What is the molecular formula of a compound whose molar mass is 88 and whose percent composition is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen?

 To determine the molecular formula, we need to follow several steps:

Step 1: Convert the percentages to grams. Since percent composition is out of 100, we can directly consider the percentages as grams for a 100-gram sample: - Carbon: 54.5g - Hydrogen: 9.1g - Oxygen: 36.4g

Step 2: Use the atomic masses to convert grams to moles for each element: - Carbon (C): Atomic mass ~ 12 g/mol \( \frac{54.5g}{12g/mol} \approx 4.54 \) moles - Hydrogen (H): Atomic mass ~ 1 g/mol \( \frac{9.1g}{1g/mol} = 9.1 \) moles - Oxygen (O): Atomic mass ~ 16 g/mol \( \frac{36.4g}{16g/mol} \approx 2.275 \) moles

Step 3: Find the simplest whole number ratio by dividing by the smallest number of moles, which in this case is for oxygen: - Carbon: \( \frac{4.54}{2.275} \approx 2 \) - Hydrogen: \( \frac{9.1}{2.275} \approx 4 \) - Oxygen: \( \frac{2.275}{2.275} = 1 \)

Step 4: This ratio gives you the empirical formula, which looks like \( C_2H_4O \) with a molar mass of approximately 44 g/mol. This is because 2 moles of carbon is \( 2 \times 12 \text{g} = 24\text{g} \), 4 moles of hydrogen is \( 4 \times 1\text{g} = 4\text{g} \), and 1 mole of oxygen is \( 16\text{g} \), adding up to 44 g/mol.

Step 5: Compare the empirical formula mass to the actual molar mass: - Empirical formula mass \( \approx 44 \) g/mol. - Actual molar mass = 88 g/mol.

Step 6: Since the actual molar mass is twice the empirical formula mass, the molecular formula contains two empirical formula units: - The molecular formula is \( C_4H_8O_2 \).