STOICHIOMETRY

STOICHIOMETRY

A. Objectives1. Examine the truth of the law of concervaton of mass,2. Find the relationship of mass of the elements in compounds,3. Determine the mol ratio of reacted substances.

B. TheoryStoichiometry is derived from the Greek stoiceon (elements) and metrein (measure). Stoichiometry means to measure the elements in this case is the ion atomic particles, molecules contained in the elements or compounds involved in a chemical reaction. Stoichiometry is the study of the relative quantities of reactants and products in chemical reactions. Stoichiometric calculations are used for many purposes. One purpose is determining how much of a reactant is needed to carry out a reaction. This kind of knowledge is useful for any chemical reaction, and it can even be a matter of life or death. Stoichiometry can be interpreted in various components of the chemical reactions and quantitative relationships between these components. Quantitative relationships in chemical reactions include the mass of reactants and reaction products, energy reactant and reaction products, as well as the volume of reactants and reaction products. There are three concepts that underlie the stoichiometry, namely conservation of mass, relative atomic mass, and concept mole.Stoichiometry of the reaction is the determination of the comparison period of the elements in a compound in the formation of compounds. On the basis of chemical stoichiometry calculations, usually takes the basic laws of chemistry, including: Law of Conservation of Mass The law of conservation of mass expressed by Antonio Lavosier Laurent (1785), which reads: mass of matter before and after the same reaction. Fixed Comparative Law Proust law or comparative law remains that reads: any compound formed from the elements with a fixed ratio.

Multiple comparisons Law/Comparative Law Multiple (Dalton's Law) "If the two types of elements can form more than one compound, then the ratio of the mass of one of the elements that are bound to other elements of the same mass, is an integer and simple." Comparative Law Volume (Gay Lussac's Law). "At the same temperature and pressure, the volume ratio of gases that react and the reaction and the result is an integer and simple." Avogadro's Law "At the same temperature and pressure, the gases containing the same volume the same number of particles as well."

but in this experiment the basic laws of chemical used is the Law of Conservation of Mass. Law of Conservation of mass states than in chemical reaction, the mass of the product equals the mass of the reactant. This experiment will examine and prove the law conservation of mass.

Law of Conservation of Mass

Lavosier (1783) was the first to conduct scientific observations of the right to study the chemical changes. He weighed substances before and after the chemical change occurs. This weighting does not only to substances in the form of solids or liquids, but also gas. Most of the observations showed that the mass of all the substances that undergo a chemical change equals the mass of substances formed in a chemical change that. Of course it does limited weighing accuracy limits of mass observations do at that time. By because it is very basic and common, then the discovery Lavosier was referred to as a law that became known as the law of conservation of mass, which up to this time the law is stated as follows:

In a chemical reaction the mass of substances before and after the chemical reaction is still

Relative atomic mass

One part of Dalton's atomic theory stated that atoms have different masses, if the element is different. determining the atomic mass of an element can be described by Dalton's atomic theory. It has been realized that it is not possible to weigh the atoms one by one and comparing the mass of an atom to the atomic mass of the other, which can be done is to use a relative scale of the mass of a single atom with an atomic mass of the other. The usefulness of the relative scale can be used to determine the composition of atoms of an element in a compound. Since 1961, it has been determined isotope C-12 as the basis for determining the relative atomic mass. Relative atomic mass is the average price of an atomic mass of an element.

Mole concept

Isotope was first used as a standard is 16O but since 1961 has agreed to take the 12C isotope as its standard. if taken exactly 12 g of pure isotope C-12, the number of atoms contained in it is called Avogadro's number (L), which to date is the price received L= 6,002.1023. Avogadro's number has great significance in the field of chemistry, because chemical reactions always related to the number of atoms, molecules, ions or electrons in a substance that many shots in the order of 1023, the amount of the mole is used as the definition of the number of particles.

Some reaction stoichiometry can be learned easily, one of them with methods JOB or Continuous variation method, which is the mechanism that is to be done observation of the reactant molar quantity change, but the same total molar. Certain of the physical properties (mass, volume, temperature, absorption) is examined, and the changes are used to predict the system stoichiometry. Physical properties of graphs groove to the quantity of reagents, will be obtained maximal or minimal point corresponding stoichiometric point of the system, stating ratio of the reactants in the compound.

Changes in the heat of the chemical reaction depends on the amount of its reagent. If the mole which reacts changed with a fixed volume, the stoichiometry can be determined from the point of maximum heat changes, with the rise of temperature on the composition of the flowing mixture.

