Question

3. A compound contains only C, H, and O. A 6.50 g sample is analyzed to show 3.54g C and 0.59 g H, and the rest is oxygen. Determine the empirical formula for this compound.

Answer 1

The empirical formula for the compound is C₂H₄O₁

Step-by-step explanation

Given:

Mass of sample = 6.50 g

Mass of C = 3.54 g

Mass of H = 0.59 g

Mass of O = Mass of sample - (Mass of C + Mass of H) = 6.50 g - (3.54 g + 0.59 g) = 2.37 g

What to find:

The empirical formula for the compound.

Step-by-step solution:

Step 1: Determine the mole of each element present.

`\begin{gathered} 3.54g\text{ }C*\frac{1mol\text{ }C}{12.01g\text{ }C}=0.2948mol\text{ }C \\ \\ 0.59g\text{ }H*\frac{1mol\text{ }H}{1.008g\text{ }H}=0.5853mol\text{ }H \\ \\ 2.37g\text{ }O*\frac{1mol\text{ }O}{15.998g\text{ }O}=0.1481mol\text{ }O \end{gathered}`

Step 2: Divide each mole by the smallest number of moles (0.1481 mol)

`\begin{gathered} C=(0.2948mol)/(0.1481mol)=2 \\ \\ H=(0.5853mol)/(0.1481mol)=4 \\ \\ O=(0.1481mol)/(0.1481mol)=1 \end{gathered}`

Step 3: Determine the empirical formula for the compound by using the mole ratio as the subscript.

Therefore, the empirical formula for the compound is:

`C_2H_4O_1`

The empirical formula for the compound is C₂H₄O₁