Question

An unknown compound contains only C, H, and O. Combustion of 8.00 g of this compound produced 16.0 g CO₂ and 6.54 gH₂O. What is the empirical formula of the unknown compound? Insert subscripts as needed.empirical formula: CHO

Answer 1

The empirical formula of the unknown compound is C₂H₂O₁

Step-by-step explanation

Given:

Mass of the unknown compound = 8.00 g

Mass of CO₂ produced = 16.0 g

Mass of H₂O produced = 6.54 g

What to find:

The empirical formula of the unknown compound.

Step-by-step solution:

Step 1: Determine the moles of CO₂ and H₂O in 16.0 g CO₂ and 6.54 g

H₂O.

The molar mass of CO₂ = 44.01 g/mol

The molar mass of H₂O = 18.02 g/mol

The moles can be calculated using the mole formula:

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Moles\text{ }of\text{ }CO₂=\frac{16.0g}{44.01g\text{/}mol}=0.364\text{ }mol \\ \\ Moles\text{ }of\text{ }H₂O=\frac{6.54g}{18.02g\text{/}mol}=0.363\text{ }mol \end{gathered}`

Step 2: Determine the mole fraction of C and H

`\begin{gathered} 0.364\text{ }mol\text{ }CO_2*\frac{1\text{ }mol\text{ }C}{1\text{ }mol\text{ }CO_2}=0.364\text{ }mol\text{ }C \\ \\ 0.363\text{ }mol\text{ }H_2O*\frac{2\text{ }mol\text{ }H}{1\text{ }mol\text{ }H_2O}=0.726\text{ }mol\text{ }H \end{gathered}`

Step 3: Determine the mass of C and H in moles of C and H in step 2.

The molar mass of C = 12.01 g/mol

The molar mass of H = 1.01 g/mol

`\begin{gathered} Mass=Moles* Molar\text{ }mass \\ \\ Mass\text{ }of\text{ }C=0.364mol*12.01g\text{/}mol=4.37\text{ }g \\ \\ Mass\text{ }of\text{ }H=0.726mol*1.01g\text{/}mol=0.733\text{ }g \end{gathered}`

Step 4: Determine the mass of O in the unknown.

Since the mass of C and H in the unknown compound have been determined to be 4.37 g and 0.733 g respectively, then the mass of O can be calculated as follows:

`\begin{gathered} Total\text{ }Mass=Mass\text{ }of\text{ }C+Mass\text{ }of\text{ }H+Mass\text{ }of\text{ }O \\ \\ 8.00g=4.37g+0.733g+Mass\text{ }O \\ \\ Mass\text{ }of\text{ }O=8.00g-5.103g \\ \\ Mass\text{ }of\text{ }O=2.897\text{ }g \end{gathered}`

Step 5: Convert the grams of O in step 4 to moles.

The molar mass of O = 16.00 g/mol

`Moles\text{ }of\text{ }O=\frac{2.897g}{16.00g\text{/}mol}=0.181\text{ }mol`

Moles ratio: 0.364 mol C, 0.363 mol H, 0.181 mol O

To get the empirical formula for the unknown compound, divide each mole of the element by the smallest mole number:

`\begin{gathered} C:\frac{0.364\text{ }mol}{0.181\text{ }mol}=2 \\ \\ H:\frac{0.363\text{ }mol}{0.181\text{ }mol}=2 \\ \\ O:\frac{0.181\text{ }mol}{.0181\text{ }mol}=1 \end{gathered}`

Therefore, the empirical formula of the unknown compound is C₂H₂O₁