Solution (Empirical Formula)
Step 1: Assume that the mass of a compound is 100 g.
[tex]\text{mass C = 44.4 g}[/tex]
[tex]\text{mass H = 6.21 g}[/tex]
[tex]\text{mass S = 39.5 g}[/tex]
[tex]\text{mass O = 9.86 g}[/tex]
Step 2: Calculate the number of moles of each element.
[tex]n \: \text{C = 44.4 g} × \frac{\text{1 mol}}{\text{12.01 g}} = \text{3.70 mol}[/tex]
[tex]n \: \text{H = 6.21 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{6.16 mol}[/tex]
[tex]n \: \text{S = 39.5 g} × \frac{\text{1 mol}}{\text{32.07 g}} = \text{1.232 mol}[/tex]
[tex]n \: \text{O = 9.86 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{0.616 mol}[/tex]
Step 3: Represent an empirical formula.
[tex]\text{empirical formula} = \text{C}_{w}\text{H}_{x}\text{S}_{y}\text{O}_{z}[/tex]
Step 4: Divide the number of moles of each element by the least number of moles.
[tex]w = \frac{\text{3.70 mol}}{\text{0.616 mol}} = 6[/tex]
[tex]x = \frac{\text{6.16 mol}}{\text{0.616 mol}} = 10[/tex]
[tex]y = \frac{\text{1.232 mol}}{\text{0.616 mol}} = 2[/tex]
[tex]z = \frac{\text{0.616 mol}}{\text{0.616 mol}} = 1[/tex]
Step 5: Write the empirical formula.
[tex]\boxed{\text{empirical formula} = \text{C}_{6}\text{H}_{10}\text{S}_{2}\text{O}}[/tex]
Solution (Molecular Formula)
Step 1: Represent a molecular formula.
[tex]\text{molecular formula} = (\text{C}_{6}\text{H}_{10}\text{S}_{2}\text{O})_{n}[/tex]
Step 2: Calculate the empirical mass.
empirical mass = (12.01 g/mol × 6) + (1.008 g/mol × 10) + (32.07 g/mol × 2) + (16.00 g/mol × 1)
empirical mass = 162.28 g/mol
Step 3: Divide the molecular mass by the empirical mass.
[tex]n = \frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
[tex]n = \frac{\text{162.28 g/mol}}{\text{162.0 g/mol}}[/tex]
[tex]n = 1[/tex]
Step 4: Multiply the subscripts by the value of n to obtain the molecular formula.
[tex]\text{molecular formula} = (\text{C}_{6}\text{H}_{10}\text{S}_{2}\text{O})_{1}[/tex]
[tex]\boxed{\text{molecular formula} =\text{C}_{6}\text{H}_{10}\text{S}_{2}\text{O}}[/tex]
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