Questions: What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

Solution Steps

Assume we have 100 grams of the compound. This means the mass of each element in the compound is equal to its percentage by mass:

• Hydrogen: \(3.25 \, \text{g}\)
• Carbon: \(19.36 \, \text{g}\)
• Oxygen: \(77.39 \, \text{g}\)

Use the molar mass of each element to convert grams to moles:

• Molar mass of Hydrogen (H): \(1.008 \, \text{g/mol}\)
• Molar mass of Carbon (C): \(12.01 \, \text{g/mol}\)
• Molar mass of Oxygen (O): \(16.00 \, \text{g/mol}\)

Calculate moles for each element:

\[ \text{Moles of H} = \frac{3.25 \, \text{g}}{1.008 \, \text{g/mol}} = 3.2262 \, \text{mol} \]

\[ \text{Moles of C} = \frac{19.36 \, \text{g}}{12.01 \, \text{g/mol}} = 1.6120 \, \text{mol} \]

\[ \text{Moles of O} = \frac{77.39 \, \text{g}}{16.00 \, \text{g/mol}} = 4.8369 \, \text{mol} \]

Divide the number of moles of each element by the smallest number of moles calculated:

\[ \text{Ratio of H} = \frac{3.2262}{1.6120} = 2.0014 \]

\[ \text{Ratio of C} = \frac{1.6120}{1.6120} = 1 \]

\[ \text{Ratio of O} = \frac{4.8369}{1.6120} = 3.0006 \]

Round the ratios to the nearest whole numbers to get the empirical formula:

• Hydrogen: 2
• Carbon: 1
• Oxygen: 3

Thus, the empirical formula is \( \text{H}_2\text{CO}_3 \).

Final Answer