Using English to Predict Rendement of Product a Reaction

Rendement 

Rendement is actually a term in the field of chemistry studies. The yield represents the inaccuracy of the reaction result, which results always lower than the mathematical calculation. For example, in a chemical reaction, should produce a substance weighing 100 grams, mathematically, but in reality the results obtained only 90 grams. Unconsciously this also often happens in our daily lives.

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. 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:

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 rendement that is by using the following equations percent rendemen = weight yield / weight of yield divided by the sample weight multiplied by 100%.

In determining the direction of a chemical reaction we must rely on an understanding based on a number of factors, and contents that are not always easy to assess. Although the assessment is prone to error, it is usually reliable where it seems reasonable to try, and that is certainly to answer ignorance.

In predicting chemical reactions there are many known factors such as, if there is no problem predicting a chemical reaction. Then if only enthalpy changes are known, then predictions usually apply to room temperature but are more or less reliable as well for higher temperatures. If the reaction occurring in the solution and the oxidation potential of the compound is relatively simple, and this oxidation potential buys a rough guide for possible reactions in the absence of a solvent. If the equilibrium constant is known, relating to ΔG0 = - RH in K gives us a change of free energy. But information on this is still lacking, so we must rely on our understanding of the preceding principles.

For chemical reactions at the temperature of the reactants and the products are produced. Here are the rules that may be useful and will be used.

Reactions tend to occur where the bonds of the orbital and some of the electrons are available and allow for attractive tensile interactions between atoms.

There is a possibility of electrons to be divided, and it always happens with energy changes. Therefore we can predict that the atoms of the atoms under the applicable conditions produce the same or different elements. The only elements of the atomic forests are not contained atoms of low external energy ie: "inert" elements or helium groups. Even this, when the circumstances are created, so through the influence of highly electronegative elements such as fluorine can be united by chemical bonds.

When that possibility exists, the tendency of atoms to form temperatures, then at the temperatures, then on the outcome there is little influence from the circumstances. In predicting reactions at ordinary temperatures, we can consider the preceding principles or bond strength and seek the total bond strength in the reactants or products. On this basis we can make prediction rules for definite possibilities or on reactions.

All chemical reactions can be classified into one of six categories:

ü Burning Reactions

The combustion reaction is when oxygen combines with other compounds to form water and carbon dioxide. These reactions are exothermic, which means they produce heat. For example naphthalene combustion reaction. C10H8 + 12 O2 -> 10 CO2 + 4 H2O

ü Reaction Synthesis

The synthesis reaction is when two or more simple compounds combine to form one more complex compound. These reactions appear in a general form:

A + B -> AB

One example of a synthesis reaction is a combination of iron and sulfur to form iron (II) sulfide:

8 Fe + S8 -> 8 FeS

ü Decomposition Reactions

The decomposition reaction is the opposite of the synthesis reaction - the complex molecule is broken down to make a simpler molecule. These reactions appear in a general form:

AB -> A + B

One example of a decomposition reaction is electrolysis of water to make oxygen and hydrogen gas:

2 H2O -> 2 H2 + O2

ü Single Displacement Reaction

This reaction is when one element alternates with another in a compound. These reactions appear in a general form:

A + BC -> AC + B

One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas:

Mg + 2 H2O -> Mg (OH) 2 + H2

ü Double displacement reaction

This is when the anions and cations of two different molecules switch places, forming two completely different compounds. These reactions appear in a general form:

AB + CD -> AD + CB

One example of a dual displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate:

Pb (NO3) 2 + 2 KI -> PbI2 + 2 KNO3

ü Acid-base Reactions

This is a special kind of double displacement reaction that occurs when acids and bases react with each other. H + ions in acid react with OH⁻ ions in the base, causing water formation. Generally, the product of this reaction is ionic and water salts:

HA + BOH -> H2O + BA

An example of an acid-base reaction is the reaction of bromide acid (HBr) with sodium hydroxide:HBr + NaOH -> NaBr + H2O

The efficiency of a chemical reaction can be determined by calculating the percentage of results. Almost in all reactions, we will get fewer results than expected. This happens because most of the reactions are equilibrium reactions, or because of some reaction conditions that cause the reaction not to run perfectly. Chemists can obtain reaction efficiency by calculating the following percentage of results:

Percentage of results = (actual results / theoretical results) x 100%

The real result is how many products are obtained after the reaction is over. The theoretical result is how many products are obtained based on stoichiometric calculations. The comparison of these two results provides an explanation of how efficient the reaction is. For example, the theoretical result of ferrous metals is 699.47 grams. While the real result is 525 grams. Therefore, the percentage of the results is:

% Yield = (525 grams / 699.47 grams) x 100% = 75.05%

The 75% yield percentage is not a too bad result.

