Ano organic compound contains 52.17%of carbon, 13% of hydrogen and the rest being oxygen. Given that the relative molecular mass is92. A, it's empirical formula is what?

Step 1: Determine the percentage of oxygen. The sum of percentages of all elements in a compound must be 100%. $Percentage of Oxygen = 100\% - (Percentage of Carbon + Percentage of Hydrogen)$ $Percentage of Oxygen = 100\% - (52.17\% + 13\%)$ $Percentage of Oxygen = 100\% - 65.17\%$ $Percentage of Oxygen = 34.83\%$

Step 2: Assume a 100 g sample and convert percentages to masses. In a 100 g sample: Mass of Carbon (C) = 52.17 g Mass of Hydrogen (H) = 13 g Mass of Oxygen (O) = 34.83 g

Step 3: Convert masses to moles using atomic masses. Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.

Moles of Carbon: $Moles of C = \frac{52.17 g}{12.01 g/mol} \approx 4.344 mol$

Moles of Hydrogen: $Moles of H = \frac{13 g}{1.008 g/mol} \approx 12.897 mol$

Moles of Oxygen: $Moles of O = \frac{34.83 g}{16.00 g/mol} \approx 2.177 mol$

Step 4: Determine the simplest whole-number ratio by dividing by the smallest number of moles. The smallest number of moles is 2.177 mol (for Oxygen).

Ratio for Carbon: $\frac{4.344 mol}{2.177 mol} \approx 1.995 \approx 2$

Ratio for Hydrogen: $\frac{12.897 mol}{2.177 mol} \approx 5.924 \approx 6$

Ratio for Oxygen: $\frac{2.177 mol}{2.177 mol} = 1$

The simplest whole-number ratio of C:H:O is 2:6:1.

The empirical formula is $\boxed{C_2H_6O}$.