What number of atoms of nitrogen are present in 5.00 g of each of the following? $\begin{array}{ll}{\text { a. glycine, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{N}} & {\text { c. calcium nitrate }} \\ {\text { b. magnesium nitride }} & {\text { d. dinitrogen tetroxide }}\end{array}$

Molar mass of: a. Glycine, \(C_2H_5O_2N\): (2 × 12.01)g/mol + (5 × 1.01)g/mol + (2 × 16.00)g/mol + 14.01g/mol = 75.07 g/mol b. Magnesium Nitride, \(Mg_3N_2\): (3 × 24.31)g/mol + (2 × 14.01)g/mol = 100.95 g/mol c. Calcium Nitrate, \(Ca(NO_3)_2\): 40.08g/mol + 2 × (14.01g/mol + 3 × 16.00g/mol) = 164.10 g/mol d. Dinitrogen Tetroxide, \(N_2O_4\): (2 × 14.01)g/mol + (4 × 16.00)g/mol = 92.02 g/mol

a. \(5.00 \,g\, C_2H_5O_2N × \frac{1\, mol C_2H_5O_2N}{75.07\,g\, C_2H_5O_2N} = 0.0666\, mol\, C_2H_5O_2N\) b. \(5.00 \,g\, Mg_3N_2 × \frac{1\, mol Mg_3N_2}{100.95\,g\, Mg_3N_2} = 0.0495\, mol\, Mg_3N_2\) c. \(5.00 \,g\, Ca(NO_3)_2 × \frac{1\, mol \,Ca(NO_3)_2}{164.10\,g\, Ca(NO_3)_2} = 0.0305\, mol\, Ca(NO_3)_2\) d. \(5.00 \,g\, N_2O_4 × \frac{1\, mol \,N_2O_4}{92.02\,g\, N_2O_4} = 0.0543\, mol\, N_2O_4\)

a. In Glycine, 1 mole of Glycine (\(C_2H_5O_2N\)) contains 1 mole of Nitrogen b. In Magnesium Nitride, 1 mole of Magnesium Nitride (\(Mg_3N_2\)) contains 2 moles of Nitrogen c. In Calcium Nitrate, 1 mole of Calcium Nitrate (\(Ca(NO_3)_2\)) contains 2 moles of Nitrogen d. In Dinitrogen Tetroxide, 1 mole of Dinitrogen Tetroxide (\(N_2O_4\)) contains 2 moles of Nitrogen

a. \(0.0666\, mol\, C_2H_5O_2N × \frac{1\, mol\, N}{1\, mol\, C_2H_5O_2N} = 0.0666\, mol\, N\) b. \(0.0495\, mol\, Mg_3N_2 × \frac{2\, mol\, N}{1\, mol\, Mg_3N_2} = 0.0990\, mol\, N\) c. \(0.0305\, mol\, Ca(NO_3)_2 × \frac{2\, mol\, N}{1\, mol\, Ca(NO_3)_2} = 0.0610\, mol\, N\) d. \(0.0543\, mol\, N_2O_4 × \frac{2\, mol\, N}{1\, mol\, N_2O_4} = 0.1086\, mol\, N\) Now, we convert moles of Nitrogen to the number of atoms using Avogadro's number: \(1\, mol\, N = 6.022 × 10^{23}\, atoms\, N\) a. \(0.0666\, mol\, N × 6.022 × 10^{23}\, atoms\, N/mol = 4.01 × 10^{22}\, atoms\, N\) b. \(0.0990\, mol\, N × 6.022 × 10^{23}\, atoms\, N/mol = 5.96 × 10^{22}\, atoms\, N\) c. \(0.0610\, mol\, N × 6.022 × 10^{23}\, atoms\, N/mol = 3.67 × 10^{22}\,atoms\, N\) d. \(0.1086\, mol\, N × 6.022 × 10^{23}\, atoms\, N/mol = 6.54 × 10^{22}\, atoms\, N\) In conclusion, the number of nitrogen atoms in \(5.00 \,g\) of each compound is: a. Glycine: \(4.01 × 10^{22}\, atoms\, N\) b. Magnesium Nitride: \(5.96 × 10^{22}\, atoms\, N\) c. Calcium Nitrate: \(3.67 × 10^{22}\, atoms\, N\) d. Dinitrogen Tetroxide: \(6.54 × 10^{22}\, atoms\, N\)