceeds 10%, so it is cheat .

Preparation of ash extract of feedstuff samples:

This operation is carried out after the separation of silica and foreign matters from the ash of the

feedstuff. As mentioned earlier, the remaining solution of the receiving flask after filtration process

using the acid contain all mineral elements of dietary origin which are important in feeding of farm

animals. The solution is diluted up to (100ml) by distil water using measuring flask after this process,

the solution is ready for determination of mineral elements of the ash using chemical method. Atomic

absorption and emission spectrophotometry are instruments used according the nature of each element

intended to be determined.


Preparation of standard solutions

It is necessary that a worker or an operator in charge of food analysis should recognize the

methodology of standard chemical solution preparation determination of dietary elements depends on

performance of chemical titration between standard solutions in order to determine the percentage of

dietary (nutritional) compounds of the sample applying certain equations and laws.

Standard solution:

It is the dissolving of equivalent weight of a chemical matter in one liter of distilled water .The

equivalent weight for any chemical matter is the molecular weight of it divided on the valence. The

molecular weight is the summation of atomic weight of all elements that contribute their composition

(Fig. No.1). The valence is a number of hydrogen atoms (power of hydrogen = pH) that join with the

chemical matter, so

equivalent weight = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡

𝑣𝑎𝑙𝑒𝑛𝑐𝑒

= 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 ∗𝑎𝑡𝑜𝑚𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 (𝑎𝑡.𝑤𝑡.)

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 𝑎𝑡𝑜𝑚𝑠 (𝑝𝐻)


Table No.(1):- Atomic weights (at.wt) and atomic number (at. No.) of related elements.

Generic name Chemical symbol Atomic weights (at.wt) atomic number (at.

No.)

Aluminum Al 26 13

Arsenic As 74 33

Barium Ba 137 56

Calcium Ca 40 20

Carbon C 12 6

Chlorine Cl 35 17

Chromium Cr 52 24

Cobalt Co 58 27

Copper Cu 63 29

Hydrogen H 1 1

Iodine I 126 53

Iron Fe 55 26

Lead Pb 207 82

Magnesium Mg 24 12

Manganese Mn 54 25

Mercury Hg 200 80

Molybdenum Mo 95 42

Nickel Ni 58 28

Nitrogen N 14 7

Oxygen O 16 8

Phosphorus P 30 15

Potassium K 39 19

Silver Ag 107 47

Sodium Na 23 11

Sulphur S 32 16

Zinc Zn 65 30

For example, to prepare a standard solution of sodium hydroxide (NaoH), forty gram of NaoH

granules should be dissolved in one liter of distilled water. This weight of NaoH granules was estimated

as follows:

The atomic weight of each H= 1 and O= 16 and Na= 23 hence, the molecular weight for this

compound is 1+16+23= 40 gm.

The valence of each one of these materials is (one) hence, the equivalent weight = 40

1= 40𝑔𝑚 So, 40gm

should be dissolved in one liter of distilled water.

To prepare a standard solution of sulphuric acid (H2So4)the weights are:


The atomic weights of each of O= 16, S=32, and H= 1. So the equivalent weight for (H2So4)

= (4∗16)+ (1∗32)+ (2∗1)

2=

98

2= 49

So, forty nine gram of sulphuric acid crystals should be dissolved in one liter of distilled water

to obtain a standard solution. It is noticed that, the available (H2So4) in the laboratory assumes a

concentrated solution of known density. Physical laws can be used in this field which is:

Density = 𝑚𝑎𝑠𝑠 (𝑤𝑒𝑖𝑔ℎ𝑡)

𝑣𝑜𝑙𝑢𝑚𝑒→ 𝑣𝑜𝑙𝑢𝑚𝑒 =

𝑚𝑎𝑠𝑠 (𝑤𝑒𝑖𝑔ℎ𝑡)

𝑑𝑒𝑛𝑠𝑖𝑡𝑦

The specific density of (H2So4) available in lab. Is a bout (1.8gm/ml). so number of cubic

centimeter can be estimated though the division of equivalent weight on known density of the acid, as

follows:

The volume = 𝑤𝑒𝑖𝑔ℎ𝑡

𝑑𝑒𝑛𝑠𝑖𝑡𝑦=

49

1.8= 27.2𝑚𝑙 of concentrated

Sulphuric acid, then it is diluted into one liter by addition of distil water.