In some chemical reactions, the reactants provided do not always correspond to the stoichiometric ratio. This means, we will run out of one of the reactants and still leave another reactant. The former reactant is known as a limiting reagent. The limiting reagent determines the amount of product to be produced by a chemical reaction. Here we will discuss how to determine the limiting reagents through the following example:

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

We start with 100 grams of ammonia gas which is reacted with 100 grams of oxygen gas. Which reactants are limiting reagents? How many grams of nitrogen monoxide (NO) gas can be produced?

To determine which reactants are limiting reagents, we can use a ratio (mole ratio) to the reaction coefficient. We calculate the number of moles each and then divide by their respective reaction coefficients based on equations of equalizing chemical reactions. The ratio of moles to the smallest coefficients is a limiting reagent.

Mol NH3 = 100 gram / 17,024 gram.mol-1 = 5,874 mol

Mol NH3 / NH3 coefficient = 5,874 / 4 = 1,468

Mol O2 = 100 gram / 32.00 gram.mol-1 = 3.125 mol

Mol O2 / coefficient O2 = 3.125 / 5 = 0.625

The ammonia gas has a ratio of mole to the coefficient of 1.468. Meanwhile, oxygen gas has a value of ratio of 0.625. Thus, oxygen gas is a limiting reagent. The calculation of the product to be produced depends on the oxygen gas mole.

The stoichiometric ratio of NO to O2 is 4: 5

Mol O2: Mol NO = reaction coefficient O2: reaction coefficient NO

3,125: Mole NO = 5: 4

Mol NO = 4/5 x Mol O2 = 4/5 x 3.125 mol = 2.5 mol NO

Mass NO = mole NO x Ar NO = 2.5 mol NO x 30.00 gram NO / mol NO = 75.00 gram NO

The value of 75.00 grams of NO is the theoretical result. If the actual result is 70.00 grams, the percentage of the reaction product is (70.00 gram / 75.00 gram) x 100% = 93.33%.

Thus, the amount of ammonia gas remaining (not used) is as much as 100 grams - 42.56 grams = 57.44 grams.

Report : Chemical practicum report in Creating Ice Cream

A.  TITLE                                                   : Chemical practicum report in Creating Ice Cream "

B.  OBJECTIVES OF EXPERIMENT     : To know the freezing point and frozen point dropping solution by                                                                          applying it in Ice Cream making.

C.  DAY / PRIVATE DATE                      : Friday , April 03, 2015

D.  THEORETICAL BASIS

v The Colligative Properties of the Solution

The solution is a homogeneous mixture of two or more substances. The interaction between solute and solvent can result in the change of physical properties of the components of the solution. One of the attributes caused by the interaction between solute and solvent is the colligative nature of the solution. The colligative nature of the solution is the nature of the solution which is only affected by the amount of solute particles in the solution, and is not influenced by the nature of the solute.

The Law of Ralout is the basis for the four properties of a dilute solution called colligative nature (from another language of colligare, meaning "to collect together") because they depend on the collective effect of the number of solute particles, not on the properties of the involved particles. The four properties are:

1)     Decrease in the vapor pressure of the solution relative to the purified solvent vapor pressure.

2)     Boiling point increase.

3)    Decrease of freezing point.

