A \(1.540-g\) sample of a liquid is subjected to combustion analysis, yielding \(40.00 \% \mathrm{C}\) and \(6.71 \% \mathrm{H}\). It may also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(30 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

Percentage of oxygen = 100% - (40.00% + 6.71%) Percentage of oxygen = 53.29% #tag_title# Step 2: Convert percentages to grams #tag_content# Now, we will convert the percentages of each element to grams using the mass of the sample, which is $1.540~g$. Grams of carbon = $1.540~g \times 40.00\% = 0.616~g$ Grams of hydrogen = $1.540~g \times 6.71\% = 0.103~g$ Grams of oxygen = $1.540~g \times 53.29\% = 0.821~g$ #tag_title# Step 3: Convert grams to moles #tag_content# Now, we will convert the grams of each element to moles using their molar masses: Moles of carbon = $\frac{0.616~g}{12.01~g/mol} = 0.0513~mol$ Moles of hydrogen = $\frac{0.103~g}{1.008~g/mol} = 0.102~mol$ Moles of oxygen = $\frac{0.821~g}{16.00~g/mol} = 0.0513~mol$ #tag_title# Step 4: Find the empirical formula #tag_content# To find the empirical formula, we will divide the moles of each element by the smallest number of moles, which in this case is 0.0513, and round to the nearest whole number: Carbon: $\frac{0.0513}{0.0513} = 1$ Hydrogen: $\frac{0.102}{0.0513} = 1.99 \approx 2$ Oxygen: $\frac{0.0513}{0.0513} = 1$ Thus, the empirical formula of this compound is \(\mathrm{CH_2O}\). #tag_title# Step 5: Find the molecular formula #tag_content# Using the empirical formula and the given molar mass of $30~g/mol$, we can find the molecular formula. First, we need to find the empirical formula mass: Empirical formula mass = $\mathrm{C} + 2 \times \mathrm{H} + \mathrm{O} = 12.01~g/mol + 2 \times 1.008~g/mol + 16.00~g/mol = 30.03~g/mol$ Since the empirical formula mass is almost equal to the molar mass of the compound, the molecular formula is the same as the empirical formula: Molecular formula = \(CH_2O\)