Notebook: concentrated acid solution should be poured (added) to the distilled water at dilution

and not the reverse to prevent reactions that lead splashing of concentrated acid.

One of the character of exact standard solutions is their reactions with each other in equal

volumes. Number of grams of Alkalines reacts with similar grams of acid of the same volume. In

chemical titration, standard solutions are usually do not use for the following reasons:

1. Highly concentrated solution.

2. The neutralization between their solutions is fast.

3. Any sample fault may lead to great fault in determination of feedstuff compounds.

So, tenth (1/10) of standard solution is used laboratory practice due to:

1. Economy in chemical materials.

2. Accuracy of titration process is great.

3. Any error has no great influence on results of calculations because it forms only (0.1) of the solution.

In titration process, only one drop may change the color of the indicator from one color to another

which is so called "balance point"


( balance point ) is defined as the point that neutralization occurs between acid and alkaline

which concurs with the change of color to another color. If the error caused by one drop of the standard

solution, then it equals to ten drops of the tenth of the standard solution.

Tenth of the standard solution is prepared by dissolving of tenth of the equivalent weight of the

chemical material in one liter of distilled water. Proper tenth of the standard solutions reacts with

volumes of each with others. Correct weighing of tenth of the standard solutions of the equivalent weight

of the chemical material require long period of time using the electrical digital scale. So, an approximate

required amount is weighed. When such amount is dissolved in one liter of distilled water, the solution

will not correctly be tenth of standard solution. However, the weight of chemical matter forming this

solution may be more or less than the correct weight which must actually be dissolved. Therefore, this

volume is multiplied by a number called the " normality". Normality can be estimated by the division

of actually dissolved weight on weight be dissolved. Normality value is usually almost whole one

(0.5-1.9)without unit.

Normality = actually dissolved weight

weight be dissolved (equivalent weight)

The tenth of the standard solutions react with their counterpart solutions according to the following

equation:

Volume of Acid * Normality = Volume of alkaline ∗ Normality

The accuracy of balance between the weights of reacted chemical substances may reach to the

accuracy of the digital scale or even exceed it. Usually, Indicator are frequently used in chemical

titration. Indicators are defined as weak organic acids or alkalines , their colours are changed in certain

PH.

The most common indicators are Methyl orange indicator, Methyl red indicator. These indicators

are usually red colour in acidic PH and yellow colour in alkaline PH. To recognize methods that

determine normality of certain solution, the following exercise is carried out:

Definition of normality of Sulphuric acid. This normality is carried out using pure sodium bicarbonate

(NaHCo3). This material can be obtained in pure structure in the lab. So, it’s used to determine the

property and vividness of standard solutions.


Procedure of experiment:

1. Find the equivalent weight of the alkaline (NaHCo3).

Equivalent weight = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡


= (3∗16)+(1∗12)+(1∗1)+(1∗23)

1= 84𝑔𝑚

2. Prepare tenth of the standard solution of the base

= 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

10=

84

10= 8.4𝑔𝑚

3. Prepare tenth of the standard dissolving in (100ml) of distil water = 8.4

10= 0.84𝑔𝑚.

4. Estimation of normality of alkaline =

𝑎𝑐𝑡𝑢𝑎𝑙𝑙𝑦 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡

𝑤𝑒𝑖𝑔ℎ𝑡 𝑏𝑒 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 (𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡)=

0.82

0.84= 0.96

5. Twenty ml. of the tenth of the standard solution of the alkaline is taken.

6. Addition of one drop of methyl red indicator to the tenth of the standard solution of the base.

7. Application of the equation:-

Volume of acid ∗normality = volume of alkaline ∗ normality.

Unknown

= 21.4 ∗ 𝑁(unknown)= 20∗ 0.96 N= 20∗0.96

21.4= 0.8

Known from burette

Volume of the acid

Burette containing

sulphuric acid (S/10) of

unknown normality


Figure explains the titration process between acid and alkaline.

