VIDEO_About Material Chemistry

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THE COLIGATIVE SOLUTION

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CHEMISTRY LESSON PLAN BASED ON CURRICULUM 2013

CHEMISTRY LESSON PLAN BASED ON CURRICULUM 2013

Steps to compile RPP

1. Specify Indicators

2. List the learning materials and complete the descriptions that have been developed in the syllabus.

3. explained what is the purpose of the learning

4. Formulate a Learning Strategy or Scenario

5. Setting Learning Facilities and Resources

6. Assessment and Follow-up

7. The development of the lesson plan should pay attention to the interests and attention of learners to the basic materials and basic competencies that are used as the study material.

Lesson plan

(R P P)

 High School

             Subject:  Chemistry

 Class / semester: X MIPA / 1

Basic Material: Atomic Model and Periodic System

Time allocation: 6 meetings x 3 hours Lesson (18 Hours Lesson)

A.  Learning objectives:

After observing the environment, reading the developmental articles and applications of chemistry students can understand and realize the nature, the importance of the role of chemistry in life, the scientific method and work safety.

B. Basic competence:

 1.1 Recognizing the orderliness of the structure of material particles as a manifestation of the greatness of God YME and the knowledge of the structure of matter particles as the result of human creative thinking whose truth is tentative.

 2.1 Shows scientific behavior (curiosity, discipline, honest, objective, open, able to distinguish facts and opinions, resilient, meticulous, responsible, critical, creative, innovative, democratic, communicative in designing and conducting experiments and discussions embodied in Everyday attitude.

 2.2 Demonstrate co-operative, courteous, tolerant, peace-loving and caring about the environment and thrifty in utilizing natural resources.

 2.3 Demonstrate responsive, and proactive and wise behavior as a form of problem-solving ability and decision making.

3.2 Analyze the development of the atomic model

3.3 Analyzing atomic structure based on Bohr's atomic theory and quantum mechanical theory.

3.4 Analyze the relationship of electron configuration and orbital diagrams to determine the location of elements in the periodic table and the periodic properties of the elements.

H. Indicators of achievement of competence:

  1. To mention the existence of the order of the structure of material particles as a manifestation of the greatness of God YME and the knowledge of the particle structure of matter as the result of human creative thinking whose truth is tentative

  2. Behave scientific (have a curiosity, discipline, honest, objective, open, able to distinguish facts and opinions, resilient, meticulous, responsible, critical, creative, innovative, democratic, communicative in designing and conducting experiments and discussions embodied in Everyday attitude.

  3. Can cooperate, courteous, tolerant, love peace and care for the environment and efficient in utilizing natural resources.

  4. Be responsive, and proactive and wise as a form of problem-solving ability and make decisions.

5. Determining elementary particles of atoms (protons, electrons and neutrons)

6. Determining relative atomic mass based on periodic table

7. Classify the elements into isotopes, isobars and isotons

8. Explain the development of atomic theory to show the weaknesses and advantages of each atomic theory

9. Explain the difference Bohr's atomic structure and quantum mechanics

10. Describe an orbital diagram

11. Determine the electron and the valence electron configuration

12. Determining the location of classes and period elements in the periodic system based on the electron configuration

13. Explain the tendency 4 The nature of the periodicity of the elements in the periodic table

C. Learning materials:

• atomic number and mass number

• Isotopes, isobars, isotons

• The development of the atomic model

• Bohr and atomic structures

• Bohr electron configuration

• Atomic structure of quantum mechanics.

• Orbital shape

• Quantum numbers

• Configuration of electrons of quantum mechanics

• Groups and periods

• Orbital diagrams

• The nature of the nature of the elements

D. Learning method: Scientific, discovery, problem solving

E. Learning media:

Learning resources: BSE chemistry book Class 

Print media: Magazines, newspapers

Electronic media: Internet articles

F. learning steps:

Meeting 1 (3 Hours Lessons = 135 minutes)

Atomic number, mass number, isotope, isobar, isoton

Stage Time activity

1. Introduction

 - Pray with students

- Saying greeting

- Checking student attendance

- Gathered assignments last week

- Listening to the Basic Material / Competence to be studied

- Listening to the Learning Method information that will be used

- Divide the group

-Give homework:

                           1. Create concept maps about the development of the atomic model

                           2. Summarize how to determine the number of electrons, protons and neutrons

                           3. Make a summary of how to determine isotope, isobar and isoton

                           (15 minutes)

2. Core 

Observing

• Observing atomic particles and their relation to mass numbers and atomic numbers.

