smc-4260-60-Empirical formula

CHEMISTRY, M2 EQ-Bank 4 MC

What is the empirical formula for a compound that is 52.1% \(\ce{Al}\) and 47.9% \(\ce{O}\)?

\(\ce{AlO}\)
\(\ce{AlO2}\)
\(\ce{Al3O2}\)
\(\ce{Al2O3}\)

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\(D\)

Show Worked Solution

Assume that there is \(100\) grams in sample.
There are \(52.1\) grams of \(\ce{Al}\) and \(47.9\) grams of \(\ce{O}\).
\(n(\ce{Al}) = \dfrac{52.1}{26.98} = 1.93\ \text{mol}\).
\(n(\ce{O}) = \dfrac{47.9}{16.00} = 2.99\ \text{mol}\).
Divide through by the smallest number of moles to determine the empirical formula:
\(\ce{Al}:\dfrac{1.93}{1.93} = 1\)
\(\ce{O}:\dfrac{2.99}{1.93} = 1.55\)
The empirical formula for the compound is \(\ce{Al2O3}\).

\(\Rightarrow D\)

CHEMISTRY, M2 EQ-Bank 3 MC

Which manganese oxide contains 69.6% manganese and 30.4% oxygen?

Manganese \(\text{(II)}\) oxide
Manganese \(\text{(III)}\) oxide
Manganese \(\text{(IV)}\) oxide
Manganese \(\text{(VII)}\) oxide

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\(B\)

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Assume that there is 100 grams in sample.
Therefore sample has 69.6 grams of \(\ce{Mn}\) and 30.4 grams of \(\ce{O}\).
\(n(\ce{Mn}) = \dfrac{69.6}{54.94} = 1.27\ \text{mol}\).
\(n(\ce{O}) = \dfrac{30.4}{16.00} = 1.90\ \text{mol}\).
Divide through by the smallest number of moles to determine the empirical formula:
\(\ce{Mn}:\dfrac{1.27}{1.27} = 1\)
\(\ce{O}:\dfrac{1.90}{1.27} = 1.5\)
The empirical formula for the compound is \(\ce{Mn2O3}\).
The empirical formula for the compound is \(\ce{Mn2O3}\).
The ionic formula equation for the compound above must be \(\ce{2Mn^{3+} + 3O^{2-}\rightarrow Mn2O3}\)
Therefore, the name of the manganese oxide compound is manganese \(\text{(III)}\) oxide.

\(\Rightarrow B\)

CHEMISTRY, M2 EQ-Bank 5

An organic compound is analysed and found to have the following percentage composition by mass:

\(\ce{C}\) - 54.5%, \(\ce{H}\) - 9.1%, \(\ce{O}\) - 36.4%

Calculate the empirical formula of this compound. (2 marks)

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If the relative molecular mass of the compound is 88.104, determine its molecular formula. (1 mark)

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a. \(\ce{C2H4O}\)

b. \(\ce{C4H8O2}\)

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a. Assume that there is \(100\) grams in sample of the organic compound.

Then there are \(54.5\) grams of \(\ce{C}\), \(9.1\) grams of \(\ce{H}\) and \(36.4\) grams of \(\ce{O}\).

\(n(\ce{C}) = \dfrac{54.5}{12.01} = 4.54\ \text{mol}\).

\(n(\ce{H}) = \dfrac{9.01}{1.008} = 8.94\ \text{mol}\).

\(n(\ce{O}) = \dfrac{36.4}{16.00} = 2.28\ \text{mol}\).

Divide through by the smallest number of moles to determine the empirical formula

\(\ce{C}: \dfrac{4.54}{2.28} \approx 2\)

\(\ce{H}: \dfrac{8.94}{2.28} \approx 4\)

\(\ce{O}: \dfrac{2.28}{2.28} = 1\)

The empirical formula for the organic compound is \(\ce{C2H4O}\).

b. The molar mass of \(\ce{C2H4O} = 2(12.01) + 4(1.008) + 16.00 = 44.052\).

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

Ratio \(= \dfrac{88.104}{44.052} = 2\).

Hence, the molecular formula for the compound is \(\ce{C4H8O2}\).

CHEMISTRY, M2 EQ-Bank 8v3

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)

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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

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)

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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 4

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

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The empirical formula of a compound is \(\ce{C4H5O2}\) and its molar mass is determined to be 340.32 g mol\(^{-1}\).
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 = \(\ce{CH2O}\)

Lactic acid and fructose 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 = \(\ce{CH2O}\)

Lactic acid and fructose 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 \(\ce{C16H20O8}\).