I LOVE CHEMIST (SPM LIKE Q)

SPM FAVOURITE QUESTIONS

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

Advanced Empirical Formulae

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.

ANSWER:

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

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.
Answer :

	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? 
Answer :

	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.

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 = CaO₂H₂ ≈ Ca(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

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]
= C_6/2 H_14/2

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)