SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
CHEM 110 – General Principles of Chemistry
TUTORIAL 2
26th & 28th February 2014
1. Name the following ionic compounds:
(a) MgS
magnesium sulfide
(b) Li3PO4
lithium phosphate
(c) Ba(ClO4)2
barium perchlorate
(d) Cu(NO3)2
copper(II) nitrate
(e)) Fe(OH)2
iron(II) hydroxide
(f) K2CrO4
(g) (NH4)2SO4
ammonium sulphate
(h) Co(NO3)2
potassium chromate
cobalt(II) nitrate
2. Write the chemical formula for the following compounds:
(a) Potassium sulfate
K2SO4
(b) Copper(I) oxide
Cu2O
(c) Iron(III) carbonate
Fe2(CO3)3
(d) Copper(II) perchlorate
Cu(ClO4)2
(e) Magnesium hydrogen carbonate
Mg(HCO3)2
(f) Nitrous acid
HNO2
(g) Dinitrogen tetroxide
N2O4
(h) Hydrogen cyanide
HCN
3. Balance the following equations:
(a) CH 4 (g) + 4 Cl 2 (g)  CCl 4 (l) + 4 HCl(g)
(b) 2 C 5 H 1 0O 2 (l) + 13 O 2 (g)  10 CO 2 (g) + 10 H 2 O(g)
(c) 2 Fe(OH) 3 (s) + 3 H 2 SO 4 (aq)  Fe 2 (SO 4 ) 3 (aq) + 6 H 2 O(l)
(d) Mg 3 N 2 (s) + 4 H 2 SO 4 (aq)  3 MgSO 4 (aq) + (NH 4 ) 2 SO 4 (aq)
4. Determine the formula masses of :
(a) CuSO4
FM: 1(63.55) + 1(32.07) + 4(16.00) = 159.62 u
(b) (NH4)3PO4
FM: 3(14.01) + 12(1.008) + 1(30.97) + 4(16.00) = 149.09 u
(c) Fe 2 (SO 4 ) 3
FM : 2(55.85) + 3(32.07) + 12(16.00) = 399.91 u
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SCHOOL OF CHEMISTRY & PHYSICS
UNIVERSITY OF KWAZULU-NATAL, WESTVILLE CAMPUS
5. Calculate the percentage by mass of the indicated element in the following
compounds:
(a)
carbon in acetylene (C 2 H 2 )
C 2 H 2 : FM = 2(12.01) + 2(1.008) = 26.04 u
%C=
(b)
(
)
X 100 = 92.2%
hydrogen in ammonium sulphate, (NH 4 ) 2 SO 4
(NH 4 ) 2 SO 4 : FM = 2(14.01) + 8(1.008) + 1(32.07) + 4(16.00) = 132.15 u
%H=
(
)
X 100 = 6.102%
6. Calculate the following quantities:
(a)
Mass in grams of 1.906 x 10-2 mol BaI2
molar mass = 1(137.33 g mol-1) + 2(126.904 g mol-1)
= 391.14 g mol-1
m = n x molar mass = 1.906 × 10-2 mol BaI2 × 391.14 g mol-1 = 7.455 g BaI2
(b)
number of moles of NH4Cl in 48.3 g of this substance
molar mass = 1(14.01 g mol-1) + 4(1.008 g mol-1) + 1(35.45 g mol-1)
= 53.49 g mol-1
48.3 g NH4Cl
= 0.903 mol NH4Cl
53.49 g mol-1
n = mass / molar mass =
(c)
number of molecules in 0.05752 mol CH2O2
no. of molecules = n x NA
0.05752 mol HCHO2 × 6.02214 × 1023 molecules mol-1
= 3.464 × 1022 HCHO2 molecules
(d)
Number of O atoms in 4.88 x 10-3 mol of Al(NO3)3
= n x NA
4.88 × 10-3 mol Al(NO3 )3 ×
9 mol O
× 6.022 × 1023 atoms mol-1
1 mol Al(NO3 )3
= 2.64 × 1022 O atoms
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