Respuesta :
Answer : The molecular formula of kethoxal is, [tex]C_6H_{12}O_4[/tex]
Solution :
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 = 48.6 g
Mass of H = 8.17 g
Mass of O = 43.2 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{48.6g}{12g/mole}=4.05moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{8.17g}{1g/mole}=8.17moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{43.2g}{16g/mole}=2.7moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.05}{2.7}=1.5[/tex]
For H = [tex]\frac{8.17}{2.7}=3.02\approx 3[/tex]
For O = [tex]\frac{2.7}{2.7}=1[/tex]
The ratio of C : H : O = 1.5 : 3 : 1
To make a whole number we multiplying the ratio by 2, we get:
The ratio of C : H : O = 3 : 6 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_6O_2[/tex]
The empirical formula weight = 3(12) + 6(1) + 2(16) = 74 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{148}{74}=2[/tex]
Molecular formula = [tex](C_3H_6O_2)_n=(C_3H_6O_2)_2=C_6H_{12}O_4[/tex]
Therefore, the molecular of the kethoxal is, [tex]C_6H_{12}O_4[/tex]
