Question

an organic compound contain 49.3% carbon, 6.84% hydrogen and also present oxygen. its vapour density is 73. find the molecular formula of the compound

Answer 1

S.No.

Element

Percentage

Atomic mass

No. of moles = Percent/Atomic                mass  

Simplest molar ratio                                      

Simplest whole no. molar ratio

   1

    C

     49.3

  12

 49.3/ 12          =4.10

4.10/2.74 = 1.49

     3

   2

    H

     6.84

   1

 6.84/1 = 6.84

6.84/2.74 = 2.49

     5

   3

     O

     43.86

   16

 43.86/ 16 = 2.74

2.74/2.74 = 1

     2


Empirical formula = C3H5O2
Empirical formula mass = 

3×12 + 5×1 + 2×16=73 

u
n = 

Molecular mass

Empirical formula mass

 = 

146

73

 = 2
So , the molecular formula is ​(C3H5O2)2  =  C6H10O4

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Answer 2

Answer: The molecular formula will be,  $C_{6}H_{10}O_4$

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 = 49.3 g

Mass of H = 6.84 g

Mass of O = 100-(49.3+6.84)= 43.86 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.3g}{12g/mole}=4.1moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.84g}{1g/mole}=6.84moles$

Moles of O = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{43.86g}{16g/mole}=2.7mole$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.1}{2.7}=1.5$

For H = $\frac{6.84}{1}=2.5$

For O =$\frac{2.7}{2.7}=1$

The ratio of C : H : : O= 1.5 : 2.5 : 1

Converting them to simple whole number ratios:  

The ratio of C : H : : O= 3 : 5 : 2

Hence the empirical formula is $C_{3}H_{5}O_2$  

The empirical weight is  $C_{3}H_{5}O_2$ = 3(12) + 5(1) + 2(16)= 73 g.

The molecular weight = $2\times {\text {vapour density}}=2\times 73=146g$

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight }}$

$n=\frac{146g/mole}{73g/eq}=2$

Thus Molecular formula is $2\times C_{3}H_{5}O_2=C_{6}H_{10}O_4$