Question

What is the molecular formula of a substance that decomposes into `1.33` `g` of `H` and `21.3` `g` of `O`, and was found to have a molar mass of `34.1` `gmol^-1`?

Answer 1

The molecular formula of the substance is `H_2O_2`.

Explanation:

`H` has a molar mass of `1` `gmol^-1`, so `1.33` `g` = `1.33` mol.

(I have assumed that we are talking about individual `H` atoms, not `H_2` molecules, and similarly for `O` versus `O_2`.

Oxygen has a molar mass of `16` `gmol^-1`, so using `n=m/M`, `n = 21.3/16=1.33` `gmol^-1`.

We see that there are the same number of moles of each element in the substance, so we can think of the molecular formula as `H_xO_x`. We need to find the value of `x`.

If `x` was equal to `1`, the molar mass would be `1+16 = 17` `gmol^-1`.

Since the given molar mass is almost exactly 2x that, at `34.1` `gmol^-1`, the value of `x` must be `2`, and the molecular formula must be `H_2O_2`.

(incidentally, this is the formula of the chemical substance hydrogen peroxide)