Questions: A compound contains 76.6% C, 6.38% H and 17.0% O. Which of the following is the correct empirical formula for the compound? a C6H6O b CH2O c C2H2O d C4H4O e CHO

Solution Steps

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

• 76.6 grams of carbon (C)
• 6.38 grams of hydrogen (H)
• 17.0 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: \[ \text{Moles of C} = \frac{76.6 \, \text{g}}{12.01 \, \text{g/mol}} = 6.3797 \, \text{mol} \] \[ \text{Moles of H} = \frac{6.38 \, \text{g}}{1.008 \, \text{g/mol}} = 6.3254 \, \text{mol} \] \[ \text{Moles of O} = \frac{17.0 \, \text{g}}{16.00 \, \text{g/mol}} = 1.0625 \, \text{mol} \]

Divide each mole value by the smallest number of moles calculated: \[ \text{Ratio for C} = \frac{6.3797}{1.0625} = 6.004 \] \[ \text{Ratio for H} = \frac{6.3254}{1.0625} = 5.953 \] \[ \text{Ratio for O} = \frac{1.0625}{1.0625} = 1.000 \]

The ratios are approximately:

• C: 6
• H: 6
• O: 1

Final Answer