• Observe the modern periodic table

Ask

1. What are the atomic particles?

2. Where is the position of electrons in the atom?

3. How to specify atomic element number?

4. How to specify the element mass number?

5. Shows what, atomic element number?

6. Shows what, elemental mass number?

7. How to determine the number of proton elements?

8. How to determine the number of element electrons?

9. How to determine the number of neutron elements?

10. What dimadsud with isotopes, isobars and isotons!

11. Why is the atomic model evolving?

Data collection

       • Observe the atomic number and mass number of several elements to determine the             number of electrons, protons and neutrons of the element.

       • Analyzing the atomic number and mass number of some case examples on elements            to understand isotopes, isobars, and isotons. 

           (90 minutes)

3. End 

Gathering assignments (5 minutes)

1. concept map about the development of atomic model

2. a summary of how to determine the number of electrons, protons and neutrons

3. a summary of how to determine isotopes, isobars and isotons

Formative Test Written  (25 minutes)

G. Evaluation of learning outcomes:

1. Affordable

With Observation

Attitude assessment sheets Attitude during discussions and presentations

2. Cognitive

Task

Create a concept map about the development of the atomic model

Summarize how to determine the number of electrons, protons and neutrons

Make a summary of how to determine isotopes, isobars and isotons

Written Format:

Question :

1. How to determine atomic number and mass number?

2. How to determine the number of electrons, protons, and neutrons of an elemental atom?

3. How to determine isotope, isobar and isoton?

3. Psychomotor

   Skills Assessment Sheet

PERTEMUAN_15. Predict Rendement of Product a Reaction

RENDEMENT CHEMISTRY

In chemistry, the chemical yield, the yield of the reaction, or only the rendement refers to the amount of reaction product produced in the chemical reaction. [1] Absolute rendement can be written as weight in grams or in moles (molar yield). The relative yield used as a calculation of the effectiveness of the procedure is calculated by dividing the amount of product obtained in moles by the theoretical yield in moles:  

The relative yield used as a calculation of the effectiveness of the procedure is calculated by dividing the amount of product obtained in moles by the theoretical yield in moles:

Rendemen fractional = rendemen actual / theoretical rendemen 

To obtain a percentage yield, multiply the fractional yield by 100%.

One or more reactants in chemical reactions are often used redundantly. The theoretical rendement is calculated based on the number of moles of the limiting reagent. For this calculation, it is usually assumed that there is only one reaction involved.

The ideal chemical yield value (theoretical rendement) is 100%, a value highly unlikely to be achieved in its practice. Calculate the percentage of rendemen that is by using the following equations percent rendemen = weight yield / weight of yield divided by the sample weight multiplied by 100%.

FACT FROM PREDICTION

Is known :

· Ρ Phenol = 1.07 gram / mL BM phenol = 94, 11 gram / mol

· Ρ Formaldehyde 37% = 1.08 gr / mL BM formaldehyde = 30.03 grams / mol

Asked:

Ø Phenol mass = V. Phenol x ρ Phenol

                        = 1mL x 1.07 grams / mL = 1, 07gram

Ø Formaldehyde Mass = V. Formaldehyde x ρ Formaldehyde

                                     = 2 mL x 1, 08 gram / mL = 2.16 grams

So:

Mol Phenol =

                        = = 0.011 mol

TEORY

We can learn with this practice :

Formaldehyde Phenol Polymers

The purpose of this pratikum is:

A. Praktikan know the process of making formaldehyde phenol

B. Praktikan able to conduct experiments of condensation polymerization

C. Praktikan able to recognize the characteristics of condensation polymerization

D. Praktikan able to study the making of phenol formaldehid by using base catalyst.

Polymers are a science that develops applicatively. Paper, plastics, tires, natural fibers, are products of polymer products. Polymers are giant molecules (macromolecules) formed by repeating simple units of monomer.

