Question

The molar mass and empirical formula of several compounds are listed below. Find the molecular formula of each compound. $$ \begin{array}{ll}{\text { a. }} & {\mathrm{C}_{4} \mathrm{H}_{9}, 114.22 \mathrm{g} / \mathrm{mol}} \\ {\text { b. }} & {\mathrm{CCl}, 284.77 \mathrm{g} / \mathrm{mol}} \\ {\text { c. }} & {\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{N}, 312.29 \mathrm{g} / \mathrm{mol}}\end{array} $$

Short answer

The molecular formulas are a. C8H18, b. C6Cl6, c. C18H12N6.

Step-by-step solution

- Calculate Molar Mass of Empirical Formula for Compound a

Calculate the molar mass of the empirical formula for compound a, which is C4H9. The atomic masses are approximately 12.01 g/mol for carbon (C) and 1.008 g/mol for hydrogen (H). Thus, the molar mass is (4 \( \times \) 12.01) + (9 \( \times \) 1.008) = 48.04 + 9.072 = 57.112 g/mol.

- Determine the Molecular Formula Multiplier for Compound a

Divide the given molar mass by the empirical formula molar mass to find the multiplier: \(\frac{114.22 \, \text{g/mol}}{57.112 \, \text{g/mol}} \approx 2\).

- Calculate the Molecular Formula for Compound a

Multiply the subscripts in the empirical formula by the multiplier to find the molecular formula: \(C_{4 \times 2}H_{9 \times 2} = C_8H_{18}\).

- Calculate Molar Mass of Empirical Formula for Compound b

Calculate the molar mass of the empirical formula for compound b, which is CCl. The atomic mass of chlorine (Cl) is approximately 35.45 g/mol. Thus, the molar mass is 12.01 + 35.45 = 47.46 g/mol.

- Determine the Molecular Formula Multiplier for Compound b

Divide the given molar mass by the empirical formula molar mass to find the multiplier: \(\frac{284.77 \, \text{g/mol}}{47.46 \, \text{g/mol}} \approx 6\).

- Calculate the Molecular Formula for Compound b

Multiply the subscripts in the empirical formula by the multiplier to find the molecular formula: \(C_{1 \times 6}Cl_{1 \times 6} = C_6Cl_6\).

- Calculate Molar Mass of Empirical Formula for Compound c

Calculate the molar mass of the empirical formula for compound c, which is C3H2N. The atomic masses are approximately 14.01 g/mol for nitrogen (N). Thus, the molar mass is (3 \( \times \) 12.01) + (2 \( \times \) 1.008) + 14.01 = 36.03 + 2.016 + 14.01 = 52.056 g/mol.

- Determine the Molecular Formula Multiplier for Compound c

Divide the given molar mass by the empirical formula molar mass to find the multiplier: \(\frac{312.29 \, \text{g/mol}}{52.056 \, \text{g/mol}} \approx 6\).

- Calculate the Molecular Formula for Compound c

Multiply the subscripts in the empirical formula by the multiplier to find the molecular formula: \(C_{3 \times 6}H_{2 \times 6}N_{1 \times 6} = C_{18}H_{12}N_6\).

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. It does not show the actual number of atoms in a molecule but reflects the ratio in the simplest form. For instance, the empirical formula for both glucose ((C_6H_{12}O_6)) and acetic acid ((C_2H_4O_2)) is CH_2O, as both have the same simplest ratio of 1 carbon to 2 hydrogens to 1 oxygen.

The determination of an empirical formula involves dividing the percentage composition or mass of each element by its atomic mass to find the relative number of moles of each element. Then, the mole ratios are simplified to the smallest whole numbers. For example, if we have a compound composed of 40% carbon and 60% oxygen by mass, the first step is to convert this to moles by the formula:
\[ \text{moles of C} = \frac{\text{mass of C}}{\text{atomic mass of C}} \]
and
\[ \text{moles of O} = \frac{\text{mass of O}}{\text{atomic mass of O}} \]
Next, divide both values by the smallest number of moles obtained to reach the empirical formula.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, and it's measured in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms present in the molecular formula of the substance, which can be found on the periodic table of elements.

For instance, to calculate the molar mass of water (H_2O), you would use the atomic masses of hydrogen (1.008 g/mol) and oxygen (16.00 g/mol), resulting in:
\[ \text{Molar mass of } H_2O = 2 \times 1.008 \text{ g/mol} + 16.00 \text{ g/mol} = 18.016 \text{ g/mol} \]
In the provided exercise, the molar masses of various empirical formulas are calculated by multiplying the atomic mass of each element by its number in the empirical formula and summing them up.

Molecular Mass Multiplier

The molecular mass multiplier is found by comparing the molar mass of a compound's empirical formula with the compound's known molar mass. It tells us how many times the empirical formula must be multiplied to achieve the actual molecular formula. This multiplier is a crucial step in determining the molecular formula from the empirical formula.

To find the molecular mass multiplier:

• Determine the molar mass of the compound's empirical formula.
• Divide the actual molar mass of the compound (usually given) by the molar mass of the empirical formula.
• Round the resulting number to the nearest whole number to get the multiplier.

For example, if a compound has an empirical formula of CH with a molar mass of 13.02 g/mol and a molecular mass of 26.04 g/mol, the multiplier is calculated as follows:
\[ \text{Multiplier} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{26.04 \text{ g/mol}}{13.02 \text{ g/mol}} \approx 2 \]
Then, to obtain the true molecular formula, each subscript in the empirical formula is multiplied by the computed multiplier.