Stoikiometri reaksi

1. PUJI LASTARI A1C312075 KHUSNUL HATIMAH A1C312021 NOVITASARI A1C312030 ALFIA RIANI A1C312080 WENNY REZKY AMELIA A1C312032 MIFTAH FARID A1C312013 GROUP 4 STOICHIOMETRY

2. After learning this chapter, you are expcted to be able to : a. Prove based on experiment that substance mass before and after reaction is constant (law of conservation of mass/Lavoisier’s law ). d. Use experimental data to prove volume ratio law (Gay Lussac’s Law ). b. Prove in accordance with experiment and interpret data concerning two elements mass that compounds (Proust’s law ). e. Estimate gases volume of reactants or products based on Gay Lussac’s law . c. Prove the applicable of ratio multiplication law (Dalton’s law) on several compound .

3. f. Find gases volume relation with number of molecules measured at the same temperature and pressure (Avogadro’s Law ). g. Explain mole definition as a unit amount of substance . i. Determine emprical formula, molecular formula and water crystal and percent composition of compounds . h. Converse total mole with number of particles, mass and substances volume . j. Determine limiting reagent in a reaction .

4. Fundamental Laws of Chemistry Chemical equation Avogadro’s Hypothesis Mole Concept Reaction Stoichiometry Compound of Stoichiometry

5. Fundamental Laws of Chemistry Law of Conservation of Mass Proust’s Law Dalton’s Law Gay Lussac’s Law 1

6. The Law of Conservation of Mass “ I n every chemical transformation, an equal quantity of matter exists before and after the reaction.” Fundamental Laws of Chemistry

7. Solution : 2Mg + O 2  2MgO Mg = = e O = 2Mg O = Ar O 2 = 32 Mass of O 2 = = 1,6 g   At 2,4 grams magnesium burning in air, magnesium oxide produce 4 grams. How many grams of oxygen are used up in the reaction?(Ar Mg = 24 Ar O = 16 ) Example: Menu

8. The Law of Definite Proportions “ In a given chemical compound the proportion by mass of the elements that compose it are fixed, independen of the origin of the compound or its mode of preparation.”

9. Example : Analysis of the salt from various regions. Check to see it meets law of proust ?

10. Solution : As shown in the calculation above, the mass ratio of Na to Cl apparently fixed. Namely 1 : 1,54. so fulfill the law of compound proust Salt of Mass of sodium (Na) Mass of chlorine (Cl) Mass Na : Cl Indra mayu Madura Impor 0,786 g 0,59 g 0,983 g 1,214 g 0,91 g 1,517 g 0,786 g : 1,214 g = 1 : 1, 54 0,59 g : 0,91 g = 1 : 1, 54 0,983 g : 1,517 g = 1 : 1, 54 Menu

11. The Law of Multiple Proportions “When two elements form a series of compounds, the masses of one that combine with a fixed mass of the other are in the ratio of small integers yo each other . ”

12. Example : Sulfur (S) and oxygen (O) formed two types of compounds. Levels of sulfur in compounds I and II in a row is 50% and 40%. Whether the law applies to the compound dalton.

13. Solution : Compound I consists of 50% sulfur, the mass oxygen is 50%. Compound II consists of 40% sulfur, the mass oxygen is 60%. Mass S : O of compound I = 50 : 50 = 1 : 1 Mass S : O of compound II = 40 : 60 = 2 : 3 or 1 : 1,5 If mass S in the compound I = compound II, as equally as 1 gram, then mass O of compound I : II = 1 : 1,5 = 2 : 3. Comparison is a simple integer and two compounds hat fulfill the law of dalton. Menu

14. The Law of Combin ing Volume s “ W hen two gases are allowed to react, such that the goes are at the same temperature and pressure, the volumes of each gas consumed will be in the ratio of small integers. Morever, the ratio of the volumes of each product gasto the volume of either reacting gas will be ratio of simple integers.”

