Solution (a)
Step 1: Calculate the number of moles of each element.
[tex]n \: \text{Al = 0.545 g} × \frac{\text{1 mol}}{\text{26.98 g}} = \text{0.0202 mol}[/tex]
[tex]n \: \text{O = 0.485 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{0.0303 mol}[/tex]
Step 2: Represent an empirical formula.
[tex]\text{empirical formula} = \text{Al}_{x}\text{O}_{y}[/tex]
Step 3: Divide the number of moles of each element by the least number of moles.
[tex]x = \frac{\text{0.0202 mol}}{\text{0.0202 mol}} = 1 × 2 = 2[/tex]
[tex]y = \frac{\text{0.0303 mol}}{\text{0.0202 mol}} = 1.5 × 2 = 3[/tex]
Step 4: Write the empirical formula.
[tex]\boxed{\text{empirical formula} = \text{Al}_{2}\text{O}_{3}}[/tex]
Solution (b)
Step 1: Assume that the mass of a compound is 100 g.
[tex]\text{mass Pt = 65.0 g}[/tex]
[tex]\text{mass Cl = 23.6 g}[/tex]
[tex]\text{mass N = 9.35 g}[/tex]
[tex]\text{mass H = 2.02 g}[/tex]
Step 2: Calculate the number of moles of each element.
[tex]n \: \text{Pt = 65.0 g} × \frac{\text{1 mol}}{\text{195.1 g}} = \text{0.333 mol}[/tex]
[tex]n \: \text{Cl = 23.6 g} × \frac{\text{1 mol}}{\text{35.45 g}} = \text{0.666 mol}[/tex]
[tex]n \: \text{N = 9.35 g} × \frac{\text{1 mol}}{\text{14.01 g}} = \text{0.6674 mol}[/tex]
[tex]n \: \text{H = 2.02 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{2.00 mol}[/tex]
Step 3: Represent an empirical formula.
[tex]\text{empirical formula} = \text{Pt}_{w}\text{Cl}_{x}\text{N}_{y}\text{H}_{z}[/tex]
Step 4: Divide the number of moles of each element by the least number of moles.
[tex]w = \frac{\text{0.333 mol}}{\text{0.333 mol}} = 1[/tex]
[tex]x = \frac{\text{0.666 mol}}{\text{0.333 mol}} = 2[/tex]
[tex]y = \frac{\text{0.6674 mol}}{\text{0.333 mol}} = 2[/tex]
[tex]z = \frac{\text{2.00 mol}}{\text{0.333 mol}} = 6[/tex]
Step 5: Write the empirical formula.
[tex]\boxed{\text{empirical formula} = \text{Pt}\text{Cl}_{2}\text{N}_{2}\text{H}_{6}}[/tex]
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