Question

What is the empirical formula of a compound comprised of 1.75% hydrogen, 56.1% sulfur and 42.15% oxygen?

Answer 1

The empirical formula for this compound is: `"H"_2"S"_2"O"_3`.

This compound is thiosulfuric acid.

Explanation:

An empirical formula represents the smallest whole number ratio of elements in a compound.

Since the percentages add up to `100%`, we can make a direct conversion to grams.

The steps for determining an empirical formula are:

1) Determine the moles of each element from the masses given and their molar masses.

2) Find the mole ratio for each element by dividing moles of each element by the smallest number of moles.

3) If the mole ratios are whole numbers, then they are the subscripts for each element.

4) If the mole ratios are not all whole numbers, multiply by a factor that will make them all whole numbers.

Moles of each element

Divide the mass in grams by the molar mass of each element. The molar mass is the element's atomic weight on the periodic table in `"g/mol"`. Since molar mass is a fraction `"g"/"mol"`, I prefer to divide by multiplying by the reciprocal of the molar mass, `"mol"/"g"`. I think it shows how to handle the units better than just dividing.

`"H":` `1.75color(red)cancel(color(black)("g H"))xx(1"mol H")/(1.008color(red)cancel(color(black)("g H")))="1.74 mol H"`

`"S":` `56.1color(red)cancel(color(black)("g S"))xx(1"mol S")/(32.06color(red)cancel(color(black)("g S")))="1.75 mol S"`

`"O":` `42.15color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="2.63 mol O"`

Mole ratios

Divide the moles of each element by the smallest number of moles, which is `"1.74 mol"`. There is no dimension for mole ratios, so I will omit "moles".

`"H":` `(color(red)cancelcolor(black)1.74)/(color(red)cancelcolor(black)1.74)=1.00`

`"S":` `1.75/1.74=1.01`

`"O":` `2.63/1.74=1.51`

Since the molar ratio for `"O"` is not a whole number, we must multiply by a factor that will make it a whole number. All mole ratios must be multiplied by the same factor. Since `1.51xx2=3.02~~3`, we will multiply all ratios by `2`.

`"H":` `1.00xx2=2.00`

`"S":` `1.01xx2=2.02~~2`

`"O":` `1.51xx1=3.02~~3`

The empirical formula for this compound is: `"H"_2"S"_2"O"_3`.

This compound is thiosulfuric acid.