4)   Symptoms of osmotic pressure.

v Decrease of Frozen Point of Solution

The freezing process of a liquid occurs when the temperature is lowered, so the distance between the particles is so close to each other and ultimately works a very strong intermolecular attraction. The presence of particles of solute will result in the movement of solvent molecules hindered, consequently to be more closer intermolecular distances required lower temperatures. So the freezing point of the solution will be lower than the freezing point of the pure solvent. The difference of freezing point due to the presence of solute particles is called the decrease of freezing point (ΔTf). The reduction of the freezing point of the solution is proportional to the product of the solution molarcity with the settlement of the solvent freezing point (Kf), expressed by the equation:

ΔTf = Kf m     or     ΔTf = Kf (n x 1000 / p)

 Where:

ΔTf = decrease of freezing point

Kf = the molal freezing drop point

N = number of moles of solvent

P = mass of solvent

The freezing point of the solution is the freezing point of the purified solvent minus its drop in freezing point or Tf = Tfa^o- ΔTf.

v Cause and Definition of Frozen Point Decline

What is a decrease in freezing point? Pure water freezes at 0 ° C, in the presence of a solute such as the added sugar into the water, the freezing point of this solution will not be equal to 0 ° C, but will fall below 0 ° C, this is what is referred to as the "freezing point drop."

So the solution will have a lower freezing point than the pure solvent. For example, the salt solution in water will have a lower freezing point than the pure solvent which is water, or the phenol solution in alcohol will have a lower freezing point than the pure solvent ie alcohol.

Why does this happen? Does the solute resist the solvent from freezing? The explanation of why this happens is easier when described from the thermodynamic point of view as follows.

For example, pure water at 0 ° C at this temperature of water is at equilibrium between the liquid phase and the solid phase. This means that the water velocity changes from liquid to solid or vice versa is the same, so it can be said to phase water and solid phase. In this condition has the same chemical potential, or in other words the energy level of both phases is the same. The magnitude of the chemical potential is affected by the temperature, so at a certain temperature the chemical potential of the solid phase or liquid phase will be lower than the other, the phase having the lower energy potential is preferably preferable, for example at 2oC the liquid phase has a higher kimi potential Lower than the solid phase so that at this temperature the water tends to lie in the liquid phase, otherwise at a temperature of -1o C the solid phase has a lower chemical potential so at this temperature the water tends to be in the solid phase.

If into pure water we dissolve the salt and then the temperature we decrease little by little, then with the cooling time slowly some of the solution will turn into a solid phase until at a certain temperature will turn into solid phase as a whole. In general, the solute preferably resides in the liquid phase compared to the solid phase, consequently in the cooling process takes place, the solution will retain its phase in the liquid state, because the energy solution preferably resides in the liquid phase compared to the solid phase. This causes the chemical potential of the solvent in the liquid phase to be lower (down) while the chemical potential of the solvent in the solid phase is not affected. Then more energy will be needed to convert the solution into a solid phase because its freezing point is lower than its pure solvent. This is why the presence of solutes will lower the freezing point of the solution. The formula for finding the freezing point of the solution is as follows:

ΔTf = Kf. M. I

Information:

ΔTf = freezing point

Δm = molalilatis solution

Kf = constant constant freezing point of solution

Do not forget to add the Van Hoff factor to the above formula if the solution in question is an electrolyte solution.

E.     TOOLS AND MATERIALS

1)     Plastic

2)    Tablespoons

3)    Water

4)   Jar

5)    Ice Stone

6)   Salt

7)    Powder flavorings

8)    Rubber

9)   Basin

F.      HOW TO WORK

1)     Preparing the Tools and Materials needed.

2)     Enter the powder into a plastic container.

3)    Beat the powder until dissolved.

4)   Tie the plastic with rubber.

5)    Break the ice cubes.

6)   Put ice cubes and salt into the jar.

7)    Insert plastic into the rock and salt.

8)    Then shake the pan ± 10 minutes.

9)   Observe the mambo ice changes.

10)  If the ice in the plastic is solid evenly then remove it from the jar

G.  DATA RESULT OF EXPERIMENT

1)   Ice cream dough after being ejected from the panic formed into ice cream that is ready to eat. With a soft texture, not solid but not too liquid.

2)  With the process of ice play on ice cubes and kitchen salt freezing ice cream happens faster.

H. DISCUSSION

Ice has an average temperature below 00 C. In this ice cream making there is the addition of salt, so that ice melting without adding heat and water formed from the melting process will also have a temperature below 00 C. This is why the addition of salt can Reducing the melting point of water, because ice can melt below its normal melting point, which is 00 C. So salt particles are the triggers of melting ice. On the other hand, smelting requires energy (absorbing heat). Well, because here the heat is not in the supply from outside, then the ice absorbs heat from itself, so the temperature becomes far down. The ice is very cold temperatures and then will absorb a lot of heat from the dough ice cream, so the ice cream dough can freeze

   I.       CONCLUSION

Ø The colligative nature of the solution is the nature of the solution which depends only on the amount of solute particles in the solution, and does not depend on the type of solute. Colligative properties of the solution include decreased vapor pressure, boiling point boost, decrease in freezing point, and osmotic pressure.

