Chem100_11. 2011/2012
Assignment #2 CHAPTERS 2 and 3|
1. {10} How many protons, neutrons, and electrons are in the following atoms?
What element symbols do Q, X, Y, Z, and @ represent?
a) 7 protons, 7 neutrons, 7 electrons, N
b) 20 protons, 21 neutrons, 20 electrons, Ca
c) 53 protons, 78 neutrons, 53 electrons, I
d) 14 protons, 16 neutrons, 14 electrons, Si
e) 8 protons, 9 neutrons, 8 electrons, O
2. {5} Silver (Ag; Z=47) has two naturally occurring isotopes, 107Ag (106.90509 amu) and 109Ag (108.90476amu). Calculate the fractional abundance of the isotopes if the average atomic mass of silver is 107.8682 amu.
107.8682amu = 106.90509 amu x X + 108.90476 amu x (1-X)
107.8682 -108.90476 = 106.90509X-108.90476X
-1.0366 = -1.9997X; X= 0.5184 ; 1-X = 0.4816
0.5184 of light isotope (107) and 0.4816 of heavy isotope (109)
53I (s), nonmetal ; 80Hg(l), metal; 35Br,
(l), nonmetal; 14Si, metalloid, (s); 18Ar, nonmetal, (g).
Note: atomic number as a superscript is also
acceptable, e.g., 53I.
1A 2A 6A 7A 8A
Alkali metals, alkaline earths metals, chalcogens, halogens, noble gases.
(a) Bromine Br-
(b) Magnesium Mg2+
(c) Sodium Na+
(d) Phosphorus P3-
(e) Aluminum Al3+
(a) sodium and chlorine NaCl sodium chloride
(b) calcium and fluorine CaF2 calcium fluoride
(c) iodine and aluminum AlI3 aluminum iodide
(d) sulfur and potassium K2S, potassium sulfide
(e) lithium and oxygen Li2O, lithium oxide.
(a) chromium (II) nitrate Cr (NO3)2
(b) FeI2 ferrous iodide or iron (II) iodide
(c) Ferric oxide Fe2O3
(d) ScS scandium (II) sulfide
(e) sodium dihydrogen phosphate NaH2PO4
(f) Ag2S Silver (I) sulfide
(g) FeCl3 iron (III) chloride or ferric chloride
(h) Mercurous fluoride Hg2F2
(i) Cu(ClO4)2 cupper (II) perchlorate or cupric perchlorate
(j) Potassium dichromate K2Cr2O7
(k) Dichlorine trioxide Cl2O3
(l) SF4 Sulfur tetrafluoride
(m) Dinitrogen pentoxide N2O5
(n) Sulfuric acid H2SO4
(o) Nitrous acid HNO2
mol C2H2O2 = 65g x 1mol/58g= 1.12 mol;
mol O = 2x1.12= 2.24mol
g O = 2.24 mol x 16g/1mol = 36g
mol C: 1.24 g x 1mol/(12.01 +32)g = 0.02818 mol C; gC: 0.02818mol x 12.01g/1mol = 0.3384g C
molH: 0.255g x 1mol /(2x1.008 +16) x2= 0.02831mol H; = 0.02853g H
mol O: 0.515 g – 0.3384g -0.02853g = 0.149g; 0.149g/16 =0.00928 mol
0.02818/0.00928 = 3.04; 0.02831/0.00928=3.05
C3H3O
S(s) + 3 F2 (g) à SF6 (g)
You run this reaction in the lab using 2.0 g of S and 2.0 g of F2 and obtained 1.5 g of SF6. What is the percent yield of this reaction?
mol S= 2.0 /32.1 = 0.0623mol; mol F2 = 2.0/38 =0.0526mol; F2 is limiting
theoretical yield: 0.0526 mol /3 = 0.0175 mol SF6 = 0.0175x (32+6x19) = 2.56g
%yield= 1.5/2.56 x100 = 58.6%= 59%
assume 100g of compound
mol Na: 29g /23g = 1.26 mol
mol S: 41/32 = 1.28 mol
mol O: 30/16 = 1.88 mol
1.28/1.26 =1.01; 1.88/1.26 =1.5 1, 1, 1.5 multiply by 2
2:2:3
Na2S2O3
GeH4 + 3GeF4 à 4GeF3H
If the reaction yield is 92.6%, how many grams of GeF4 are needed to produce 8.00 moles of GeF3H?
theoretically, for 8 mol of GeF3H, 6 mol GeF4 are needed
actually: 6 mol/0.926 = 6.48 mol of GeF4 ; 6.48mol x (72.6 + 4x19)g/mol = 962.9 g