             Polymers can be grouped by source, ie natural polymers and synthetic polymers. In addition, the polymer may also be grouped from chain arrangement, polymerization reaction, monomer type, thermal properties and application. Synthetic polymers are materials that are widely used in our daily life in various applications. One type of synthetic polymer widely used commercially in both the plastic and paint industries (surface coatings) is phenol formaldehyde.

             Phenol formaldehyde is a thermosetting synthetic resin, produced from the condensation polymerization reaction between phenol and formaldehyde. One application of phenol formaldehyde resin is for varnish.

          Phenol is also known as carbolic acid or benzenol is a colorless crystalline substance that has a distinctive odor. Phenol is an organic compound whose chemical formula is C6H5OH and its structure has a hydroxyl group (-OH) attached to a phenyl ring.

METHODELOGY

1.1 Tools and Materials

A. The tools used in the experiment are:

1. Analytical balance sheet 6. Pipette drops

2. Test tube 7. Bunsen lamp

3. Mixers 8. Tweezers

4. Pipette volume 9. Mold (porcelain cup)

5. Pro pipette

B. The materials used in the experiment are:

Ø Phenol solution

Ø Formaldehyde solution 37%

Ø 2.5 N NaOH solution

Ø Rock didh

Ø Aluminum foil

1.2 Working steps

· Prepared tools and materials to be used.

· Dipipet 1 cc of phenol solution then put in the test tube.

· Added 2mL of 37% formaldehyde solution, then shaken the test tube to homogeneous solution.

· Added 2 drops of 2.5N NaOH solution and inserted sufficient boiling stones into the test tube, then the reaction tube clamped and then heated to a white milk-colored solution

· Stopped warming, allowed to separate mixture into two phases. The top layer is taken with a dropper pipette (the top layer is mostly water, and the bottom layer is yellow thick liquid)

Then the bottom layer is reheated until it becomes brown and then expands and eventually turns into a solid like a reddish-brown glass, after being poured into a mold that has been coated with aluminum foil.

1.3 Anali sis Data

Table 1. Comparison of polymer preparation results from all groups

No

Group

B.Mold print

B. Molds + polymers

Polymer weight

Polymerization Time

1.

I.

129.369 grams

131,282 grams

1.913 grams

2.

II

34,081 grams

35.705 grams

1,624 grams

3.

III

143,811 grams

146,681 grams

2,870 grams

15 minutes

4.

IV

132,553 grams

135.156 grams

2,002 grams

5.

V

125.320 grams

128.210 grams

2,890 grams

3.50 minutes

6.

VI

41, 125 grams

42.752 grams

1,627 grams

9 minutes

7.

VII

134, 709 grams

135,270 grams

0.561 grams

13 Minutes

Type:

Ü Volume of phenol: 1 ml

Ü Formaldehyde volume 37%: 2 ml

Ü V NaOH: 22 drops

Color Changes during Polymerization

1. Phenol + Formaldehyde = brown

2. Phenol + formaldehyde + NaOH 2.5 N = Old Chocolate

3. Phenol + Formaldehyde + NaOH 2.5 N after heated = Young white sediment chocolate and then reheated to remove water reddish-brown solution poured on a mold coated with aluminum foil and let it harden and freeze.

3.4 Calculations

Is known :

· Ρ Phenol = 1.07 gram / mL BM phenol = 94, 11 gram / mol

· Ρ Formaldehyde 37% = 1.08 gr / mL BM formaldehyde = 30.03 grams / mol

Asked:

Ø Phenol mass = V. Phenol x ρ Phenol

                        = 1mL x 1.07 grams / mL = 1, 07gram

Ø Formaldehyde Mass = V. Formaldehyde x ρ Formaldehyde

                                     = 2 mL x 1, 08 gram / mL = 2.16 grams

So:

Mol Phenol =

                        = = 0.011 mol

DISCUSSION

Theoretically, the production of phenol formaldehyde by using a base catalyst with a mole ratio of reactant P / F> 1, the resulting product is a thermoset resol type. Resol is the result of a reaction between phenol and formaldehyde excess by the presence of an alkaline catalyst. The most commonly used types of base catalysts are sodium hydroxide (NaOH) and ammonium hydroxide (NH₄OH) at pH = 8-11. The product of phenol formaldehyde produced with a sodium hydroxide catalyst will have a water solubility and if the catalyst used ammonium hydroxide will give insoluble in water due to the formation of bis and tris hydroxylbenzylamin (Martin, 1956).

