Problem Set 13
1.
How many g of oxygen are in 5.2 g iron(II) chlorate?
Formula: Fe(ClO3 )2
Molar mass Fe(ClO3 )2
element
Fe
# moles
Molar mass
1
55.85 g Fe
Cl
2
70.90 g Cl
O
6
96.00 g O
Molar mass Fe(ClO3)2
Given 5.2 g Fe(ClO3)2
Target: g O
Path: g Fe(ClO3 )2 6 mol Fe(ClO3 )2 6 mol O 6 g O
Factors:
Computation:
2.
How many atoms of oxygen are in 0.256 mol [SO3]2-? Name the compound.
The compound is sulfite ion. There are 3 moles of oxygen in one mole of sulfite ion
Given: 0.256 mol [SO3]2Target # O atoms
Path: 0.256 mol [SO3]2- 6 mol O 6 # O atoms
Factors:
Computation:
3.
Draw Lewis diagrams of nitrate ion and (ClO2)- (chlorine is less electronegative than oxygen). When drawing Lewis
diagrams follow the procedure that you have learned in class.
nitrate ion (NO3)n = 4 x 8 = 32
v = 5 + 3 x 6 + 1= 24
s = 32 - 24 = 8
b=4
lp = (32 - 2 x 8)/2 = 8
chlorite ion (ClO2)n = 3 x 8 = 24
v = 7 + 2 x 6 + 1 = 20
s=4
b=2
lp = (24 - 2 x 4)/2 = 8
4.
Write a symbol for each of the following elements, ions, isotopes or compounds that exist in nature:
a.
5.
sulfur-34
b.
nitrogen
N2
c.
potassium hydroxide
KOH
d.
phosphorus pentachloride PCl5
e.
chloric acid
HClO3
f.
calcium sulfide
CaS
g.
fluoride ion
Fh.
copper(II) nitride
Cu3N2
i.
hydrofluoric acid
HF
j.
sodium phosphate
Na3PO4
k.
barium ion
Ba2+
l.
calcium carbonate
CaCO3
m.
ammonium ion
(NH4)+
n.
phosphorous acid
H3PO3
Write the name for each of the following elements, ions, isotopes or compounds that exist in nature:
a.
O2
oxygen
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
6.
magnesium-25
IP2O5
BaS
Fe(OH)3
Na3N
CaF2
Be2+
N2
Mg3 (PO4 )2
HClO2
S2(OH)-
iodide ion
diphosphorus pentoxide
barium sulfide
iron(III) hydroxide
sodium nitride
calcium fluoride
beryllium ion
nitrogen
magnesium phosphate
chlorous acid
sulfide ion
hydroxide ion
Fill in the table:
formula
acid or base
name the two ions
write formulas for ions
+
(SO3)
2-
name the compound
H2SO3
acid
hydrogen ion, sulfite ion
H
sulfurous acid
Ca(OH)2
base
calcium ion; hydroxide ion
Ca2+ (OH)-
calcium hydroxide
NaOH
base
sodium ion; hydroxide ion
Na+
(OH)-
sodium hydroxide
HBrO3
acid
hydrogen ion, bromate ion
H+
(BrO3)-
+
3-
bromic acid
H3PO4
acid
hydrogen ion, phosphate ion
H
(PO4)
phosphoric acid
CuOH
base
copper(I) ion; hydroxide ion
Cu+
(OH)-
Mn(OH)2
base
manganese(II) ion; hydroxide
ion
Mn2+ (OH)-
manganese(II)
hydroxide
HF
acid
hydrogen ion, fluoride ion
H+
F-
hydrofluoric acid
H2CO3
acid
hydrogen ion, carbonate ion
H+
(CO3)2-
carbonic acid
copper(I) hydroxide
7.
Fill in the table:
name of the compound
acid or base
name the two ions
write formula of the two
ions
write the formula of the
compound
hydroiodic acid
acid
hydrogen ion; iodide ion
H+
I-
HI
sulfurous acid
acid
hydrogen ion; sulfite ion
H+
(SO3)2-
H2SO3
magnesium hydroxide
base
magnesium ion; hydroxide ion
Mg2+
(OH)-
Mg(OH)2
iron(III) hydroxide
base
ion(III) ion; hydroxide ion
Fe3+
(OH)-
Fe(OH)3
hydrobromic acid
acid
hydrogen ion; bromide ion
H+
Br-
HBr
carbonic acid
acid
hydrogen ion; carbonate ion
H+
(CO3)2-
H2CO3
chromium(II) hydroxide
base
chromium(II) ion; hydroxide ion
Cr2+
(OH)-
Cr(OH)2
+
bromous acid
acid
hydrogen ion; bromite ion
H
(BrO2)
iron(II) hydroxide
base
iron(II) ion; hydroxide ion
Fe2+
(OH)-
8.
-
HBrO2
Fe(OH)2
Calculate the molarity of the following solutions:
a.
0.025 mol of sodium hydroxide is dissolved in 12 mL of solution
2 significant digits
b.
0,354 mol of barium chloride is dissolved in 293 mL of solution
3 significant digits
c.
0.020 mol of calcium nitrate is dissolved in 1.00 mL of solution
3 significant digits
9.
Calculate the number of moles of solute in each of the two following solutions:
a.
349 mL of 0.0010 M KOH
Given: 349 mL of 0.0010 M KOH solution
Target: mol KOH
Path: mL of 0.0010 M KOH solution 6 L of 0.0010 M KOH solution 6 mol KOH
Factors:
Computation:
Note that the molarity 0.0010 has only two significant digits and so the answer has also two significant digits
b.
5.000 x 103 mL of 3.100 M CoCl2
We note that 5.000 x 103 mL is equal to 5.000 L
Given: 5.000 L 3.100 M CoCl2 solution
Target: mol CoCl2
Path: L 3.100 M CoCl2 solution 6 mol CoCl2
Factors:
Computation:
Note that there are four significant digits in the given data and therefore, the answer has 4 significant digits.
10.
Calculate the grams of solute in each of the following solutions:
a.
0.035 L of 10.0 M HCl
Molar mass HCL
element
# moles
H
Molar mass
1
1.008 g H
Cl
1
35.45 g Cl
Molar mass HCl
Given: 0.035 L 10.0 M HCl solution Target: g HCl
Path: L 10.0 M HCl solution 6 mol HCl 6 g HCL
Factors:
Computation:
b.
8.00 mL of 8.00 M Na2SO4
Molar mass Na2SO4
element
Na
# moles
Molar mass
2
45.98 g Na
S
1
32.07 g S
O
4
64.00 g O
Molar mass Na2SO4
Given: 8.00 mL 8.00 M Na2SO4 solution
Target: g Na2SO4
Path: mL 8.00 M Na2SO4 solution 6 L 8.00 M Na2SO4 solution 6 mol Na2SO4 6 g Na2SO4
Factors:
Computation:
There are 3 significant digits in the answer (rule of multiplication).