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Chapter 3: Problem 23

Determine the formula weights of each of the following compounds: (a) lead (IV) chloride; (b) copper(II) oxide; (c) iodic acid, $\mathrm{HIO}_{3} ;(\mathbf{d})\( sodium perchlorate, \)\mathrm{NaClO}_{4} ;$ (e) indium nitride, (f) phosphorus pentoxide, \(\mathrm{P}_{4} \mathrm{O}_{10} ;(\mathbf{g})\) boron trichloride.

Short Answer

Expert verified
The formula weights for the given compounds are: (a) Lead(IV) chloride: \(349 \, g/mol\) (b) Copper(II) oxide: \(79.546 \, g/mol\) (c) Iodic acid: \(175.912 \, g/mol\) (d) Sodium perchlorate: \(122.44 \, g/mol\) (e) Indium nitride: \(128.825 \, g/mol\) (f) Phosphorus pentoxide: \(283.896 \, g/mol\) (g) Boron trichloride: \(117.17 \, g/mol\)

Step by step solution

01

(1) Lead(IV) chloride#

: First, we will write the chemical formula for lead(IV) chloride. The Roman numeral (IV) indicates the oxidation state of lead, which is +4. Chlorine has an oxidation state of -1. Thus, we need to balance the charges to get PbCl4. Now, we will calculate the formula weight: \(PbCl_4 = 1 \times (Pb) + 4 \times (Cl) \\ = 1 \times 207.2 + 4 \times 35.45 \\ = 207.2 + 141.8 \\ = 349 \, g/mol\)
02

(2) Copper(II) oxide#

: First, we will write the chemical formula for copper(II) oxide. The Roman numeral (II) indicates the oxidation state of copper, which is +2. Oxygen has an oxidation state of -2. Thus, we have CuO. Now, we will calculate the formula weight: \(CuO = 1 \times (Cu) + 1 \times (O) \\ = 1 \times 63.546 + 1 \times 16 \\ = 63.546 + 16 \\ = 79.546 \, g/mol\)
03

(3) Iodic acid (\(HIO_3\))#

: The chemical formula for iodic acid is given as \(HIO_3\). We need to calculate its formula weight: \(HIO_3 = 1 \times (H) + 1 \times (I) + 3 \times (O) \\ = 1 \times 1.008 + 1 \times 126.904 + 3 \times 16 \\ = 1.008 + 126.904 + 48 \\ = 175.912 \, g/mol\)
04

(4) Sodium perchlorate (\(NaClO_4\))#

: The chemical formula for sodium perchlorate is given as \(NaClO_4\). We need to calculate its formula weight: \(NaClO_4 = 1 \times (Na) + 1 \times (Cl) + 4 \times (O) \\ = 1 \times 22.99 + 1 \times 35.45 + 4 \times 16 \\ = 22.99 + 35.45 + 64 \\ = 122.44 \, g/mol\)
05

(5) Indium nitride#

: First, we write the chemical formula for indium nitride. Indium typically has an oxidation state of +3 and nitrogen has an oxidation state of -3, so we have InN. Now, we will calculate the formula weight: \(InN = 1 \times (In) + 1 \times (N) \\ = 1 \times 114.818 + 1 \times 14.007 \\ = 114.818 + 14.007 \\ = 128.825 \, g/mol\)
06

(6) Phosphorus pentoxide (\(P_4O_{10}\))#

: The chemical formula for phosphorus pentoxide is given as \(P_4O_{10}\). We need to calculate its formula weight: \(P_4O_{10} = 4 \times (P) + 10 \times (O) \\ = 4 \times 30.974 + 10 \times 16 \\ = 123.896 + 160 \\ = 283.896 \, g/mol\)
07

(7) Boron trichloride#

: First, we write the chemical formula for boron trichloride. Boron typically has an oxidation state of +3 and chlorine has an oxidation state of -1, so we have BCl3. Now, we will calculate the formula weight: \(BCl_3 = 1 \times (B) + 3 \times (Cl) \\ = 1 \times 10.81 + 3 \times 35.45 \\ = 10.81 + 106.35 \\ = 117.17 \, g/mol\) Formula weights for the given compounds are as follows: (a) Lead(IV) chloride: 349 g/mol (b) Copper(II) oxide: 79.546 g/mol (c) Iodic acid: 175.912 g/mol (d) Sodium perchlorate: 122.44 g/mol (e) Indium nitride: 128.825 g/mol (f) Phosphorus pentoxide: 283.896 g/mol (g) Boron trichloride: 117.17 g/mol

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Most popular questions from this chapter

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{NaClO}_{3}(s) \longrightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\) (b) $\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)$ (c) \(\mathrm{K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)\) (d) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l)$

(a) Ibuprofen is a common over-the-counter analgesic with the formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2} .\) How many moles of \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\) are in a 500-mg tablet of ibuprofen? Assume the tablet is composed entirely of ibuprofen. (b) How many molecules of $\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}$ are in this tablet? (c) How many oxygen atoms are in the tablet?

Balance the following equations: (a) $\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(l)$ (b) $\mathrm{Au}_{2} \mathrm{~S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Au}(s)+\mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{Ba}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(a q) \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)$ (d) $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$

Determine the formula weights of each of the following compounds: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) which is responsible for the rotten smell of spoiled food; (b) sodium perborate, \(\mathrm{NaBO}_{3}\), a substance used as bleach; (c) calcium carbonate, \(\mathrm{CaCO}_{3},\) a substance found in marble. (c) $\mathrm{CF}_{2} \mathrm{Cl}_{2},\( a refrigerant known as Freon; \)(\mathbf{d}) \mathrm{NaHCO}_{3},$ known as baking soda and used in bread and pastry baking; \((\mathbf{e})\) iron pyrite, \(\mathrm{FeS}_{2}\) which has a golden appearance and is known as "Fool's Gold."

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is $0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\( If the HCN is formed by reaction of \)\mathrm{NaCN}$ with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains $850 \mathrm{~g}\( of Acrilan \)^{\circledast}$ fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.
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