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

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 2

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}}= (55.80g)/(12g/mole)=4.65moles`

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

`Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= (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 =
`(4.65)/(2.32)=2`

For H =
`(7.04)/(2.32)=3`

For O =
`(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=(301.35g/mole)/(43g/eq)=7`

Thus molecular formula will be
`7* C_(2)H_(3)O=C_(14)H_(21)O_7`