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Determine via calculations, the molecular formula of ethanoic acid with a molecular mass of 60g/mol and consisting out of 39,9% carbon, 6,7% hydrogen and 53,4% oxygen.

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To determine the molecular formula of ethanoic acid (also known as acetic acid) with a molecular mass of 60 g/mol and given percentage composition of carbon (C), hydrogen (H), and oxygen (O), we can follow these steps:

  1. Convert the percentages to grams: Assume we have 100 grams of the substance. This means we have:

    • Carbon: 39.9 g
    • Hydrogen: 6.7 g
    • Oxygen: 53.4 g

  2. Convert grams to moles using the molar mass of each element:

    • Molar mass of Carbon (C) = 12.01 g/mol
    • Molar mass of Hydrogen (H) = 1.008 g/mol
    • Molar mass of Oxygen (O) = 16.00 g/mol

    Now, calculate the number of moles of each element: \[ \text{Moles of C} = \frac{39.9 , \text{g}}{12.01 , \text{g/mol}} \approx 3.32 , \text{mol} \] \[ \text{Moles of H} = \frac{6.7 , \text{g}}{1.008 , \text{g/mol}} \approx 6.64 , \text{mol} \] \[ \text{Moles of O} = \frac{53.4 , \text{g}}{16.00 , \text{g/mol}} \approx 3.34 , \text{mol} \]

  3. Calculate the simplest mole ratio: To find the simplest molar ratio, divide each value by the smallest number of moles obtained.

    • Smallest number of moles ≈ 3.32 (C moles)

    Now, calculate the ratios: \[ \text{Ratio of C} = \frac{3.32}{3.32} = 1 \] \[ \text{Ratio of H} = \frac{6.64}{3.32} \approx 2 \] \[ \text{Ratio of O} = \frac{3.34}{3.32} \approx 1 \]

  4. Identify the empirical formula: The simplest whole-number ratio is \( C_1H_2O_1 \), which gives the empirical formula: \[ \text{Empirical formula} = CH_2O \]

  5. Determine the molar mass of the empirical formula: \[ \text{Molar mass of CH}_2\text{O} = 12.01 + (2 \times 1.008) + 16.00 = 30.03 , \text{g/mol} \]

  6. Calculate the number of empirical units in the molecular formula: \[ \text{Molecular mass} = \frac{\text{Molecular mass of ethanoic acid (60 g/mol)}}{\text{Molar mass of empirical formula (30.03 g/mol)}} \] \[ = \frac{60}{30.03} \approx 2 \]

  7. Determine the molecular formula: Since we have two empirical units in the molecular formula, we can multiply each subscript in the empirical formula \( CH_2O \) by 2: \[ \text{Molecular formula} = C_{2}H_{4}O_{2} \]

Therefore, the molecular formula of ethanoic acid is Câ‚‚Hâ‚„Oâ‚‚.