Respuesta :
The empirical formula of the compound will be C₁₂ H₁₈ O
1) Mass of carbon (C) in 3.190 g of carbon dioxide (CO₂)
- atomic mass of C: 12.0107 g/mol
- molar mass of CO₂: 44.01 g/mol
Proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 3.190 g of CO₂
Solving for x:
x = (12.0107 g of C / 44.01 g of CO₂ ) × 3.190 g of CO₂ = 0.87053 g of C
2) Mass of hydrogen (H) in 0.9360 g of water (H₂O)
- atomic mass of H: 1.00784 g/mol
- molar mass of H₂O: 18.01528 g/mol
Proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.9360 g of H₂O
Solving for x:
x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.9360 g of H₂O = 0.10472 g of H
3) Mass of oxygen (O) in 1.0857 g of pure sample
Mass of O = mass of pure sample - mass of C - mass of H
Mass of O = 1.0857 g - 0.87053 g - 0.10472 = 0.09392 g O
Round to four decimals: Mass of O = 0.0939 g
4) Mole calculations
Dividing the mass in grams of each element by its atomic mass:
C: 0.87053 g / 12.0107 g/mol = 0.07247 mol
H: 0.10472 g / 1.00784 g/mol = 0.10390 mol
O: 0.0939 g / 15.999 g/mol = 0.00586 mol
5) Dividing every amount by the smallest value to find the mole ratios :
C: 0.07247 mol / 0.00586 mol = 12.36 ≈ 12
H: 0.10390 mol / 0.00586 mol = 17.7 ≈ 18
O: 0.00586 mol / 0.00586 mol = 1
Thus the mole ratio is 12 : 18 : 1, and the empirical formula is:
C₁₂ H₁₈ O
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