GTF112 Inorganic Chemistry lab report EXP 1 EMPIRICAL FORMULA PDF

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Find the empirical formula of zinc chloride and copper sulfide...

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Name: Hazirah binti Hashim Matrix Number: 157667 Title: Experiment 1 Chemical Formulas A. Zinc Chloride 1. Mass of evaporating dish and zinc

: 40.7114 g

2. Mass of evaporating dish

: 40.2948 g

3. Mass of zinc

: 0.4766 g

4. Mass of evaporating dish and zinc chloride First determination

: 41.3245 g

Second determination:

-

Third determination

-

:

5. Mass of zinc chloride

: 1.0297 g

6. Mass of chlorine in zinc chloride

: 0.5531 g

7. Empirical formula for zinc chloride Molar mass of Zn is 65.38 g/mol,

Number of moles of Zn=0.4766 g Zn ×

1 mol Zn 65.38 g Zn

¿ 0.0073 mol Zn

Molar mass of Cl is 35.453

g/mol,

Number of moles of Cl=0.5531 g Cl ×

Ratio of moles,

Zn :

0.0073 =1.0 0.0073

1mol Cl 35.453 g Cl Cl :

¿ 0.0156 mol Cl

0.0156 ≈ 2.0 0.0073

The ratio of Zn:Cl in integer of 1:2. Therefore, the empirical formula of zinc chloride is ZnCl2. 8. Balance chemical equation for the formation of zinc chloride from zinc and HCl.

Zn (s)+2 HCl (aq )→ Zn Cl2 ( aq )+ H 2(g )

B. Copper sulfide 1. Mass of crucible, cover and copper

: 37.5267 g

2. Mass of crucible and cover

: 35.9797 g

3. Mass of copper

: 1.5497 9

4. Mass of crucible, cover and copper sulfide : First determination

: 38.0122 g

Second determination : 37.9656 9 Third determination

: 37.9229 g

5. Mass of copper sulfide

: 1.9432 g

6. Mass of sulphur in copper sulpfide

: 0.3935 g

7. Empirical formula for copper sulfide Molar mass of Cu is 63.546 g/mol,

Number of moles of Cu=1.5497 g Cu ×

1mol Cu 63.546 g Cu

¿ 0.0244 mol Cu Molar mass of S is 32.065 g/mol,

Number of moles of S=0.3935 g S ×

1 mol S 32.065 g S

¿ 0.0123 mol S Ratio of moles,

Cu :

S:

0.0244 ≈ 2.0 0.0123

0.0123 =1.0 0.0123

The ratio of Cu:S in integer of 2:1. Therefore, the empirical formula of copper sulfide is Cu2S. 8. Balanced chemical equation for the formation of copper sulfide from copper and sulphur.

2Cu (s)+ S(s) →Cu2 S (s )

QUESTIONS (Practical – Chemical Formula) 1. Can you determine the molecular formula of a substance from its percent composition?

Answer: No. We can only deduce the empirical formula of the substance from its percent composition. However, the molecular formula of the substance can be determined from the empirical formula if the approximate molar mass is known.

2. Given that zinc chloride has a formula weight of 136.28 amu, what is its formula? Answer: Zn= 65.38 amu

Cl= 35.45 amu

1 Zn + 1 Cl = 1(65.38 amu) + 1(35.45 amu) = 100.83 amu 136.28 amu – 100.83 amu = 35.45 amu The remaining mass left is approximately the mass of Cl. Therefore, zinc chloride has 1 zinc and 2 chlorines. The formula of zinc chloride is ZnCl2.

3. Can you determine the atomic weights of zinc or copper by the methods used in this experiment? If so, how? What additional information is necessary to do this? Answer: Yes. The additional information needed are the molar mass of sulphur, hydrogen and chlorine. Find the mole ratio of the two reaction and let the atomic weight of zinc as x g/mol and copper as y g/mol.

4. How many grams of zinc chloride could be formed from the reactions of 3.57 g of zinc with excess HCl? Answer: Zn + 2HCl → ZnCl2 + H2 Mass of Zn= 3.57 g

Molar mass of Zn= 65.38 g/mol

Mass of ZnCl2=?

Molar mass of ZnCl2= 136.285 g/mol

Mass of Zn Cl2=3.57 g Zn×

1mol Zn 1 mol ZnCl 2 136.285 g Zn Cl2 × × 1 mol ZnCl 2 1 mol Zn 65.38 g Zn

¿ 7.44 g Zn Cl2

5. Magnesium reacts directly with nitrogen gas to form magnesium nitride. In an experiment conducted, it was found that 0.36 g of magnesium produces 0.50 g of magnesium nitride. (a) How many moles of nitrogen combined with 0.36 g of magnesium? (b) What is the ratio of nitrogen atoms to magnesium atoms in magnesium nitride? (c) What is the empirical formula of magnesium nitride? Answer: a) Mass of nitrogen= 0.50 g – 0.36 g = 0.14 g Molar mass of nitrogen gas is 28 g/mol,

Moles of nitrogen=0.14 g nitrogen×

1 mol nitrogen 28 g nitrogen

¿ 0.005 mol nitrogen b) Percentage of magnesium and nitrogen in magnesium nitride

0.36 % magnesium= ×100=72 % 0.50 % nitrogen=

0.14 ×100=28 % 0.50

Molar mass of magnesium is 24 g/mol,

Magnesiumatom :72 g ×

1 mol =3 mol 24 g

Molar mass of nitrogen is 14 g/mol,

Nitrogen atom :28 g ×

1 mol =2 mol 14 g

The ratio of nitrogen atoms to magnesium atom in magnesium nitride is 2:3. c) Empirical formula of magnesium nitride is Mg3N2

6. When copper(I) sulfide is partially roasted in air (reaction with

O 2 ), copper(I)

sulfite is formed first. Subsequently, upon heating, the copper sulfite thermally

decomposes to copper(I) oxide and sulfur dioxide. Write the balanced chemical equations for these two reactions. Answer:

2Cu2 S (s)+ 3O 2 (g)→2 Cu2 SO 3 (s ) Cu2 SO 3 (s) heat Cu2 O (s )+ SO 2 (g) →...