Answer
There are $2.77 \times 10^{-19} CCl_2F_2$ molecules and 326 mg of chlorine in 5.56 mg of Freon-12.
Work Step by Step
- Calculate or find the molar mass for $ CCl_2F_2 $:
$ CCl_2F_2 $ : ( 35.45 $\times$ 2 )+ ( 12.01 $\times$ 1 )+ ( 19.00 $\times$ 2 )= 120.91 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 5.56 \space mg \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole}{ 120.91 \space g} = 4.60 \times 10^{-5} \space mole$$
- Using avogadro's constant calculate the amount of molecules
$$4.60 \times 10^{-5} \space mol \times \frac{6.02 \times 10^{23} \space molecules}{1 \space mol} = 2.77 \times 10^{19} \space molecules$$
- Each mole of Freon-12 has 2 moles of chlorine:
$$4.60 \times 10^{-5} \space mol \space CCl_2F_2 \times \frac{2 \space moles \space Cl_2}{1 \space mol \space CCl_2F_2} = 9.20 \times 10^{-5} \space mol Cl$$
- Calculate the mass:
$$9.20 \times 10^{-5} \space mol \space Cl \times \frac{35.45 \space g \space Cl}{1 \space mol \space Cl} = 3.26 \times 10^{-3} \space g \space Cl = 326 \space mg \space Cl$$
