# Question 0fe92

Jun 17, 2015

Molecular formula: ${C}_{12} {H}_{8} C {l}_{6} O$

#### Explanation:

So, you know that you have a 2.416-g sample of dieldrin that undergoes combustion to produce 3.350 g of carbon dioxide and 0.458 g of water.

Assuming that all the carbon the compound contained is now in trhe carbon dioxide and all the hydrogen is now in the water, you can determine the number of moles of c arbon and of hydrogen the initial sample contained.

Use carbon dioxide and water's molar amsses to see how many moles were produced by the combustion reaction

3.350cancel("g") * ("1 mole "CO_2)/(44.01cancel("g")) = "0.0761 moles" $C {O}_{2}$

Since 1 mole of $C {O}_{2}$ contains 1 mole of carbon, the initial sample contained

${n}_{C} = {n}_{C {O}_{2}} = \text{0.0761 moles}$ $C$

0.458cancel("g") * ("1 mole "H_2O)/(18.02cancel("g")) = "0.02542 moles" ${H}_{2} O$

Since you get 2 moles of hydrogen for every 1 mole of water, the initial sample contained

0.02542cancel("moles"H_2O) * "2 moles H"/(1cancel("mole"H_2O)) = "0.0508 moles" $H$

Use the fact that you have a $2 : 1$ atom ratio between carbon and chlorine to determine how many moles of chlorine the initial sample contained.

0.0761cancel("moles C") * "1 mole Cl"/(2cancel("moles C")) = "0.03805 moles Cl"

To get the number of moles of oxygen in the sample, you need to subtract from the mass of the sample the masses of carbon, hydrogen, and chlorine. The sample contained

0.0761cancel("moles C") * "12.01 g"/(1cancel("mole C")) = "0.9140 g C"

0.0508cancel("moles H") * "1.01 g"/(1cancel("mole H")) = "0.05131 g H"

0.03805cancel("moles Cl") * "35.45 g"/(1cancel("mole Cl")) = "1.349 g Cl"

This means that the mass of oxygen in the sample was

${m}_{O} = {m}_{\text{sample}} - {m}_{C} - {m}_{H} - {m}_{C l}$

${m}_{O} = 2.416 - 0.9140 - 0.05131 - 1.349 = \text{0.1017 g O}$

The number of moles of oxygen are

0.107cancel("g") * "1 mole O"/(16.0 cancel("g")) = "0.006356 moles O"

Now, divide the number of moles each element had in the sample by the smallest one to get the mole ratio that exists in the sample between the elements

"For C":(0.0761cancel("moles"))/(0.006356cancel("moles")) = 11.97 ~= 12

"For H": (0.0508cancel("moles"))/(0.006356cancel("moles")) = 7.99 ~= 8

"For Cl": (0.03805cancel("moles"))/(0.006356cancel("moles")) = 5.99 ~= 6

"For O": (0.006356cancel("moles"))/(0.006356cancel("moles")) = 1#

The empirical formula of the compound is

${\left({C}_{12} {H}_{8} C {l}_{6} O\right)}_{\textcolor{b l u e}{n}}$

To get the molecular formula, you need to determine the value of $\textcolor{b l u e}{n}$. This is where the molar mass of the compound comes in handy. You know that

${\left({C}_{12} {H}_{8} C {l}_{6} O\right)}_{\textcolor{b l u e}{n}} = \text{381 g/mol}$

This is equivalent to

$\left(12 \cdot 12.01 + 8 \cdot 1.01 + 6 \cdot 35.45 + 1 \cdot 16.0\right) \cdot \textcolor{b l u e}{n} = 381$

$\textcolor{b l u e}{n} = \frac{381}{380.9} = 1$

Therefore, the molecular formula of dieldrin is

${\left({C}_{12} {H}_{8} C {l}_{6} O\right)}_{1} \iff \textcolor{g r e e n}{{C}_{12} {H}_{8} C {l}_{6} O}$