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CHEMISTRY, M2 EQ-Bank 8v3

  1. Consider the compounds ethanol (\(\ce{C2H6O}\)), formaldehyde (\(\ce{CH2O}\)), and acetic acid (\(\ce{C2H4O2}\)).
    Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  2. The empirical formula of a compound is \(\ce{C4H5N2}\) and its molar mass is determined to be 243.3 g mol\(^{-1}\).
    Calculate the molecular formula of this compound.    (3 marks)
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a.    Formaldehyde (\(\ce{CH2O}\)) and acetic acid (\(\ce{C2H4O2}\)) have the same empirical formula of \(\ce{CH2O}\).

b.    The molecular formula of the compound is \(\ce{C12H15N6}\).

Show Worked Solution

a.    Determine the empirical formula of each compound:

Ethanol (\(\ce{C2H6O}\)):
\[\text{Empirical formula} = \ce{C2H6O}\]

Formaldehyde (\(\ce{CH2O}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetic acid (\(\ce{C2H4O2}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Formaldehyde and acetic acid have the same empirical formula of \(\ce{CH2O}\).

b.     Calculate the molar mass of the empirical formula \(\ce{C4H5N2}\):

\[\text{Molar mass of} \ \ce{C4H5N2} = 4 \times 12.01 + 5 \times 1.01 + 2 \times 14.01 = 81.1 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{243.3 \ \text{g/mol}}{81.1 \ \text{g/mol}} = 3\]

Multiply the subscripts in the empirical formula by 3 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C4H5N2)} \times 3 = \ce{C12H15N6}\]

Thus, the molecular formula of the compound is \(\ce{C12H15N6}\).

CHEMISTRY, M2 EQ-Bank 8v2

  1. Consider the compounds glyceraldehyde (\(\ce{C3H6O3}\)), glycolic acid (\(\ce{C2H4O3}\)), and ribose (\(\ce{C5H10O5}\)).
    Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  2. The empirical formula of a compound is \(\ce{C3H5O}\) and its molar mass is determined to be 114.14 g mol\(^{-1}\).
    Calculate the molecular formula of this compound.    (3 marks)
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a.    glyceraldehyde (\(\ce{C3H6O3}\)) and ribose (\(\ce{C5H10O5}\)) have the same empirical formula of \(\ce{CH2O}\).

b.    The molecular formula of the compound is \(\ce{C6H10O2}\).

Show Worked Solution

a.    Determine the empirical formula of each compound:

Glyceraldehyde (\(\ce{C3H6O3}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Glycolic acid (\(\ce{C2H4O3}\)):
\[\text{Empirical formula} = \ce{C2H4O3}\]

Ribose (\(\ce{C5H10O5}\)):
\[\text{Empirical formula} = \ce{CH2O}\]

Acetone and ribose have the same empirical formula of \(\ce{CH2O}\).

b.    Calculate the molar mass of the empirical formula \(\ce{C3H5O}\):

\[\text{Molar mass of} \ \ce{C3H5O} = 3 \times 12.01 + 5 \times 1.008 + 1 \times 16.00 = 57.07 \ \text{g/mol}\]

Determine the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[\text{Ratio} = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{114.14 \ \text{g/mol}}{57.07 \ \text{g/mol}} = 2\]

Multiply the subscripts in the empirical formula by 2 to get the molecular formula:

\[\text{Molecular formula} = \ce{(C3H5O)} \times 2 = \ce{C6H10O2}\]

Thus, the molecular formula of the compound is \(\ce{C6H10O2}\).

CHEMISTRY, M2 EQ-Bank 8

  1. Consider the compounds butyraldehyde \(\ce{(C4H8O)}\), lactic acid \(\ce{(C3H6O3)}\), and fructose \(\ce{(C6H12O6)}\).
  2. Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  3. The empirical formula of a compound is \(\ce{C4H5O2}\) and its molar mass is determined to be 340.32 g mol\(^{-1}\).
  4. Calculate the molecular formula of this compound.     (3 marks)

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a.    Determine the empirical formula of each compound.

Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)

Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)

Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)

→ Butyraldehyde and lactic acid have the same empirical formula of \(\ce{CH2O}\).
 

b.    \(\ce{C16H20O8}\)

Show Worked Solution

a.    Determine the empirical formula of each compound.

Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)

Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)

Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)

→ Butyraldehyde and lactic acid have the same empirical formula of \(\ce{CH2O}\).

 

b.    Calculate the molar mass of the empirical formula \(\ce{C4H5O2}\):

  \(\text{Molar mass}\ \ce{(C4H5O2)} = 4 \times 12.01 + 5 \times 1.008 + 2 \times 16.00 = 85.08\ \text{g/mol}\)
 

Ratio of the molar mass of the compound to the molar mass of the empirical formula:

  \(\text{Ratio} = \dfrac{\text{Molar mass}}{\text{Empirical formula mass}} = \dfrac{340.32\ \text{g mol}^{-1}}{85.08\ \text{g mol}^{-1}} =4\)
 

Multiply the subscripts in the empirical formula by 4 to get the molecular formula:

  \(\text{Molecular formula} = \ce{(C4H5O2)} \times 4 = \ce{C16H20O8}\)

→ Thus, the molecular formula of the compound is \(\ce{C16H20O8}\).