An unknown organic compound containing carbon, hydrogen, and oxygen, is subjected to combustion analysis. When 3.498 g of the unknown compound is combusted, 7.950 g of CO2 and 3.255 g of H2O is formed. What is the molecular formula of the unknown compound if its molar mass is 174.27 g/mol?

Molecular formula for the organic compound is:  C₉H₁₈O₃

Explanation:

We determine the combustion reaction:

CₓHₓOₓ  +  O₂  →  CO₂  +  H₂O

We don't know the subscrip of the organic compound, that's why we put x.

Notice we have C in the compound and in CO₂. Let's find out the mass of C.  1 mol of CO₂ contains 1 mol of C

Then (7.950 g /44g/mol) of CO₂ may contain ?

(7.950 g /44g/mol) represents the moles of dioxide → 0.181 moles of C

These moles came from the mass of the organic compound, the 3.498 g.

So now, we can know the subscript for C.

If 3.498 g of the compound contain 0.181 moles of C

174.27 g of the compound may contain (174.27 .  0.181)/ 3.498 = 9

Let's do the same for the H.

1 mol of water contain 2 moles of H

(3.255 g /18 g/mol) of water, may contain  ?

3.255 g /18 g/mol = 0.181 moles . 2 = 0.362 moles of H

These moles came from the mass of the organic compound, the 3.498 g.

Let's find out the subscript for H

If 3.498 g of the compound contain 0.362 moles of H

174.27 g of the compound may contain (174.27 .  0.362)/ 3.498 = 18

Organic compound is C₉H₁₈Oₓ. If the molar mass is 174.27 g we can easily determine the moles of O

12 g/mol . 9 + 1 g/mol . 18 + 16 g/mol . X = 174.27 g/mol

(174.27 g/mol - 12 g/mol . 9 - 1 g/mol . 18 ) / 16 = 3

Molecular formula for the organic compound is:  C₉H₁₈O₃.  Therefore the complete combustion reaction is:

C₉H₁₈O₃  +  12O₂  →  9CO₂  +  9H₂O

IDK sorry

Explanation: