Question

use the references to access important values if needed for this question. a. how many atoms of nitrogen are present in 2.42 grams of dinitrogen tetrafluoride? b. how many grams of fluorine are present in 6.66×10^22 molecules of dinitrogen tetrafluoride? grams of fluorine.

Explanation:

Step1: Calculate moles of $N_2F_4$

The molar mass of $N_2F_4$ is $M = 2\times14.01+4\times19.00=104.02\ g/mol$. For part a, the mass of $N_2F_4$ is $m = 2.42\ g$. The number of moles of $N_2F_4$, $n=\frac{m}{M}=\frac{2.42\ g}{104.02\ g/mol}\approx0.0233\ mol$.

Step2: Determine number of nitrogen atoms

In one molecule of $N_2F_4$, there are 2 nitrogen atoms. Using Avogadro's number $N_A = 6.022\times 10^{23}\ mol^{-1}$, the number of nitrogen - atoms $N = 2\times n\times N_A=2\times0.0233\ mol\times6.022\times 10^{23}\ mol^{-1}\approx2.80\times 10^{22}$ atoms.

Step3: Calculate moles of $N_2F_4$ for part b

The number of molecules of $N_2F_4$ is $N_{molecules}=6.66\times 10^{22}$. The number of moles of $N_2F_4$, $n=\frac{N_{molecules}}{N_A}=\frac{6.66\times 10^{22}}{6.022\times 10^{23}\ mol^{-1}}\approx0.111\ mol$.

Step4: Determine moles of fluorine

In one molecule of $N_2F_4$, there are 4 fluorine atoms. So the number of moles of fluorine, $n_F = 4\times n = 4\times0.111\ mol = 0.444\ mol$.

Step5: Calculate mass of fluorine

The molar mass of fluorine is $M_F=19.00\ g/mol$. The mass of fluorine, $m_F=n_F\times M_F=0.444\ mol\times19.00\ g/mol = 8.44\ g$.

Answer:

a. $2.80\times 10^{22}$ atoms
b. $8.44$ g