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Unformatted text preview: 4 (b) CHgCHzOH (d) CH3(CH2)3COZH (e) CH2(OH)CH2C1
” tr ttttﬁ 3H
H—(il—(IZ—OH H-—(|3—-(l3-—([3-—Cll——C-—OH Ho—c—o—CI
H H H H H H H H 4mol P 30.974g P 12. (a) P mass = 6.25 x I0"2mol R, x
lmol P4 lmol P =‘ 7.74gP (b) Stearic acid mass = 4.03 x 102‘1 molecules x + x 6.022 x10 molecules 1 mole CmH3602 = 1.90>< 103 g stearic acid. lmol cﬁnmNzo2 X 146.19g CGHHNzoz (c) mass = 3.03 mole
2 molN lmol C‘SHmNzO2 = 221.6 g 06111461202 g. (a) Cu(U02)2(PO4)2-8H20 has 1 Cu, 2 U, 2 P, 20 O, 16 H or a total of 41 atoms. (b) By number, Cu(U02)2(P04)2-8H20 has a H to 0 ratio of 16:20 or 4:5
or 0.800 H atoms/0 atom. (c) By number CuCUOz)2(PO4)2-8H20 has a Cu to P ratio of 1:2 The mass ratio of Cu:P is
63.546 g Cu lmol Cu
30.9738 g P lmolP _ lmol Cux = 1.026
2 [1101 P x (d) With a mass percent slightly greater than 50%, U has the largest mass percent
with oxygen coming in at ~34%. 1 mol P 1 mol CuGJOz)2(PO4)2-8HZO X 937.666 g Cu(U02)2 (1)04)2 431120
(e) 1'00 g P x 30.9738 g P X 2 me] P 1 mol Cu(U02)2(PO4)2-8H20 = 15.1 g of Cu(UOg)2(PO4)2-8H20 22. Determine the mass of a mole of Cr(NO3 )3 -9H20 , and then the mass otlwater in a mole. molar mass Cr(No, )3-9H10 = 51.9961g Cr +(3><14.0067g N)+(18x15.9994g 0)
+(1 8x1.00794g H) = 400.148g/mol Cr(N03)3 X9H20 9molH O 18.0153gH20
H o = —————L—x—M-—— = 162.14gH o molCr NO -9H 0
mass 2 1moICr(N03)3>_<9H10 lmoIHZO 1 l ( 3)3 _ 2
162.14g HZO/molC1-(NO3 )3 .9Hzo x100%=40.52 %H20
400.148g/m01Cr(N03 )3 -9H20 3_5_. Conyen each percentage into the mass in 100.00 g, and then to the moles of that element. 94-34 a Cx—lﬂ = 7.854 mo] 0 +5.615 = 1.40m01 C x5 = 7.00
12.011 g c "
5.66 g Hx—--———— = 5.61§m01 H +5.615 = 1.00 mol RX 5 = 5.00
1.00794g H - Multiply by 5 to achieve whole nurnber ratios. The empirical formula is C5 H7 , and formula mass [(7 x 12.01] g C)+(5 x 1.00794 g H)] = 89.1 17 11. Since this empirical molar mass is
one—half of the 178 u, the correct molecular mass, the molecular formula must be twice the empirical formula. Molecular formula: CMHm ﬂ First determine the mass of carbon and hydrogen present in the sample. lmol CO2 lmol C = 0.0104molcx 12.011 3C 0.458 C0 -—x
g 2x44.01g co2 imolco2 1:116ch =0.125 gC lmol H20 2mol H = 0041511101 Hx 1.00794g H ——-x = 0.04135 g H
13.0153 gH20 lmol H20 lmol H ' 0.374 g HIOx Then, the mass of N that this sample would have produced is determined.
(Note that this is also the mass of N2 produced in the reaction.) 0.312g lst sample
0.486 g 2nd sample
From which we can calculate the mass of N in the sample. lrnolN2 x 2molN x14.00671gN'
28.0134gN2 ImolN2 lmolN 0.226 g N2 x = 0. 145 g N2 0.145 g Nzx = 0.145 gN .1; We may alternatively determine the mass of N by difference:
0.312 g— 0.125 g C— 0.0418; g H = 0.145 g N Then, we can calculate the relative number of moles of each element. 0.145ngL01N—= 0.01035 molN +0.0103§->1.0_Q_m01N 14.0067g N . .
Thus, the empirical 0.0104 11101 C +0.0103§ —> 1.00mol C ,
' formula is CH4N 0.0415 mo] H +0.0103§ —> 4.01mol H a -, The oxidation state (0.8.) is given ﬁrst, followed by the explanation for its assignment. (a) C = —4 in CH 4 H has an oxidation state of +1 in its non-metal compounds
(Remember that the sum of the oxidation states in a neutral compound equals 0.)
(b) S = +4 in SE1 F has 0. S. = -—1 in its compounds.
(c) O = —1 in NaIO2 Na has 0. S. = +1 in its compounds. ((1) C —_- 0 in (121130; H has 0. S. = +1 in its non-metal compounds; that of O = —2
(usually). (Remember that the sum of the oxidation states in a
polyatomic ion equals the charge on that ion.) (e) Fe = +6 in F6042" O has .0. S. = -2 in most of its compounds (especially metal
containing compounds). 56. (a) 3510403)2 barium nitrate (b) HINO2 nitrous acid
(c) CrO2 chromitun(IV) oxide (d) K103 potassium iodate
(e) LiCN lithium cyanide (1') K10 potassium hypoiodite
(g) 174011): iron(II) hydroxide (h) Can—[21304 )2 calcium dihydro gen
- q ‘ phosphate
(0 HgPO}. Phosphoric add (i) NaHSO4 j sodium hydrogen
(k) Nazcr207 sodium dichromate (1) NH 4 C2 HJO2 ammonium acetate
(111) Mng4 magnesium oxalate (n) Nazczo4 sodium oxalate 68. The increase in mass of the solid is the result of each mole of the solid absorbing ten moles of water.
lmolNa2304 x lOmol H20 added X 18.0153 g H20 142.043 g NaZSO4 lmol NaZSO4 lmol H20
= 30.50 g H20 added increase in mass = 24.05 g Nazso4 x 74. Molecules (a), (b) and (c) are structural isomers. They share a common formula, namer
C5HHCI, but have different molecular structures. Molecule (d) has a diﬁ'erent“chemlcal formula (C5H13C1) and hence cannot be classiﬁed as an isomer. ...
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This note was uploaded on 10/25/2009 for the course CHEM 2090 at Cornell.