C. Apparatus and Reagent1. Apparatus :a. Erlenmeyer flask 100 mLe. Spatulab. Test tube 75x12 mmf. Graduated cylinderc. Analytical balanceg. Beaker glassd. Burner methylatedh. Thermometer2. Reagent :a. NaOH 0,2 M solutione. Pb(NO3)2 0,1 M solution b. H2SO4 0,2 M solutionf. Cooper metal c. CuSO4 0,1 M solutiong. Sulfur powderd. KI 0,1 M solutionD. Procedur

E. Observation Data1. The conservation of mass in chemical reactiona. The reaction between NaOH and CuSO4 solutionErlenmeyer mass before reaction : 135,9082 gramsErlenmeyer mass after reaction : 135,9082 gramsThe change of substance states after filtrationbefore reaction after reactionNaOH(aq)+ CuSO4(aq) Na2SO4 (aq)+ Cu(OH)2(s)transparent light blue transparent blue precipitate

b. The reaction between KI and Pb(NO3)2Erlenmeyer mass before reaction : 136,122 gramsErlenmeyer mass after reaction : 136,122 gramsThe change of substance states after filtrationbefore reaction after reaction 2KI(aq) + Pb(NO3)2(aq) 2KNO3 (aq) + PbI2(s)transparent transparent transparent yellow precipitate 2. The relationship between the mass element in compoundAmount of sulfur (soatula tip)12345

The length of remained Cu (mm)811101212

Graph 1. The relationship between among of sulfur precipitate and the length of remained Cu

3. Determination of reactant moles ratioTest tube123456

Vol Pb(NO3)2 (ml)123579

Vol KI (ml)987531

Height of precipitate (mm)876543

Graph.2 relationship between the height of precipitate and the compostion of solution4. Acid-base stoichiometryVolume of NaOH (ml)Volume of H2SO4 (ml)TMTAT

03029,029,00

52529,030,21,2

102029,231,22,0

151531,031,00

201031,030,01,0

25527,029,02,0

30029,029,00

Graph.3 relationship between T and the compostion of acid base solutionF. Data Analysis 1. The conservation of mass in chemical reactionNaOH(aq) + CuSO4(aq) Na2SO4 (aq)+ Cu(OH)2(s) 135,9082gram 135,9082gram2KI(aq) + Pb(NO3)2(aq) 2KNO3 (aq) + PbI2(s) 136,122 grams 136,122 grams2. The relationship between the mass element in compoundAmount of sulfur (spatula tip )12345

The length of remained Cu (mm)811101212

length (mm)30111

3. Determination of reactant moles ratio

a. Reaction in the test tube 1

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0,1mmol0,9mmolR0,1mmol0,2mmol0,1mmol0,2mmolS-0,7mmol0,1mmol0,02mmol

Mass of PbI2 = 0,1 mmol . 461= 46,1 mg

b. Reaction in the test tube 2

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0,2 mmol0,8mmolR0,2 mmol0,4 mmol0,2mmol0,4mmolS -0,4mmol0,2mmol0,4mmol


c. Reaction in the test tube 3

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0,3mmol0,7mmolR0,3mmol0,6mmol0,3mmol0,6 mmolS -0,1mmol0,3mmol0,6mmolMass of PbI2 = 0,3 mmol . 461= 138,3 mg

d. Reaction in the test tube 4

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0,5mmol0,5mmolR0,25mmol0,5mmol0,25mmol0,5mmolS0,25mmol -0,25mmol0,5mmol


e. Reaction in the test tube 5

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0,7 mmol0,3mmolR0,15mmol0,3mmol0,15mmol0,3mmolS0,55mmol - 0,15mmol0,3mmolMass of PbI2 = 0,15 mmol . 461= 69,15 mg

f. Reaction in the test tube 6

Pb(NO3)2 (aq)+2KI(aq)PbI2(s)+2KNO3(aq)M0.9mmol0,1mmolR0,05mmol0,1mmol0,05mmol0,1mmolS0,85mmol - 0,05mmol0,1mmol


4. Acid-base stoichiometrya. Reaction between 0 NaOH and 30 mL H2SO4(T=0) mol NaOH = M.Vmmol H2SO4 = M.V = 1.0 =1.30 =0 =30 mmolThe ratio between NaOH and H2SO4 = 0 : 30

b. Reaction between 5 NaOH and 25 mL H2SO4(T=1,2)mol NaOH = M.Vmmol H2SO4 = M.V = 1.5 =1.25 =5mmol =25 mmolThe ratio between NaOH and H2SO4 = 5 : 25 = 1 : 5

c. Reaction between 10 NaOH and 20 mL H2SO4(T=2,0)mol NaOH = M.Vmmol H2SO4 = M.V = 1.10 =1.20 =10mmol =20 mmolThe ratio between NaOH and H2SO4 = 10 : 20 = 1 : 2d. Reaction between 15 NaOH and 15 mL H2SO4(T=0)mol NaOH = M.Vmmol H2SO4 = M.V = 1.15 =1.15 =15mmol =15 mmolThe ratio between NaOH and H2SO4 = 15 : 15 = 1 : 1

