# Help with question 11 c guys ?

Aug 20, 2017

$\text{C"_4"H"_4"S}$ is both the molecular and empirical formula of thiophene: The idea here is to use two situations and conservation of mass to separately determine the mass percent of two out of three components in thiophene, which we denote as ${\text{C"_x"H"_y"S}}_{z}$. Main steps:

1. Find mass percent of $\text{S}$ in thiophene.
2. Find mass percent of $\text{C}$ in thiophene and use that to deduce mass percent of $\text{H}$ in thiophene.
3. Assume $\text{100 g}$ of thiophene to find the mol ratios.
4. Check with the molecular mass to see if the empirical formula is any different from the molecular formula.

In the first reaction:

${\text{7.96 g C"_x"H"_y"S"_z harr "16.65 g CO}}_{2}$

In the second set of reactions:

${\text{4.31 g C"_x"H"_y"S"_z harr "11.96 g BaSO}}_{4}$

where we assume thiophene was the limiting reagent...

This begins part $\left(i\right)$:

STEP 1: MASS PERCENT OF S

As the quantity of sulfur must obey conservation of mass, here is a nice trick we can use: we can convert from the mass of one substance to another using the ratio of their molar masses.

"11.96 g " cancel("BaSO"_4) xx (32.065 cancel"g/mol" " S")/(233.38 cancel"g/mol" cancel("BaSO"_4))

$=$ $\text{1.643 g S}$ that came from thiophene,

assuming no other reactants contained sulfur.

That means we can use the thiophene mass from the second set of reactions to find the mass percent of $\text{S}$ in thiophene.

$\left(1.643 \cancel{\text{g" " S")/(4.31 cancel("g") " thiophene") xx 100% = color(green)ul(38.13%"w/w S}}\right)$ in thiophene

STEP 2: MASS PERCENT OF C & H

From the first reaction, notice that thiophene is the only source of carbon for ${\text{CO}}_{2}$. Thus, we can similarly find the mass of $\text{C}$ that is in ${\text{CO}}_{2}$, and by conservation of mass, that was the mass in thiophene.

"16.65 g " cancel("CO"_2) xx (12.011 cancel"g/mol" " C")/(44.009 cancel"g/mol" cancel("CO"_2))

$=$ $\text{4.544 g C}$ that came from thiophene.

This means the mass percent of $\text{C}$ in thiophene is:

$\left(4.544 \cancel{\text{g" " C")/(7.96 cancel"g" " thiophene") xx 100% = color(green)ul(57.09% "w/w C}}\right)$ in thiophene

This leaves a hydrogen atom mass percent of:

100.00% - 38.13% - 57.09%

= color(green)ul(4.78_7% "w/w H") in thiophene

(where the subscript indicates the digit past the last significant digit.)

STEP 3: MOL RATIOS & EMPIRICAL FORMULA

Currently our mass percents are only relative. If we choose a $\text{100 g}$ mass of thiophene, then in that mass, we have absolute masses:

• ${4.78}_{7}$ $\text{g H}$
• $\text{57.09 g C}$
• $\text{38.13 g S}$

Now, if we have everything in mols, they can be compared in a ratio. So, divide by their molar masses to get the mol quantities of each:

(4.787 cancel"g H")/(1.0079 cancel"g H""/""mol H") = "4.749 mols H"

(57.09 cancel"g C")/(12.011 cancel"g C""/mol C") = "4.753 mols C"

(38.13 cancel"g S")/(32.015 cancel"g S""/mol S") = "1.191 mols S"

We see that the mols of $\text{H}$ and $\text{C}$ are about equal, and are also about four times the mols of $\text{S}$. To proceed, divide by the smallest mol quantity:

$\left(1.191 \cancel{\text{mols" "S")/(1.191 cancel"mols}}\right) = \boldsymbol{1}$ equiv. of $\text{S}$

$\left(4.753 \cancel{\text{mols" "C")/(1.191 cancel"mols") ~~ (4.749 cancel"mols" "H")/(1.191 cancel"mols}}\right) = 3.991$

$\approx \boldsymbol{4}$ equivs. of $\text{C}$ and $\text{H}$ each

So, we construct an empirical formula of:

$\textcolor{b l u e}{\text{C"_4"H"_4"S}}$

This is currently the most "reduced" form of thiophene's chemical formula, i.e. the mol ratios are as small as they can be with integer subscripts.

And now for part $\left(i i\right)$.

STEP 4: MOLECULAR FORMULA

Knowing that its molecular mass is $\text{84 amu}$, or $\text{84 g/mol}$, we compare with the molecular mass we would get using its empirical formula to see if we have the proper molecular formula yet.

M_("C"_4"H"_4"S") = 4 xx "12.011 g/mol" + 4 xx "1.0079 g/mol" + "32.065 g/mol"

$=$ $\text{84.141 g/mol}$ $\approx$ $\text{84 g/mol}$

So, our empirical formula IS the molecular formula.