Question

What number of atoms of nitrogen are present in \(5.00 \mathrm{~g}\) of each of the following? a. glycine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}\) b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide

Short answer

In 5.00 grams of each compound, there are approximately: a. Glycine: \(4.01 \times 10^{22}\) nitrogen atoms b. Magnesium nitride: \(5.96 \times 10^{22}\) nitrogen atoms c. Calcium nitrate: \(3.68 \times 10^{22}\) nitrogen atoms d. Dinitrogen tetroxide: \(6.54 \times 10^{22}\) nitrogen atoms

Step-by-step solution

a. Glycine, C₂H₅O₂N

Step 1: Calculate molar mass of glycine The molar mass of glycine is the sum of the molar masses of all its elements: \(M_{glycine} = 2 \times M_{C} + 5 \times M_{H} + 2 \times M_{O} + M_{N}\) Using the standard atomic masses, \(M_{glycine} = 2 \times 12.01 + 5 \times 1.01 + 2 \times 16.00 + 14.01 \approx 75.07 \, g/mol\) Step 2: Calculate moles of glycine Use the given mass and molar mass of glycine to find the number of moles: \(n_{glycine} =\frac{5.00 \, g}{75.07 \, g/mol} \approx 0.0666 \, mol\) Step 3: Calculate number of nitrogen atoms Number of molecules of glycine = moles × Avogadro's number \(\text{Molecules} = 0.0666 \, mol \times 6.022 \times 10^{23} \, mol^{-1} \approx 4.01 \times 10^{22}\, \text{molecules}\) Since there is 1 nitrogen atom in a glycine molecule, Number of nitrogen atoms = number of molecules × nitrogen atoms per molecule \(N_{N} = 4.01 \times 10^{22} \times 1\) So, there are approximately \(4.01 \times 10^{22}\) nitrogen atoms in 5.00 grams of glycine.

b. Magnesium nitride, Mg₃N₂

Step 1: Calculate molar mass of magnesium nitride \(M_{Mg_{3}N_{2}} = 3 \times M_{Mg} + 2 \times M_{N}\) \(M_{Mg_{3}N_{2}} = 3 \times 24.31 + 2 \times 14.01 \approx 100.95 \, g/mol\) Step 2: Calculate moles of magnesium nitride \(n_{Mg_{3}N_{2}} = \frac{5.00 \, g}{100.95 \, g/mol} \approx 0.0495 \, mol\) Step 3: Calculate number of nitrogen atoms Number of formula units of magnesium nitride = moles × Avogadro's number \(\text{Formula units} = 0.0495 \, mol \times 6.022 \times 10^{23} \, mol^{-1} \approx 2.98 \times 10^{22} \, \text{formula units}\) Since there are 2 nitrogen atoms in a formula unit of magnesium nitride, Number of nitrogen atoms = formula units × nitrogen atoms per formula unit \(N_{N} = 2.98 \times 10^{22} \times 2\) So, there are approximately \(5.96 \times 10^{22}\) nitrogen atoms in 5.00 grams of magnesium nitride.

c. Calcium nitrate, Ca(NO₃)₂

Step 1: Calculate molar mass of calcium nitrate \(M_{Ca(NO_{3})_{2}} = M_{Ca} + 2 \times ( M_{N} + 3 \times M_{O} )\) \(M_{Ca(NO_{3})_{2}} = 40.08 + 2 \times ( 14.01 + 3 \times 16.00 ) \approx 164.09 \, g/mol\) Step 2: Calculate moles of calcium nitrate \(n_{Ca(NO_{3})_{2}} = \frac{5.00 \, g}{164.09 \, g/mol} \approx 0.0305 \, mol\) Step 3: Calculate number of nitrogen atoms Number of formula units of calcium nitrate = moles × Avogadro's number \(\text{Formula units} = 0.0305 \, mol \times 6.022 \times 10^{23} \, mol^{-1} \approx 1.84 \times 10^{22} \, \text{formula units}\) Since there are 2 nitrogen atoms in a formula unit of calcium nitrate, Number of nitrogen atoms = formula units × nitrogen atoms per formula unit \(N_{N} = 1.84 \times 10^{22} \times 2\) So, there are approximately \(3.68 \times 10^{22}\) nitrogen atoms in 5.00 grams of calcium nitrate.

d. Dinitrogen tetroxide, N₂O₄

Step 1: Calculate molar mass of dinitrogen tetroxide \(M_{N_{2}O_{4}} = 2 \times M_{N} + 4 \times M_{O}\) \(M_{N_{2}O_{4}} = 2 \times 14.01 + 4 \times 16.00 \approx 92.02 \, g/mol\) Step 2: Calculate moles of dinitrogen tetroxide \(n_{N_{2}O_{4}} = \frac{5.00 \, g}{92.02 \, g/mol} \approx 0.0543 \, mol\) Step 3: Calculate number of nitrogen atoms Number of molecules of dinitrogen tetroxide = moles × Avogadro's number \(\text{Molecules} = 0.0543 \, mol \times 6.022 \times 10^{23} \, mol^{-1} \approx 3.27 \times 10^{22} \, \text{molecules}\) Since there are 2 nitrogen atoms in a molecule of dinitrogen tetroxide, Number of nitrogen atoms = molecules × nitrogen atoms per molecule \(N_{N} = 3.27 \times 10^{22} \times 2\) So, there are approximately \(6.54 \times 10^{22}\) nitrogen atoms in 5.00 grams of dinitrogen tetroxide.