CONCLUSION

From the results of the discussion above can be summed up as follows:

  Phenol formaldehyde is a synthetic polymer prepared by the process of condensation with a by-product of water.

PERTEMUAN_14. Using English to report

 Practical Chemistry Report of Decreasing Freezing Point A Solution

Chemical Practicum Report

Decrease of Freezing Point A Solution

A. Test Topics

Colligative nature.

B. Purpose of Experiment

Determine the reduction of freezing point of a solution.

C. Basic Theory

The freezing point is the temperature at which the vapor pressure of the liquid is equal to the vapor pressure in a solid state. The freezing point of the solution is lower than the freezing point of pure solvent. In freezing a solution, the freezing is solely the solvent, while the solute does not freeze. The freezing point is a fixed temperature in which an appropriate substance changes from a liquid to a solid state. Every substance that has freezing has a pressure of 1 atm.

The freezing point of a liquid will change if the vapor pressure changes, usually caused by the introduction of a solute or in other words, if the liquid is not pure, the freezing point changes (the freezing point will decrease).

The presence of solute causes a more difficult solvent to freeze, consequently the freezing point of the solution will be lower than the freezing point of the pure solvent. The difference between the freezing point of the solution and the freezing point of the pure solvent is called the decrease of the freezing point of the solution.

Experiments also show that the drop in freezing does not depend on the type of solute, but depends only on the concentration of the solution.

The decrease in the molasses freezing point is the value of the freezing drop if the concentration of the solution is one mol.

Electrolyte ΔTf = Kf x m x i

Non Electrolyte ΔTf = Kf x m

D. Tools And Materials

Tool:

1. Balance Sheet

2. Test tube

3. Spoon

4. Mixer Stem

5. Glasses of chemicals

6. Thermometer

7. The reaction tube shelf

Material:

1. Distilled water

2. Ice cubes

3. Urea 1 m & 2 m

4. NaCl 1 m & 2 m

E. Introduction to Experiments

Freezing point, a pure liquid has a certain freezing point value. If the pure liquid has been mixed with other substances, the freezing point may change and the change is dependent on the mixing agent. At each temperature, the vapor pressure of the solution is lower than the vapor pressure of the solvent thereby causing the freezing point of the solution to be lower than that of the solvent freezing point. The increase in the boiling point and the decrease in freezing point is one of the colligative properties of the solution.

F. Safety

1. Materials used, not to be swallowed.

2. Ingredients of asthma / bases used, do not get exposed to skin, if exposed to the skin immediately rinse with running water.

3. In each observation made, keep your eye distance to the observed material, be careful not to expose the effects of the substance / reaction that occurs. If exposed immediately rinse with water.

4. Clean (wash) the equipment every time after the experiment is used. The reaction tube is cleaned with a reaction tube brush.

G. Experimental Procedure

The unit of concentration of molality (m) represents 1 mol of solute in 1000 grams of solvent.

1. Prepare a 250 ml glass of water and fill it with ice cubes that have been crushed until the volume reaches approximately 3/4 tall glass of chemistry. Sprinkle the ice cubes with salt.

2. Insert 5 ml of water into the test tube, then insert the test tube into a glass containing the ice. Adjust the position of the tube soaked in ice in the beaker.

3. Stir the contents of the test tube by slowly stirring the stir bar (not stirring in a circle) until the liquid in the tube completely frozen.

4. Remove the test tube from inside the beaker and let the ice in the tube melt slightly.

5. Remove the stir bar and insert the thermometer. Stir back the water by throwing the thermometer (be careful not to stop the thermometer and then break) and then read the temperature (the temperature will decrease and then increase again, take the lowest temperature).

6. Repeat steps 1-5 using urea, 1 m and 2 m solutions, and a solution of NaCl, 1 m and 2 m.

I. Discussion

A solution will freeze at a lower temperature than the freezing point of water. To learn this further needs to be understood about freezing. What is meant by freezing is the temperature at which the liquid phase and the solid phase are together (in equilibrium).

