Question

una muestra de un compuesto desconocido tiene una composición porcentual de 52,14% de carbono, 13,13% de hidrógeno y 34,73% de oxígeno. ¿qué compuestos podrían ser la muestra? ch₃ch₃ch₂oh₂ do₂h₅oh do₄h₁₀oh₂ do h oh

Explanation:

Step1: Calculate molar - mass ratios

Assume 100g of the compound. So, we have 52.14g of C, 13.13g of H, and 34.73g of O.
The molar - mass of C is 12.01 g/mol, of H is 1.008 g/mol, and of O is 16.00 g/mol.
The number of moles of C: $n_C=\frac{52.14g}{12.01g/mol}\approx4.342mol$.
The number of moles of H: $n_H=\frac{13.13g}{1.008g/mol}\approx13.026mol$.
The number of moles of O: $n_O=\frac{34.73g}{16.00g/mol}\approx2.171mol$.

Step2: Find the mole - ratio

Divide each number of moles by the smallest number of moles (2.171mol in this case).
For C: $\frac{4.342mol}{2.171mol} = 2$.
For H: $\frac{13.026mol}{2.171mol}\approx6$.
For O: $\frac{2.171mol}{2.171mol}=1$.
The empirical formula is $C_2H_6O$.

Step3: Analyze the options

The molecular formula $C_2H_6O$ can represent ethanol ($CH_3CH_2OH$).
The formula $CH_3CH_3CH_2OH$ is incorrect as written. But if we consider the correct structure of ethanol which has the formula $C_2H_6O$.

Answer:

The compound could be ethanol (assuming the first - option was meant to be $CH_3CH_2OH$). The other options with "do" seem to be mis - written and not valid chemical formulas in the context.