Question

A compound with molecular mass of 34 g mol is known to contain

Answer 1

Answer:

$H_{2}O_{2}$

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

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 = $\frac{molecular  weight}{equivalent  weight}$= $\frac{34}{17}$= 2

The molecular formula will be= 2 X HO=$H_{2}O_{2}$