### PS_CHEM7_ch2 - WordPress.com

```CHAPTER 2
CH7 PROBLEM SOLVING CLASS
R.D. A. BOLINAS
2.11 State the mass law(s) demonstrated
by the following experimental
• 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 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.
• 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 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
• 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 the
following elements:
(a) cesium and bromine;
(b) sulfur and barium;
(c) calcium and fluorine.
• a) CsBr
• b) BaS
• c) CaF2
cesium bromide
barium sulfide
calcium fluoride
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
• 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 of
the 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 molecules
consist 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
•
•
b) 2 atoms of nitrogen
c) 12 atoms of oxygen
139.15 amu
130.14 amu
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
```