Respuesta :
55.80 C/12= 4.65/2.3225= 2
7.04 H/1= 7.04/2.3225= 3
37.16 O/16= 2.3225/2.3225= 1
Empirical Formula= C2H3O
301.35/43= 7
(C2H3O)7= C14H21O7= Molecular Formula
Answer:The empirical formula is [tex]C_{2}H_{3}O[/tex] and molecular formula is [tex]C_{14}H_{21}O_7[/tex]
Solution : Given,
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 = 55.80 g
Mass of H = 7.04 g
Mass of O = 37.16 g
Step 1 : convert given masses into moles.
[tex]Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{55.80g}{12g/mole}=4.65moles[/tex]
[tex]Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.04g}{1g/mole}=7.04moles[/tex]
[tex]Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{37.16g}{16g/mole}=2.32moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.65}{2.32}=2[/tex]
For H = [tex]\frac{7.04}{2.32}=3[/tex]
For O =[tex]\frac{2.32}{2.32}=1[/tex]
The ratio of C : H : O= 2:3:1
Hence the empirical formula is [tex]C_{2}H_{3}O[/tex]
The empirical weight of[tex]C_{2}H_{3}O[/tex] = 2(12)+3(1) +1(16)= 43g.
The molecular weight = 301.35 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}[/tex]
[tex]n=\frac{301.35g/mole}{43g/eq}=7[/tex]
Thus molecular formula will be [tex]7\times C_{2}H_{3}O=C_{14}H_{21}O_7[/tex]
