Solution (Empirical Formula):
Step 1: Assume that the mass of a compound is 100 g.
[tex]\text{mass C = 75.0 g}[/tex]
[tex]\text{mass H = 5.05 g}[/tex]
[tex]\text{mass O = 20.0 g}[/tex]
Step 2: Calculate the number of moles of each element.
[tex]n \: \text{C = 75.0 g} × \frac{\text{1 mol}}{\text{12.01 g}} = \text{6.24 mol}[/tex]
[tex]n \: \text{H = 5.05 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{5.01 mol}[/tex]
[tex]n \: \text{O = 20.0 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{1.25 mol}[/tex]
Step 3: Represent an empirical formula.
[tex]\text{empirical formula} = \text{C}_{x}\text{H}_{y}\text{O}_{z}[/tex]
Step 4: Divide the number of moles of each element by the least number of moles.
[tex]x = \frac{\text{6.24 mol}}{\text{1.25 mol}} = 5[/tex]
[tex]y = \frac{\text{5.01 mol}}{\text{1.25 mol}} = 4[/tex]
[tex]z = \frac{\text{1.25 mol}}{\text{1.25 mol}} = 1[/tex]
Step 5: Write the empirical formula.
[tex]\boxed{\text{empirical formula} = \text{C}_{5}\text{H}_{4}\text{O}}[/tex]
Solution (Molecular Formula):
Step 1: Represent a molecular formula.
[tex]\text{molecular formula} = (\text{C}_{5}\text{H}_{4}\text{O})_{n}[/tex]
Step 2: Calculate the empirical mass.
empirical mass = (12.01 g/mol × 5) + (1.008 g/mol × 4) + (16.00 g/mol × 1)
empirical mass = 80.082 g/mol
Step 3: Divide the molar mass by the empirical mass.
[tex]n = \frac{\text{molar mass}}{\text{empirical mass}}[/tex]
[tex]n = \frac{\text{240.28 g/mol}}{\text{80.082 g/mol}}[/tex]
[tex]n = 3[/tex]
Step 4: Multiply the subscripts by the value of n to obtain the molecular formula.
[tex]\text{molecular formula} = (\text{C}_{5}\text{H}_{4}\text{O})_{3}[/tex]
[tex]\boxed{\text{molecular formula} = \text{C}_{15}\text{H}_{12}\text{O}_{3}}[/tex]
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