Determine Weight Percent given molecular weight....

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hdp12
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Homework Statement


So for my homework assignment in 'Materials Science and Engineering', We were given the following problem.

1. Li and Ra both are both BCC metals. For this problem assume they form a complete solid solution (even though the very large difference in their atomic size tell us they will not form a complete solid solution). We want to make a solid solution in which half the atoms in the solid solution (by counting) are Li and the other half of the atoms (by counting) are Ra.

a. Calculate the wt% of Li and Ra needed to make this alloy. This corresponds to weighing out the Li and Ra to make the alloy.

Useful data:
Li: MW = 6.941 g/mole r = 0.53 g/cm3
Ra: MW = 226.0254 g/mole r = 5.00 g/cm3

Homework Equations


wt%=(m1 / m1 + m2)(100 wt%)

The Attempt at a Solution


I can't figure out how to do it because we were given molecular weight but we need literal mass. My first instinct was that you simply halve both molecular weights and plug those numbers into the formula.
Then I started analyzing the wording and think the fact that it's BCC is important as well as how it explicitly states (by counting)
Could someone just lead me in the right direction please? I appreciate it
 
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hdp12 said:

Homework Statement


So for my homework assignment in 'Materials Science and Engineering', We were given the following problem.

1. Li and Ra both are both BCC metals. For this problem assume they form a complete solid solution (even though the very large difference in their atomic size tell us they will not form a complete solid solution). We want to make a solid solution in which half the atoms in the solid solution (by counting) are Li and the other half of the atoms (by counting) are Ra.

a. Calculate the wt% of Li and Ra needed to make this alloy. This corresponds to weighing out the Li and Ra to make the alloy.

Useful data:
Li: MW = 6.941 g/mole r = 0.53 g/cm3
Ra: MW = 226.0254 g/mole r = 5.00 g/cm3

Homework Equations


wt%=(m1 / m1 + m2)(100 wt%)

The Attempt at a Solution


I can't figure out how to do it because we were given molecular weight but we need literal mass. My first instinct was that you simply halve both molecular weights and plug those numbers into the formula.
Then I started analyzing the wording and think the fact that it's BCC is important as well as how it explicitly states (by counting)
Could someone just lead me in the right direction please? I appreciate it
You are given the molar masses for Li and Ra. The molar mass for each element contains the same number of atoms. What is this famous number?

Suppose you wish to make a sample of Li-Ra alloy which contains N atoms total. You know that N/2 atoms are Li and N/2 atoms are Ra.
 
SteamKing said:
You are given the molar masses for Li and Ra. The molar mass for each element contains the same number of atoms. What is this famous number?

Suppose you wish to make a sample of Li-Ra alloy which contains N atoms total. You know that N/2 atoms are Li and N/2 atoms are Ra.

That was actually my other attempt before I decided to post. I divided the atoms in a BCC cell by Avogadro's number and got 3.321*10^-24 Mol
Then I was going to multiply that by each molecular weight, respectively, to get their masses
I just didn't want to follow that thread and be sure of it when I wasn't positive