Calculating Mass Percentage of Mg in Mg-Al Alloy

AI Thread Summary
To calculate the mass percentage of magnesium (Mg) in a Mg-Al alloy with a total mass of 0.263 g and a density of 2.40 g/cm3, the volume of the alloy must first be determined. The participant initially attempted to find the mass of Mg by assuming additive volumes, which is incorrect since volumes are not additive in mixtures. The correct approach involves using the mass and density relationships to derive two equations, one for mass and one for volume. After some confusion, the participant received clarification from their teacher and is now working towards a solution. Understanding the non-additivity of volumes is crucial for accurately calculating the mass percentage of Mg in the alloy.
Lancelot59
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Homework Statement


A Mg-Al alloy has a mass of 0.263 g and a density of 2.40 g/cm3. The density of pure Mg =
1.74 g/cm3 and for pure Al = 2.70 g/cm3. Calculate the % Mg in the alloy.

Homework Equations


D=M/V

The Attempt at a Solution


Well I'm not really sure where to do. I solved for the volume of the alloy, then tried to use it to determine the mass of the Mg. Once there I divided by the total mass and multiplied by 100 to get the wrong answer.

How can I go about getting the mass of Mg in the alloy?
 
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Assume that volume of the alloy is the sum of volumes and mass is the sum of masses. That gives two equations in two unknowns.

In reality only one of these equations is correct - that is, mass is sum of masses. Volumes are not additive. But that's not your fault.

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methods
 
Could you please elaborate on why volumes aren't additive in this case?4

Also I'm getting nowhere fast.

I wound up with 0.263g = MassMg + MassAl

Then solved for MassAl like so:

MassAl = 0.263 - MassMg

Then tried to substitute it back in:

0.263 = MassMg + 0.263 - MassMg

Which just cancels itself out.
 
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Where is the equation for sum of volumes? Note, that volumes can be expressed using mass and density.

Volumes are not additive in general. If you mix 50 mL of water and 50 mL of ethanol you will get 96.4 mL of mixture.

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Ok. I finally got word back from my teacher, and I think I see what to do now. I'll post my resulting solution when I figure it out.
 
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