Respuesta :
The solution would be like this for this specific problem:
Moles of carbon = 58.8 /
12 = 4.9
Moles of hydrogen = 9.8 / 1 = 9.8
Moles of oxugen = 31.4 / 16 m= 1.96
Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1
I hope this helps and if you have any further questions, please don’t hesitate to ask again.
Answer: Thus the molecular formula is [tex]C_{5}H_{10}O_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 = 58.8 g
Mass of H = 9.8 g
Mass of O = 31.4 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{58.8g}{12g/mole}=4.9moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.8g}{1g/mole}=9.8moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{31.4g}{16g/mole}=1.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.9}{1.96}=2.5[/tex]
For H = [tex]\frac{9.8}{1.96}=5[/tex]
For O =[tex]\frac{1.96}{1.96}=1[/tex]
The ratio of C : H : O= 2.5 : 5 : 1
Converting them into whole number ratios:
Hence the empirical formula is [tex]C_{5}H_{10}O_2[/tex]
The empirical weight of[tex]C_{5}H_{10}O_2[/tex] = 5(12) + 10(1) + 2(16)= 102g.
Given :The molecular weight = 102 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{102g/mole}{102g/eq}=1[/tex]
Thus the molecular formula is [tex]C_{5}H_{10}O_2[/tex]
