Question

A compound with molar mass of 34 g/mol is known to contain 5.88% hydrogen and 94.12% oxygen. find molecular formula of compound

Answer 1

Answer: $H_2O_2$

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 H= 5.88 g

Mass of O = 94.12 g

Step 1 : convert given masses into moles.

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.88g}{1g/mole}=5.88moles$

Moles of O =$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{94.12g}{16g/mole}=5.88moles$

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

For H = $\frac{5.88}{5.88}=1$

For O =$\frac{5.88}{5.88}=1$

The ratio of H : O= 1:1

Hence the empirical formula is $HO$

The empirical weight of $HO$ = 1(1)+1(16)= 17g.

The molecular weight = 34 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{34}{17}=2$

The molecular formula will be=$2\times HO=H_2O_2$