Respuesta :
Answer: Empirical formula and molecular formula are [tex]CH_2O[/tex] and [tex]C_2H_4O_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 40.002 g
Mass of H= 6.7135 g
Mass of O = 53.284 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{ 40.002g}{12g/mole}=3.3335moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{ 6.7135g}{1g/mole}=6.7135moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.284 g}{16g/mole}=3.3302moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.3335}{3.3302}=1[/tex]
For H = [tex]\frac{6.7135}{3.3302}=2[/tex]
For O =[tex]\frac{3.3302}{3.3302}=1[/tex]
The ratio of C : H : O= 1 : 2 : 1
Hence the empirical formula is [tex]CH_2O[/tex]
The empirical weight of [tex]CH_2O[/tex] = 1(12)+2(1)+1(16)= 30 g.
The molecular weight = 60.052 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{ 60.052 }{30}=2[/tex]
The molecular formula will be=[tex]2\times CH_2O=C_2H_4O_2[/tex]
Thus empirical formula and molecular formula are [tex]CH_2O[/tex] and [tex]C_2H_4O_2[/tex]
The empirical formula and molecular formula of the compound with a molecular mass of 60.052 amu containing 40.002% C, 6.7135% H, and 53.284% O are:
1. The empirical formula of the compound is CH₂O
2. The molecular formula of the compound is C₂H₄O₂
1. Determination of the empirical formula of the compound.
C = 40.002%
H = 6.7135%
O = 53.284%
Divide by their molar mass
C = 40.002 / 12 = 3.3335
H = 6.7135 / 1 = 6.7135
O = 53.284 / 16 = 3.33025
Divide by the smallest
C = 3.3335 / 3.33025 = 1
H = 6.7135 / 3.33025 = 2
O = 3.33025 / 3.33025 = 1
Thus, the empirical formula of the compound is CH₂O
2. Determination of the molecular formula of the compound
Empirical formula = CH₂O
Molar mass = 60.052 amu
Molecular formula =?
Molecular formula = Empirical × n = molar mass
[CH₂O]n = 60.052
[12 + (2×1) + 16]n = 60.052
30n = 60.052
Divide both side by 30
n = 60.052 / 30
n = 2
Molecular formula = [CH₂O]n
Molecular formula = [CH₂O]₂
Molecular formula = C₂H₄O₂
Therefore, the molecular formula of the compound is C₂H₄O₂
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