Chapter One Scientific Measurements
Pra cti ce Exer cises
I. I The scie ntific n1ethod is an iterative process of gath ering information through making observations and collecting data and then for mula ting expla na tions that lead to a conclusion.
1. 2 (a) elen1ent (d) mixture , heterogeneous
1.3 (a) chen1ical c han ge (c) phys ical cha nge
1.4 (a) intensive (c) intensive
(b) mixture, hon1ogeneous (e) element
(b) physical change
(d) phy sical change
(b) extensive ( d) extensive (c) compoun d
1.5 V = 4 ; tr3 , the S I unit for rad ius, r, is n1eters, the nun1bers and 1t do not have units. Therefore, 3 3 the SI unit for volun1e is meter3 or m 3
1.6 Force equals n1ass x acceleration (F = ,na), and acceleratio n eq ua ls chan ge in velocity divided by h ( change in v ) d I I d d d d b ( d ) p h c ange 111 ttme a = . , an ve oc1ty equa s 1stance 1v 1 e y time v = - . ut t e change 111 t t equations together:
F= 111 (change in v) change 111 t c ange 1111 change in d h d ( ) F= ,n ---~ = ,n change int changeint 2
The unit for mass is kilogran1 (kg); the unit fo r di stance is n1eter (m) and the unit for time is second (s) Substitute t he units into the equatio n above:
Unit for force in S I base units= kg (;) or kg m s2 J. 7 (F =( : :~)(C+32°f=( : :~)(3 55 °C) +32 °f=67 JOf
1. 8 f c = (t F-32 °F)(! :~)=(ss°F-32 °F)(! :~)= 13°C
To convert fro m ° F to K we first convert to °C (C = (t f-32 Of)( !:~)=(68 °f-32 Of)(! :~)=20 oc
TK = (2 73 °C + tc)( / 0 ) = (27 3 ° C + 20 °C)( i~) = 293 K
1.9
(a) (b) 21.0233 g + 2 1. 0 g = 42.0233 g: round ed co rrect ly to 42.0 g 10.0324 g / 11.7 mL = 0 8 574 g I mL: rounded correctly to 0.857 g/n1L
(c)
{4.25 cm x 12. 334 c) = 148.57 cm : rou nded correctly to 149 c111 2.223 cm - 1.04 cm
1. 10 (a) 54. 183 g - 0.0278 g = 54. 155 g (b) I 0.0 g + 1.03 g + 0.2 4 3 g = 11. 3 g (roun ded after addi ng) (c) 43.4 in x ( 1 ) = 3.62 ft. ( I and 12 a re exact numbers) 12 m
(d) J.03 m x 2.074 m x 2.9 m O36 2 = 111 12.46 m + 4.778 m
I. I I m2 = ( 124ft2}(30.48cm)2( I m )2 = J l 5m2 I ft 100 cm
1. 12 (a) in . 3 =(3.00yd3 )( 1 3 y:) 3 ( 1 ~:) 3 =140,000 in. 3 = l.4 0x J05 in 3 cm = ( 1. 25 km)( l ~Oimm )( 10 1°;m) = 1. 25 x 105 cm
13
14
15 (b) (c) (d) g= (3.27 02)( 2~·!! g) =92.7 g km = ( 20.2 mile)( 1.609 km)( I gal ) = 8 59 km/ L I gal I mi le 3.785 L /L
Density= mass0 244 g = 0.0 163 g/111L = 0.0 163 g cm-3 vo lume 15.0 mL D 1 mass mass 0.547 g 2 2 L-' 2 2 -3 ens, y = - = = gm = . g cm vo lume v1 -v; (5.95mL - 5.70mL) . mass Density= volume 365 " 3
Density of the obJect = "' = 16.5 g/cm 22.12cm 3
The obj ect is not co111posed of p ure gold s ince the density of gold is 19 .3 g/c111 3
1. 16 The density of th e alloy is 12.6 g/c m 3 To determ ine the mass of the 0.822 ft 3 samp le of the a ll oy, first convert the de nsity fro111 g/c m 3 to lb ./ft3, the n fi nd t he weight. 3
Density in lb ./ ft 3 = 12 6 g ( 1 lb )( 30 48 cm) = 787 lb ./ft 3 cm3 453.6 g I ft
Mass of sa111 p le alloy= (0.822 ft3) (787 lb ./ ft 3 ) = 647 lb. ·fi1 density of substance 1. 17 spec , ,c gr av1 y = -~. density of water
1 090 = density of wine ' 62.4 lb ft ' density of w ine= 1.090 x 62 .4 lb. ft3 = 68.02 lb . ft3
1. 18 fi 1 density of substance s pec, 1c g r av1 y = -----''-density of water
d . f 1. 00 g( I oz )(29 574 mL) I 043 / I. ·d ens1ty o water= = . oz. 1qu1 oz.
The specific gravity of urine is be low the norma l range.
Revi e, v Pro bl e ms
1.39 ( a) Physica l change. Copper does not change chemically when electricity flows through it: It
1. 41 ren1a1ns copper.
(b) Physical change. Galliun1 is changes its state, not its chen1ical composition when it melts.
(c) Chen1ical change. Th is is an exan1ple of the Maillard reaction describing the chemical reaction of sugar mo lecules and amino acids.
( d ) Chen1ical change. W i ne contains ethano l which can be converted to acetic acid.
(e) Chen1ical change. Concrete is composed of many di ffe rent substances that undergo a chen1ical process called hydration when water is added to it.
(a) (b) (c) ( d )
Hydrogen is a gas at roon1 ten1peratu re. A l un1inum is a solid at room temperature. N itrogen is a gas at roon1 ten1perature.
Mercury is a li quid at roon1 temperatu re. 1.43 (a) 0.0 I 0. I (b)
1.45 ( a) tp = (tc) + 32 °f = ( 57 °C) + 32 ° f = 135 °f when rounded to the p roper (9°F) (9 °F) 5 °c 5 °c nun1ber of s ignificant figures.
(b) le=(!:~) (tp- 32 °f) = (! :~) (-25.5 Of - 32 °f) = -31.9 oc (1°c) (1°c) (c) tc = (TK - 273 K) lK = ( 378 K- 273 K ) lK = 105 °C
(d) TK=(tc + 273°C)(/ 0~) =(-31 + 273) (i'~) =242K
1. 47 Ten1pe rature in °C: (1°c) 6 ( 1°c) 6 tc=(TK-273 K) lK =( 15.7 x 10 K-273 K) lK = 15.7 x 10 °C
Ten1pe rature in ° f :
tp = (: :~) (°C) + 32 ° f = (: :~) ( 15.7 x I 0 6 °C) + 32 ° f = 2.83 x I 0 7 0 f
1. 49 'c - {tF - 32 °F)( ! :~) - (103.S °F- 32 °F )(! :~) · 39.7 ·c
This dog has a fever; the ten1perature above is out of non11al canine range.
