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Chemistry 9 Lesson 47: Fat

08/08/2021 by admin Leave a Comment

Chemistry 9 Lesson 47: Fat

1. Theoretical Summary

1.1. Where is fat?

foods high in fat

Fat is the main component of fats, oils, etc. found in the body of animals and plants.

1.2. Physical properties of fats

Fats are lighter than water, insoluble in water, soluble in benzene, gasoline, kerosene, etc.

Fat solubility

a) Water b) Benzene

1.3. Composition and structure of fats

Fats are mixtures of many esters of glycerol with fatty acids and have the general formula (RCOO)3C3H5

Glycerol with CTCT is: .

Abbreviated as C3H5(OH)3

Fatty acids have a common CT of RCOOH. The general formula of fat is (RCOO)3OLD3H5

โ€‹ Fats are mixtures of many esters of glycerol with fatty acids and have the general formula (RCOO).3OLD3H5

1.4. Chemical properties

a) Hydrolysis reaction in acidic medium

(RCOO)3OLD3H5 + 3 HOURS2O \(\xrightarrow[axit]{t^{0}}\)C3H5(OH)3 + 3RCOOH

b) Hydrolysis in alkali (saponification reaction)

(RCOO)3OLD3H5 + 3NaOH \(\overset{t^{0}}{\rightarrow}\) C3H5(OH)3 + 3RCOONa

1.5. Application

Energy of fat

  • Fat is a basic ingredient in human and animal food
  • Fat is rich in energy and helps the body absorb vitamins A, D, E, K…

  • Used to make glycerol and soaps

1.6. summary

Diagram of fat metabolism in the body

post fat mind map

2. Illustrated exercise

2.1. Type 1: Saponification reaction

Calculate the mass of salt obtained from complete hydrolysis of 17.16 kg of a fat. Knowing the hydrolysis process, it is necessary to use 2.4 kg of NaOH and obtain 0.736 kg of Glycerol. Calculate the mass of soap obtained from the above mass of salt. Know that the salts of Fatty Acids make up 60% of the soap mass.

Solution guide

The general equation is: Fat + NaOH โ†’ Salt + Glycerol

Applying the law of conservation of mass to the reaction we have:

mfat + mNaOH = mmsalt + mGlixerol

โ‡’ mm salt = mfat + mNaOH – mGlixerol = 17.16 + 2.4 – 0.736 = 18.824 (kg)

Since the salts of Fatty Acids make up 60% of the soap’s mass, then:

msoap = \(\frac{{18,824,100}}{{60}} = 31.37(kg)\)

2.2. Form 2: Structural formula of fat

When carrying out the saponification reaction of a type of fat A with NaOH solution, glycerol and a mixture of two salts C are obtained.17H35COONa and OLD15Hthirty firstCOONa with a corresponding molar ratio of 2 : 1.

Determine the possible structural formula of this fat.

Solution guide

Because fat A, when hydrolyzed, produces only two salts of the acid:

It’s C17H35COONa and OLD15Hthirty firstCOONa with the corresponding molar ratio of 2 : 1. So ester A contains two acid radicals C17H35COO- and an acid radical C15Hthirty firstCOO– So the structure of ester A is:

or

3. Practice

3.1. Essay exercises

Question 1: For m kg of fat, react with NaOH to get 17.72 kg of salt mixture and 1.84 kg of glycerol. Calculate m and mass of NaOH used.

Verse 2: To completely hydrolyze 8.58 kg of a type of fat, 1.2 kg of NaOH is needed, to obtain 0.368 kg of glycerol and m kg of a salt mixture of fatty acids.

a) Calculate m.

b) Calculate the mass of bar soap that can be obtained from m kg of the above salt mixture. Know that salts of fatty acids make up 60% of the mass of soap.

Question 3: Let glycerol react with a mixture of fatty acids consisting of C17H35COOH and C15Hthirty firstHow many triesters will COOH get?

3.2. Multiple choice exercises

Question 1: Fat reacts with alkali to yield glycerol and

A. a salt of fatty acids.

B. two salts of fatty acids.

C. three salts of fatty acids.

D. a salt mixture of fatty acids

Verse 2: Completely saponification of 17.24 grams of fat requires just enough 0.06 mol of NaOH. The solution is depleted after the reaction, yielding a mass of soap of

A. 17.80g B. 18.24g C. 16.68g D. 18.38

Question 3: Cooking oil is:

A. an ester

B. ester of glycerol

C. esters of glycerol and fatty acids

D. a mixture of many esters of glycerol and fatty acids

Question 4: Which of the following is not a fat:

A. Coconut oil

B. Sesame oil (sesame oil)

C. Peanut oil (peanut)

D. Lubricant oil

4. Conclusion

After the lesson to know:

Know the definition of fat.

– Understand the natural state, physical properties, chemical properties and applications of fats.

– Write CTPT of glycerol, CT of fat.

.

=============

Filed Under: Grade 9 Chemistry Tagged With: Chapter 5 Chemistry 9, Chemistry 9, Chemistry Lecture 9, Derivatives of hydrocarbons, Polymer

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