Question

1. given 10.00 gram samples of licl, libr, lif and lii, place the samples in order of least to greatest number of atoms of lithium, li.

Explanation:

Step1: Calculate moles of each compound

Molar mass of \( \text{LiCl} \): \( 6.94 + 35.45 = 42.39 \, \text{g/mol} \)
Moles of \( \text{LiCl} \): \( \frac{10.00}{42.39} \approx 0.236 \, \text{mol} \) (1 mol Li per mol LiCl, so moles of Li = 0.236 mol)

Molar mass of \( \text{LiBr} \): \( 6.94 + 79.90 = 86.84 \, \text{g/mol} \)
Moles of \( \text{LiBr} \): \( \frac{10.00}{86.84} \approx 0.115 \, \text{mol} \) (moles of Li = 0.115 mol) Wait, no: Wait, 10/86.84 ≈ 0.115? Wait, 86.840.115≈10? No, 86.840.115≈10.0? Wait, 86.840.115 = 86.840.1 + 86.84*0.015 = 8.684 + 1.3026 = 9.9866 ≈10. So moles of LiBr is ≈0.115 mol? Wait, no, 10g / 86.84g/mol ≈0.115 mol. Then moles of Li is 0.115 mol.

Molar mass of \( \text{LiF} \): \( 6.94 + 19.00 = 25.94 \, \text{g/mol} \)
Moles of \( \text{LiF} \): \( \frac{10.00}{25.94} \approx 0.385 \, \text{mol} \) (moles of Li = 0.385 mol)

Molar mass of \( \text{LiI} \): \( 6.94 + 126.90 = 133.84 \, \text{g/mol} \)
Moles of \( \text{LiI} \): \( \frac{10.00}{133.84} \approx 0.0747 \, \text{mol} \) (moles of Li = 0.0747 mol)

Step2: Compare moles of Li (since 1 mol compound has 1 mol Li)

Moles of Li:

• LiI: ~0.0747 mol
• LiBr: ~0.115 mol
• LiCl: ~0.236 mol
• LiF: ~0.385 mol

Number of atoms is proportional to moles (Avogadro's number). So order from least to greatest number of Li atoms:

LiI (least) < LiBr < LiCl < LiF (greatest)

Answer:

Order (least to greatest number of Li atoms): \( \text{LiI} < \text{LiBr} < \text{LiCl} < \text{LiF} \)