The normal freezing point of a liquid is a freezing point at a pressure of 760 mmHg or 1 atm. For example pure water freezes at a fixed temperature, ie 0 ˚C at 1 atm pressure. Decrease in freezing point is proportional to the amount of solute concentration is greater then the bigger the frozen point is also greater. Thus, in the presence of solutes in water, the freezing point of water becomes less than 0˚C at 1 atm pressure.

If we pay attention to the making of the swivel ice, to obtain a lower temperature and 0 ˚C then the rotating ice dough is placed in a vessel submerged in ice cubes and water that has been given the salt of the kitchen, while rotated and stirred then the ice batter in the vessel will freeze, Where the freezing point of the rotating ice dough is a few degrees below the freezing point of pure water. This happens because there is a process of heat transfer from the ice batter into a mixture of ice cubes, water and salt. For more details can be seen and the following picture:

Images change the pure water ice cubes and the process of freezing ice

Information:

○ = pure solvent particles

● = solute particles

If the pure water in a container is immersed in ice cubes and the pure water salt will freeze at a certain temperature (normally 0 C measured at 1 atm pressure). While at the same temperature, the ice dough has not been perfectly frozen or has not even frozen. The presence of substances or solutes added in the rotary ice dough prevents the purge of the purified solvent molecules to freeze normally, so that the freezing point of the solution decreases (a decrease in freezing), resulting in a lower temperature to freeze it.

Thus, it is clear that the solution will freeze at a lower temperature than the freezing point of water. The difference between the freezing point of the pure solvent and the freezing point of the solution is called the decrease of the freezing point of the solution denoted by ΔTf.

          ΔTf = Tºf - Tf

Information:

ΔTf = decrease of freezing point

Tºf = freezing point of solution

Tf = freezing point of solvent

The freezing point does not depend on the type of solute, but depends only on the concentration or the amount of solute particles in the solution. So, the greater the concentration of the solution the greater the freezing point. Mathematically writable:

Information:

ΔTf = decrease of freezing point

Kf = the molal freezing drop point

M = concentration of solution

G = mass dissolved in grams

P = solvent mass in grams

Mr. = the relative molecular mass of the solute

Where, Kf is the same as the molecular freezing point constant, ie the value of the freezing point of the solution as much as 1 mole of solute in 1000 grams of solvent (Kf). The price of Kf depends on the properties of the liquid used as the solvent, so the price of Kf for each solvent varies.

Application in Everyday Life

.

A. Creating a Cooling Blend

The coolant is an aqueous solution that has a freezing point well below 0oC. Coolant liquids are used in ice factories, also used to make ice turns. The coolant is made by dissolving different types of salt into the water. In the manufacture of ice, the cooling liquid is made by mixing the kitchen salt with ice cubes in a wood-lined vessel. At the mixing, the ice cubes will melt while the mixed temperature falls. Meanwhile, the mixture of swivel ice maker was put into another vessel made of stainless steel. The vessel is then fed into the coolant, while continuously stirring so the mixture freezes.

B. Anti-freeze on Car Radiators

In cold climates, the water of the radiator is usually added ethylene glycol. In areas with cold climates, the radiator water is easily frozen. If this situation is allowed, then the vehicle radiator will be quickly damaged. With the addition of ethylene glycol into the water the radiator is expected to freeze the water in the radiator decreases, in other words the water does not easily freeze.

C. Antibeku in Animal Body

Animals living in cold climates, such as kutup bears, utilize the principle of the colligative nature of frost reduction solutions to survive. The blood of sea fish contains antifreeze substances capable of lowering the freezing point of water up to 0.8oC. Thus, the latter can survive in winter where the temperature reaches 1.9oC because the antifreeze it contains can prevent the formation of ice crystals in tissues and cells. Other animals whose bodies contain antifreeze substances among insects, amphibians, and nematodes. Insect body contains glycerol and metal sulphoxide, amphibians containing glucose and blood glycerol while nematode contains glycerol and trilose.

