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A compound contains only carbon, hydrogen and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg of $C{O_2}$ and 4.37 mg ${H_2}O$. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

Last updated date: 13th Jun 2024
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Hint: First find out the moles of $C{O_2}$ and ${H_2}O$ released due to combustion. Using them to find out the moles of carbon, hydrogen and oxygen involved in the reaction, this stoichiometry in which they are present will give us the empirical formula. Now check how many times the empirical weight gives us the molecular weight. This will lead us to the molecular formula.

Complete Solution :
-Since this compound contains only carbon, hydrogen and oxygen its molecular formula would be somewhat like: ${C_x}{H_y}{O_z}$. During combustion such compounds release only carbon dioxide and water, so the combustion reaction for this compound will be:
${C_x}{H_y}{O_z} + {O_2} \to xC{O_2} + \dfrac{y}{2}{H_2}O$

-According to the question this compound releases 16.01mg of $C{O_2}$ and 4.37mg ${H_2}O$. We will first find out the number of moles associated with them.
For $C{O_2}$: Molecular weight = 44 g/mol
moles = given weight / molecular weight
= $\dfrac{{16.01 \times {{10}^{ - 3}}}}{{44}}$
= 0.364 m moles

For ${H_2}O$: Molecular weight = 18 g/mol
moles = given weight / molecular weight
= $\dfrac{{4.37 \times {{10}^{ - 3}}}}{{18}}$
= 0.243 m moles

 -0.364 m moles of $C{O_2}$, contains 0.364 m moles of carbon, which can be written as:
Weight of carbon = 0.364 × 12
= 4.368 mg

-And in 0.243 m moles of ${H_2}O$, amount of hydrogen present will be = 0.243 × 2 m moles
 Weight of hydrogen = 0.243 × 2 × 1
= 0.486 mg

-Since the total weight of the unknown compound is given in the question to be 10.68 mg, out of which 4.368mg is carbon and 0.486mg is hydrogen. We can calculate the amount of oxygen present.
Weight of oxygen = 10.68 – (4.368 + 0.486)
 = 10.68 – 4.854
 = 5.826 mg
So, moles of oxygen = $\dfrac{{5.826}}{{16}}$ = 0.364 m moles
-Now to find out the empirical formula of the compound let us take the ratio of moles of all the constituent elements.

0.364 0.486 0.364
1 1.34 1
0 4 3

Hence the empirical formula of this unknown compound will be: ${C_3}{H_4}{O_3}$
And the weight of this empirical formula is = 36 + 4 + 48 = 88 g/mol
-The question gives us the molecular weight of the molecular formula = 176.1 g/mol
This molecular weight is twice the weight of the empirical formula, so to find out the molecular formula we will multiply 2 to the stoichiometry of the empirical formula.
So, the molecular formula would be: ${C_6}{H_8}{O_6}$
Finally, the empirical formula is ${C_3}{H_4}{O_3}$ and the molecular formula is ${C_6}{H_8}{O_6}$.

Note: The empirical formula is the simplest whole number ratio in which the elements are present. Molecular formulas can be any multiple of this whole number ratio, but always remember that the constituents will always remain in the stoichiometry of the empirical formula.