e. Reaction between 20 NaOH and 10 mL H2SO4(T=1,0)mol NaOH = M.V mmol H2SO4 = M.V = 1.20 =1.10 =20mmol =10 mmolThe ratio between NaOH and H2SO4 = 20 : 10 = 2 : 1

f. Reaction between 25 NaOH and 5 mL H2SO4(T=2,0)mol NaOH = M.V mmol H2SO4 = M.V = 1.25 =1.5 =25mmol =5 mmolThe ratio between NaOH and H2SO4 = 25 : 5 = 5 : 1

g. Reaction between 30 NaOH and 0 mL H2SO4(T=0)mol NaOH = M.V mmol H2SO4 = M.V = 1.30 =1.0 =30mmol =0The ratio between NaOH and H2SO4 = 30 : 0

G. Discussion1. The conservation of mass in chemical reactiona. The reaction between NaOH and CuSO4 solutionThis experiment used a solution of 5ml NaOH and CuSO4 in a sealed container. NaOH initially included in the erlenmeyer flask while CuSO4 put into a test tube. The test tube is inserted into the erlenmeyer then closed using plastic and rubber tied then weighed and the results was 135,9082 grams. After weighed, the two solutions are mixed by tilting the flask erlenmeyer so the two solutions can be mixed. After all the solution has been mixed, the erlenmeyer flask weighed again and the results was 135,9082 grams. In this situation, the chemical reactions is occured it characterized by the color change before and after the mixed. Before reactant the colour of solution is transparent, but after the reaction the colour become blue precipitate, with the following reaction: before reaction after reaction2NaOH(aq) + CuSO4(aq) Na2SO4 (aq) + Cu(OH)2(s)transparenttransparenttransparent blue precipitateThen the mixture was filtered using filter paper and the clear filtrate results will be obtained and a blue precipitate. The clear filtrate is Na2SO4 while blue precipitate is Cu(OH)2

b. The reaction between KI and Pb(NO3)2This experiment used a solution of 5ml KI and Pb(NO3)2 in a sealed container. KI initially included in the erlenmeyer flask while Pb(NO3)2 put into a test tube. The test tube is inserted into the erlenmeyer then closed using plastic and rubber tied then weighed and the results was 136,122grams. After weighed, the two solutions are mixed by tilting the flask erlenmeyer so the two solutions can be mixed. After all the solution has been mixed, the erlenmeyer flask weighed again and the results was 136,122 grams. In this situation, the chemical reactions is occured it characterized by the color change before and after the mixed. Before reactant the colour of solution is transparent, but after the reaction the colour become yellow precipitate, with the following reaction: before reaction after reaction2KI(aq) + Pb(NO3)2(aq) 2KNO3 (aq)+ PbI2(s)transparent transparent transparent yellow precipitateThen the mixture was filtered using filter paper and the clear filtrate results will be obtained and a blue precipitate. The clear filtrate is KNO3 while blue precipitate is PbI2.From the two experiments show that the law of conservation of mass applied in this experiment, where the mass of substances before chemical reaction (before mixed) will be equal to the mass of matter after chemical reaction (after mixed).

2. The relationship between the mass element in compoundThis experiment to prove the Proust Law. In this experiment after copper and sulfur was burned, copper and sulfur reacted. Seen that the length of the copper after reacted with sulfur decreased, with increasing sulfur powder reacted the diminishing length of copper. in this case the sulfur acts as a limiting reactant is completely reacted with the copper. the reaction is complete when the sulfur has reacted completely discharged not remained. But our data not like that. Maybe when we had been measuring the presipitate copper, its still warm. So that happened expansion.

