Question

An elemental analysis is performed in an unknown compound. It is found to contain 40.0 % mass in Carbon, 6.71% mass in Hydrogen, and the remaining mass in Oxygen. Determine its empirical formula. The formula mass of the unknown is independently determined to be 90.08 g/mol, determine the unknown’s molecular formula

Answer 1

Answer: Molecular formula will be
`C_3H_6O_3`

Step-by-step 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= 40 g

Mass of H = 6.71 g

Mass of O = 100 - (40+6.71 ) = 53.29 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (40g)/(12g/mole)=3.33moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (6.71g)/(1g/mole)=6.71moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (53.29g)/(16g/mole)=3.33moles`

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

For C =
`(3.33)/(3.33)=1`

For H =
`(6.71)/(3.33)=2`

For O =
`(3.33)/(3.33)=1`

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

Hence the empirical formula is
`CH_2O`

The empirical weight of
`CH_2O` = 1(12)+2(1)+1(16)= 30g.

The molecular weight = 90.08 g/mole

Now we have to calculate the molecular formula:

`n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=(90.08)/(30)=3`

The molecular formula will be=
`3* CH_2O=C_3H_6O_3`

Molecular formula will be
`C_3H_6O_3`