1.51 9 2 cm, 2 significant figures; 9.1 S cm, 3 significant figures
1.53 ( a) (c)
(e)
1.55 ( a) (c) ( d )
1.57 ( a)
(b) (c) ( d )
1.59 ( a)
(b)
(c) ( d)
(e)
(t)
1.61 ( a)
(b)
(c) ( d )
4 significant figures
4 significant figures
2 significant figures (b) (d) 5 significant figures 2 significant figures
0.72 m 2 (b) 84.24 kg
4 19 g/cm 3 (dividing a nun1ber with 4 sig. figs by one with 3 sig. figs) 19.42 g/mL (e) 858 .0 cm 2
finite nun1ber of significant figures
exact nun1ber
finite number of sign ificant figures
finite nun1ber of sign ificant figures
km/hr = {32 .0 dm!s)( 1 m )( 1 km )( 3600 s) = I LS km/h !Odm 1000 m I h
µg/L = (8 2 mg/mL)( I O~Ogmg)( I x / ~6 µg )( IO~OLmL) = 8.2 x 106 µg/L
kg= (75.3 mg)( I g )( I kg ) = 7.53 x 10- 5 kg 1000 mg 1000 g
L = (137.S mL)( IL ) = 0.1375 L 1000 mL
mL = (0.025 L)( JO~OLmL) = 25 mL
cm = {36 in.) ( 2 · 5 ~ cm) = 9 1 cm I m .
kg = (s.o Ib)( 1 kg ) = 2.3 kg 2.205 lb
mL= (3.0qt)( 946 4 mL) = 2800 mL I qt
mL = (8 oz)( 29 6 mL) = 200 mL I oz
km/hr = (55 mi/hr)( 1 609 ~m ) = 88 km/hr I m1
= (50.0 mi)( 1 609 ~ ) = 80.4 km I m1
4 0 x 103 mL + 0.9 mL = 4,000 pistachios (don't try this at home) m = (200 mi)(5280 ft)(30.48 cm ) [Jx J02 m ) ( I
mi = (2230ft)( I mi )( 60s )(60min) = 1520 mi hr I s 5280 ft I min I hr hr I light year= I Y(36~-~5d)(214dh )(3~0is)(3.00~~os m) =9.47 x 10 1s 111 m1 es = . 1g t years = . x m1 .1 871 . h (9.47 x J0 15 m)( I km)( I mi ) 5 1 10 u I light year I 000 m 1 609 km dens ity= 111ass/ volu111e = 36.4 g/45.6 mL = 0.798 g/mL
mL - 25.0g( I mL ) - 3 1 6mL 0.79 1 g g - 185 mL( l .492 g) - 276 g I mL mass of si lve r = 62.00 g - 27.35 g = 34.65 g volu me of silver= 18.3 mL-15 mL = 3 3 111L o r 3.3 cm 3 dens ity of silver= ( mass of silver) /(vol u111e of si lve r)= (3 4 .65 g) /(3.3 c111 3 ) = 11 g/ cm 3
d ·1 ( 227,641 lb) O59 1 lb 1- • ens1 y =. ga 385, 265 gal
density= 0.59 1 lb. gar 1( 453 6 g )( 3785 ~7-1 )= 0.0708 ginL- 1 I lb I gal
Chapter Two Elements, Compounds, and the Periodic Table
Pra c ti ce Exer c ises
2.1
(b)
(c) ( d ) ( a)
2.2 ( a)
(b)
(c)
SF 6 contains I S and 6 F atoms per molecu le
(C2 Hs)2 N2H2 conta ins 4 C, 12 H, and 2 N per n1olecu le
CaJ(P0 4) 2 contains 3 Ca, 2 P, and 8 0 atoms per formula un it
Co(N0 3)2 ·6H20 conta ins I Co , 2 N , 12 0, and 12 H per formu la unit
NH4N03 contains 2 nitrogen , 4 hydrogen , 3 oxygen aton1s per formula unit
FeNH 4(S04) 2 contains l iron, l nitrogen , 4 hydrogen , 2 su lfur, 8 oxygen atoms per fon11ula unit
Mo(N0 3) 2·SH20 contains I n1ol ybdenum , 2 nitrogen, 11 oxygen, and 10 hydrogen atoms per forn1ula unit
(d) C6H4C lN02 contains 6 carbon, 4 hydrogen , I ch lo rine , I nitrogen , and 2 oxygen atoms per mo lecule
2.3 Reactants: 4 N, 12 H, and 6 O; Product s : 4 N , 12 H, and 6 0.
2. 4 Reactants : 6 N, 42 H, 2 P, 20 0, 3 Ba , 12 C; Products: 3 Ba, 2 P, 20 0 , 6 N, 42 H, 12 C; The reaction is balanced.
2.5 The nun1ber of protons is equal to the aton1ic number of an elen1ent. The number of e lectrons is equa l to the aton1ic nun1ber for a neutral aton1. If the atom has a positive charge , the nun1ber of electrons is deten11ined by subtracting the charge from the aton1ic number. If the aton1 has a negative charge the nun1ber of electrons is determined by adding the charge to the atomic nun1ber.
(a) Fe ha s 26 protons , and 26 electrons
(b) Fe 3 • has 26 protons , and 23 electrons
(c) N3- has 7 protons, and l O electrons
(d) N has 7 proton s, and 7 e lectrons.
2.6 The nun1ber of protons is equal to the aton1ic number of an e len1ent. The number of e lectrons is equa l to the atomic nun1ber for a neutral aton1. If the aton1 has a positive charge, the number of electrons is detern1ined by subtracting the charge from the aton1ic number. If the aton1 has a negative charge the nun1ber of electrons is determined by adding the charge to the atomic
2.7
2.8 nun1ber.
