Question

The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas. Empirical formula:

Answer 1

Answer: Empirical formula and molecular formula are $CH_2O$ and $C_2H_4O_2$

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= 40.002 g

Mass of H= 6.7135 g

Mass of O = 53.284 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{ 40.002g}{12g/mole}=3.3335moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{ 6.7135g}{1g/mole}=6.7135moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.284 g}{16g/mole}=3.3302moles$

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

For C = $\frac{3.3335}{3.3302}=1$

For H = $\frac{6.7135}{3.3302}=2$

For O =$\frac{3.3302}{3.3302}=1$

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)= 30 g.

The molecular weight = 60.052 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{ 60.052 }{30}=2$

The molecular formula will be=$2\times CH_2O=C_2H_4O_2$

Thus empirical formula and molecular formula  are $CH_2O$ and $C_2H_4O_2$