## Sunday, March 27, 2011

### RELATIONSHIP MOLE AND MOLECULE

1 mol "molecule" = 6.02 x 1023 molecule
number of molecule = 6.02 x 1023/mol

1 mole of substances - "amount" / "number of particles" in substances.

Example:
1 mole of Zn - number of Zn atom in Zinc

## Friday, March 25, 2011

Combustion 0.202 g of a sample of organic compound containing elements carbon, hydrogen and oxygen to produce carbon dioxide 0.361 g and 0.147 g of water. If the relative molecular mass is 148, determine the molecular formula.

mass of carbon in carbon dioxide = 12 x 0.361 / 44 = 0.0984 g
mass of hydrogen in water = 2 x 0.147 / 18 = 0.0163 g
mass  of oxygen in compound = 0.202 - (0.0984 + 0.0163) = 0.0873 g

 Carbon, C Hydrogen , H Oxygen , O Mass (g) 0.0984 0.0163 0.0873 Number of mole (mol) 0.0082 0.0163 0.0055 Mole ratio 0.0082/0.055 = 1.5 0.0163/0.0055 = 3 0.0055/0.0055 = 1 Simplest ratio 3 6 2 Empirical formula C3H6O2

Molecular formula =
(C3H6O2)n = 148
3(12) + 6(1) + 2(16) = 148
74n = 148
n = 2

molecular formula = C6H12O4

## Saturday, March 19, 2011

### PERINGATAN : TIPS UNTUK CALON SPM

Bila anda menjawab soalan mengenai chemical formula.

Selalunya soalan yang akan di tanya.
[SOALAN LAZIM | FOVOURITE QUESTION | SPOT QUESTION]

1. Definisi "empirical formula".
Question : What is meant by empirical formula?
Answer : Formula shows the simplest ratio of each elements in a chemical compound.

* ingat : simplest ratio of each elements - "nisbah terkecil bagi setiap element dalam sebatian kimia."

2. Menentukan empirical formula untuk suatu bahan dengan jisim yang diberi.
Question : Determine the empirical formula of magnesium oxide with mass of magnesium ribbon is 2.4 g and oxygen in magnesium oxide is 1.6 g.

 Magnesium, Mg Oxygen, O Mass (g) 2.4 1.6 Number of mole (mol) 2.4 / 24 = 0.1 1.6 / 16 = 0.1 Mole ratio 0.1/0.1 0.1/0.1 Simplest ratio 1 1 Empirical formula MgO

*ingat :
1. Buat jadual untuk setiap element(unsur) yang terkandung dalam sebatian. [ Penting ! Pengiraan melibatkan satu atom sahaja].
2. Cari bilangan mol setiap element(unsur) . [ Bahagi dengan jisim atom relatif // divide with RAM]
3. Cari nisbah mol (mole ratio) setiap element (unsur) . [ Bahagi dengan bil. mol terkecil antara unsur yang terlibat // choose smallest no. of mole between no of mole in step 2.
4. Bandingan nisbah mol kepada nisbah yang paling kecil. [ Simplify and round off to the simplest whole number ]
5. Tuliskan empirikal formula mengikut bilangan nisbah tadi. [ Write empirical formula by the simplest ratio ]

3. Menentukan empirikal formula daripada peratus bahan yang terdapat dalam sebatian kimia.
Question : Determine the empirical formula of substance X consists 82.75% of carbon and 17.25% of hydrogen by mass?

 Carbon, C Hydrogen , H Assume percentage(%) as Mass (g) 82.75 17.25 Number of mole (mol) 82.75/12 = 6.9 17.25 / 1 = 17.25 Mole ratio 6.9 / 6.9 = 1 17.25 / 6.9 =  2.5 Simplest ratio 2 5 Empirical formula C2H5
*ingat:
1. Buat jadual dengan column mengikut bilangan element yang terdapat dlm sebatian. [ kira one atom jek]
2. Anggap 100% sama dengan 100 g. Oleh itu kita boleh anggap peratus yang diberi adalah jisim bagi setiap element yang terdapat dalam sebatian.
3. Baru boleh cari bilangan mol setiap element.
4. Cari nisbah mol (mole ratio) setiap element.
5. Cari nisbah terkecil untuk bandingkan element.
6. Tulis empirical formula berdasarkan nisbah terkecil yang diperolehi.