(a) 0 has 8 protons , and 8 electrons (b) (c) At 3 • had 13 proton s, and I O e lectrons (d) ( a)
2.9 ( a) CrCIJ and CrCh, Cr203 and CrO CuCI, CuC1 2, Cu20 and CuO (b)
2.10 (a) Au 2S and Au2S3, Au3N and AuN (b) TiS , Ti 2S3 and TiS 2; Ti 3N2, Ti N and Ti3N 4
2.1 I ( a) (b)
0 2 - has 8 protons , and I Oelectrons Al has 13 proton s, and 13 electrons ( d) ( d)
2.12
2.13
2.14 (a) (a) (e) (a) (c) (e)
KC2H 302 (b)
K2S (b) Ca3P2 alu111inum chloride so dium brom ide potassium phosphide
Chapter 2
Sr(N03)2 (c) Fe (C2 H302)3 BaBr2 (c) NaCN (d) Al(OH) 3 (b) bariu111 hydroxide (d) calciu111 fluoride
2.15 lithium sulfide, magnesium phosphide , nickel ( ll) chloride , titaniu111(1I) chloride, iron(III) oxide
2.16 (a) Al2S 3 (b) SrF 2 (c) Ti02 (d) CoO (e) Au20 3
2.17 (a) lithium carbonate (b) potassium permanganate (c) iron ( III) hydroxide
2.18 (a) KC I03 (b) N aOC I (c) Nh(P04)2
2.19 The term "octa' 111eans eight, therefore there are 8 carbon atoms in octane. The fon11ula for an alkane is C.H1,, +2, so octane has 8 carbons and ((2 x 8) + 2) = 18 H. The mo le cu lar for111ula is CsH 1s and the condensed structural formu la is CH3CH2CH2CH2CH2CH2CH2CH 3
2.20 ( a) (b)
2.21 ( a) (c)
2.22 ( a)
Propano l : C 3HsO, CH3CH2CH20H
Butanol: C4Hio0, CH3CH2CH2CH20H phosphorus trich loride dich lorine heptoxide
AsC ls (b) (b) (d)
2.23 di iodine pentoxide
2.24 chro111ium(III) acetate
Re v i e, v Probl e ms
2.67 I Cr, 6 C, 9 H, 6 0
2.69 MgS04
2.71 CH3COOH or C2H402
2.73 NH 3
2.75 (b)
2.77 (a) H I H N H
2 K potassium, 2 C carbon, 4 0 oxygen
2 H hydrogen , I S sulfur, 3 0 oxygen
12 C carbon, 26 H hydrogen
4 H hydrogen , 2 C carbon , 2 0 oxygen
sulfur dioxide hydrogen sulfide (c) (b) (c) ( d ) (e) 9 H hydrogen , 2 N nitrogen , I P phosphorus , 4 0 oxygen.
( d)
2.79 (a) (b) (c) ( d) (e)
2.81 (a) (b) (c)
2.83 (a)
I Ni nicke l, 2 C l ch lorine , 8 0 oxygen
I C car bon , I O oxygen, 2 Cl chlor ine
2 K potassium, 2 Cr chron1ium, 7 0 oxygen
2 C car bon , 4 H hydrogen , 2 0 oxygen
2 N nitrogen , 9 H hydrogen , I P phosphorus, 4 0 oxygen.
14C ,28 H, 140
4 N, 8 H, 2 C, 2 0
15 C, 40 H, 15 0
6 N a (b) 3C (c) 270 ( d)
2.85 No , S(CH3)2, 02 , S02, C02, H20
2.87 Reaction is not balanced as written.
2.89 (a) (d)
2 91 (a) (d)
2.93 (a) (b) (c)
2.95 (a) ( d )
2.97 (a) (d)
2 .99 (a) (c) (e)
2.10 1 (a) (c)
2.103 (a) (c)
2.105 (a) (c)
2.107 (a) (b) (c) (d) (e) (t) (g) ( h) (i) (b) (e) (b) (e)
CoCb and CoCb
TiCb, TiCh and TiC14 MnCb and MnCh
KN03
Fe2(C03)3
PbO and Pb0 2 FeO and FeiOJ (b) (e) (b) (e)
ca lc ium sulfide
sodi um phosph ide
Ca(C2H302)2 Mg3(P04)2
SnO and Sn0 2 Cu20 and CuO
rub idium sulfide (b) (d)
si licon dioxide (b) tetraphosphorus decoxide (d)
Mg2+ BaO Baf2 (c) (c)
a luminum brom ide bar ium arsenide
xenon tetrafluoride dichlori ne heptoxide
iron(II) sulfide (b) copper(II) oxide
tin(! V) oxide ( d) coba lt(II) chloride hexahydrate
so dium nitrite (b)
magnesium s ulfate heptahydrate (d)
1on 1c
mo lecular
mo lecular
1on 1c
1on 1c
molecu lar
1on1c
1on 1c
1on 1c
chron1ium(U) ch loride
disu lfu r dichloride
sulfur trioxide
amn1onium acetate
potassium iodate
tetraphosphorus hexoxide
ca lciun1 sulfite
silver cyanide
zinc(II) b romide
potassium permanganate potas s ium thiocyanate
mo lec ular
sele n ide
2.109 (a) Na2HP04 (b) Li2Se (c) Cr (C2 H302)3 (d) S2 F JO
(e) Ni (CN)2 (t) Fe203 (g) Sbfs
2. I I I (a) ( NH4)2S (b) Cr2(S04)3 6H20(c) Sif4 (d) MoS2 (e) S nCl4 (t) H2Se (g) P4S1
2.1 13 (a) In correct Hg2Ch
(b) Correct
(c) Correct
(d) In correct: Ah(C0 3)3
(e) Incorrect: KMn04
(t) In correct: CaF2
(g) Incorrect: N a20
2.1 15 (a) In correct: iron ( II ) su lfide, iron(III) s ul fide
(b) In correct: sodiu111 ch loride
(c) In correct: ca lc ium bromide
(d) Correct: Ah(C0 3)3
(e) Correct: Ca(OC l)2
(t) Incorrect: magnesium ph osphate
(g) In correct: potass ium sulfide
2.1 17 dise lenium hexasulfide and diselenium tetrasulfide
Chapter Three
The Mo le and Stoich iometry
mo l A l =3 47gA l x( lmolAI ) =0.129mo l Al 26.98 g Al
MM h = 2 x 126.9 1 g mor' h = 253 82 g 111or1 h
mass I, = g I,= 0 .023 mol 12 x 2 = 5.84 g I, ( 253.82 g I ) - - I mol 1 2 -
Uncertainty in mo les = ± 0.002 g ( 1 mol K2SO 4 ) = ± 1.1 5 x I 0-5 mo! K 2S04 174.25 g K2S04
atOlllS of Au= I oz . Au x (28.35 g Au) x ( I mol Au ) x (6.022 x 10 23 atoms Au) I oz Au 196.967 g Au I mol Au = 8.67 x 1022 atoms Au
Mass of one molecular of table sugar, C,2H22011 MM C12H22011 = 12 x 12.0 I g/11101 + 22 x 1.0 I g/mol + 11 x 16.00 g/11101 = 342.34 g mor'
Mass of one mo lecular of tab le sugar, C,2H22011 = I ato111 C,2H220 11 x
( I m2: I C12H22012 )x(342J4 g C12H 22012) = 5.68 x 10-22 g
6.022x 10 atoms C12 H 220 12 I mol c,2H 22012
Find the mass of 5.64 x I 0 18 111olec ules of C 18H38 (MW = 254.50 g/ mol)
g = 5 64 x I 0 ,s ( I mol C18 H38 ) (254.50 g C18 H38 ) = 2 38 x I o-3 g 6 022 x I023 mo lec ule s C H I mol C18 H38 · 18 38
g = 2.38 x I 0-3 g = 0.00238 g
The mass is 2 mg, and whi le a balance can measure I 111g , the uncertai nty is ± I 111g , so the 111ass cannot be measured accurately .
