Answer to Question #334030 in Chemistry for Isa

Question

A 0.784-g sample of an alloy containing Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated, providing a mixed precipitate of Al(C9H6NO)3 and Mg(C9H6NO)2 that weighs 8.215 g. Igniting the precipitate converts it to a mixture of Al2O3 and MgO that weighs 0.932 g. Calculate the %w/w Al and %w/w Mg in the alloy.

Answer 1

Solution:

For Al we have:

Al → Al(C₉H₆NO)₃ → Al₂O₃

According to chemical scheme above:

2 × Moles of Al₂O₃ = Moles of Al(C₉H₆NO)₃

Converting from moles to grams and solving yields an equation relating the grams of Al₂O₃ to the grams of Al(C₉H₆NO)₃:

2 × (Mass of Al₂O₃ / Molar mass of Al₂O₃) = (Mass of Al(C₉H₆NO)₃ / Molar mass of Al(C₉H₆NO)₃)

Molar mass of Al₂O₃ is 101.96 g/mol

Molar mass of Al(C₉H₆NO)₃ is 459.4317 g/mol

Therefore,

2 × (Mass of Al₂O₃ / 101.96) = (Mass of Al(C₉H₆NO)₃ / 459.4317)

918.8634 × Mass of Al₂O₃ = 101.96 × Mass of Al(C₉H₆NO)₃

Mass of Al₂O₃ = 0.11096 × Mass of Al(C₉H₆NO)₃

For Mg we have:

Mg → Mg(C₉H₆NO)₂ → MgO

According to chemical scheme above:

Moles of MgO = Moles of Mg(C₉H₆NO)₂

Converting from moles to grams and solving yields an equation relating the grams of MgO to the grams of Mg(C₉H₆NO)₂:

(Mass of MgO / Molar mass of MgO) = (Mass of Mg(C₉H₆NO)₂ / Molar mass of Mg(C₉H₆NO)₂)

Molar mass of MgO is 40.3044 g/mol

Molar mass of Mg(C₉H₆NO)₂ is 312.605 g/mol

Therefore,

(Mass of MgO / 40.3044) = (Mass of Mg(C₉H₆NO)₂ / 312.605)

312.605 × Mass of MgO = 40.3044 × Mass of Mg(C₉H₆NO)₂

Mass of MgO = 0.12893 × Mass of Mg(C₉H₆NO)₂

Mass of Al₂O₃ + Mass of MgO = 0.932 g

Therefore,

0.11096 × Mass of Al(C₉H₆NO)₃ + 0.12893 × Mass of Mg(C₉H₆NO)₂ = 0.932 g

Mass of Al(C₉H₆NO)₃ + Mass of Mg(C₉H₆NO)₂ = 8.215 g

0.11096 × Mass of Al(C₉H₆NO)₃ + 0.12893 × Mass of Mg(C₉H₆NO)₂ = 0.932 g

Multiplying the first equation by 0.11096 and subtracting the second equation gives:

−0.01797 × Mass of Mg(C₉H₆NO)₂ = −0.020464

Mass of Mg(C₉H₆NO)₂ = 1.1388 g

The mass of Al(C₉H₆NO)₃ can then be calculated using the known combined mass of the two original precipitates:

Mass of Al(C₉H₆NO)₃ = 8.215 g − Mass of Mg(C₉H₆NO)₂ = 8.215 g − 1.1388 g = 7.0762 g

Mass of Al(C₉H₆NO)₃ = 7.0762 g

For Al we have:

Al → Al(C₉H₆NO)₃ → Al₂O₃

According to chemical scheme above:

Moles of Al = Moles of Al(C₉H₆NO)₃

Converting from moles to grams and solving yields an equation relating the grams of Al to the grams of Al(C₉H₆NO)₃:

(Mass of Al / Molar mass of Al) = (Mass of Al(C₉H₆NO)₃ / Molar mass of Al(C₉H₆NO)₃)

Molar mass of Al is 26.9815 g/mol

Molar mass of Al(C₉H₆NO)₃ is 459.4317 g/mol

Mass of Al(C₉H₆NO)₃ = 7.0762 g

Therefore,

Mass of Al = (Mass of Al(C₉H₆NO)₃ × Molar mass of Al) / (Molar mass of Al(C₉H₆NO)₃)

Mass of Al = (7.0762 g × 26.9815 g/mol) / (459.4317 g/mol) = 0.4156 g

Mass of Al = 0.4156 g

%Al = (Mass of Al / Mass of sample) × 100% = (0.4156 g / 0.784 g) × 100% = 53.01%

%Al = 53.01%

For Mg we have:

Mg → Mg(C₉H₆NO)₂ → MgO

According to chemical scheme above:

Moles of Mg = Moles of Mg(C₉H₆NO)₂

Converting from moles to grams and solving yields an equation relating the grams of Mg to the grams of Mg(C₉H₆NO)₂:

(Mass of Mg / Molar mass of Mg) = (Mass of Mg(C₉H₆NO)₂ / Molar mass of Mg(C₉H₆NO)₂)

Molar mass of Mg is 24.305 g/mol

Molar mass of Mg(C₉H₆NO)₂ is 312.605 g/mol

Mass of Mg(C₉H₆NO)₂ = 1.1388 g

Therefore,

Mass of Mg = (Mass of Mg(C₉H₆NO)₂ × Molar mass of Mg) / (Molar mass of Mg(C₉H₆NO)₂)

Mass of Mg = (1.1388 g × 24.305 g/mol) / (312.605 g/mol) = 0.08854 g

Mass of Mg = 0.08854 g

%Mg = (Mass of Mg / Mass of sample) × 100% = (0.08854 g / 0.784 g) × 100% = 11.29%

%Mg = 11.29%

Answer:

The %w/w Al in the alloy is 53.01%

The %w/w Mg in the alloy is 11.29%