## Friday, March 18, 2011

### QUESTIONS : EMPIRICAL / MOLECULAR FORMULA

Question 1 :
A compound was analyzed and found to contain 13.5 g calcium, Ca; 10.8 g oxygen, O and 0.675 g H. What is the empirical formula of the compound?
[ RAM : Ca=40 , O=16 , H=1 ]

Solution:

Step 1 : Given ...
Mass of calcium, Ca = 13.5 g
Mass of oxygen, O = 10.8 g
Mass of hydrogen, H = 0.675 g

Step 2 : Find number of mole each elements.
Mole of Ca = 13.5 / 40 = 0.3375
Mole of O = 10.8 / 16 = 0.675
Mole of H = 0.675 / 1  = 0.675

Step 3 : Simplest mole ratio each elements.
Mole ratio of Ca = 0.3375 / 0.3375 = 1
Mole ratio of O = 0.675 / 0.3375 = 2
Mole ratio of H = 0.675 / 0.3375 = 2

Step 4 : Write the empirical formula
Empirical formula = CaO2H2Ca(OH)2

Question 2 : From table result for an experiment below. Find empirical formula of magnesium oxide?
Result of experiment:
 Mass of crucible + lid 24.0 g Mass of crucible + lid + magnesium ribbon 26.4 g Mass of crucible + lid + magnesium oxide 28.0 g
[ RAM : Mg=24 ; O=16]

Solution:
 Magnesium, Mg Oxygen, O Mass (g) 26.4 - 24.0 = 2.4 28.0 - 26.4 = 1.6 Number of mole (mol) 2.4 / 24 = 0.1 1.6 / 16 = 0.1 Mole ratio 0.1/0.1 0.1/0.1 Simplest ratio 1 1 Empirical formula MgO

## Thursday, March 17, 2011

### Relationship between molecular formula and empirical formula

Empirical formula the simplest ratio of an element in molecular formula
(Empirical formula)n = Molecular compound

Example of RELATIONSHIP:

 COMPOUND MOLECULAR FORMULA EMPIRICAL FORMULA VALUE OF n Glucose C6H12O6 CH2O 6 Carbon dioxide CO2 CO2 1 Water H2O H2O 1 Benzene C6H6 CH 6 Hydrogen Peroxide H2O2 HO 2 Butene C4H8 CH2 2 Butane C4H10 C2H5 2 Ethane C2H6 CH3 2

Example Questions :

(a) What empirical formula of hexane, C6H14?

Solution :
Step 1:
Find simplest ratio between carbon and hydrogen in compound.
[ Use fractional of each element]
= C6/2 H14/2

Step 2:
Divisible is value of n; n=2   ⇒ (empirical formula) n = molecular formula
= (C3H7)2

Step 3:

Empirical formula
= C3H7

### CHEMICAL FORMULA

CHEMICAL FORMULA - (FORMULA KIMIA) : is a way to representation chemical compound.

EXAMPLE :

Hydrogen atom - H
Hydrogen gas - H2
Hydrogen ion - H+

Letter-shows an elements. Each elements started with Capital letter. If have two letter, follow with small letter.

EXAMPLE :
Chlorine - Cl
Helium - He
Magnesium - Mg
Nickle - Ni
Neon - Ne

Subscript numbering-shows the number of elements in an compound.

The number of elements shows actual number ratio and exactly in chemical compound called molecular formula.
The number of elements shows simplest number ratio in chemical compound is called empirical formula.

EXAMPLE:
MOLECULAR FORMULA
carbon dioxide, CO2
nitrogen dioxide, NO2
carbon tetrachloride, CCl4
glucose, C6H12O6

EMPIRICAL FORMULA
carbon dioxide, CO2
nitrogen dioxide, NO2
carbon tetrachloride, CCl4
glucose, CH2O

Note : empirical formula like as simplify fractions.
Simplify molecular formula to empirical formula(simplest ratio).

TO BE CONTINUE... (relationship between molecular formula with empirical formula)