CHAPTER 2CH7 PROBLEM SOLVING CLASSR.D. A. BOLINAS

2.11 State the mass law(s) demonstrated by the following experimentalresults, and explain your reasoning:

• Experiment 1: A student heats 1.27 g of copper and 3.50 g of iodine to produce 3.81 g of a white compound, and 0.96g of iodine remains.

• Experiment 2: A second student heats 2.55 g of copper and 3.50 g of iodine to form 5.25 g of a white compound, and 0.80 g of copper remains.

• 2.11 These two experiments demonstrate the Law of Definite Composition. In both cases, the ratio of

• (g Cu reacted)/(g I reacted) = 0.50,

so the composition is constant regardless of the method of preparation. They also demonstrate the Law of Conservation of Mass, since in both cases the total mass before reaction equals the total mass after reaction.

2.13 Galena, a mineral of lead, is a compound of the metal withsulfur. Analysis shows that a 2.34g sample of galena contains2.03 g of lead. Calculate the:

• (a) mass of sulfur in the sample;• (b) mass fractions of lead and sulfur in galena; • (c) mass percents of lead and sulfur in galena.

• 2.13 a) 2.34 g compound – 2.03 g lead = 0.31 g sulfur•

• b) Mass fraction Pb = 2.03/2.34 = 0.86752 = 0.868• Mass fraction S = 0.31/2.34 = 0.13248 = 0.13•

• c) Mass % Pb = 0.868 fraction x 100 = 86.752 = 86.8%Mass % S = 0.13 fraction x 100 = 13.248 = 13%

2.17 Show, with calculations, how the following data illustrate thelaw of multiple proportions:

•Compound 1: 77.6 mass % xenon and 22.4 mass % fluorine

•Compound 2: 63.3 mass % xenon and 36.7 mass % fluorine

• 2.17 Compound 1: 77.6% Xe/ 22.4% F = 3.4643 = 3.46• Compound 2: 63.3% Xe / 36.7% F = 1.7248 = 1.72

• Ratio: 3.46/1.72 = 2.0116 = 2.01 / 1.00

• The ratios are in a 2:1 ratio, supporting the Law of Multiple Proportions.

2.33 Magnesium has three naturally occurring isotopes, 24Mg (isotopic mass 23.9850 amu, abundance 78.99%), 25Mg (isotopic mass 24.9858 amu, abundance 10.00%), and 26Mg (isotopic mass 25.9826 amu, abundance 11.01%).

Calculate the atomic mass of magnesium.

• Atomic mass of Mg = • (0.7899)(23.9850 amu)• + (0.1000)(24.9858 amu)• + (0.1101)(25.9826 amu)• ---------------------------------• = 24.3050 • = 24.31 amu

2.61 Give the name and formula of the compound formed from thefollowing elements:

(a) cesium and bromine; (b) sulfur and barium;(c) calcium and fluorine.

• a) CsBr cesium bromide• b) BaS barium sulfide • c) CaF2 calcium fluoride

2.65 Give the systematic names for the formulas or the formulasfor the names:

(a) Na2HPO4

(b) potassium carbonate dihydrate(c) NaNO2(d) ammonium perchlorate

• 2.65• a) sodium hydrogen phosphate • b) K2CO3 2H2O • c) sodium nitrite • d) NH4ClO4

2.67 Correct each of the following names:

(a) CuI is cobalt(II) iodide.(b) Fe(HSO4)3 is iron(II) sulfate.(c) MgCr2O7 is magnesium dichromium heptaoxide.

• 2.67• a) copper(I) iodide, Cu is copper, and since iodide is I–,

this must be copper(I).

• b) iron(III) hydrogen sulfate, HSO4– is hydrogen sulfate,

and this must be iron(III) to be neutral.

• c) magnesium dichromate, Mg forms Mg2+ and Cr2O72– is

named dichromate ion.

2.69 Give the name and formula for the acid derived from each ofthe following anions:

(a) perchlorate, ClO4-

(b) nitrate, NO3-

(c) bromite, BrO2-

 (d) fluoride, F-

• 2.69

• a) perchloric acid, HClO4 • b) nitric acid, HNO3 • c) bromous acid, HBrO2 • d) hydrofluoric acid, HF

2.71 Give the name and formula of the compound whose moleculesconsist of two chlorine atoms and one oxygen atom.

• 2.71

dichlorine monoxide Cl2O

2.73 Give the number of atoms of the specified element in a formula unit of each of the following compounds, and calculate the molecular (formula) mass:

(a) Hydrogen in ammonium benzoate, C6H5COONH4

(b) Nitrogen in hydrazinium sulfate, N2H6SO4

(c) Oxygen in the mineral leadhillite, Pb4SO4(CO3)2(OH)2

• 2.73 a) 9 atoms hydrogen 139.15 amu• b) 2 atoms of nitrogen 130.14 amu• c) 12 atoms of oxygen 1078.9 amu

2.75 Write the formula of each compound, and determine its molecular (formula) mass:

(a) sodium dichromate; (b) ammonium perchlorate; (c) magnesium nitrite trihydrate.

• 2.75 a) Na2Cr2O7 261.98 amu• b) NH4ClO4 117.49 amu• c) Mg(NO2)2•3H2O 170.38 amu