Questions: A compound contains 62.1% C, 10.4% H and 27.5% O. What is the empirical formula of the compound? CH2O C2H4O C3H4O C3H6O

Solution Steps

Assume we have 100 grams of the compound. This means we have:

• 62.1 grams of carbon (C)
• 10.4 grams of hydrogen (H)
• 27.5 grams of oxygen (O)

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

• Molar mass of C = 12.01 g/mol
• Molar mass of H = 1.008 g/mol
• Molar mass of O = 16.00 g/mol

Calculate moles for each element: \[ \text{Moles of C} = \frac{62.1 \, \text{g}}{12.01 \, \text{g/mol}} = 5.1715 \, \text{mol} \] \[ \text{Moles of H} = \frac{10.4 \, \text{g}}{1.008 \, \text{g/mol}} = 10.3175 \, \text{mol} \] \[ \text{Moles of O} = \frac{27.5 \, \text{g}}{16.00 \, \text{g/mol}} = 1.7188 \, \text{mol} \]

Divide each mole value by the smallest number of moles calculated: \[ \text{Ratio for C} = \frac{5.1715}{1.7188} = 3.008 \] \[ \text{Ratio for H} = \frac{10.3175}{1.7188} = 6.005 \] \[ \text{Ratio for O} = \frac{1.7188}{1.7188} = 1 \]

The ratios are approximately whole numbers:

• C: 3
• H: 6
• O: 1

Final Answer