Alu111inum su lfate: Al 2(S04)3, the alu111 ·num is Al 3+ ) 2 mol Al 3+
mo le A l3+ = 0.0774 11101 SO/- 3 mol soi - = 0.0516 mol A l3+
The for111ula of di nitroge n pentoxide is N 20 5
mo l N = 8.60 mo l O = 3.44 mol N ato111s ( )( 2mo l N) 5 mo l O
The fon11ula fo r sodiun1 bica rbonate is NaHC0 3.
mo l O = 1.29 mol NaHC03( 3 mol O ) = 3.87 11101 0 atoms I mo l NaHC03
gFe= (25 6 go)( I mo l O )(2 mo l Fe)(55.85 g Fe)= 59 6 g Fe 16 00gO 3mo l 0 lmolFe
3.11 gF=0.163gBr ( I molBr )( 3molF)( l 9.00gF) = o . 116gF
79 90 g Br I mol Br I mo l F
3.12 %N = 0. 20l 2 gN x 100% =36.84 % N
0.5462 g sample
% 0= OJ 45 0gO x 100% =63.16% 0 0.5462 g sample
Since these two values constitute I 00 %, there are no other elements pre s ent.
3.13 % H=( massH ) x t00%= ( O.OS 70gH } t00 %,=13.04 % H total mass 0.66 72 g total
% C=( massC ) x l00%= ( OJ4SlgC } 1oo o;,,=52.17 % C total mass 0.667 2 g total
I t is like ly that the compound contain s another e len1ent since the percentages do not add up to 100 %
3.14 We first determine the nun1ber of grams of each element that are present in one mo l of N 20:
2 mol N x 14.0 1 g/mol = 28.02 g N
I mo! 0 x 16.00 g/mol = 16.00 g O
The percentages b y mass are then obtained u s ing the fon11ula mas s of N20 ( 44.02 g ) :
% N = 2 8 02 g N x 100% = 63.65 % N
44 02 g sample
% 0 = 16 · 00 g O x 100% = 36.35 % 0
44 02 g sample
3.15 NO: Forn1ula n1ass = 30.0 I g/ mol
I mol N x 14.01 g/mol = 14.01 g N o/o N = ( 14.0 1/ 30.01) x 100% = 46.68 % N
I mol O x 16.00 g/11101 = 16.00 g O % 0 = ( 16.00/ 30.0 I ) x I 00% = 53.32 % 0
N0 2: Fon11ula mass= 46.0 I g/mol
I mol N x 14.01 g/mol = 14.01 g N % N = ( 14.0 1/ 46.01 ) x 100% = 30.45 % N
2 mol O x 16.00 g/11101 = 32.00 g O % 0 = ( 32.00/ 46.0 I) x I 00% = 69.55 % 0
N 20 f Fon11ula mass= 76.02 g/mo l
2 mol N x 14.0 1 g/11101 = 28.02 g N % N = ( 2 8 .02 / 76.02 ) x 100% = 36.86% N
3 mol O x 16.00 g/11101 = 48.00 g O % 0 = ( 48.00/ 76.02 ) x I 00% = 63. 14% 0
N 20 4 : Fon11ula mass= 92.02 g/mol
2 mol N x 14.0 I g/mol = 28.02 g N % N = (28.02 /92.02) x I 00% = 30.45 % N
4 mol O x 16.00 g/11101 = 64.00 g O % 0 = ( 64.00/ 92.02 ) x I 00% = 69.55 % 0
The con1pound N 20 3 corresponds to the data in Practice Exercise 3.12.
3 .16 We first determine the nun1ber of 11101 of each e lement as follows:
mo l N=l.333gN x lmolN =0.095 167mo l N 14.007 g N
We need to determine the mas s of H in the s an1ple. S ince there is a tota l n1as s of 1. 525 g of compound and 1.333 g of H , the n1ass of H is :
mass H = 1.525 g sample - 1.333 g N = 0.192 g H
mo l H=0.192gH x lmo l H =0.1905mo l H 1.00794 g H
Empirical formu la: N00951 61Ho. 190 = N 0095 167 H 0 190 = NH 2 0.095 167 0.095 167
3.17 First, find the nu111ber of mo les of each element, then determine the empirical fon11ula by co111par ing the ratio of the numbe r of moles of each element
Start with the number of moles ofS:
mo t S = 0.7625 g S ( 1 mol S ) = 0.02378 mot S 32.066 g S
Then find the numbe r of moles of 0: since there are only two ele111ents in the compound, S and 0 , the remaining mass is O g O = 1. 525 g compound- 0.7625 g S = 0.7625 g O
mo t O = 0.7625 g O ( 1 mo l O ) = 0.04766 mot O 15.9994 g O
The empirical formula is So 0237s0o.4766
The empirical for111ula 111ust be in whole numbers , so divide by the smalle r subscript: S0 02378 0 0 04766 which becomes 80 2. 0.02378 0 .02378
3.18
mo t 0.259 g N = 0.259 g N x 1 mol N = 0.0185 mot N 14 007 g N
mo t I= 7.04 g I x I mol I = 0.0555 mot I 126.9 g I
E111pirical formu la: No. 01s sio.osss = N 0 0 185 I0 0555 = NI 3 0 .0 185 0 .0 185
3.19 mo l A l =5.68tonsAl( 20001 bAIX 454 gAIX lmolAI J=l.91 x 105 1110IAI I ton Al I lb Al 26.98 g Al
mo t O = 5.04 tons 0( 2000 lb O )( 454 g O ) ( 1 mol O J = 2.86 x 105 11101 0 I ton O I lb O 16.00 g O
Empirical For111ula: AJl. 9 1XI0' 0 2 86xlo'
In who le nu111be rs : Al1.9 1x,o'0 2 86xtO' which is AI01.s and mu ltipl y the subscripts by 2: Al 20 3 1.9 lxl o' 1. 9 lx l O '
3.20 It is convenient to assu111e that we have I 00 g of the sa111ple, so that the percentage by mas s va lues 111ay be taken directly to represent masses . Thus there is 32.4 g of Na, 22.6 g of S and ( 100.00- 32.4- 22.6 ) = 45.0 g of 0. Now , convert these masses to a nu111ber of 11101:
mo ! Na= (32.4 g Na)( 1 mo ! Na ) = l .40 mo ! Na 23.00 g Na
mo l S = (22.6 gs) ( 1 mol S J= 0.705 mol S 3206gS
mo ! 0 = (45.0 go)( l mo ! O ) = 2.8 1 mo ! 0 16.00 g O
Next , we divide each of these mole amounts by the smallest in order to deduce the simplest whole nu111ber ratio:
For Na: 1.40 mol = 1.99
0.705 mol
For S: 0.705 mol = 1.00 0.705 mol
For 0: 2 8 1 mo l = 3.99 0.705 mol
The empirical formula is Na2S04
3 .2 1 It is convenient to assun1e that we have I 00 g of the san1ple, so that the percentage by mass va lues may be taken direct ly to rep resent masses. Thus there is 81.79 g of C, 6. 10 g of H and ( 100.00-8 1.79-6. 10) = 12.1 1 g of 0. Now , convert these n1asses to a number of mo le:
mo l C = (8 1. 79 g c)( 1 mol C ) = 6.8 1 mo l C 12.0 l gC
mo l H = {6 10 g H)( 1 mo l H ) = 6 05 mol H 1. 008 g H
mo l O = {12 11 g O)( 1 mol O ) = 0 757 mo l O 16.00 g O
Next , we divide each of these 11101 an1ounts by the sn1allest in order to deduce the simplest whole nun1ber ratio:
For C: 6·8 1 mo l = 9.00 0.757 mo l
For H: 6.05 mol = 7 .99 0.757 mol
For 0: 0.757 mol = 1.00 0.757 mol
The emp irical formu la is C9HsO.