D. Anti-freeze to Liquid Snow

In an area that has winter, every snowfall occurs, the streets are filled with snow ice. This of course makes the vehicle difficult to drive. To overcome this, the snowy streets are sprinkled with a mixture of salt NaCl and CaCl2. The sowing of salt is sown, the more snow will melt.

E. Determining Relative Molecular Mass (Mr)

The measurement of the colligative properties of the solution can be used to determine the relative molecular mass of the solute. It can be used because the colligative nature depends on the concentration of the solute. Knowing the dissolved mass can be determined.

J. Conclusions

The conclusions of the experiments that have been done are as follows:

1. What is the function of adding salt to ice cubes?

Answer: The function of adding salt to ice cubes is as a substance that lowers the freezing point of ice cubes so that ice cubes will not freeze at 0 ° C.

2. What is the decrease in freezing point for each solution?

In the urea solution the molality of 1 m of the freezing point - 3oC and molalitas 2 m of freezing point - 6 oC. While in NaCl solution molalitas 1 m freezing point - 5 oC and molalitas 2 m freezing point -9 oC.

3. What is the relationship between the molality of the solution and the decrease in its freezing point?

Answer: The relationship between the molality of the solution and the decrease in the freezing point is the greater the molality the greater the freezing point. This relationship can be formulated as follows:

ΔTf = Tºf - Tf

Information:

ΔTf = decrease of freezing point

Tºf = freezing point of solution

Tf = freezing point of solvent

4. How does the freezing point drop for urea and NaCl solutions in the same molality? Explain?

Answer:

At the same molality, the decrease of freezing point of NaCl (electrolyte) solution is higher than that of urea (non-electrolyte) solution. Because the electrolyte substances decompose into ions so that the number of particles more than non-electrolyte substances.

5. The greater the molality of the solution, the higher the decrease in the freezing point of the solution.

6. Decrease of the freezing point of the solution (Tf) is directly proportional to the molality of the solution.

7. The freezing point of pure solvent is higher than the freezing point of the solution.

8. The freezing point of the electrolyte solution is lower than the non-electrolyte solution in the same kemolalan.

9. The smaller the concentration of the solution, the greater the antarion distance and the more free ions.

10. For the same concentration, the electrolyte solution contains more number of particles than the non-electrolyte solution.

11. Electrolyte solution has a greater colligative properties than non-electrolyte colligative properties.

12. The higher the kemolalan the lower the freezing point.

13. The higher the kemolalan the greater the decrease in the freezing point.

K. Follow Up

1. Look for examples of events in everyday life that indicate a decrease in freezing?

Answer: Examples of events in everyday life that indicate the occurrence of a decrease in freezing is as follows:

The presence of solutes in the solution will cause the freezing point of the solution to be smaller than the freezing point of the solvent. For example salt can melt the snow. Another example is that the snow on the road in the winter country is easy to clean by adding salt, thereby lowering the freezing point of the solution, so the freezing point of the solution (salt + snow) will be lower than the freezing point of the snow melting ice on the road by sprinkling salt. As well as on the ice cream mixture does not freeze due to the decrease in freezing point. Additionally Prevention of water freezing of car radiators during winter in Europe also uses applications of the colligative nature. Even some animals that have poles or sea with cold temperatures also use a chemical compound (salt) in his blood so as not to freeze to death.

