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% ?

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Answer 1 · 2018-02-22T12:46:29

$"H"_2"SO"_4$

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.