Ø The difference of freezing point due to the presence of soluble zar particles is called the decrease of freezing point (ΔTf). The freezing point of the solution is proportional to the product of solution molality with the solvent freezing solvent (Kf) settlement, expressed by the equation:

ΔTf = Kf m or ΔTf = Kf (n x 1000 / p)

Ø Normal temperature of ice and water mixture is 0oC but not cold enough to freeze ice cream. The temperature required to freeze the ice cream is -3 oC or lower. To reach the temperature it is necessary to add salt in the process of making ice cream. Salt serves to lower the freezing point of the solution. When ice is mixed with salt, ice melts and dissolves to form salt water and lowers its temperature. This process requires heat from the outside. The mixture gets hot from the ice cream dough then the result is solid ice cream as you wish. 

J. DOCUMENTATION 

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 (stoichiometry 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 Inter atom (Ar) and inter molecular (Mr) masses can be determined. Knowledge of relative atomic mass and chemical formula of compounds is the basis of chemical calculations.

A. Atomic Mass

1. Average atomic mass

Atoms of the same element do not always have the same mass. This is known as isotopes. The atoms in nature can have different masses, then the atomic mass is calculated on the average mass of all the atoms in nature.

Example:

The chlorine atoms in nature are present in two isotopes, 75% as Cl-35 with mass 35 s.m.a, and 25% as Cl-37 with a mass of 37 s.m.a. Then the average mass of chlorine atoms is:

2. relative atomic mass (Ar)

Measuring the mass is comparing the mass of an object to another, in which the mass of the reference object is called the standard mass.

Example:

If the mass of 1 atom C-12 is 2.04 x 10^-27 kg. What is the average mass of 1 magnesium atom, if Ar Mg = 24?

Answer:

An average mass of 1 atom Mg = 4.08 x 10^-27kg

3. Relative Molecular Mass and Mass Relative Formula (Mr)

The mass of a molecule of a compound is called the relative molecular mass (Mr). The magnitude of the relative molecular mass of a compound is the sum of the relative atomic mass (Ar) of its constituent elements.

M_r A_xB_y = (x  A_r A + y A_r B)

example

Calculate the relative molecular mass of H2O

Mr= (2 × atomic mass of hydrogen) + (1 × mass of oxygen atom)

Mr= (2 × 1) + (1 × 16) = 18.

The multiplier number in front of the atomic mass of each element is the index of the element in the compound.

B. Mol

For practical reasons, in determining the particle size, chemists agree to look for units called moles. One mole is a number of particles contained in a substance equal to the number of atoms present in 12.00 grams of C-12.

From the experiments performed by Joseph Loschmidt and then justified by Avogadro, it turns out that the number of carbon atoms contained in 12.00 grams of C-12 is 6.02 x 1023butir atom. This number is hereinafter called Avogadro number or Avogadro constant and given the symbol L.

1 mol of substance = 6.02 x 10²³particles

1. Molar Mass

The mass relationship with the number of particles is expressed in the molar mass. The molar mass is the mass of the substance that is equal to the atomic mass or the mass of the formula of the substance expressed in grams.

Molar mass (M) = Mass 1 mol of substance X = (ArX) gram

Since Mr of a molecule or unit of chemical formula of a compound is the amount of Ar of its constituent atoms, then:

Molar mass (M) = Mass 1 mol of substance AxBy = (MrAxBy) gram

By means of the molar mass (M), the number of moles of a substance can be calculated by:

With      n = the number of moles of the substance

A = the mass of matter in a gram

M = molar mass = Mr in grams

Example:

Calculate how many mol molecules are contained in 6gram glucose (C₆H₁₂O₆) if known Ar C = 12, O = 16, and H = 1.

