A compound has a molar mass of 180 g/mol and the following composition by mass: 40.0% C, 6.7% H, and 53.3% O. What is the molecular formula of the compound?

• A. C12H22O11
• B. C2H4O2
• C. C3H6O3
• D. C6H12O6

Answer: (D)

Verified step by step guidance

1. Convert the percentage composition of each element to grams assuming a 100 g sample. This means you have 40.0 g of C, 6.7 g of H, and 53.3 g of O.
2. Calculate the number of moles of each element by dividing the mass of each element by its atomic mass: use 12.01 g/mol for C, 1.008 g/mol for H, and 16.00 g/mol for O. The formulas are: \(\text{moles C} = \frac{40.0}{12.01}\), \(\text{moles H} = \frac{6.7}{1.008}\), \(\text{moles O} = \frac{53.3}{16.00}\).
3. Determine the simplest whole number mole ratio by dividing each mole value by the smallest number of moles calculated in the previous step. This gives the empirical formula ratio.
4. Write the empirical formula using the mole ratios as subscripts for each element. If necessary, multiply all ratios by the same factor to get whole numbers.
5. Calculate the empirical formula mass by summing the atomic masses multiplied by their subscripts in the empirical formula. Then, divide the given molar mass (180 g/mol) by the empirical formula mass to find the multiplier. Multiply the subscripts in the empirical formula by this multiplier to get the molecular formula.