The mineral spodumene has the empirical formula \(\mathrm{LiAlSi}_{2} \mathrm{O}_{6} .\) Given that the percentage of lithium- 6 atoms in naturally occuring lithium is \(7.40 \%,\) how many lithium- 6 atoms are present in a 518 g sample of spodumene?

Short Answer

Expert verified
The number of lithium- 6 atoms present in a 518 g sample of spodumene is approximately \(1.883 \times 10^{23}\) atoms.

Step by step solution

01

Determine gram-molecular weight of Spodumene

Using the periodic table, we find the atomic weights of Li, Al, Si, and O: \[Li = 6.94\, g/mol\]; \[Al = 26.98\, g/mol\]; \[Si = 28.09\, g/mol\]; \[O = 16.00\, g/mol\]. Using the empirical formula \(LiAlSi_{2}O_6\), the gram-molecular weight calculates to \[6.94\, g/mol + 26.98\, g/mol + 2(28.09\, g/mol) + 6(16.00\, g/mol) = 154.22\, g/mol\].
02

Calculate the grams of Lithium in 518 g sample

The fractional part by weight of Lithium in the empirical formula equals to 6.94 g out of the total 154.22 g. Hence, the grams of Lithium in a 518 g sample of Spodumene calculates to \[(6.94\, g / 154.22\, g) * 518\, g = 29.37\, g\].
03

Calculate the moles of Lithium

From the knowledge of basic Chemistry, we can say that, one mole equals the atomic weight expressed in grams. Hence, the moles of Lithium in a 29.37 g is \[29.37\, g / 6.94\, g/mol = 4.23\, mol\].
04

Calculate the total number of Lithium atoms

Using Avogadro's number, \(6.022 \times 10^{23}\), which states that there are \(6.022 \times 10^{23}\) atoms/ions in one mole of any substance. So, the total number of Lithium atoms in 4.23 mol is \[4.23 \times 6.022 \times 10^{23} = 2.545 \times 10^{24}\].
05

Calculate the number of Lithium-6 atoms

Since only 7.40% of all naturally occurring Lithium atoms are Lithium-6, the total number of Lithium-6 atoms in our Spodumene sample will be \[0.0740 * (2.545 \times 10^{24}) = 1.883 \times 10^{23}\].

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