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Cumene is a compound containing c and h that is used in the production rnof acetone and phenol in the chemical industry. Combustion of 47.6 mg rnof cumene produced some CO2 and 42.8 mg water. The molar mass rncumene is 125 g/mol. Determine the empirical and molecular formulas of cumene
Accepted Answer
Combustion of cumene:
C9H12 + 12O2 → 9CO2 + 6H2O
From the given data:
Mass of cumene = 47.6 mg = 0.0476 g
Mass of water produced = 42.8 mg = 0.0428 g
1. Determining the empirical formula:
- Convert the mass of cumene to moles:
Moles of cumene = 0.0476 g / 125 g/mol = 0.000381 mol
- Convert the mass of water produced to moles:
Moles of water = 0.0428 g / 18 g/mol = 0.00238 mol
- Calculate the mole ratio of C to H:
Mole ratio of C : H = Moles of C : Moles of H
= (0.000381 mol x 9) : (0.00238 mol x 12)
= 0.00343 : 0.0286
- Divide the mole ratios by the smallest mole ratio to get the simplest whole-number ratio:
Simplest mole ratio = 0.00343 : 0.0286 ÷ 0.00343
= 1 : 8
- Therefore, the empirical formula of cumene is CH8.
2. Determining the molecular formula:
- The molar mass of the empirical formula (CH8) is 125 g/mol, which is the same as the given molar mass of cumene.
- Therefore, the empirical formula and the molecular formula of cumene are the same: CH8.
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