It is compose d of 40.92 percent carbon (C),4.58 percent hydrogen (H) , and 54.50 percent oxygen (O) by mass. Determine its empirical formula.

To determine the empirical formula, we need to find the ratio of the number of atoms of each element in the compound.<br /><br />Given:<br />- 40.92% carbon (C)<br />- 4.58% hydrogen (H)<br />- 54.50% oxygen (O)<br /><br />Step 1: Convert the percentages to grams.<br />- Carbon (C): 40.92 g<br />- Hydrogen (H): 4.58 g<br />- Oxygen (O): 54.50 g<br /><br />Step 2: Calculate the number of moles of each element.<br />- Carbon (C): 40.92 g / 12.01 g/mol = 3.40 mol<br />- Hydrogen (H): 4.58 g / 1.008 g/mol = 4.53 mol<br />- Oxygen (O): 54.50 g / 16.00 g/mol = 3.41 mol<br /><br />Step 3: Find the ratio of the number of moles of each element.<br />- Carbon (C): 3.40 mol<br />- Hydrogen (H): 4.53 mol<br />- Oxygen (O): 3.41 mol<br /><br />Step 4: Divide each ratio by the smallest number of moles.<br />- Carbon (C): 3.40 mol / 3.40 mol = 1<br />- Hydrogen (H): 4.53 mol / 3.40 mol = 1.33<br />- Oxygen (O): 3.41 mol / 3.40 mol = 1<br /><br />Step 5: Round the ratios to the nearest whole number.<br />- Carbon (C): 1<br />- Hydrogen (H): 1.33 (rounded to 1)<br />- Oxygen (O): 1<br /><br />Therefore, the empirical formula of the compound is C1H1O1.