Step 1: Assume that the mass of a compound is 100 g.
[tex]\text{mass C = 45.90 g}[/tex]
[tex]\text{mass H = 2.75 g}[/tex]
[tex]\text{mass O = 26.20 g}[/tex]
[tex]\text{mass S = 17.50 g}[/tex]
[tex]\text{mass N = 7.65 g}[/tex]
Step 2: Calculate the number of moles of each element.
[tex]n \: \text{C = 45.90 g} × \frac{\text{1 mol}}{\text{12.01 g}} = \text{3.822 mol}[/tex]
[tex]n \: \text{H = 2.75 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{2.728 mol}[/tex]
[tex]n \: \text{O = 26.20 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{1.638 mol}[/tex]
[tex]n \: \text{S = 17.50 g} × \frac{\text{1 mol}}{\text{32.07 g}} = \text{0.5457 mol}[/tex]
[tex]n \: \text{N = 7.65 g} × \frac{\text{1 mol}}{\text{14.01 g}} = \text{0.5460 mol}[/tex]
Step 3: Represent an empirical formula.
[tex]\text{empirical formula} = \text{C}_{v}\text{H}_{w}\text{O}_{x}\text{S}_{y}\text{N}_{z}[/tex]
Step 4: Divide the number of moles of each element by the least number of moles.
[tex]v = \frac{\text{3.822 mol}}{\text{0.5457 mol}} = 7[/tex]
[tex]w = \frac{\text{2.728 mol}}{\text{0.5457 mol}} = 5[/tex]
[tex]x = \frac{\text{1.638 mol}}{\text{0.5457 mol}} = 3[/tex]
[tex]y = \frac{\text{0.5457 mol}}{\text{0.5457 mol}} = 1[/tex]
[tex]z = \frac{\text{0.5460 mol}}{\text{0.5460 mol}} = 1[/tex]
Step 5: Write the empirical formula.
[tex]\text{empirical formula} = \text{C}_{7}\text{H}_{5}\text{O}_{3}\text{SN}[/tex]
Step 6: Represent a molecular formula.
[tex]\text{molecular formula} = (\text{C}_{7}\text{H}_{5}\text{O}_{3}\text{SN})_{n}[/tex]
Step 7: Calculate the empirical weight.
empirical weight = (12.01 g/mol × 7) + (1.008 g/mol × 5) + (16.00 g/mol × 3) + (32.07 g/mol × 1) + (14.01 g/mol × 1)
empirical weight = 183.19 g/mol
Step 8: Divide the molar mass by the empirical mass.
[tex]n = \frac{\text{molecular weight}}{\text{empirical weight}}[/tex]
[tex]n = \frac{\text{183.18 g/mol}}{\text{183.19 g/mol}}[/tex]
[tex]n = 1[/tex]
Step 9: Multiply the subscripts by the value of n to obtain the molecular formula.
[tex]\text{molecular formula} = (\text{C}_{7}\text{H}_{5}\text{O}_{3}\text{SN})_{1}[/tex]
[tex]\boxed{\text{molecular formula} = \text{C}_{7}\text{H}_{5}\text{O}_{3}\text{SN}}[/tex]
[tex]\\[/tex]
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