Data Tabel & Lampiran

Experiment’s Data:

No.

Procedure Hypothesis Observation Conclusion

1. Preparation of double salt CuSO4.5H2O (s) + (NH4)2SO4 (s) + H2O (l) CuSO4.(NH4)2SO4.6H2O (s)

Before:CuSO4.5H2O (s) = blue, 1.24 grams(NH4)2SO4 (s) = white, 0.66 gramsAfter:Both mixed + H2O become clear blue (+++++)After placed in ice bath, blue crystal is formed.Mass of crystal formed = 2.218 gramsMass of crystal after dried:w1 = 0.9908 gramsw2 = 0.9781 gramsw3 = 0.9776 grams

Double salt: CuSO4.(NH4)2SO4.6H2O has mass 0.98213 gthe percentage rendement is 49.17%

2. Preparation of complex salt NH4OH (aq) + CuSO4.5H2O (s) Cu(NH3)4SO4.5H2O (s)Complex salt is formed from two substances which one has role as central atom and the other as ligand.

Before:CuSO4.5H2O (s) = blue, 1.24 gramsAmmonia = colorless, 5mLEthanol = colorlessAfter:CuSO4.5H2O + ammonia = blue turbid (+)Added ethanol = blue

Complex salt: Cu(NH3)4SO4.5H2Ohas mass 0.7226 gthe percentage rendement is 41.35%

turbid (++)After 30 minutes, there are two layers, upper is colorless and bottom is blue (++)Washed by ammonia = blue solid (+++)Mass of crystal formed = 2.4076 gramsMass of crystal after dried:w1 = 0.7288 gramsw2 = 0.7195 gramsw3 = 0.7194 grams

3. Test for double salt With aquadestCuSO4.(NH4)2SO4.6H2O + H2O Cu2+ + 2SO4

2- + 2NH4

+ + 7H2OWith NaOHCuSO4.(NH4)2SO4.6H2O + NaOH NH4OH + Na2SO4

Before:Double salt crystal = blue (+)Aquadest = colorlessHCl = colorlessNaOH = colorlessAfter:Double salt crystal + aquadest = clear blue (+++)Double salt + aquadest = clear blue (++)Double salt + HCl = clear blue (++)Double salt + NaOH =

With aquadest, double salt is ionized well.With HCl, double salt is ionized well.With NaOH, double salt forms new compound.

turbid blue (++)Test for complex salt With aquadest

Cu(NH3)4SO4.5H2O + H2O NH4OH (aq) + CuSO4 (s) + 5H2O (l)

Before:Complex salt crystal = blue (+)Aquadest = colorlessHCl = colorlessNaOH = colorlessAfter:Complex salt crystal + aquadest = turbid blue (+++)Complex salt + aquadest = turbid blue (+)Complex salt + HCl = colorlessComplex salt + NaOH = turbid blue (+)

Addition of aquadest, HCl, and NaOH can cause the changing of ligand in complex salt.

Double salt:CuSO4(NH4)2SO4.6H2O CuSO4 + (NH4)2SO4 + 6H2O ↑Complex salt:Cu(NH3)4SO4.5H2O → CuSO4 + 5H2O + NH3↑

Flame test will give green color because it has cupric anhydrate.Double salt has ionic bond, the melting point is 285°CComplex salt has coordination of covalent bond, the melting point is 360°C

Double:After heated = tosca.After tested by litmus paper = blue color (base)After tested by spatula with HCl = greenMelting point = 250°CComplex:After heated = green.After tested by litmus paper = red color (acid)After tested by spatula with HCl = greenMelting point = 310°C

Flame test of complex salt give green color (cupric).Double salt has base characteristic.Melting point of double salt is 250°C.Complex salt has acid characteristic.Melting point of complex salt is 310°C.

ATTACHMENTCalculation

1. Preparation of double saltn CuSO4.5H2O = 0.005 moles

molar mass CuSO4.5H2O = 249.5 g/molemass CuSO4.5H2O = molar mass CuSO4.5H2O x mole CuSO4.5H2Omass CuSO4.5H2O = 249.5 g/mole x 0.005 molesmass CuSO4.5H2O = 1.2475 grams

n (NH4)2SO4 = 0.005 molesmolar mass (NH4)2SO4 = 132 g/molemass (NH4)2SO4 = molar mass CuSO4.5H2O x mole CuSO4.5H2Omass (NH4)2SO4 = 132 g/mole x 0.005 molesmass (NH4)2SO4 = 0.66 grams

v H2O = 5mLmass H2O = v H2O . ρ H2O = 5mL x 1g/mL = 5 gramsmolar mas H2O = 18 g/molemol H2O = mass H2O /molar mass H2Omol H2O = 5/18 = 0.2778 moles

Reaction:

CuSO4.5H2O + (NH4)2SO4 + H2O CuSO4.(NH4)2SO4.6H2OInitial 0.2778 mol 0.005 mol 0.005 molReaction 0.005 mol 0.005 mol 0.005 mol 0.005 molRemains 0.2728 mol - - 0.005 mol

