Question

What is the molecular formula of each compound? (a) Empirical formula \(\mathrm{CH}_{2}(\mathscr{A}=42.08 \mathrm{~g} / \mathrm{mol})\) (b) Empirical formula \(\mathrm{NH}_{2}(\mathscr{M}=32.05 \mathrm{~g} / \mathrm{mol})\) (c) Empirical formula \(\mathrm{NO}_{2}(\mathscr{M}=92.02 \mathrm{~g} / \mathrm{mol})\) (d) Empirical formula CHN \((\mathscr{M}=135.14 \mathrm{~g} / \mathrm{mol})\)

Short answer

(a) \(\mathrm{C}_3\mathrm{H}_6\), (b) \(\mathrm{N}_2\mathrm{H}_4\), (c) \(\mathrm{N}_2\mathrm{O}_4\), (d) \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}_5\)

Step-by-step solution

Calculate the Molar Mass of Empirical Formula (a)

For empirical formula \(\mathrm{CH}_{2}\), calculate the molar mass: - Carbon (C): 12.01 \mathrm{g/mol} - Hydrogen (H): 1.01 \mathrm{g/mol} - Total molar mass: \[\mathscr{M}_{\text{empirical}}=12.01 + 2(1.01) = 14.03 \mathrm{~g/mol}\]

Determine the Ratio for Empirical Formula of (a)

Divide the given molecular mass \(\mathscr{M}=42.08 \mathrm{~g/mol}\) by the empirical molar mass calculated: \[\frac{42.08 \mathrm{~g/mol}}{14.03 \mathrm{~g/mol}}\approx 3\]

Find the Molecular Formula for (a)

Since the ratio is 3, multiply the subscripts in \(\mathrm{CH}_{2} \) by 3: - Molecular formula: \[\mathrm{C}_3\mathrm{H}_6\]

Calculate the Molar Mass of Empirical Formula (b)

For empirical formula \(\mathrm{NH}_{2}\), calculate the molar mass: - Nitrogen (N): 14.01 \mathrm{g/mol} - Hydrogen (H): 1.01 \mathrm{g/mol} - Total molar mass: \[\mathscr{M}_{\text{empirical}}=14.01 + 2(1.01) = 16.03 \mathrm{~g/mol}\]

Determine the Ratio for Empirical Formula of (b)

Divide the given molecular mass \(\mathscr{M}=32.05 \mathrm{~g/mol}\) by the empirical molar mass calculated: \[\frac{32.05 \mathrm{~g/mol}}{16.03 \mathrm{~g/mol}}\approx 2\]

Find the Molecular Formula for (b)

Since the ratio is 2, multiply the subscripts in \(\mathrm{NH}_{2}\) by 2: - Molecular formula: \[\mathrm{N}_2\mathrm{H}_4\]

Calculate the Molar Mass of Empirical Formula (c)

For empirical formula \(\mathrm{NO}_{2}\), calculate the molar mass: - Nitrogen (N): 14.01 \mathrm{g/mol} - Oxygen (O): 16.00 \mathrm{g/mol} - Total molar mass: \[\mathscr{M}_{\text{empirical}}=14.01 + 2(16.00) = 46.01 \mathrm{~g/mol}\]

Determine the Ratio for Empirical Formula of (c)

Divide the given molecular mass \(\mathscr{M}=92.02 \mathrm{~g/mol}\) by the empirical molar mass calculated: \[\frac{92.02 \mathrm{~g/mol}}{46.01 \mathrm{~g/mol}}\approx 2\]

Find the Molecular Formula for (c)

Since the ratio is 2, multiply the subscripts in \(\mathrm{NO}_{2}\) by 2: - Molecular formula: \[\mathrm{N}_2\mathrm{O}_4\]

Calculate the Molar Mass of Empirical Formula (d)

For empirical formula \(\mathrm{CHN}\), calculate the molar mass: - Carbon (C): 12.01 \mathrm{g/mol} - Hydrogen (H): 1.01 \mathrm{g/mol} - Nitrogen (N): 14.01 \mathrm{g/mol} - Total molar mass: \[\mathscr{M}_{\text{empirical}} = 12.01 + 1.01 + 14.01 = 27.03 \mathrm{~g/mol}\]

Determine the Ratio for Empirical Formula of (d)

Divide the given molecular mass \(\mathscr{M}=135.14 \mathrm{~g/mol}\) by the empirical molar mass calculated: \[\frac{135.14 \mathrm{~g/mol}}{27.03 \mathrm{~g/mol}}\approx 5\]

Find the Molecular Formula for (d)

Since the ratio is 5, multiply the subscripts in \(\mathrm{CHN}\) by 5: - Molecular formula: \[\mathrm{C}_5\mathrm{H}_5\mathrm{N}_5\]

Key concepts

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element present in the compound. It does not give the exact number of atoms but instead expresses the relative proportions. To determine the empirical formula, follow these steps:

• Determine the moles of each element in the compound by using their masses and atomic weights.
• Find the simplest whole-number ratio by dividing all the moles by the smallest value among them.

For example, consider the empirical formula \(\text{CH}_2\): this indicates there is 1 Carbon atom for every 2 Hydrogen atoms in the simplest form of this compound.

Molar Mass Calculation

Calculating the molar mass of a compound is essential for progressing from the empirical to the molecular formula. The molar mass is the sum of the atomic masses of all atoms in a molecule.

• Identify the atomic masses of each element using the periodic table.
• Multiply the atomic mass of each element by the number of atoms of that element in the empirical formula.
• Add these values together to get the total molar mass of the empirical formula.

For instance, for the empirical formula \(\text{CH}_2\), the molar masses are:
\(\mathcal{M}_{\text{empirical}} = 12.01 + 2(1.01) = 14.03 \text{ g/mol}\).

Chemical Ratios

The ratio between the molar mass of the molecular formula and the empirical formula is a critical step. This ratio helps determine how many times the empirical formula's subscripted units must be multiplied to achieve the molecular formula.

• Divide the given molecular mass by the calculated empirical molar mass.
• Ensure that this ratio is a whole number or close to it, as it represents the factor by which each element's subscript in the empirical formula will be scaled.

For example, if the given molecular mass is 42.08 g/mol for a compound with the empirical formula \(\text{CH}_2\), the ratio is established using:
\[\frac{42.08 \text{ g/mol}}{14.03 \text{ g/mol}} \approx 3\].

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a compound. It can be found by scaling up the subscripts in the empirical formula by the ratio calculated.

• Multiply each subscript in the empirical formula by the integral ratio determined.
• Ensure the resulting formula accurately reflects the given molecular mass.

For the empirical formula \(\text{CH}_2\) with a ratio of 3, the molecular formula becomes:
\[\text{C}_3\text{H}_6\].
This process ensures that the molecular formula provides the precise composition of the compound based on empirical data and given molecular mass.