15. Example : Two researchers independently studied the decomposition reaction of dinitrogen pentaoksida be nitrogen dioxsyde and oxygen. The researcher found that the decompotion of 50mL (100°C, 1 atm) nitrogen pentaoksida produce 100mL (100°C, 1 atm) nitrogen dioxyde and 25mL (100°C, 1 atm) oxygen. Whether the result of these trials fulfill the law of combining volumes?

16. Solution : 50mL N 2 O 5  100mL NO 2 + 25mL O 2 2 4 1 Then 2N 2 O 5  4NO 2 + O 2 This statement is meet Menu

17. Chemical Equations Lavoisier, 1788 Because of the principle of the law conservation of mass , an equation must be balanced . It must have the same number of atoms of the same kind on both sides. 2

18. Balancing Equations Depict the kind of reactants and products and their relative amounts in a reaction. ___ Al(s) + ___ Br 2 ( liq ) ---> ___ Al 2 Br 6 (s) 2 Al(s) + 3 Br 2 (g) ---> Al 2 Br 6 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

19. Coefficients Index Physical state reactants product Equation is said equivalent if the type and number of atoms reacted substances (reactants) together with the type and number of atoms of the reaction (products) Reactants are written on the left followed by an arrow and then the product 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l )

20. AVOGADRO’S HYPOTHESIS "At the same temperature and pressure, all gases with the same volume will contain the same number of molecules as well." 3

21. Exercise Problem A hydrocarbon compounds (C XHY) by burning gaseous reaction: C x H y (g) + O 2 (g) → CO 2 (g) + H 2 O (g) (not equal) Of the experiment is known that 2 liters of gas to burn C x H y (T, P) required 5 liters of oxygen gas (T, P) and produced 4 liters of carbon dioxide gas (T, P). Determine the molecular formula of the hydrocarbon?

22. Answer Due to volume ratio is a reaction coefficient, then the equation becomes: 2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + .... H 2 O (g) (not equal) For equality oxygen atom, then the coefficient of H 2 O is 2 (10-8), thus the equations become: 2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O (g) For C and H atoms equality as follows. Thus, the hydrocarbon molecular formula is C 2 H 2.

23. Mole Concept Mole definition R elation with the number of moles of particles Molar mass and Molar volume Calculating Mass and volume of Product Kemolaran Solution 4

24. M o le Concept Mole Definition In Chemistry, number of atomic particles or elements that involved in chemical reaction explained by mole. Mole is a unit to express the number of particles . Menu

25. Relationship Between M oles and N umber of P articles T he number of particles in the 12 gram 12 C atom specified based experiment result is 6,02 x 10 23 Relationship B etween the number of moles ( n ) and the number of particles ( x ) can be formulated as follows : x = n × 6,02 × 10 23

26. E A X M P L E How many particles are found in 3 mol of Iron metal? Answer : 1 mol of Iron metal (Fe) = 6,02 × 10 23 3 mol of Iron metal = 3 × 6,02 × 10 23 = 1,806 × 10 24 Menu

27. M O A L R Mass & Volume Molar mass in one mol mass of substance expressed in grams. Molar Mass m = mass (g) m = n × M m n = number of mole (mol) M m = molar mass(g/mol)

28. E A X M P L E Known A r Fe = 56. Calculate the molar mass of Fe ? Answer : Mass of 1 mol of Fe = 1 × 56 = 56 g Thus, Molar Mass of Fe = 56 g/mol.

29. Molar volume expresses volume for each 1 mol of gas. The gas molar volume is 22,4 L at STP and 24,5 L at room temperature. Molar Volume If expressed at STP molar volume is symbolized by V m . Relationship between volume, number of moles, and the gas molar volume is as follows V = Volume of the gas V = n × V m n = Number of moles V m = The gas molar volume

30. When volume of the gas is measured at certain temperature and pressure, then equation used is is ideal gas equation. Mathematically, the ideal gas equation is as follows : P = pressure of the gas (atm) PV = nRT V = volume of the gas (L) n = number of moles V = nRT R = gas constant (0,0821 atm L mol -1 K -1 P T = absolute temperature (K)

31. E A X M P L E At certain temperature and pressure,5 mole of SO 2 gas have 100 L volume. At the same temperature and pressure . What is the volume of 3 mole of No 2 gas?

