Respuesta :
The empirical formula of the compound is C3H4O as the mole ratio is 3 : 4 : 1
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
Setting up proportion as:
12.0107 g of C / 44.01 g of CO₂ = x / 3.190 g of CO₂
x = 0.87057 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
Setting up proportion as:
2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.9360 g of H₂O
x = 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 = 0.38397 g O ≈ 0.3840 g
4) Mole calculations
C: 0.87057 g / 12.0107 g/mol = 0.07248 mol
H: 0.10472 g / 1.00784 g/mol = 0.10390 mol
O: 0.3840 g / 15.999 g/mol = 0.02400 mol
So, to find empirical formula the mole ratios are:
C: 3
H: 4
O: 1
Thus the mole ratio is 3 : 4 : 1, and the empirical formula is: C3 H4 O
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