CHEMISTRY FORM 4

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EMPERICAL FORMULA AND MOLECULAR FORMULA Emperical Formula : The chemical formula that shows the simplest whole number ratio of atoms of each

element in the compound

Element A B

Mass of atom x y

Mole of atom x÷Relative atomic mass of A y÷Relative atomic mass of B

Mole Ratio p q

Emperical formula : ApBq Molecular Formula : The chemical formula that shows the actual number of atoms of each element that are

present in a molecule of the compound.

Molecular formula = (Emperical Formula)n

n is a positive integer 1. 6.0 g of carbon combined with 1.0 g of hydrogen to form organic compound with its relative molecular mass

28. Find emperical formula and the molecular formula of the compound. [Relative atomic mass, C=12 ; H=1] [EF=CH2][MF= C2H4] 2. 9.6 g of carbon combined with 1.6 g of hydrogen to form organic compound with its relative molecular mass

56. Find emperical formula and the molecular formula of the compound. [Relative atomic mass, C=12 ; H=1] [EF=CH2][MF= C4H8]

CHEMISTRY FORM 4

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3. The information below is regarding an organic compound. The percentage of each element according to mass

in the compound are as followed: [Relative atomic mass, C=12 ; H=1]

Carbon = 85.7% Hydrogen = 14.3 % Relative molecular mass = 42

Base on the information, determine the emperical formula and molecular formula for the compound. [EF=CH2][MF= C3H6] 4. An organic compound R, has a relative molecular mass of 46. The percentage of composition by mass of each

element are: Carbon = 52.2%, Hydrogen = 13.0 % and oxygen = 34.8 %. Find the emperical formula and the molecular formula of the molecule.

[Relative atomic mass, C=12 ; H=1 ; O=16] [EF=C2H6O][MF= C2H6O] 5. 3.12 g of element X combines with 0.64 g of oxygen to form a compound with the emperical formula X2O.

What is the relative atomic mass of X? [Relative atomic mass, O=16] [Relative atomic mass= 39]

CHEMISTRY FORM 4

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6. It is found that y g of metal M combines with 0.96 g of oxygen to form a compound with emperical formula

M2O3. Find the value of y. [Relative atomic mass, O=16 ; M=27] [y = 1.08 g] 7. 25 cm3 nitrogen gas and 50 cm3 of oxygen gas which are measures at room temperature and pressure reacted

completely to produced and oxide of nitrogen. Determine the emperical formula of the oxide nitrogen. [Relative atomic mass, O=16 ; N=14 ; 1 mole gas occupies a volume 24 dm3 at room temperature and

pressure] [EF = NO2] 8. A compound CxHyOz contains 40.0% of carbon, 6.7% of hydrogen and 53.3% of oxygen by mass. If the relative

molecular mass of the compound is 180, finds its emperical formula and molecular formula. [Relative atomic mass, C=12 ; H=1 ; O=16] [EF=CH2O][MF= C6H12O6]

CHEMISTRY FORM 4

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9. A organic compound contains 12.77% of carbon, 2.13% of hydrogen and 85.10% of bromine by mass. Its

molecular mass is 188. Find its emperical formula and molecular formula. [Relative atomic mass, C=12 ; H=1 ; Br=80] [EF=CH2Br][MF= C2H4Br2] 10. Nicotine has molar mass 162. Its contains 74.1 % of carbon, 8.64 % of hydrogen and the rest is nitrogen.

Determine the emperical formula and molecular formula of nicotine. [Relative atomic mass, C=12 ; H=1 ; N=14] [EF=C5H7N][MF= C10H14N2]