Question

 A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it has a molar mass of 301.35 grams​

Answer 1

Answer:The empirical formula is $C_{2}H_{3}O$   and molecular formula is  $C_{14}H_{21}O_7$

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.

$Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{55.80g}{12g/mole}=4.65moles$

$Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.04g}{1g/mole}=7.04moles$

$Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{37.16g}{16g/mole}=2.32moles$

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

For C = $\frac{4.65}{2.32}=2$

For H = $\frac{7.04}{2.32}=3$

For O =$\frac{2.32}{2.32}=1$

The ratio of C : H : O= 2:3:1

Hence the empirical formula is $C_{2}H_{3}O$  

The empirical weight of$C_{2}H_{3}O$ = 2(12)+3(1) +1(16)= 43g.

The molecular weight = 301.35 g/mole

Now we have to calculate the molecular formula.

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

$n=\frac{301.35g/mole}{43g/eq}=7$

Thus molecular formula will be $7\times C_{2}H_{3}O=C_{14}H_{21}O_7$

Answer 2

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