3.22 Find the n1ole s of S and C using the stoichiometric ratios, and then find the empirical formu la fron1 the ratio of n1oles of S and C.
Mo lar n1ass ofS02 = 64.06 g mor' Molar mass ofC02 = 44.01 g n1or 1
mo l S = 0.640 g S02( 1 mol S02 )( 1 mol S )= 9.99 x 10-3 11101 64.06 g S02 I mol S02
mo l C = 0.220 g C0 2( I mo l C02 ) ( I mo l C ) = 5.00 x I 0-3 mo l 44.0 1 g C0 2 I mo l C02
Emp irical Forn1ula C5 00 x 10 , S 9 99 x 10 , divide both subscripts by 5.00 x I 0-3 to get CS 2
3 .23 Since the entire an1ount of car bon that was present in the original san1ple appears among the products only as C02, we calculate the amount of carbon in the san1ple as follows :
gC=( 7.40 6 gCO )( l molC0 2 )( l molC )( 12 0lgC)= 2 02 l gC 2 44.0 1 g C02 I mol C02 I mo l C
Sin1ilarly, the entire mass of hydrogen that was present in the origina l sample appears an1ong the products only as H20. Thus the mass of hydrogen in the san1ple is:
g H = (3 027 g H o) ( I mo l H20 )( 2 mol H ) ( 1. 008 g H ) = 0 3386 g H 2 18.02 g H20 I mo l H20 I mo l H
The mass of oxygen in the origi na l san1ple is determined by difference:
5.048 g - 2.021 g - 0.3386 g = 2.688 g O
Next , these mass an1ounts are converted to the co rr esponding mole amounts:
mo l C= (2.021 gc)( 1 molC) =0 . 1683 mo l C 12.0 1 gC
mo l H = (03386 g H)( 1 mot H ) = 03359 mo l H 1. 008 g H
mo l O = (2.688 go)( 1 mo l O ) = 0. 1680 mot O 16 00 g O
The sin1plest formula is obtained by dividing each of these n1ole amounts by the smallest:
For C: 0.1683 mo l = 1.002
For H: 0 3359 mo l = 1.999 0. 1680 mol 0 1680 mo l
For O: 0.1680 mol = LOOO 0.1680 mol
These va lues give us t he simplest fon11ula direct ly, name ly CH20
3.24 The forn1ula n1ass of the empirical unit is I N + 2 H = 16.03. Since this is half of the mo lecular mass , the mo lecular formula is N2H4.
32.0 g n1or 1 hydrazine x I mo l NH2 = 2 mol NH2/ mo l hydrazine 1603 g
To determine the molecular formu la, mu ltiply the subscripts ofNH 2 by 2 to obtain N2H4
Fon11ula mass ofCHCl: 48.47 g n1or 1
For CH,C l IOO = 2.02 and 289 = 5.84 - 49.48 49.48 For CHCl:
3.25 To fi nd the n1olecu la r forn1ula, divide t he molecular mass by the formu la mass of the en1pirica l 3.28
3.30 fon11ula, then n1ultip ly the subscripts of the empirical formula by that value . Fon11ula mass ofCH 2Cl: 49.48 g n1or 1
The CH 2C l rounds better using the molecular n1ass of l 00, ther efore multiply the subscripts by 2 and the forn1ula is C2H4Cl 2.
For CH Cl, the mo lecular mass of289 gives a multiple of 6, a nd t herefore the fonnula is C6H6Cl6. I mol O mo l 0 2 = 6.76 mol S03 x 2 = 338 mol 0 2 2 mo l S03 I mot H SO
mo l H2S04 = 0366 mot NaOH x 2 4 = 0. 183 mol H 2S04 2 mo l NaOH
Fei03(s) + 2A l(s) 2Fe(l) + Ab0 3(s)
g Al O = (86 0 g Fe)( I mol Fe )( I mot AI 2 0 3 )( I 02.0 g Al2 0 3 ) = 78 5 g Al ,0 3 2 3 55.85 g Fe 2 mot Fe I mol Al20 3 mo l Al 20 3 -
3 .31 g C02 = ( 1.50 x I 02 g CaO) ( I mo! CaO )( I mo! C02 )( 44.0 I g C02) = 1.18 x I02 g C02 56 .08 g CaO I mo! CaO I mol C02
3.32 First deter111ine the number of grams of CaC03 that would be required to react co111pletely with the given a111ount of HCI:
g CaC0 3 = ( 125 g HCI) ( I mol HC I )( I mo l CaC03 )( I00.088 g CaC03) 36.461 g HCI 2 mol HC I I mo l Ca03 = 171.57 g CaC0 3
Since this is more than the a111ount that is availab le , we conclude that CaC0 3 is the limiting reactant. The rest of the calculation is therefore based on the availab le amount of CaC03:
g C0 2 = (! 25 g CaC03)( I mol CaC03 )( I mol C0 2 )(44 .01 g C0 2 ) = 55 0 g C0 2 I00.088 g CaC0 3 I mo! CaC03 I mol C02
For the nu111ber of grams of left ove r HCI, the excess reagent, find the a111ount of HCI used and then subtract that from the amount of HC I started with, 125 g.