Note book: when the solution (exact 𝒔

𝟏𝟎 )it means that normality = 1

Example:

Prepare one liter of tenth of standard solution from ammonium sulphate (NH4)2so4 . the atomic

weight of O= 16, S=32, N= 14 and H= 1.

Equivalent weight = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡


= 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 ∗𝑎𝑡𝑜𝑚𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 𝑎𝑡𝑜𝑚𝑠

= (4∗16)+ 32+8+(2∗14)

2=

132

2= 66𝑔𝑚

66

10= 6.6 𝑔𝑚 is dissolved in (1 liter) of distilled water.

Indirect method for preparation of chemical standard solution (use of alkaline as a

medium in titration process).

In this method, unknown normality of the alkaline (NaoH) can be determined using sulphuric

acid (H2So4) of unknown normality. In this case, an alkaline is used , sodium bicarbonate (NaHCo3)

of known normality is used as a medium for finding the normality of the acid and then to find the

normality of the alkaline (NaoH).

The first stage:

1. Normality of sodium bicarbonate is determined following equivalent weight.


2. Ten ml. of sodium bicarbonate solution is put in conical flask. One drop of red methyl indicator is

added to the flask.

3. Unknown normality acid (H2so4) is put in the burette .

4.Find the unknown normality of the acid following the equation.

Volume of Acid ∗normality of the acid= Volume of alkaline ∗ Normality of alkaline

(taken from burette) (to be found) = (10ml) ( known )

The second stage:

1- Ten ml of unknown Normality of sodium hydroxide is taken in conical flask . Later , a drop of

red methyl indicator is added .

2- The previous acid of known Normality is put in a burette .

3- The tap of the burette is opened with consequent dropping until the reaching of balance point

.The Normality of the alkaline (NaoH) can be calculated using the following equation :

Volume of Acid ∗ Normality of the Acid= Volume of alkaline ∗ Normality of alkaline

( taken from burette ) ( taken from the 1st step ) (10 ml) ( to be found )

Example:

The volume of (H2so4) (N 𝑠

10 )at titration with NaHCo3 (N

𝑠

10 exact) equals to (11ml).At titration,

with the base NaoH is (12.2ml). Find the normality of the alkaline NaoH? Ten ml. Was taken from each

base at titration.

Solution:

Volume of acid ∗normality of the acid= volume of alkaline (NaHCo3) ∗ normality of alkaline.

11 ∗N= 10∗1

Normality of H2so4 = 10

11 = 0.9

Volume of acid ∗normality of the acid= volume of alkaline (NaoH) ∗ normality of alkaline.

12.2 ∗0.9 = 10 ∗N of NaoH

Normality of alkaline NaoH = 12.2∗0.9

10 = 1.09


Determination of Crude protein in feedstuffs

Crude protein is considered as an important compound which is essential in the nutrition of farm

animals. The benefits of protein is great particularly for non-ruminant farm animal such as poultry.

Therefore, any deficiency of protein leads nutritional deficiency and pathological cases in poultry. Crude


protein is defined as all nitrogen present in feedstuff sample. The importance of crude protein lies in the

recognition of the nutritional value of the feedstuff with subsequent determination of the suitable price

of those feedstuffs. The prices of feedstuffs vary according to their contents of crude proteins. For

example the meals such as soya bean meal contains high percentage of crude protein which may reach

up to (30-40%), the alfalfa may contain 17% crude protein which is more expensive than the yellow

corn (maize) which contains 9% of protein. Consequently, corn is more expensive than the straw which

contains 1-2% protein. The prices of the some feedstuff itself vary from on kind to another according

to their contents of crude proteins. First class alfalfa contains 22% protein which is more expensive than

second class alfalfa which contains 17% protein . The 3rd class of alfalfa is the worst because it contains

about 11% protein.

Nutritionists found out that in nitrogen of feedstuffs used in feeding of farm animals. This refers that

(6.25gm) of protein contains (1 gram) of nitrogen (it indicates that each 1 gm of nitrogen must be

multiplied with the factor " 6.25" to estimate number of grams of protein found in that sample as shown

in the following figure:

Protein Nitrogen

100 gm. 16 gm.

𝑿 1 gm.

