Respuesta :
Answer:
C18H22O2
Explanation:
The following were obtained from the question:
Mass of the unknown compound = 1.893g
Mass of CO2 produced = 5.545g
Mass of H2O produced = 1.388g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol
Now let us obtain the various masses of C, H and O present in the unknown compound.
This is illustrated below:
Mass of C in the unknown compound = 12/44 x 5.545 = 1.512g
Mass of H in the unknown compound = 2/18 x 1.388 = 0.154g
Mass of O in the unknown compound = 1.893 — (1.512+0.154) = 0.227g
Now, let us calculate the empirical formula of the compound.
C = 1.512g
H = 0.154g
O = 0.227g
Divide by their molar Mass
C = 1.512/12 = 0.126
H = 0.154/1 = 0.154
O = 0.227/16 = 0.014
Now, divide by the smallest
C = 0.126/0.014 = 9
H = 0.154/0.014 =11
O = 0.014/0.014 = 1
The empirical formula is C9H11O
from the question,
Number of mole the unknown compound = 7.002x10^-3 mol
Mass of the unknown compound = 1.893g
Molar Mass of the unknown compound = Mass /number of mole = 1.893g/7.002x10^-3 mol = 270.35g/mol
Now, we can easily find the molecular formula of the unknown compound as follows
[C9H11O]n = 270.35
[(12x9) + (11x1) +16 ]n = 270.35
135n = 270.35
Divide both side by the coefficient of n i.e 135
n = 270.35/135 = 2
Therefore, the molecular formula is => [C9H11O]n
=> [C9H11O]2
=> C18H22O2
Answer:
The molecular formula is C18H22O2
Explanation:
Step 1: Data given
Mass of the sample = 1.893 grams
Mass of CO2 produced = 5.545 grams
Mass of H2O produced = 1.388 grams
The 1.893-gram sample contained 7.002 * 10^-3 moles
Molar mass CO2 = 44.01 g/mol
Molar mass H2O = 18.02 g/mol
Molar mass C = 12.01 g/mol
Molar mass H = 1.01 g/mol
Molar mass O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = 5.545 grams / 44.01 g/mol
Moles CO2 = 0.1260 moles
In 1 mol CO2 we have 1 mol C
In 0.1260 moles CO2 we have 0.1260 moles C
Step 3: Calculate mass C
Mass C = 0.1260 moles * 12.01 g/mol
Mass C = 1.513 grams
Step 4: Calculate moles and H2O
Moles H2O = 1.388 grams / 18.02 g/mol
Moles H2O = 0.07703 moles
In 1 mol H2O we have 2 moles H
In 0.07703 moles H2O we have 2*0.07703 = 0.15406 moles H
Step 5: Calculate mass H
Mass H = 0.15406 moles * 1.01 g/mol
Mass H = 0.1556 grams
Step 6: Calculate mass O
Mass O = 1.893 - 1.513 - 0.1556
Mass O = 0.2244 grams
Step 7: Calculate moles O
Moles O = 0.2244 grams / 16.0 g/mol
Moles O = 0.01403 moles
Step 8: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.1260 moles/ 0.01403 moles = 9
H: 0.15406 moles / 0.01403 moles = 11
O: 0.01403 moles / 0.01403 moles = 1
The empirical formula is C9H11O
The molecular mass of this formula is 135.2 g/mol
Step 9: Calculate the molar mass of this compound
Molar mass = 1.893 grams / 0.007002 moles
Molar mass = 270.4 g/mol
Step 9: Calculate the molecular formula
We have to multiply the empirical formula by n
n = 270.4 / 135.2 = 2
The molecular formula is 2*(C9H11O) = C18H22O2
