1. CH7 PROBLEM SOLVING CLASS R.D. A. BOLINAS
Chapter 2
CH7 PROBLEM SOLVING CLASS
R.D. A. BOLINAS

2. 2.11 State the mass law(s) demonstrated by the following experimental results, 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.

3. 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.

4. 2.13 Galena, a mineral of lead, is a compound of the metal with sulfur. Analysis shows that a 2.34g sample of galena contains 2.03 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.

5. 2.13 a) 2.34 g compound – 2.03 g lead = 0.31 g sulfur
b) Mass fraction Pb = 2.03/2.34 = = 0.868
Mass fraction S = 0.31/2.34 = = 0.13
c) Mass % Pb = fraction x 100 = = 86.8%
Mass % S = 0.13 fraction x 100 = = 13%

6. 2.17 Show, with calculations, how the following data illustrate the law 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

7. 2.17 Compound 1: 77.6% Xe/ 22.4% F = = 3.46
Compound 2: 63.3% Xe / 36.7% F = = 1.72
Ratio: 3.46/1.72 = = 2.01 / 1.00
The ratios are in a 2:1 ratio, supporting the Law of Multiple Proportions.

8. 2.33 Magnesium has three naturally occurring isotopes, 24Mg (isotopic mass amu, abundance 78.99%), 25Mg (isotopic mass amu, abundance 10.00%), and 26Mg (isotopic mass amu, abundance 11.01%). Calculate the atomic mass of magnesium.

9. Atomic mass of Mg =
(0.7899)( amu)
+ (0.1000)( amu)
+ (0.1101)( amu) =
= amu

10. 2.61 Give the name and formula of the compound formed from the following elements: (a) cesium and bromine; (b) sulfur and barium; (c) calcium and fluorine.

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

12. 2.65 Give the systematic names for the formulas or the formulas for the names: (a) Na2HPO4 (b) potassium carbonate dihydrate (c) NaNO2 (d) ammonium perchlorate

13. 2.65
a) sodium hydrogen phosphate
b) K2CO3  2H2O
c) sodium nitrite
d) NH4ClO4

14. 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.

15. 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.

16. 2.69 Give the name and formula for the acid derived from each of the following anions: (a) perchlorate, ClO4- (b) nitrate, NO3- (c) bromite, BrO2-  (d) fluoride, F-

17. 2.69
a) perchloric acid, HClO4
b) nitric acid, HNO3
c) bromous acid, HBrO2
d) hydrofluoric acid, HF

18. 2.71 Give the name and formula of the compound whose molecules consist of two chlorine atoms and one oxygen atom.

19. dichlorine monoxide Cl2O
2.71
dichlorine monoxide Cl2O

20. 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

21. 2.73 a) 9 atoms hydrogen amu
b) 2 atoms of nitrogen amu
c) 12 atoms of oxygen amu

22. 2.75 Write the formula of each compound, and determine its molecular (formula) mass: (a) sodium dichromate; (b) ammonium perchlorate; (c) magnesium nitrite trihydrate.

23. 2.75 a) Na2Cr2O amu
b) NH4ClO amu
c) Mg(NO2)2•3H2O amu