Respuesta :
The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
= 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon : Hydrogen : Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996 : 3.995 : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
= 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon : Hydrogen : Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996 : 3.995 : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O
The empirical formula of the compound is C₂H₄O
How to determine the mass of Carbon
- Mass of CO₂ = 1.6004 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.6004
Mass of C = 0.4365 g
How to determine the mass of H
- Mass of H₂O = 0.6551 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 0.6551
Mass of H = 0.0728 g
How to determine the mass of O
- Mass of compound = 0.8009 g
- Mass of C = 0.4365 g
- Mass of H = 0.0728 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 0.8009 – (0.4365 + 0.0728)
Mass of O = 0.2916 g
How to determine the empirical formula
- C = 0.4365 g
- H = 0.0728 g
- O = 0.2916 g
- Empirical formula =?
Divide by their molar mass
C = 0.4365 / 12 = 0.036
H = 0.0728 / 1 = 0.0728
O = 0.2916 / 16 = 0.018
Divide by the smallest
C = 0.036 / 0.018 = 2
H = 0.0728 / 0.018 = 4
O = 0.018 / 0.018 = 1
Thus, the empirical formula of the compound is C₂H₄O
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