Mass of reaction is 1.9975 gramsRemain substance of H2O

n CuSO4.(NH4)2SO4.6H2O = 0.005 molesn H2Oremain = 0.2728 molesmass H2Oremain = mol H2O x molar mass H2Omass H2Oremain = 0.2728 moles x 18 g/molemass H2Oremain = 4.9104 gramsv H2Oremain = mass/ρ = 4.9104/1 = 4.9104 mL

Mass of double salt (rendement):w1 = 0.9908 grams w2 = 0.9781 grams w3 = 0.9776 grams average of w = 0.9822 grams% = (average of mass / mass of reaction) x 100%% = (0.9822/1.9975) x 100% % = 49.17%

2. Preparation of complex saltn CuSO4.5H2O = 0.005 moles

molar mass CuSO4.5H2O = 249.5 g/molemass CuSO4.5H2O = molar mass CuSO4.5H2O x mole CuSO4.5H2O

mass CuSO4.5H2O = 249.5 g/mole x 0.005 molesmass CuSO4.5H2O = 1.2475 grams

v NH4OH = 5mLρ NH4OH = 0.91 g/mLmass NH4OH = v NH4OH . ρ NH4OH = 5mL x 0,91 g/mL = 4.55 gramsmolar mas NH4OH = 35 g/molemol NH4OH = mass NH4OH /molar mass NH4OH = 4.55/35 = 0.13 moles

Reaction:

CuSO4.5H2O + NH4OH Cu(NH3)4SO4.5H2OInitial 0.005 mol 0.13 molReaction 0.005 mol 0.005 mol 0.005 molRemains - 0.125 mol 0.005 mol

Mass of reaction is 1.7475 gramsRemain substance of NH4OH

n CuSO4.(NH4)2SO4.6H2O = 0.005 molesn NH4OH remain = 0.125 molesmass NH4OH remain = mol NH4OH x molar mass NH4OHmass NH4OH remain = 0.125 x 35 = 4.375 grams

Mass of complex salt (rendement):w1 = 0.7288 grams w2 = 0.7195 grams w3 = 0.7194 gramsaverage of w = 0.7226 grams% = (average of mass / mass of reaction) x 100%% = (0.7226/1.7175) x 100% % = 41.35%

ATTACHMENTAnswer of Question

1. Calculate the percentage result from experiment 1 and 2!Answer:The percentage result from experiment 1 is:% = (average of mass / mass of reaction) x 100%% = (0.9822/1.9975) x 100% % = 49.17%The percentage result from experiment 2 is:% = (average of mass / mass of reaction) x 100%% = (0.7226/1.7175) x 100% % = 41.35%

2. Write the reactions from experiment 1, 2 and 3!Answer:Experiment 1:CuSO4.5H2O (s) + (NH4)2SO4 (s) + H2O (l) CuSO4.(NH4)2SO4.6H2O (s)Experiment 2:NH4OH (aq) + CuSO4.5H2O (s) Cu(NH3)4SO4.5H2O (s)Experiment 3:For double salt + aquadestCuSO4.(NH4)2SO4.6H2O + H2O Cu2+ + 2SO4

2- + 2NH4+ + 7H2O

For complex salt + NaOHCuSO4.(NH4)2SO4.6H2O + NaOH NH4OH + Na2SO4For double salt + NaOH

3. Explain the different characteristics between double salt and complex salt according to 3rd experiment!Answer:Adding aquadest in complex salt will make the complex salt become its origin

compound (back reaction), while adding aquadest in double salt will make the compound containing is ionized well.

Adding HCl in complex salt will make the ligand of complex salt’s solution change, while adding HCl in double salt will make the compound containing is ionized well.

Adding NaOH in complex salt will make the ligand of complex salt’s solution change, while adding NaOH in double salt will make the compound containing is forming new compound.

By heating both salts, we know that the complex salt has base characteristic while the double salt has acid characteristic.

By testing spatula that contains of HCl, both of salts give green color, means that both salt has cupric for its contain.

The melting point of complex salt is higher than the melting point of double salt, as complex salt has coordination covalent bond while double salt has ionic bond.

4. What is the melting point of double salt and complex salt from your synthesis? Compare the melting point of double salt and complex salt according to theory! If it has differences, what is the reason? Explain!Answer:The melting point of double salt is 250°C and the melting point of complex salt is

310°C. both of them is appropriate with the theory that say the melting point of complex salt is higher than the melting point of double salt. This is because the bonding of complex salt makes the more attractive bond than the bonding of double salt. So, when both of them is melted, the salt that has stronger bond will be difficult to be broken. It make the melting point of complex become higher as to broke its bond needs higher energy.

ATTACHMENTCalculation

The substances that using:

CuSO4.5H2O (s) dan (NH4)2SO4 (s)

Preparation of complex and double salt:

Complex salt

Double salt

Test for complex salt’s solution

Before: After:

Test for double salt’s solution

Before: After:

Heating process: Test the melting point:

Before Heating

After tested by litmus paper