Compound A has the following percentage composition by mass: Carbon 26.7 %, Oxygen 71.1%, Hydrogen 2.2%. Determine the empirical formula of A (H=1, C=12, O=16). If the relative molecular mass of A is 90, what is the molecular formula of A?

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Solution

Step 1: Find the empirical formula:
Element Percentage weight Atomic mass Relative no. of moles Simple ratio of atoms
C 26.7 12 26.7/12= 2.2 2.2/2.2=1
H 2.2 1 2.2/1= 2.2 2.2/2.2=1
O 71.10 16 71.10/16=4.4 4.4/ 2.2= 2
Therefore, the empirical formula is CHO₂.
Step 2: Calculate the molecular formula
Empirical formula mass = 12 + 2 x 16 + 1 x 1
= 12 + 32 + 1 = 45
Molecular mass of A = 90
n = molecular mass / Empirical formula mass
= 90 / 45 = 2
Therefore, Molecular formula = n x Empirical formula
= 2 x C₁H₁O₂
= C₂H₂O₄
Step 3: Result
Hence, the empirical formula is CHO₂_, and the molecular formula is C₂H₂O₄.

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(a) The compound A has the following percentage composition by mass:

Carbon $26.7 %$ , oxygen $71.1 %$ , hydrogen $2.2 %$ . Determine the empirical formula for A (Answer to one decimal place)

$[H = 1, C = 12, O = 16]$

A hydrocarbon has the following percentage composition:

Hydrogen $= 2.2 %$ , Carbon $= 26.6 %$ and oxygen $= 71.2 %$ .

Calculate the empirical formula of the compound. If its molecular mass is $90$ , find its molecular formula.

$[C = 12, H = 1, O = 16]$

The molecular mass of an organic compound is 90 and its %age composition is C- 26.6%; O=71.1%

and H=2.2%. Determine the molecular formula of the compound.

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Element	Percentage weight	Atomic mass	Relative no. of moles	Simple ratio of atoms
C	26.7	12	26.7/12= 2.2	2.2/2.2=1
H	2.2	1	2.2/1= 2.2	2.2/2.2=1
O	71.10	16	71.10/16=4.4	4.4/ 2.2= 2