X = 𝟏 ∗𝟏𝟎𝟎

𝟏𝟔

= 𝟔. 𝟐𝟓 𝒈𝒎

Crude protein can be determined in the nutrition laboratory applying kjeldahl method which was

invented by a scientist kjeldahl in Denmark in 1883 A.C.

This method is still followed with some alterations modifications. However, the base of the

method is still constant until now. The opinion of the method is summarized by the process that all

nitrogen present in the feedstuff sample with convert into ammonium sulphate (NH4)2 So4 using

concentrated sulphoric acid (H2So4) with boiling. Then, estimation of numbers of nitrogen grams of the

sample and multiplication with ( 6.25) to find number of grams of crude protein.


There are two methods of Kjeldahl:-

1. Macro Kjeldahl method:-

In this method, large weights of feedstuff sample and large amounts of acid and base are used

for the determination of protein. In this technique, large equipment is used.

2. Micro- Kjeldahl method:-

In this technique, small weight of the sample and little amounts of acid and alkaline are used. At

the termination of the experiment, the final value is multiplied by dilution factor.

Preparation of the required solutions:

1. Concentrated sulphuric acid (H2So4).

2. Sodium hydroxide (45% NaoH).

3. Sulphuric (𝑠

10 ) of known normality.

4. alkaline: Sodium hydroxide (s/10) of known normality.

5. Red methyl indicator pigment: it is prepared by dissolving of (250mgm) (0.25gm) of red methyl

powder in 500 ml of ethyl alcohol (ethanol) (C2H5oH), then it is dissolved in 100 ml of distil water.

Notebook: solutions are prepared by dissolving of 1 kg, 900 gm, 450gm, in (1liter) of distilled

water. The concentration, consequently, are 100%, 90% and 45%, respectively. Small amounts of

solutions can be prepared i.e. 45 gm dissolved in (100 ml) of distilled water and the concentration

is 45%.

Stages of determination of crude protein:

Crude protein determination can be divided in the laboratory into three principal stages which

are: digestion, distillation and titration. In the lab, micro- Kjeldahl method is used. The serial steps of

determination of crude protein are:


First: Digestion stage:

1. Half gram of (0.5gm) of feedstuff sample is weighed in a dry clean watch glass. In case of meat or

any substance containing high ratio of moisture, The sample's weight is about five grams. Then, the

sample is added to Kjeldahl flask.

2. one and half gram (1.5 gm) of catalyst is added into the sample present in the Kjeldahl flask. Catalyst

is a substance that makes a chemical reaction happen faster without being changed itself. Catalyst can

be prepared by a mixture of potassium sulphate and copper sulphate " Kcl- pack No.2) Also, the catalyst

can be prepared by mixting of mixture potassium sulphate and mercuric oxide "Kcl- pack No. 5).

3.Concentrated sulphuric acid about (10ml) is added to Kjeldahl flask.

4. Kjeldahl flask with its content is put inside " Digestion heater" for an hour with gradual elevation of

heater temperature from low to high temp. The Kjeldahl flask remains in the heater until the colour

change into clear transparent green colour. The flask is lift for time equal to one and half time that it

required to be transparent. Such time is necessary to ensure the complete conversion of nitrogen of the

feedstuff sample into ammonium sulphate as in the following equation:

NH3+H2O (NH4)2OH

H2So4 + (NH4)2oH (NH4)2So4 + 2H2o

Time: 10 o'clock at locating the flask in the equipment.

Time: 11 alteration the colour of the solution into clear and transparent liquid.

Time 12:30: completion of digestion stage.

Digestion flask containing feedstuff

Sample+ catalyst + concentrated sulphuric acid

Boiling

Catalyst


5. The content (the solution) is transferred into a measuring flask (100ml). the solution is then diluted

by distilled water which is called " digestion extract".

Second: Distillation stage:

1. Connection of the flask with the distillation equipment and opening water flow of the condenser.

2. (300) ml of distilled water and some glass beads are added to the boiling flask. Glass beads are added

in order to prevent crackling and boiling.