Daftar Pustaka

http://chemistranger.blogspot.com/2012/07/sifat-koligatif-larutan.html

http://fathur30rahman.blogspot.com/2014/04/laporan-praktikum-kimia-penurunan

titik.html

http://riski-seftiani.blogspot.com/2012/11/laporan-pratikum-kimia.html

http://www.slideshare.net/Andirahim/kimia-kelas-12-ariharnanto

http://worldofanimeducation.blogspot.com/2012/10/laporan-penurunan-

titik-beku.html

www.google.com

PERTEMUAN13_ STOICHIOMETRY

STOICHIOMETRY

Stoichiometry comes from the Greek word stoicheion which means element and metron which means measure. Stoichiometry discusses the relation of mass between elements in a compound (stoichiometric compound) and interactivity in a reaction (reaction stoichiometry). Mass measurements in chemical reactions were initiated by Antoine Laurent Lavoisier (1743 - 1794) who found that in chemical reactions there was no mass change (mass conservation law). Furthermore Joseph Louis Proust (1754 - 1826) found that the elements form compounds in certain comparisons (fixed comparison law). Furthermore, in order to construct his atomic theory, John Dalton discovered the third basic chemical law, called the law of multiples of comparison. These three laws are the basis of the first chemical theory, the atomic theory proposed by John Dalton around 1803. According to Dalton, every material consists of atoms, elements composed of similar atoms, whereas compounds composed of different atoms in certain comparisons . However, Dalton has not been able to determine the ratio of the atoms in the compound (the chemical formula of the substance). Determination of chemical formula of substances can be done thanks to the discovery of Gay Lussac and Avogadro. Once the chemical formula of the compound can be determined, then the ratio of antaratome (Ar) and intermolecular (Mr) masses can be determined. Knowledge of relative atomic mass and chemical formula of compounds is the basis of chemical calculations.

Basic Laws of Chemistry

1. The Law of Conservation of Mass (Lavoisier Law)

"The mass of the substance before the reaction equals the mass of the substance after the reaction" Example:

S (s) + O2 (g) → SO2 (g)

1 mol of S reacts with 1 mole O2 to form 1 mole of SO2. 32 grams of S reacts with 32 grams of O2 forming 64 grams of SO2. The total mass of the reactants is equal to the mass of the resulting product.

H2 (g) + ½ O2 (g) → H2O (l)

1 mole of H2 reacts with ½ moles of O2 forming 1 mole of H2O. 2 grams of H2 reacts with 16 grams of O2 forming 18 grams of H2O. The total mass of the reactants is equal to the mass of the product formed.

2. Comparable Law (Proust Law)

"The mass ratio of the constituent elements is always fixed, even if it is made in a different way" Example:

S (s) + O2 (g) → SO2 (g)

The ratio of mass S to mass of O2 to form SO2 is 32 grams S to 32 grams O2 or 1: 1. This means that every gram of S just reacts with one gram of O2 forming 2 grams of SO2. If 50 grams of S is required, it takes 50 grams of O2 to form 100 grams of SO2.

H2 (g) + ½ O2 (g) → H2O (l)

The ratio of mass of H2 to mass of O2 to form H2O is 2 gram H2 to 16 gram of O2 or 1: 8. This means, every one gram of H2 precisely reacts with 8 gram of O2 forming 9 gram H2O. If provided 24 grams of O2, it takes 3 grams of H2 to form 27 grams of H2O.

3.The Law of Multiple Comparisons (Dalton's Law)

 The Law of Multiple Comparisons: If an element reacts with other elements, then the ratio of the weight of the element is a simple integer

 So from persmaaan:

2Na (s) + 2HCl (aq) → 2NaCl (aq) + H2 (g)

We can know that 2 moles of HCl react with 2 moles of Na to form 2 moles of NaCl and 1 mole of H2. By equalizing this reaction, it can be known the quantity of each substance involved in the reaction.

Hence the equalization of this reaction is very important in solving stoichiometric problems.

Example:

Lead (IV) Hydroxide reacts with Sulfuric Acid, by reaction as follows:

Pb (OH) 4 + H2SO4 → Pb (SO4) 2 + H2O

If we look good either:

Reactant Element

(Number of moles) Product

(Number of moles)

Pb 1 1

O 8 9

H 6 2

S 1 2

Then this equation is not equivalent. Therefore we need to equate this equation. In the reactant there are 16 atoms, but in its product there are only 14 atoms. This equation needs to add coefficients so that the number of atoms of the elements is the same.

