Page 269 - Modern Analytical Chemistry

1400-CH08 9/9/99 2:18 PM Page 252

252 Modern Analytical Chemistry

containing four unknown terms. To solve this pair of equations, we must find
two additional equations relating the four unknowns to one another. These
additional equations describe the stoichiometric relationships between the two
compounds containing aluminum and the two compounds containing
magnesium and are based on the conservation of Al and Mg. Thus, for Al we
have

2 ´moles Al 2 O 3 = moles Al(C 9 H 6 NO) 3
Converting from moles to grams and solving yields an equation relating the
grams of Al 2 O 3 to the grams of Al(C 9 H 6 NO) 3
2
9
6
2 ´g Al O 3 g Al(C H NO) 3
=
FW Al O 3 FW Al(C H NO) 3
2
6
9
9
g Al(C H NO) 3 ´FW Al O 3 g Al(C H NO) 3 ´ 101 .96 g/mol
9
6
6
2
= =
g Al O 3
2
9
2 ´FW Al(C H NO) 3 2 ´ 459 .45 g/mol
6
= . 0 11096 ´g Al(C H NO) 3
9
6
For Mg we have
Moles MgO = moles Mg(C 9 H 6 NO) 2
g MgO g Mg(C H NO) 2
9
6
=
FW MgO FW Mg(C H NO ) 2
6
9
g Mg(C H NO) 2 ´ FW MgO g Mg(C H NO) 2 ´40 .304 g /mol
9
6
9
6
g MgO = =
FW Mg(C H NO) 2 312.61 g /mol
9
6
= . 0 12893 ´ g Mg(C H NO) 2
6
9
Substituting the equations for g MgO and g Al 2 O 3 into the equation for the
combined weights of MgO and Al 2 O 3 leaves us with two equations and two
unknowns.
g Al(C 9 H 6 NO) 3 + g Mg(C 9 H 6 NO) 2 = 7.8154
0.11096 ´g Al(C 9 H 6 NO) 3 + 0.12893 ´g Mg(C 9 H 6 NO) 2 = 1.0022
Multiplying the first equation by 0.11096 and subtracting the second equation
gives
–0.01797 ´g Mg(C 9 H 6 NO) 2 = –0.1350
which can be solved for the mass of Mg(C 9 H 6 NO) 2 .
g Mg(C 9 H 6 NO) 2 = 7.5125 g
The mass of Al(C 9 H 6 NO) 3 can then be calculated using the known combined
mass of the two original precipitates.
7.8154 g – g Mg(C 9 H 6 NO) 2 = 7.8154 g – 7.5125 g = 0.3029 g Al(C 9 H 6 NO) 3
Using the conservation of Mg and Al, the %w/w Mg and %w/w Al in the
sample can now be determined as in Example 8.1, where AW is an atomic
weight.