Question

1. how many molecules of sulfur hexafluoride are present in 6.22 grams of this compound? use the references to access important values if needed for this question. molecules. 2. how many grams of sulfur hexafluoride are present in 5.51x10^22 molecules of this compound? grams. 3 item attempts remaining. submit answer try another version

Explanation:

Step1: Calculate molar mass of sulfur hexafluoride ($SF_6$)

The molar mass of $S$ is approximately $32.07\ g/mol$ and the molar mass of $F$ is approximately $19.00\ g/mol$. For $SF_6$, the molar mass $M = 32.07+6\times19.00=32.07 + 114.00=146.07\ g/mol$.

Step2: Solve for number of molecules in part 1

We know that $n=\frac{m}{M}$ and $N = n\times N_A$, where $n$ is the number of moles, $m$ is the mass, $M$ is the molar - mass, $N$ is the number of molecules and $N_A=6.022\times 10^{23}\ mol^{-1}$. Rearranging for $N$, we get $N=\frac{m}{M}\times N_A$. Given $m = 6.22\ g$, then $n=\frac{6.22\ g}{146.07\ g/mol}\approx0.0426\ mol$. And $N=0.0426\ mol\times6.022\times 10^{23}\ mol^{-1}\approx2.57\times 10^{22}$ molecules.

Step3: Solve for mass in part 2

Given $N = 5.51\times 10^{22}$ molecules, first find the number of moles $n=\frac{N}{N_A}=\frac{5.51\times 10^{22}}{6.022\times 10^{23}\ mol^{-1}}\approx0.0915\ mol$. Then find the mass $m=n\times M=0.0915\ mol\times146.07\ g/mol\approx13.4\ g$.

Answer:

1. $2.57\times 10^{22}$ molecules
2. $13.4$ g