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Chapter 3: Problem 6

Determine the mass, in grams, of (a) \(7.34 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}\) (b) \(3.16 \times 10^{24} \mathrm{O}_{2}\) molecules; (c) \(18.6 \mathrm{mol} \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (d) \(4.18 \times 10^{24}\) molecules of \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
The masses are (a) 675.47 g \(N_{2}O_{4}\), (b) 168.00 g \(O_{2}\), (c) 4644.25 g \(CuSO_{4} \cdot 5H_{2}O\), and (d) 431.05 g \(C_{2}H_{4}(OH)_{2}\)

Step by step solution

01

Calculate the mass of \(7.34 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}\)

First, calculate the molar mass of \(N_{2}O_{4}\) by adding the molar masses of its constituent atoms. The molar mass is \(2(14.01 \, g/mol) + 4(16.00 \, g/mol) = 92.02 \, g/mol\). Multiply the given number of moles by the molar mass to find the mass: \(7.34 \, mol \times 92.02 \, g/mol = 675.47 \, g\)
02

Calculate the mass of \(3.16 \times 10^{24}\) O2 molecules

First, convert the number of molecules to moles by dividing by Avogadro's number, \(6.022 \times 10^{23} mol^{-1}\): \(3.16 \times 10^{24} molecules / 6.022 \times 10^{23}\, mol^{-1} = 5.25 \, mol\). The molar mass of \(O_{2}\) is \(2(16.00 g/mol) = 32.00 g/mol\). Multiply the number of moles by the molar mass to find the mass: \(5.25 \, mol \times 32.00 \, g/mol = 168.00 \,g\)
03

Calculate the mass of \(18.6 \, mol CuSO_{4} \cdot 5H_{2}O\)

First calculate the molar mass of \(CuSO_{4} \cdot 5H_{2}O\): \((63.55 g/mol) + (32.07 g/mol) + 4(16.00 g/mol) + 5((2.02 g/mol) + (16.00 g/mol)) = 249.68 g/mol\). Then multiply the given number of moles by the molar mass to find the mass: \(18.6 mol \times 249.68 g/mol = 4644.25 g\)
04

Calculate the mass of \(4.18 \times 10^{24}\) C2H4(OH)2 molecules

First convert the number of molecules to moles: \(4.18 \times 10^{24} molecules / 6.022 \times 10^{23} \, mol^{-1} = 6.94 mol\). Then calculate the molar mass of \(C_{2}H_{4}(OH)_{2}\): \(2(12.01 g/mol) + 4(1.01 g/mol) + 2((16.00 g/mol) + (1.01 g/mol) + (1.01 g/mol)) = 62.07 g/mol\). Multiply the number of moles by the molar mass to find the mass: \(6.94 mol \times 62.07 g/mol = 431.05 g\)

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Most popular questions from this chapter

A public water supply was found to contain 1 part per billion (ppb) by mass of chloroform, \(\mathrm{CHCl}_{3}\) (a) How many \(\mathrm{CHCl}_{3}\) molecules would be present in a \(225 \mathrm{mL}\) glass of this water? (b) If the \(\mathrm{CHCl}_{3}\) in part (a) could be isolated, would this quantity be detectable on an ordinary analytical balance that measures mass with a precision of ±0.0001 g?

Name the following compounds and specify which ones are best described as ionic: (a) \(\mathrm{OF}_{2} ;\) (b) \(\mathrm{XeF}_{2}\) (c) \(\operatorname{CuSO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\)

Determine the empirical formula of (a) the rodenticide (rat killer) warfarin, which consists of \(74.01 \% \mathrm{C}\) \(5.23 \% \mathrm{H},\) and \(20.76 \%\) O, by mass; (b) the antibacterial agent sulfamethizole, which consists of \(39.98 \%\) C. \(3.73 \% \mathrm{H}, 20.73 \% \mathrm{N}, 11.84 \% \mathrm{O},\) and \(23.72 \% \mathrm{S},\) by mass.

A hydrate of copper(II) sulfate, when heated, goes through the succession of changes suggested by the photograph. In this photograph, (a) is the original fully hydrated copper(II) sulfate; (b) is the product obtained by heating the original hydrate to \(140^{\circ} \mathrm{C}\) (c) is the product obtained by further heating to \(400^{\circ} \mathrm{C}\) and (d) is the product obtained at \(1000^{\circ} \mathrm{C}\) A \(2.574 \mathrm{g}\) sample of \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) was heated to \(140^{\circ} \mathrm{C},\) cooled, and reweighed. The resulting solid was reheated to \(400^{\circ} \mathrm{C},\) cooled, and reweighed. Finally, this solid was heated to \(1000^{\circ} \mathrm{C},\) cooled, and reweighed for the last time. $$ \text {Original sample } \quad \quad\quad\quad\quad \text {\(2.574 \mathrm{g}\) } $$ $$ \text {After heating to \(140^{\circ} \mathrm{C}\) } \quad \quad\quad\quad\quad \text {\(1.833 \mathrm{g}\) } $$ $$ \text {After reheating to \(400^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(1.647 \mathrm{g}\) } $$ $$ \text {After reheating to \(1000^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(0.812 \mathrm{g}\)} $$ (a) Assuming that all the water of hydration is driven off at \(400^{\circ} \mathrm{C},\) what is the formula of the original hydrate? (b) What is the formula of the hydrate obtained when the original hydrate is heated to only \(140^{\circ} \mathrm{C} ?\) (c) The black residue obtained at \(1000^{\circ} \mathrm{C}\) is an oxide of copper. What is its percent composition and empirical formula?

Name the acids: (a) \(\mathrm{HClO}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3} ;\) (c) \(\mathrm{H}_{2} \mathrm{Se}\) (d) HNO \(_{2}\).
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