# 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% ?

Feb 22, 2018

${\text{H"_2"SO}}_{4}$

#### Explanation:

We must first find the empirical formula.

We can say that, in a $100 \text{g}$ sample, we have $2 \text{g}$ hydrogen, $\text{33g}$ sulfur, and $65 \text{g}$ oxygen.

Converting to moles:

$2 \text{g H"*(1 "mol H")/(1 "g H")= 2 "mol H}$

$33 \text{g S"*(1 "mol S")/(32 "g S")=1.03"mol S}$

$65 \text{g O"*(1 "mol O")/(16"g O")=4.06"mol O}$

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

For hydrogen: $\frac{2}{1.03} = 1.94 \approx 2$
For sulfur: $\frac{1.03}{1.03} = 1$
For oxygen: $\frac{4.06}{1.03} = 3.94 \approx 4$

Putting them all together, we get:

${\text{H"_2"SO}}_{4}$, sulfuric acid.

We must check if the molar mass of sulfuric acid is $98 \text{g}$:

$\left(1 \cdot 2\right) + 32 + \left(16 \cdot 4\right) = 98$

So the molecular formula is also ${\text{H"_2"SO}}_{4}$, sulfuric acid.