3. Determination of reactant moles ratioV This experiment was carried out by mixing a solution of Pb(NO3)2 0.5 M and 0.5 M KI and will result a mixture of 10 ml volume. This mixture is made in six test tubes . After the solution is mixed and then allowing the mixture to settle. After all the sediment settles, then measure the height.Of the six test tubes are used, the highest sediment contained in the tube with the composition 1 mL Pb(NO3)2 and 9 mL KI. This is not in accordance with the laws contained in the solution stoichiometry, the reaction coefficient is the ratio of moles of substances involved in the reaction, if the substance reacted was solution then determined by measuring the quantity of a substance volume.The precipitate which should be the highest mixed results with 7 ml Pb(NO3)2 3ml KI, because in this volume, the volume ratio of Pb(NO3)2 and KI solution is = 1: 2 and it is in accordance with the ratio of the coefficients in the equation. In addition to the conditions very effective to be used as the reaction barrier. An error in this experiment is because the tube used has a large surface area so it affecting the accuracy in measuring the height of the precipitate.4. Acid-base stoichiometryThis experiment was carried out by reacting 1 M NaOH and 1 M H2SO4 by volume of each solution composition varied so as to produce a mixture of 30 ml volume. Data obtained from this experiment with the relationship between the volume of acid-base with T. Comparison of the number of molecules reacting NaOH and H2SO4 is 0 : 30, 1: 5, 1: 2, 1: 1, 2: 1, 5: 1, 30 : 0.The temperature changes can also be used as a reference for determining the acid-base reaction stoichiometry. The temperature changes depending in the amount reagent varied when the volume of the mixture is fixed. Stoichiometry can be determined from the change of maximum temperature. The largest temperature change in the composition 20 mL NaOH and 10 mL H2SO4 . Stoichiometric reaction occurs at a volume ratio of NaOH and H2SO4 is 2 : 1 the composition can be seen from the calculations that show that NaOH/ H2SO4 completely reacted.Data that our group gained are not like the theory because there are any error in measuring temperature.

H. Conclusion1. Mass of substance before and after chemical reaction is same. This is suitable with the law of conservation of mass,2. The proportion of elements that form a pure particular compound is fixed,3. If there are two substances that are mixed, it will cause a change in temperature, color and form (precipitate),4. The numbers ratio of moles that react determines the amount of reaction product.I. Suggestion1. The presipitate copper in determination of reactant moles ratio should be weighed by analytical balance not be measured by mistar. So that the change of copper is more seen,2. Experiment with copper and sulfur should be done outdoors, because the smell of sulfur was overpowering and it is advisable to wear a mask,3. Experiments measure the height of the sediment mixture of KI and Pb (NO3)2, it takes patience to wait until the mixture is completely settles,4. For acid-base stoichiometry experiment takes precision in reading the thermometer to determine the temperature scale.

J. BibliographyAhmad, Hiskia, dkk . 1995. Materi Pokok Kimia Dasar I. Jakarta : Universitas Terbuka.IS, Kasmadi,dkk .2004. Kimia Dasar I. Unnes .Semarang : Unnes Press.Ebbing, D.D. dan Gammon, D.S.,2009.General Chemistry.Amerika: Houghton Mifflin Company.

Question1. On the experiment of conservation of, is there any chemical phenomena ? mention it if any!How does the mass of substances after the reaction and before reaction?2. Refer to graph 1. Show the relationship between the mass of copper ad the reacted sulfur?3. Refer to graph 2. 1. Indicate the volume of reactant composition which produced the highest sediment?1. Count the moles of Pb(NO3)2 and the moles of KI from the volume of solution which produces the highest precipite !1. The moles of Pb(NO3)2 and KI that reacted have certain ratio. What is the ratio?4. From the graph 3.1. How many volume of NaOH and H2SO4 mixture which showed the highest T ?1. Count the moles of NaOH and H2SO4 from the volume that produces the highest T !1. The moles of NaOH and H2SO4 that reacted have certain ratio. What is the ratio ?Answer1. Yes there is. in the mass determination experiments are substances that change color chemical events that occur in mixing NaOH and CuSO4. Where the first colored transparent and blue after mixed change to blue (precipite), as well as the mixture of KI and Pb(NO3)2 premises which the color changes from transparent to yellow (precipite). After the second mass is considered before and after the solution was not mixed mass changes, which means the mass before and after the reaction is still.2. The relationship between the mass of copper and sulfur which react mass that is more and more the amount of sulfur that reacted sulfur was reduced length of copper or otherwise.3. The higher presipitate in the tube with the composition 3 mL Pb(NO3)2 and 7 mL KI. 2KI(aq) + Pb(NO3)2(aq) 2KNO3 (aq) + PbI2(s)Initial : 0,7 mmol0,3 mmol - -Reaktion : 0,6 mmol0,3 mmol 0,6 mmol 0,3 mmolremain : 0,1 mmol - 0,6 mmol 0,3 mmolthe ratio of reactant mole 1 mmol Pb(NO3)2 : 2 mmol KI = 1 : 24. The largest T in the composition 20 mL NaOH and 10 mL H2SO4Mol NaOH = M . Vmol H2SO4 = M . V= 1 . 20 = 1 . 10= 20 mmol = 10 mmol

2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O(aq)Initial : 20 mmol10 mmol - -Reaktion : 10 mmol10 mmol 10 mmol 20 mmolremain : - - 10 mmol 20 mmol

the mol ratio of NaOH and H2SO4 = 20 : 10 = 2 : 1

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STOICHIOMETRY