How many grams of Cl are in 38.0 g of each sample of chlo- rofluorocarbons (CFCs)?(a)

Step 1 of 4

First, you need to find how many moles of chlorine are in 38.0g of each of these compounds:

Cl is 35.453 grams/mol

C is 12.011 grams/mol

F is 18.988 grams/mol

(a)   \(\mathrm{CF}_{2} \mathrm{Cl}_{2}=1 \text { Carbon }+2 \text { Fluorine }+2 \text { Chlorine }\)

\([1 \times(12.011 \mathrm{~g} / \mathrm{mol})]+[2 \times(18.988 \mathrm{~g} / \mathrm{mol})]+[2 \times(35.453 \mathrm{~g} / \mathrm{mol})]=120.893 \mathrm{~g} / \mathrm{mol}\)

So for this particular compound, one mole weighs 120.893 grams

(or in other words \(6.0221413 \times 10^{23}\) of these molecules would weight 120.893 grams)

Now find how many moles of the compound it takes to weigh 38.0 grams:

\(38.0 \text { grams } \times(1 \mathrm{~mol} / 120.893 \text { grams })=\) grams cancel and we get 0.314 moles

( 0.314 moles means we have \(0.314 \times\left(6.0221413 \times 10^{23}\right)\) molecules of compound

In each molecule of compound \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right)\) we have 2 Chlorine molecules

therefore...

In each mole of compound \(\left(6.0221413 \times 10^{23} \text { molecules of } \mathrm{CF}_{2} \mathrm{Cl}_{2}\right)\)

we have 2 moles of Chlorine \(\left(2 \times\left[6.0221413 \times 10^{23}\right] \text { molecules of } \mathrm{Cl}\right)\)

So,

In 0.314 moles of \(\mathrm{CF}_{2} \mathrm{Cl}_{2} \text { there are }(2 \times 0.314) \text { moles of Chlorine }\)

\(2 \times 0.314=0.628 \text { moles of Chlorine in } 38.0 \text { grams of } \mathrm{CF}_{2} \mathrm{Cl}_{2}\)

Chlorine = 35.453 grams/mole

So...

\(0.628 \text { moles } \times(35.453 \text { grams } / 1 \text { mole })=22.3 \text { grams }\) of chlorine in 38.0 grams of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\)

Just to help clarify...

Because there are \(0.314 \text { moles of } \mathrm{CF}_{2} \mathrm{Cl}_{2} \text {, there are } 2 \times 0.314 \text { moles }\) of Fluorine and \(1 \times 0.314 \text { moles }\) of carbon in the compound as well...

\(2 \times 0.314 \text { moles }=0.628 \text { moles }\) of Fluorine

Fluorine = 18.988 grams/mol

So… 

\(\begin{array}{l}\text{0.628 moles } \times(18.988 \text{ grams } / 1 \text{ mole } )=11.92 \text{ grams of fluorine in 38.0 grams of}\\ \mathrm{CF}_{2} \mathrm{Cl}_{2}\end{array}\)

And \(1 \times 0.314 \text { moles }=0.314\) moles of Carbon

Carbon is 12.011 grams/mol

So...

\(0.314 \text { moles } \times(12.011 \text { grams } / 1 \text { mole })=3.77\) grams of carbon in 38.0 grams of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\)

In total, in 38 grams of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) there are:

22.3 grams of chlorine

11.92 grams of fluorine

3.77 grams of carbon

= 38.0 grams total