g HCI used= ( ! 25 g CaC0 3) ( I mol CaC03 )( 2 mol HCI ) ( 36.461 g HCI) I00.088 g CaC03 I mol CaC0 3 I mol HC I = 91.1 g HCI
g HCI remaining= 125 g - 91.1 g = 34 g HCI remaining
3.33 First determine the number of grams of 02 that would be required to react completely w ith the given a111ount of ammon ia:
g O = (30.00 g NH ) ( I mol NH 3 J( 5 mol 0 2 J( 32.00 g 0 2 J 2 3 17.03 g NH 3 4 mo l NH 3 I mol 0 2 = 70.46 g 02
Since this is 111ore than the amount that is avai lable, we conclude that oxygen is the li,niting reactant. The rest of the calculation is therefore based on the available amount of oxygen:
g NO= (40.00 g O )( I mol 0 2 )( 4 mo l N0) ( 30.0 1 g NO ) 2 32.00 g 0 2 5 mol 0 2 I mol NO = 30.0 1 g NO
3.34 First determine the numbe r of gr ams of sa licylic acid , HOOCC6H40H that would be required to react co111plete ly with the given amount of acetic anhydride, C4H60 3: g HOOCC6H40H = (15.6 g C4H60 J) x ( 1 mo! C4H603 1( 2 mo! HOOCC6H40H 1( 138.12 g HOOCC6H40H I l l 02.09 g C4H603 Jl 1 mo! C4H603 Jl I mo! HOOCC6H40H J = 42.2 g HOOCC 6H40H
Since more salicylic acid is required than is availab le , it is the li111iting reagent. Once 28.2 g of sa licylic acid is reacted the reaction will stop, even though there are 15.6 g of acetic anhydride present. Therefore the sa licylic acid is the limiting reactant. The theoretical yield of aspirin HOOCC6H402C2H 3 is therefore based on the amount of salicy lic acid added. This is calcu lated: g HOOCC6 H402C2H3 = (28.2 g HOOCC6H40H) x ( I mo! HOOCC6 H40H )(2 mo! HOOCC6 H40 2C2H3)( 180.16 g HOOCC 6 H40 2C2H3 )
138. 12 g HOOCC6 H40H 2 mol HOOCC 6 H40H 1 mol HOOCC 6 H4 0 2C2H3
= 36.78 g HOOCC 6H402C2H3
Now the percentage yield can be calculated from the amount of acetyl salicy lic acid actually produced, 30.7 g:
percentage y1e - x I Vo - x I , o, o ld _ ( actual yield ) ooo ( 30 7 g HOOCC6 H 40 2C 2H 3 ) 000 , _ 83 50 , theoretica l yie ld 36 78 g HOOCC6 H 4 0 2C 2H 3
3.35 First determine the number of grams of C2H50H that would be required to react con1plete ly with the given an1ount of sodiwn dichromate:
g C2H50H = (90 0 g Na2Cr207 )( I mol Na2Cr207 )( 3 mo l C2H50H )(46.08 g C2H50H) 262.0 g Na 2Cr20 7 2 mol Na 2Cr20 7 I mol C 2H 50H = 23.7 g C 2H 50H
Once this an1ount ofC2 H50H is reacted the reaction will stop , even though there are 24.0 g C2HsOH p resent, because the Na 2Cr2 0 7 will be used up. Therefore Na 2 Cr2 0 7 is the lin1iting reactant. The theoretical yield of acetic acid (HC2HJ02) is therefo re based on the amount of Na 2 Cr20 7 added. This is ca lculated below:
g HC2 H302 = ( 90.0 g NaiCr20 7 ) ( I mol Na2Cr207 x3 mol HC2H 30 2 160.06 g HC 2H 30 2 ) 262.0 g Na2Cr207 2 mo l Na2Cr207 I mol HC2H302 = 30.9 g HC2H302
Now the percentage yie ld is calculated from the amount of acetic acid actually produced , 26.6 g:
percent yield = ( actual yield ) x I 00% - ( 26 · 6 g HC2 H 302 ) x I 000/o - 86 I% theoretica l yie ld 30.9 g HC2H 30 2
3.36 Three step synthesis overall yield= (0.872 x 0.9 11 x 0.863) x 100% = 68.6%
Two step synthesis overall yield= (0.855 x 0.843) x I OOo/o = 72. 1%
Therefore, the two-step process is the preferred process.
Re v i e, v P rob l e ms
3.31 (a)
(a)
g Fe= (1.35 mo l Fe) ( 55 85 g Fe) = 75.4 g Fe I mole Fe
g O = (24.5 mol 0) ( 16 0 g O) = 392 g O I mole O
g Ca= (0.876)( 40 08 gCa) = 35. 1 g Ca I mo le Ca
NaHC03 = I Na + I H + I C + 30 = (22.98977) + (1.00794) + ( 12.0 107) + (3 x 15.9994) = 84.0066 1 g/n1ole = 84 0066 g/mol
(NH4 )2C03 = 2N + 8H + C + 30 = (2 x 14.0067) + (8 x 1.00794) + ( 12.0107) + (3 x 15.9994) = 96.08582 g/n1ole = 96.0858 g/mol
CuS04·5H20 = !Cu + IS + 90 + JOH = 63.546 + 32.065 + (9 x 15.9994) + ( JO x 1.00794) = 249.685 g/ mole
( d ) (e)
Potassium d ichromate: K2Cr201 = 2K + 2Cr + 70 = (2 x 39.0983) + (2 x 5 1.9961) + (7 x 15.9994) = 294 .1 846 g/111ole
Altu11inum su l fate: Ab(S04) 3 = 2Al + JS + 120 = (2 x 26.98154) + (3 x 32.065) + ( 12 x 15.9994) = 342 .1 5088 g/111ole = 342.151 g/mo l
3.35 ( a) ( 3 10.18 gCa 3 (P0 4 ) l g Ca 3( P 0 4) 2 = (1.25 11101 Ca3(P04) 2) ( ) 2 = 388 g Ca3(P04) 2 I mol Ca3 P0 4 2
(b)
(c) ( d )
gC4H1o=(0 600µ molC4H10)( lmolC4Hio )( 5812 gC 4H i o) =3.49 x 10- 5 gC 4H 10 106 µmo l C4H10 I mol C4H10 ( 241.86 mg Fe( N03} ) mg Fe(N0 3) 3 = (0 625 mmol Fe(N03) 3) ( } 3 = 15 1 mg Fe(N03) 3 I mmol Fe N03 3 = 0 .15 1 g Fe(N0 3)3 ( 96.09 g (NH 4 ) C03 ) g (N H4)2C03 = (1 .45 11101 (N H4)2C0 3) ( ) 2 = 139 g ( NH4)2C03 I mo l NH 4 2 C0 3
3.37 ( a) mo l NH 3 = l. 56ngNH3( l gNH 3 )(lmole NH 3 ) =9.16 x 10- 11 molNH 3 109 ng NH 3 17 03 g NH 3
(b) (c) (d)
mo l Na2Cr04 = 6 .98 µg Na2Cr0 4( 1 g Na2Cr04 )( 1 mole Na2Cr04) 106 µg Na2Cr04 162 0 g Na2Cr04 = 4.3 1 x I o-s mo l N a2Cr0 4
mo l CaC0 3 = 21.5 g CaCO, 3 = 0 .2 15 mo l CaC0 3 ( )( I mol CaCO ) , JOO 09 g CaC0 3
mo l Sr(N03)2 = 16 8 g S r (N03)2 x ( 1 mole Sr~N03\ 2 ) = 7.94 x 10- 2 11101 Sr(N03)2 21 1. 6 g Sr N0 3 2
mo l Ni= 17 7 g Ni ( 1 mol Ni ) = 0 .3 02 mo l Ni 58 69 g Ni
1. 56 x I 0 21 ato111s Ta ( 1 mol Ta ) = 2.59 x I 03 mo le Ta 6.022 x I 0 23 atoms Ta
gK=2.00x 10 12 atoms K ( l mo! K )( 39.JOgK)= I .JO x 10- 10 g K 6.022 x I 02' atoms K I mol K
mo l C- 12 = 6 g x = 0.5 mo l C- 12 ( I mol C- 12 ) 12 .00 gC- 12
Nu111ber of atoms C- 12 = 0 5 11101 = 3.0 1 x 10 - atoms C- 12 ( 6.022 x I 023 atoms C- 12) 03 I mol C- 12
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obtaining a reprieve, there being great endeavours tho’ in vain used for that purpose, is yet unknown.”