32. Comparison of mole of SO 2 gas and NO 2 gas = 5 : 3 Volume of 5 mole of SO 2 gas is 100 L then thevolume of 3 mol of NO 2 is Thus, volume of NO 2 gas is 60 L V NO 2 = x 100 L = 60 L   A S N W E R Menu

33. Calculating Mass and Volume of Product There following steps can determine mass or volume of product . a. Calculate the number of moles of elementary unit (atoms, molecules or ions) from elements, compounds, or ions from known mass or volume of subtance . b . Calculate the number of moles of the unknown subtances using subtance coefficient in balanced equation. c. Determine mass of volume of the unknown subtances based on the number of moles calculated.

34. Calculate mass of carbon dioxide gas (CO 2 ) produced when 108 g ethane (C 2 H 6 ) burned in O 2 gas. Answer : Chemical equation : 2C 2 H 6(g) + 7O 2(g) 4CO 2(g) + 6H 2 O (g) known 108 g of C 2 H 6 , Mr C 2 H 6 = 30 and Mr CO 2 = 44 E A X M P L E

35. Three following steps can determine mass of carbon dioxide : Step I : Calculate moles of 108 g of ethane 1 mol of C 2 H 6 = 30 g then 108 g C 2 H 6 = 108 g = 3,6 mol 30 g/mol Step II : Calculate the number of moles of carbon dioxide produced. Based on equation : 2 mol C 2 H 6 produced 4 mol CO 2 . 3,6 mol C 2 H 6 produce 4 x 3,6 mol = 7,2 mol CO 2 2 Step III : Calculate the number of carbon dioxide produced in gram, 1 mol CO 2 = 44 g 7,2 mol CO 2 = 7,2 x 44 g = 316,8 g Thus, the mass of carbon dioxide gas produced in the combustion of 108 g ethane is 316,8 g. Menu

36. Kemolaran solution O ne way to express concentration of solution used in chemistry is kemolaran (M). kemolaran expressed in number of moles of solute per liter of solution, or the amount of solute in mmol per mL of solution . M = Kemolaran solution M = n = number of moles of solute V = volume of solution n V

37. C alculate the number of moles and the mass of urea ( Mr = 60) were present in 200 mL of 0.4 M urea . Answer : determine the mass of solute ,n = V x M number of moles of urea (n) = V x M = 0,2 L x 0,4 Mol L -1 = 0,08 Mol mass urea (m ) = n x M m = 0,08 mol x 60 g mol -1 ` = 4,8 gram E A X M P L E Menu

38. Reaction Stoichiometry S toichiometry coefficients of the reaction as the basis of the reaction Limiting Reagent Chemical Formula of hydrates 5

39. A As the reaction stoichiometry coefficients Elementary Reactions Reaction coefficient is the ratio of particles of substances involved in the reaction. Therefore 1 mole of any substance contains the same number of particles, the reaction coefficient is also a comparison of the number of moles of substances involved in the reaction. Number of moles = (coefficient of substances were asked) / (coefficient of known substances) x number of substances known Reaction stoichiometry

40. Aluminum dissolves in sulfuric acid produce aluminum sulfate and hydrogen gas. 2AL (S) + 3H2SO4 (aq) AL2 (SO4) 3 (aq) + 3H2 (g) How many moles of hydrogen gas can be produced when 0.5 moles of aluminum used EXAMPLE ANSWER

41. In the know: Coefficient substances were = 3 Coefficient of substances known to = 2 The number of moles of Al = 0.5 mol In question : the number of moles of H2 = ...? Number of moles = (coefficient of substances were) / (coefficient of unknown substances) x number of substances known The number of moles of H2 = (coefficient H2) / (coefficient Al) x number of moles of Al = 3/2 x 0.5 mol = 0.75 mol ANSWER Menu

42. Limiting reagent B T he reactans used up first a rection and restricts the running reaction so that no further one called limiting reagent.