3. ( 20 ) ml. of digestion extract is added to the reaction tube as well as (10ml) of concentrated (45%) of

NaoH. With the presence of high temperature, ammonia gas is released as in the following equation:

(NH4)2So4 +2 NaoH Na2So4 + 2NH3+ 2H2O

( concentrated )

4. Preparation of a receiving flask containing (20ml) of (s/10) H2So4 previously prepared and of known

normality with a drop of red methyl (the indicator). The formed and released ammonia will react to

sulphuric acid and will lead to formation of ammonium sulphate as in the following equation:

2NH3+ H2So4 (NH4)2So4 + H2O

Heat

Diluted (S/10)


5. The process is lift for half an hour or to a time sufficient to gather about (100ml). Then, the heat

source is closed. The receiving flask is lifted and is replaced by a flask containing distilled water. This

operation is called (back- suction) which occurs due to difference in the pressure of the tubes. At

termination of back suction process, distillation stage terminates.

Third : Titration stage:

The solution is gathered in a receiving flask which previously contains (S/10) H2So4 is titrated

with (S/10) NaoH of known normality. The titration is continuous until reaching to the balance point

and colour disappearance and the colour becomes light yellow. alkaline volume is recorded from the

burette. The volume of the alkaline is the volume that descends from the burette. The volume is recorded

at the balance point.

Calculation:

1. Volume of the

acid present in

receiving flask.

= volume of acid

∗ normality of

acid

= 20 ∗1 = 20 ml.

2. Volume of the

alkaline NaoH

reacted with the

remaining of acid

Boiling flask

Reaction tube

Condenser

Receiving

flask

contains

H2so4

It reacts with the

remaining of the acid

Burette contains

(S/10) NaoH of

known normality

Receiving flask

contains (s./10) H2So4

of known normality

Alteration the

colour from

pinkish to

yellowish


= volume of alkaline ∗ normality of alkaline

= 18 ∗ 0.98

= 17.64 ml.

3. The correct volume of the acid reacts with ammonia

= (volume of the acid ∗ normality of acid)-( volume of the alkaline ∗ normality of alkaline).

= 20 - 17.64

= 2.36

4. Important Note :

Each (1ml) of (S/10) H2So4 reacts with (0.0014) gm. of nitrogen. So, the amount of nitrogen =

2.36 ∗ 0.0014

= 0.0033 gm. of nitrogen

5. Basic Note:

Number of grams of crude protein of the sample

= number of grams of nitrogen ∗ 6.25

= 0.003 ∗ 6.25

= 0.02gm

6. Percentage of crude protein of the sample

= 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑟𝑢𝑑𝑒 𝑝𝑟𝑜𝑡𝑒𝑖𝑛

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 ∗ 100 ∗ dilution factor

= 0.02

0.5 ∗ 100 ∗ 5

= 20 %

Example:


Half gram of feedstuff sample was taken for determination of crude protein by micro- Kjeldahl

method. (20 ) ml. of (S/10) . (1.11 normality) of H2So4 was put in a receiving flask for distillation

process. Titration process was carried out using NaoH (exactly standard of S/10). The volume of NaoH

needed for titration is (18ml). Estimate the percentage of crude protein of the sample.(twenty ml of

digestion extract diluted up to 100 ml )

The solution:

The correct volume of acid reacting with ammonia

= (volume of acid ∗ normality of acid) – (volume of alkaline ∗normality of alkaline)

= (20 ∗ 1.11) – (18∗1)

= 22.2 – 18

= 4.2 ml.

Percentage of crude protein = 4.2 ∗0.0014 ∗6.25

0.5∗ 100

= 7.35 * dilution factor

= 7.35 * 100

20

= 7.35 * 5

= 36.75 %

Example:


Twenty five ml of digestion extract (diluted up to 100ml) was taken. Find the percentage of crude

protein of the sample. The required volume of the alkaline for titration in the receiving flask was 18ml

of 0.9 normality. The receiving flask contains 20ml (of exactly correct S/10 normality) of H2So4. The

original weight of the sample is 0.5gm.

The solution:

The correct volume of the acid reacting with ammonia

= (volume of acid ∗ normality of acid) – (volume of alkaline ∗normality of alkaline)

= (20∗1)- (18∗0.9)

= 20- 16.2

= 3.8ml

The percentage of crude protein of the sample =

3.8 ∗0.0014 ∗6.25

0.5∗ 100 ∗ 4

= 26.2%

Handout of practical Lectures in Animal Nutrition Directions ...

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