- #1

nineteen

- 68

- 12

## Homework Statement

In an organic compound "A" only C,H and O is present. 1.22g of A is completely burned and it gives 0.84g of CO

_{2}and 0.54g of H

_{2}O. If the relative molecular mass of "A" is 123. Find the molecular formula of A.

[/B]

*I have showed my attempt at solving this problem, but I think it is wrong. Please help me out here. After what I have done last, there are only a few more steps to find the empirical formula and then the molecular formula itself, but I think what I have done is wrong. Please try and correct me, I'd appreciate it a lot.*

2. Homework Equations

2. Homework Equations

no. of moles = Mass/Molar mass

n of the molecular formula = Molar mass/ Formula mass of the empirical formula

(I guess that those are the only equations)

n of the molecular formula = Molar mass/ Formula mass of the empirical formula

(I guess that those are the only equations)

## The Attempt at a Solution

[/B]

Mass of C in CO

_{2}produced: (12/44) x 0.84g = 0.22g

Mass of O in CO

_{2}produced: (32/44) x 0.84g = 0.61g

Mass of H in H

_{2}O produced: (2/18) x 0.54g = 0.06g

Mass of O in H

_{2}O produced : (16/18) x 0.54g = 0.48g

Total masses;

C = 0.22g

O = 1.09g

H = 0.06g

Molar ratios of elements;

C ----> (0.22/12) = 0.01

H ----> (0.06/1) = 0.06

O ----> (1.09/16) = 0.06

Dividing all by the smallest value;

C ----> 0.01/0.01 = 1

H ----> 0.06/0.01 = 6

O ----> 0.06/0.01 = 6

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