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Chapter 3: Problem 67

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

Expert verified
The number of nitrogen atoms in 5.00 g of each compound are as follows: a. Glycine: \(4.01\times10^{22}\) N atoms b. Magnesium Nitride: \(5.96\times10^{22}\) N atoms c. Calcium Nitrate: \(3.67\times10^{22}\) N atoms d. Dinitrogen Tetroxide: \(6.54\times10^{22}\) N atoms

Step by step solution

01

Determine the molar mass of each compound

We will use the periodic table to find the molar mass of each element in the compound, then add them together to find the total molar mass of each compound. a. Glycine: C₂H₅O₂N Molar Mass = (2 × C) + (5 × H) + (2 × O) + N Molar Mass = (2 × 12.01) + (5 × 1.01) + (2 × 16.00) + 14.01 Molar Mass = 75.07 g/mol b. Magnesium Nitride: Mg₃N₂ Molar Mass = (3 × Mg) + (2 × N) Molar Mass = (3 × 24.31) + (2 × 14.01) Molar Mass = 100.95 g/mol c. Calcium Nitrate: Ca(NO₃)₂ Molar Mass = Ca + (2 × (N + (3 × O))) Molar Mass = 40.08 + (2 × (14.01 + (3 × 16.00))) Molar Mass = 164.10 g/mol d. Dinitrogen Tetroxide: N₂O₄ Molar Mass = (2 × N) + (4 × O) Molar Mass = (2 × 14.01) + (4 × 16.00) Molar Mass = 92.02 g/mol
02

Determine the molar mass of nitrogen

From the periodic table, the molar mass of nitrogen (N) is 14.01 g/mol.
03

Determine the moles of each compound

We will now use the 5.00 g of each compound and divide it by their respective molar masses to find the moles of each compound. a. Glycine: Moles = 5.00 g / 75.07 g/mol = 0.06664 mol b. Magnesium Nitride: Moles = 5.00 g / 100.95 g/mol = 0.04953 mol c. Calcium Nitrate: Moles = 5.00 g / 164.10 g/mol = 0.03046 mol d. Dinitrogen Tetroxide: Moles = 5.00 g / 92.02 g/mol = 0.05433 mol
04

Determine the number of nitrogen atoms in each mole of the compound

a. Glycine: 1 nitrogen atom per mole b. Magnesium Nitride: 2 nitrogen atoms per mole c. Calcium Nitrate: 2 nitrogen atoms per mole d. Dinitrogen Tetroxide: 2 nitrogen atoms per mole
05

Calculate the total number of nitrogen atoms in 5.00 g of each compound

We will multiply the moles of each compound by the number of nitrogen atoms in each mole of the compound, and then by Avogadro's number (6.022 x 10²³) to find the total number of nitrogen atoms. a. Glycine: \(0.06664\text{ mol } \times 1 \times 6.022\times10^{23}\) atoms = \(4.01\times10^{22}\) N atoms b. Magnesium Nitride: \(0.04953\text{ mol } \times 2 \times 6.022\times10^{23}\) atoms = \(5.96\times10^{22}\) N atoms c. Calcium Nitrate: \(0.03046\text{ mol } \times 2 \times 6.022\times10^{23}\) atoms = \(3.67\times10^{22}\) N atoms d. Dinitrogen Tetroxide: \(0.05433\text{ mol } \times 2 \times 6.022\times10^{23}\) atoms = \(6.54\times10^{22}\) N atoms

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Most popular questions from this chapter

An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right)\). Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\mathrm{X}\).

You take \(1.00 \mathrm{~g}\) of an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen), burn it in air, and collect \(2.20 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.400 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). You know that the molar mass of aspirin is between 170 and \(190 \mathrm{~g} / \mathrm{mol}\). Reacting 1 mole of salicylic acid with 1 mole of acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) gives you 1 mole of aspirin and 1 mole of acetic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\right.\) ). Use this information to determine the molecular formula of salicylic acid.

Bacterial digestion is an economical method of sewage treatment. The reaction \(5 \mathrm{CO}_{2}(g)+55 \mathrm{NH}_{4}^{+}(a q)+76 \mathrm{O}_{2}(g) \stackrel{\text { hacteria }}{\longrightarrow}\) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}(s)+54 \mathrm{NO}_{2}^{-}(a q)+52 \mathrm{H}_{2} \mathrm{O}(t)+109 \mathrm{H}^{+}(a q)\) bacterial tissue is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is produced in a treatment plant for every \(1.0 \times 10^{4} \mathrm{~kg}\) of wastewater containing \(3.0 \% \mathrm{NH}_{4}{ }^{+}\) ions by mass? Assume that \(95 \%\) of the ammonium ions are consumed by the bacteria.

In the production of printed circuit boards for the electronics industry, a \(0.60\) -mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{~cm}\) in area. An average of \(80 . \%\) of the copper is removed from each board (density of copper \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

When \(\mathrm{M}_{2} \mathrm{~S}_{3}(s)\) is heated in air, it is converted to \(\mathrm{MO}_{2}(s)\). A \(4.000-\mathrm{g}\) sample of \(\mathrm{M}_{2} \mathrm{~S}_{3}(s)\) shows a decrease in mass of \(0.277 \mathrm{~g}\) when it is heated in air. What is the average atomic mass of M?
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