The author of this account also informs us that Kidd “could hardly be brought to a charitable reconciliation with those persons, who were evidences against him alleging that they deposed many things that were inconsistent with truth and that much of their evidence was by hearsay: and in the general part of his discourse seemed not only to reflect on them buton severalothers,whoinsteadofbeing his friends as theyprofessed, had traitorously been instrumental in hisruin!” “He further declared that as to the death of William Moore, his gunner, the blow that he gave him, it was in a passion, as being provoked by him to do so, but not with an intention of any manifest injury, much less to kill or murder him. Nay, he was so far from bearing any malice against him, that he freely gave £200 for his ransom, and further said that all his sailors knew he always had a great love and respect for him; adding that if any one concerned in his tryal had acted contrary to the dictates of his or their own conscience he heartily forgave them, and desired that God would do the like.” “He expressed abundance of sorrow for leaving his wife and children without having the opportunity of taking leave of them, they being inhabitants in New York. So that the thoughts of his wife’s sorrow at the sad tidings of his shameful death was more occasion of grief to him than that of his own sad misfortunes.” “He desiredallseamen ingeneral,moreespeciallyCaptainsinparticular to take warning by hisdismalunhappinessandshamefuldeathand that they would avoid the means and occasions that brought him thereto, and also that they would act with more caution and prudence, both in their private and public affairs by sea and land, addingthatthiswas a very fickleandfaithlessgeneration.” (He had undoubtedly found it so.) “After he had ended his discourse to the people, he spent the rest of his time in Prayer and other pious and Godly exercises with the Ordinary of Newgate and other ministers: and at last seemed very devout and penitent, expressing his hearty sorrow for his manifest transgressions, especially the unhappy and sudden death of William Moore his gunner—but would not call it
murder to the very last, esteeming it rather an accidental misfortune than a murder by reason that there was but one blow given and that in passion without any premeditated malice.”
No reference is made in this account to Kidd’s being “inflamed with drink.” It is clear from it that whether or not he had been given a drop of whiskey on his way to execution, he was to the end in the full possession of his faculties.
The only member of his crew who was hung with him was poor Darby Mullins, the remainder being at the last moment reprieved. Why Mullins, who had surrendered himself to the Governor of East Jersey along with two others, relying on the King’s proclamation, was selected as Kidd’s fellow-sufferer, is not clear. It is true that he was an Irishman, and in the opinion of the chaplain in a better frame of mind to meet his death than any of his companions: but neither of these circumstances in itself seems quite a satisfactory justification for hanging him. He had no doubt joined Culliford, unquestionably by far the most guilty of all the seamen implicated, but for whose presence at Madagascar, when the Adventure Galley arrived there, Kidd in all probability would have been able to bring his prizes home before the hue and cry had been raised against him. But Culliford, though indicted for several piracies about the same time as Kidd, apparently escaped scot free, having been clever enough to save his neck by surrendering to the right persons under the King’s proclamation, and to secure the services of a counsel who did not fail to put in an appearance on his behalf, when his case came on for hearing; the result of which was that “his case” (according to a note in the State Trials) “being particular and argued by Counsel he was respited.”
To come now to the last painful incident in this disgraceful tragedy. The day after Kidd’s corpse had been hung aloft in chains on the gallows, Somers dared at last to break the silence he had so long maintained and to put in his reply to the Articles of Impeachment brought against him by the Commons. The allegations he had to meet were that in the grant of the goods of the pirates to
the co-adventurers, the name of Samuel Newton, one of the Grantees, had been “used in trust and for the sole benefit of” himself: that “the grant manifestly tended to the obstruction of trade and navigation, the great loss and prejudice of merchants and others, His Majesty’s subjects, and the dishonour of the King and his Kingdom:” and that “by procuring and passing it,” he had been guilty of a notorious breach of his duty. In his reply he was forced to admit that Newton had been named in the grant, “by and in trust for him,” and was apparently unable to give any excuse whatever for this discreditable deception. He pleaded that the grant “did not in any way tend to the obstruction or discouragement of trade or navigation, or to the loss or prejudice of His Majesty’s subjects, nor to the dishonour of His Majesty or His Kingdom.” He denied (and the denial implied what would be considered in these days a very low estimate of official honesty) that the passing of the grant was any breach of duty, inasmuch as it “was formed as a recompense to the grantees, who at their own charge had provided and fitted out the said Ship” (the Adventure Galley) to enable Kidd “to execute the powers in the said grant mentioned, whereby thepublicmighthave receivedgreatbenefithadthesaidWilliamKiddfaithfullydischarged thetrustreposedinhimby HisMajesty andtheGrantees,whichhe failing to do, the owners of the said ship had lost their expenses, andhadnotreceivedanybenefitfromthegrant.”
