Question

A 0.6113-g sample of Dow metal, containing aluminum, magnesium, and other metals, was dissolved and treated to prevent interferences by the other metals. The aluminum and magnesium were precipitated with 8-hydroxyquinoline. After filtering and drying, the mixture of Al(C9H6NO)3 and Mg(C9H6NO)2 was found to weigh 7.8154 g. The mixture of dried precipitates was then ignited, converting the precipitate to a mixture of Al2O3 and MgO. The weight of this mixed solid was found to be 1.0022 g. Calculate the %w/w Al and %w/w Mg in the alloy​

Answer 1

Answer:

95.55% w/w Mg; 2.89% w/w Al

Explanation:

We can solve this question using both precipitates mixture. for example, for the oxide:

Mass MgO + Mass Al2O3 = 1.0022g

Moles MgO*40.3044g/mol + Moles Al2O3*101.96g/mol = 1.0022g

Where we are writting the molar mass of each oxide.

We can write, thus:

40.3044X + 50.98Y = 1.0022 (1)

Where X are moles of Mg and Y moles of Al

Because 1 mole of Al2O3 are 2 moles of Al

And for the other mixtue:

312.605X + 459.4317Y = 7.8154 (2)

Replacing (2) in (1):

312.605(1.0022-50.98Y / 40.3044) + 459.4317Y = 7.8154

312.605(0.024866-1.26487Y) + 459.4317Y = 7.8154

7.7732 - 395.406Y + 459.4317Y = 7.8154

64.0257Y = 0.0422

Y = 6.591x10⁻⁴ moles = Moles Al

The moles of Mg are:

312.605X + 459.4317*6.591x10⁻⁴ moles = 7.8154

312.605X + 0.3028 = 7.8154

312.605X = 7.5126

X = 0.02403 Moles = Moles Mg

The mass of each metal and its mass percent is:

Mg: 0.02403moles * (24.305g/mol) = 0.5841g / 0.6113g * 100 =

95.55% w/w Mg

Al: 6.591x10⁻⁴ moles * (26.98g/mol) = 0.01767g / 0.6113g * 100 =

2.89% w/w Al