43. Methane burns (reacts with oxygen) by the equation: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g ) In an experiment, a total of 8 grams of methane burned with 40 grams of oxygen. (Ar C = 12; H = 1; O = 16) Determine the limiting reagent. EXAMPLE ANSWER

44. In the know : Mass CH4 = 8 grams Mass O2 = 40 grams In question: limiting reagent = ...? The number of moles of methane (CH4) = (8 g) / (16 g / mol) = 0.5 mol The number of moles of oxygen (O2) = (40 g) / (32 g / mol) = 1.25 mol When compared with the coefficient of the reaction, methane is multiplied by the number 0.5 / 1 or 0.5, while the number of oxygen with 1.25 / 2 or 0.625. Because pengalinya smaller, the limiting reagent is methane. ANSWER Menu

45. Chemical formula Hydrant C. Hydrants are solids that bind several molecules of water as part of its crystal structure. British salt, MgSO4, 7H2O: Magnesium sulphate pentahydrate If a hydrate in the heat, some or all of the water crystals can be separated (yawn). CuSO4. 5H2O (S)  CuSO4. (S) + 5H2O (g) If a hydrate is dissolved in water, it will loose its crystalline water. CuSO4. 5H2O (S)  CuSO4. (Aq) + 5H2O (l)

46. A total of 10 grams of hydrated iron (II) sulphate crystals are heated so that all water is evaporated. The remaining solid mass was 5.47 grams. How is the formula that hydrates ? (Ar H = 1; O = 16 S = 32; Fe = 56) EXAMPLE ANSWER

47. Differences existing mass is the mass of the crystal water. For example the amount of water crystals is x, so it is FeSO4.xH2O formula hydrates . FeSO4.xH2O mass = 10 grams   FeSO4 mass = 5.47 grams Mass of water = 10 to 5.47 = 4.53 grams. ANSWER

48. FeSO4.xH2O (s) FeSO4 (aq) + xH2O (g) The number of moles of FeSO4 = (5.47 g) / (152 g / mol) = 0.036 mol The number of moles of H2O = (4.53 g) / (18 g / mol) = 0.252 mol FeSO4 mol: mol of water = 0.036: 0.252 = 1: 7 Meaning, one molecule binds FeSO4 7 water molecules. The formula hydrates is FeSO4.7H2O. Menu

49. Compound of Stoichiometry Composition Percentage Empirical and Molecular Formulas 6

50. Composition percent is determined by dividing mass of each element in one mol of compound with molar mass of compound and multiply by 100 percent. Composition Percentage A Compound stoichiometry Example

51. Answer : Total mass of 1 mol e of C 4 H 10 O derived from its relative molecular mass. Mr C 4 H 10 O = (4 x 12) + (10 x 1) + (1 x 16) = 74 % Mass of X = X 100 % % Mass of C = X 100 % = 64,9 %   Determine percentage of each element in ether anaesthesia , C 4 H 10 O. ( Ar C = 12, H = 1, O = 16).

52. % Mass of H = X 100 % = 13,5 % % Mass of C = X 100 % = 21,6 % Th o s e , percentage of each element in ether anaesthesia is 64, 9% C, 13,5 H, and 21,6 % O   Menu

53. Empirical and Molecular Formulas B A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. Empirical formula is the simplest ratio of atoms that compose a molecule, hence, empirical formula is also called comparison formula.

54. A compound consists of 84 % carbon and 16 % hydrogen. If Ar C = 12, Ar H = 1, and Mr = 100, determine the empirical and molecular formulas of the compound. The number of C atom : the number of H atoms = : = 7 : 16 Th o s e , that compound empirical formula is C 7 H 16 . Its molecular mass is (C 7 H 16 ) n = 100 100 n = 100 n = 1 That compound molecular formula is C 7 H 16   Answer: Menu

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