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 do₄h₁₂oh₂

Explanation:

Step1: Calculate molar - mass of carbon in each compound

For a compound \(C_xH_yO_z\), the mass of carbon in the compound is \(12x\) (molar - mass of \(C = 12\ g/mol\)).

Step2: Calculate molar - mass of hydrogen in each compound

The mass of hydrogen in the compound is \(1y\) (molar - mass of \(H=1\ g/mol\)).

Step3: Calculate molar - mass of oxygen in each compound

The mass of oxygen in the compound is \(16z\) (molar - mass of \(O = 16\ g/mol\)).

Step4: Calculate the mass - percentage of carbon, hydrogen and oxygen in each compound

Let's take the first compound \(C_3H_8O_2\) as an example.
The molar - mass of \(C_3H_8O_2\) is \(M=(3\times12)+(8\times1)+(2\times16)=36 + 8+32 = 76\ g/mol\).
The mass - percentage of carbon is \(\frac{3\times12}{76}\times100\%=\frac{36}{76}\times100\%\approx47.37\%\).
The mass - percentage of hydrogen is \(\frac{8\times1}{76}\times100\%=\frac{8}{76}\times100\%\approx10.53\%\).
The mass - percentage of oxygen is \(\frac{2\times16}{76}\times100\%=\frac{32}{76}\times100\%\approx42.11\%\).
For the compound \(C_2H_6O\):
The molar - mass of \(C_2H_6O\) is \(M=(2\times12)+(6\times1)+(1\times16)=24 + 6+16 = 46\ g/mol\).
The mass - percentage of carbon is \(\frac{2\times12}{46}\times100\%=\frac{24}{46}\times100\%\approx52.17\%\).
The mass - percentage of hydrogen is \(\frac{6\times1}{46}\times100\%=\frac{6}{46}\times100\%\approx13.04\%\).
The mass - percentage of oxygen is \(\frac{1\times16}{46}\times100\%=\frac{16}{46}\times100\%\approx34.78\%\).

Answer:

The compound \(C_2H_6O\) (ethanol) could be the sample as its percentage composition of carbon (\(\approx52.17\%\)), hydrogen (\(\approx13.04\%\)) and oxygen (\(\approx34.78\%\)) is very close to the given percentages \(52.14\%\) of carbon, \(13.13\%\) of hydrogen and \(34.73\%\) of oxygen. Among the options, the closest - matching compound is likely the one with the correct empirical or molecular formula based on these percentage calculations. If we assume the options are written in a non - standard way and the second option is \(C_2H_6O\), then that is the answer.