As a matter of fact, it may well be doubted whether any of the grantees, excepting Kidd and Livingstone, lost any part of their expenses. As has already been shown, one of the conditions on which their legal advisers had been careful to insist had been that if the prize moneys were insufficient to make good the full amount advanced by the grantees, other than Kidd and Livingstone, the deficiency was to be made good by Kidd and Livingstone, both of them men of substance. We have seen with what eagerness, and with what disastrous results to Kidd, Livingstone had endeavoured to get his bond restored to him by Bellamont. That Kidd’s estate of itself, notwithstanding the fact that he was unable in Newgate to get funds for his defence until the night before his trial, was sufficient to
have covered any loss sustained by the great men, who had exploited him, is clear from the fact that of his effects forfeited to the Crown, six thousand four hundred and seventy-one pounds were afterwards given by Queen Anne towards the establishment of Greenwich hospital.[14] But whether or not these great men found it inconvenient to reclaim their one thousand pounds apiece, it is impossible to doubt that when making this cruelly unjust charge of faithlessness against Kidd the day after his death, Somers was fully acquainted with the essential facts of the case. It is incredible that he had not read Kidd’s narrative, the depositions of his men, and Bellamont’s correspondence, and that he was not cognizant of all the proceedings at Kidd’s trial, the keeping back of the French passes by the Admiralty officials: the failure of Kidd’s counsel to put in an appearance on the critical day when he was tried for piracy; the break-down of the most material parts of the King’s evidence; and the manner in which the trials had been conducted throughout by the Lord Chief Baron. It is to be feared that he not only knew all this, but that his was the unseen master hand that had held the strings, which had been so skilfully and ruthlessly manipulated as to bring about Kidd’s death so opportunely by the verdicts of London juries. If this be so, what is to be said of the Whig historians, who have dealt with Kidd’s case? Is it possible to believe in the face of indisputably recorded facts, that Somers really was the immaculate politician of his day depicted for us by Macaulay, “whose integrity,” we have been assured, “was ever certain to come forth bright and pure from the most severe investigation”? In the foregoing pages an attempt has been made, it is believed for the first time, to allow the personages who took part in this melancholy business to speak for themselves, so far as the extant records permit. Hitherto by a conspiracy of silence, their voices have been hushed, and the facts of the case studiously suppressed or perverted by eminent advocates, who have thought it necessary, if the memories of Somers and his colleagues were to be cleansed from the stigma which clung to them in their own day from the part they took in it, that Kidd’s reputation should be blackened, and that he should be
depicted as a villain of the deepest dye, whom, on account of his unexceptionable antecedents, these great men were fully justified in employing, but whose character underwent so rapid a deterioration after he had once come into contact with them, that he betrayed them for the purpose of enriching himself with spoils, of which as a matter of fact he stood in little need and which he made no effort to secure for himself. He has been represented by Macaulay not only as a rapacious pirate, but also as a monster of cruelty, who for his own ends depraved his crew and led them into every kind of wickedness. To quote but one passage from Macaulay’s indefensible and inexcusable travesty, “With the rapacity he had the cruelty of his odious calling. He burnt houses: he massacred peasantry. His prisoners were tied up and beaten with naked cutlasses, in order to obtain information about their concealed hoards. One of his crew, whom he had called a dog, was provoked into exclaiming in an agony of remorse, ‘Yes, I am a dog: but it is you that have made me so.’ Kidd in a fury struck the man dead.”
These accusations have obtained ready credence; but their absurdity will be evident to any one who will take the pains to examine the records. There is no reason whatever for believing that Kidd was cruel or rapacious. The only ground for suggesting that “he massacred peasantry” is the one case, when his cooper’s throat having been cut by the natives, he retaliated by ordering one native to be shot. This was the only time when it was ever alleged in his own day that he had burnt houses: and we have it on the authority of Palmer, the King’s evidence against him, that on this occasion Kidd had given express orders to his men to spare the houses that had white flags hoisted on them, because their inmates had helped to water his ship. The episode on the strength of which Macaulay accuses him of causing his prisoners to be beaten with cutlasses, in order to extort from them information as to their concealed hoards, has already been explained. The men in question were not his prisoners. He allowed them to proceed peacefully on their voyage, and their ship was not taken from them. Kidd never went on board of her, much less did he give directions to his crew to ill-use them.
Questioned as to whether any gold had been taken from them, Palmer freely admitted that he did not see any. Asked further by Kidd, whether it was not the case that a parcel of rogues had gone on board and done the deed complained of, he virtually admitted that it was so by making no reply. In the matter of cruelty there is a marked difference between the reported doings of Kidd and of the pirates of whom the East India Company were repeatedly complaining. In these complaints mention is often made of the outrages committed: but in the case of Kidd the Company made no complaint of similar misdeeds. From all that can be learned of him, he seems to have been a kind-hearted man. There is no reason to doubt the truth of his dying statement that he had paid two hundred pounds for his gunner Moore’s ransom, probably on the occasion when the natives had cut his cooper’s throat. One of the reasons which led Bellamont to employ him is stated by Bellamont’s apologist to have been Kidd’s well-known affection for his wife and family, which was also relied on by Bellamont as being strong enough to prevent him from attempting to escape by forsaking them on his return. And we have it on the record of a witness who certainly had no bias in his favour that his chief solicitude in Newgate after he had been sentenced to death was for them and not for himself.
The suggestion that Moore, when knocked on the head by Kidd, was “in an agony of remorse” for acts of piracy which Kidd had led him to commit, is almost too ludicrous to call for comment. It is absolutely clear from the evidence of every witness of the occurrence that so far from Kidd having led Moore astray, Moore had vainly endeavoured to induce Kidd to become a pirate, and that it was his failure to succeed in this endeavour that led to the altercation which ended in his death.
But the most flagrant fiction fabricated by a Whig historian in relation to poor Kidd, is not to be found in Macaulay’s history, but in the pages of a grave historical work, compiled by an eminent lawyer, who in his day had filled not only the office of Lord Chief Justice, but also that of Lord Chancellor. That great legal luminary, Lord Campbell, in his “Life of Somers” has not hesitated to insert a
circumstantial fable to the effect that Kidd was caught red-handed on the high seas in the midst of his criminal career. In the fifth volume of his “Lives of the Chancellors,” pages 126 and 127, he tells the tale thus: “A noble vessel called the Adventure Galleywas fitted out, and the command of her given to William Kidd, a naval officer, esteemed for honour as well as for gallantry. On arriving in the Indian Seas, he turned pirate himself, and cruised against the commerce of all nations indiscriminately, till after a sharp engagement with an English frigate in which several fell on both sides, he was captured and brought home in irons.” To such depths can history sink when written by political partisans of the highest rank and respectability.