Answer:

2. Molar Volume

The molar volume of the gas is the volume of 1 mole of gas at a given temperature and pressure. To determine the molar volume of the gas in the standard state we are weighing a certain amount of gas volume in the already known empty mass tube at 0⁰ and the gas pressure of 1 atm. The standard state (0⁰C, 1 atm) volume of 1 mol of gas is 22.4 liters.

V = n mol x 22.4 L / mol

With    V = gas volume at 0⁰C, 1 atm

n = number of moles of gas

Example:

Calculate the volume of 4 grams of SO₃ gas if it is known Ar S = 32, O = 16.

Answer:

C. The basic laws of chemistry

The basic laws of chemistry are:

1. Law eternity mass

2. Law comparison permanent

3. Dalton's law or law comparison Multiple

4. Gay-Lussac Law or law Volume comparison

5. Avogadro's law

GIVE EVIDANCE ~ Chemical Reactions In Daily Life

1.      Photosynthesis

Photosynthesis is a process used by plants and other organisms to convert light energy, usually from the Sun, into a chemical energy that can then be liberated to fuel organism activity. This chemical energy is stored in carbohydrate molecules, such as sugars, which are synthesized from carbon dioxide and water. In most cases, oxygen is also produced as a waste product. Most plants, mostly algae, and cyanobacteria do photosynthesis, and these organisms are called photoautotrophs. Photosynthesis maintains atmospheric oxygen levels and supplies all organic compounds and most of the energy needed for life on Earth.

Briefly, plants use a chemical reaction called photosynthesis to convert carbon dioxide and water into food (glucose) and oxygen. It is one of the most common daily chemical reactions and also one of the most important, because this is how plants produce food for themselves and animals and convert carbon dioxide into oxygen.

6 CO2 + 6 H2O + light → C6H12O6 + 6 O2

2.     Aerobic Cell Respiration

Aerobic cellular respiration is the reverse process of photosynthesis in molecular energy coupled with the oxygen we breathe to release the energy needed by our cells plus carbon dioxide and water. The energy used by cells is the chemical energy in the form of ATP (adenosine triphosphate).

Aerobic respiration requires oxygen to produce ATP. Although carbohydrates, fats, and proteins are consumed as reactants, it is the preferred method of breaking pyruvate in glycolysis and requiring pyruvate to enter the mitochondria to be completely oxidized by the Krebs cycle. The products of this process are carbon dioxide and water, but the energy transferred is used to break strong bonds in ADP as a third phosphate group is added to form ATP, by substrate level phosphorylation, NADH and FADH2

Here is the overall equation for aerobic cell respiration:

C6H12O6 + 6O2 → 6CO2 + 6H2O + energy (36 ATPs)

3.     Anaerobic Respiration

In contrast to aerobic respiration, anaerobic respiration represents a set of chemical reactions that allow cells to gain energy from complex molecules without oxygen. Cell muscles perform anaerobic respiration every time we get rid of oxygen which then reaches them, like during intense or prolonged exercise. Anaerobic respiration by yeast and bacteria used for fermentation, to produce ethanol, carbon dioxide, and other chemicals that make cheese, wine, beer, yogurt, bread, and many other common products. The overall chemical equation for one form of anaerobic respiration is:

C6H12O6 → 2C2H5OH + 2CO2 + energy

4.    Burning

Every time we light a match, burn a candle, make a fire, or light a grill, we will see a burning reaction. Combustion combines energetic molecules with oxygen to produce carbon dioxide and water.

For example, propane combustion reactions, found in gas grills and some fireplaces, are:

C3H8 + 5O2 → 4H2O + 3CO2 + energy

5.     Rust

Rust is iron oxide, usually a red oxide formed by redox reactions of iron and oxygen in the presence of water or air humidity. Some forms of rust are distinguished both visually and by spectroscopy, and form under different circumstances. Rust consists of hydrated iron (III) oxide Fe2O3 · nH2O and iron (III) oxides-hydroxides (FeO (OH), Fe (OH) 3).

In sufficient time, oxygen, and water, every iron mass will eventually convert entirely of rust and crumbling. The rust surface is flaked and brittle, and does not provide protection to the base iron, such as the formation of patina on the surface of the copper. Rust is a general term for corrosion of iron and its alloys, such as steel. Many other metals have equal corrosion, but the resulting oxide is not commonly called rust.

Here is the chemical equation for iron rust:

Fe + O2 + H2O → Fe2O3. XH2O