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 $C_3H_6O_3$

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 =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles$

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

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

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

For C = $\frac{3.33}{3.33}=1$

For H =$\frac{6.71}{3.33}=2$

For O = $\frac{3.33}{3.33}=1$

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:

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{90.08}{30}=3$

The molecular formula will be=$3\times CH_2O=C_3H_6O_3$

Molecular formula will be $C_3H_6O_3$