Question

Find the molecular formula of an acid with a molar mass of 98 grams per moles and the percentage composition of the elements in the acid is as such: hydrogen=2%, sulphur=33%, oxygen=65% ?

Answer 1

`"H"_2"SO"_4`

Explanation:

We must first find the empirical formula.

We can say that, in a `100"g"` sample, we have `2"g"` hydrogen, `"33g"` sulfur, and `65"g"` oxygen.

Converting to moles:

`2"g H"*(1 "mol H")/(1 "g H")= 2 "mol H"`

`33"g S"*(1 "mol S")/(32 "g S")=1.03"mol S"`

`65 "g O"*(1 "mol O")/(16"g O")=4.06"mol O"`

We must divide each mole number by the smallest number, which, here, is `1.03`. Then round to the nearest whole number.

For hydrogen: `2/1.03=1.94~~2`
For sulfur: `1.03/1.03=1`
For oxygen: `4.06/1.03=3.94~~4`

Putting them all together, we get:

`"H"_2"SO"_4`, sulfuric acid.

We must check if the molar mass of sulfuric acid is `98"g"`:

`(1*2)+32+(16*4)=98`

So the molecular formula is also `"H"_2"SO"_4`, sulfuric acid.