In front of H2SO4 it is necessary to add coefficient 2 so that the number of sulfur atoms corresponds, then in front of H2O it is necessary to add coefficient 4 so that the number of oxygen atoms is appropriate. Then the equivalent reaction is:

Pb (OH) 4 + 2H2SO4 → Pb (SO4) 2 + 4H2O

Reactant Element

(Number of moles) Product

(Number of moles)

Pb 1 1

O 12 12

H 8 8S 2 2

The condition in which the equation of the reaction is equal is when it satisfies the following two criteria:

1. The number of atoms of each element on the left and right sides of the equation has been the same.

2. The number of ions on the left and right has been the same (using redox reaction equation)

3. Comparative Law of Volume (Gay Lussac Law)

Applies only to chemical reactions that involve the gas phase

"At the same temperature and pressure, the ratio of reactant gas volume to the gas volume of the reaction product is a simple integer (equal to the ratio of the reaction coefficient)" Example: N2 (g) + 3 H2 (g) → 2 NH3 (g)

The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of N2 gas exactly reacts with 3 mL of H2 gas to form 2 mL of NH3 gas. Thus, to obtain 50 L of NH 3 gas, it takes 25 L of N2 gas and 75 L of H2 gas.

CO (g) + H2O (g) → CO2 (g) + H2 (g)

The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of CO gas reacts exactly with 1 mL of H2O gas to form 1 mL of CO2 gas and 1 mL of H2 gas. Thus, as much as 4 L of CO gas requires 4 L of H2O gas to form 4 L of CO2 gas and 4 L of H2 gas.

4. Avogadro's Law

Applies only to chemical reactions that involve the gas phase

"At the same temperature and pressure, the same volumes of gas contain the same number of moles" .Avogadro's law is closely related to Gay Lussac's Law : Example:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

The mole ratio is equal to the ratio of the reaction coefficient. This means that every 1 mole of precise N2 gas reacts with 3 moles of H 2 gas to form 2 moles of NH 3 gas. The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 L of N2 gas precisely reacts with 3 L of H 2 gas to form 2 L of NH3 gas. Thus, if at a certain temperature and pressure, 1 mole of gas is equivalent to 1 L of gas, then 2 moles of gas is equivalent to 2 L of gas. In other words, the mole gas ratio is equal to the ratio of gas volume.

Here are some examples of problems and chemical calculations that use the basic laws of chemistry:

1. 20 gram calcium powder (Ar Ca = 40) is reacted with 20 grams of sulfur (Ar S = 32) according to the Ca + S → CaS reaction equation. What substance is left after the reaction is completed? How much mass of the substance is left after the reaction is complete?

Resolution:

The ratio of Ca mole to S is 1: 1. This means that every 40 grams of Ca exactly reacts with 32 grams of S to form 72 grams of CaS. The ratio of mass of Ca to S is 40: 32 = 5: 4.

If 20 grams of S just exhausts, it takes (5/4) x 20 = 25 grams Ca, to form 45 grams of CaS. Unfortunately, the amount of Ca provided is insufficient.

Therefore, 20 grams of Ca will be properly reacted. The required S mass of (4/5) x 20 grams = 16 grams. Thus, the remaining substance is sulfur (S). The remaining sulfur mass is 20-16 = 4 grams.

Equalization of Chemical Reaction

Chemical reactions are often written in bentu equations using element symbols. The reactants are the substances that are on the left, and the product is the substance that is on the right, then both are separated by arrows (can be one or two alternating arrows). Example:

2Na (s) + HCl (aq) → 2NaCl (aq) + H2 (g)

The equation of a chemical reaction is like a prescription in the reaction, thus indicating everything associated with the reaction, whether it is an ion, an element, a compound, a reactant or a product. All.

Then as in the recipe, there is a proportion of the equation shown in the figures in front of the molecular formula.

When considered again, the number of H atoms on the reactant (left) is not equal to the number of H atoms on the product (right). Then this reaction needs to be synchronized. The equalization of chemical reactions must satisfy some chemical laws of matter.

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Siahaan, P., Radiman, C.L., Rahayu, S.I., Martoprawiro, M.A.,  and Ziessow, D., 2008, Kajian Interaksi Molekul Benzena Tersubstitusi melalui Studi Relaksasi  T₁ NMR ¹³C dan Perhitungan Mekanika Kuantum  ab initio, Disertasi, ITB.

Siahaan, P., Radiman, C.L., Rahayu, S.I., Martoprawiro, M.A.,  and Ziessow, D., 2009, Molecular Interaction Between Benzonitrile and Hexamethylphosphoric Triamide  by  ¹³C NMR T₁ Relaxation Time Studies and ab initio QM calculations: Extended  Investigation, Indo. J. Chem., Vol. 9 (2), 292-296.