Respuesta :
Answer
The empirical formula for the compound is C₂H₄O₁
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.
[tex]\begin{gathered} 3.54g\text{ }C\times\frac{1mol\text{ }C}{12.01g\text{ }C}=0.2948mol\text{ }C \\ \\ 0.59g\text{ }H\times\frac{1mol\text{ }H}{1.008g\text{ }H}=0.5853mol\text{ }H \\ \\ 2.37g\text{ }O\times\frac{1mol\text{ }O}{15.998g\text{ }O}=0.1481mol\text{ }O \end{gathered}[/tex]Step 2: Divide each mole by the smallest number of moles (0.1481 mol)
[tex]\begin{gathered} C=\frac{0.2948mol}{0.1481mol}=2 \\ \\ H=\frac{0.5853mol}{0.1481mol}=4 \\ \\ O=\frac{0.1481mol}{0.1481mol}=1 \end{gathered}[/tex]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:
[tex]C_2H_4O_1[/tex]The empirical formula for the compound is C₂H₄O₁
