Question

Dixanthogen, an herbicide, is composed of \(29.73 \%\) carbon, \(4.16 \%\) hydrogen, \(13.20 \%\) oxygen, and \(52.91 \%\) sulfur. It has a molar mass of \(242.4 \mathrm{~g} / \mathrm{mol}\). Determine the empirical and the molecular formulas. (f) \(3.99 \% \mathrm{P}, 82.3 \% \mathrm{Br}, 13.7 \% \mathrm{Cl}\)

Short answer

Empirical formula: \( C_3 H_5 O_1 S_2 \). Molecular formula: \( C_5 H_8 O_2 S_3 \).

Step-by-step solution

- Convert percentage to grams

Assume a 100 g sample of the compound. Hence, there will be 29.73 g of carbon, 4.16 g of hydrogen, 13.20 g of oxygen, and 52.91 g of sulfur.

- Convert grams to moles

Use the molar masses of each element to convert the mass in grams to moles. The molar masses are: Carbon (\text{C}) = 12.01 g/mol, Hydrogen (\text{H}) = 1.008 g/mol, Oxygen (\text{O}) = 16.00 g/mol,Sulfur (\text{S}) = 32.07 g/mol.Moles of C: \ \( \frac{29.73}{12.01} \approx 2.48 \). Moles of H: \ \( \frac{4.16}{1.008} \approx 4.13 \). Moles of O: \ \( \frac{13.20}{16.00} \approx 0.825 \). Moles of S: \ \( \frac{52.91}{32.07} \approx 1.65 \).

- Find the simplest mole ratio

Divide each mole value by the smallest number of moles calculated: 0.825. C: \ \( \frac{2.48}{0.825} \approx 3.01 \). H: \ \( \frac{4.13}{0.825} \approx 5.01 \). O: \ \( \frac{0.825}{0.825} = 1 \). S: \ \( \frac{1.65}{0.825} = 2 \).

- Write Empirical Formula

Use the ratios obtained to write the empirical formula. Here, round off to the nearest whole number to get the empirical formula: \( C_3 H_5 O_1 S_2 \).

- Determine molecular formula

Calculate empirical formula molar mass by summing the molar masses of all atoms in the empirical formula: \ C (12.01 g/mol) * 3 + H (1.008 g/mol) * 5 + O (16.00 g/mol) * 1 + S (32.07 g/mol) * 2 = 145.25 g/mol. \Given molar mass / empirical formula molar mass: \ \( \frac{242.4}{145.25} \approx 1.67 \).\Multiply empirical formula by this factor to obtain molecular formula: \ \( C_3 H_5 O_1 S_2 \) * 1.67 = \( C_5 H_8 O_2 S_3 \).

Key concepts

molar mass

Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It's important because it helps convert between grams and moles, which is crucial for solving chemical problems.
To find molar mass, simply add the atomic masses of all atoms in the molecule. For example, in carbon dioxide (CO₂), the molar mass is calculated as follows:

• Carbon: 12.01 g/mol (1 atom)
• Oxygen: 16.00 g/mol (2 atoms, hence 16*2 = 32 g/mol)

The molar mass of CO₂ is 12.01 + 32 = 44.01 g/mol.

mole ratio

A mole ratio is a comparison of the moles of different elements in a compound, giving insight into the proportions in which elements combine. It's found by dividing the number of moles of each element by the smallest number of moles of any element present.
For example, let's take a look at the mole calculations from our exercise:

• Moles of Carbon (C) = 2.48
• Moles of Hydrogen (H) = 4.13
• Moles of Oxygen (O) = 0.825
• Moles of Sulfur (S) = 1.65

Here, the smallest mole value is 0.825 (for Oxygen). Therefore, the mole ratios are:

• C: 2.48 / 0.825 ≈ 3.01
• H: 4.13 / 0.825 ≈ 5.01
• O: 0.825 / 0.825 = 1
• S: 1.65 / 0.825 = 2

This ratio forms the basis of the empirical formula.

empirical formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. It doesn't necessarily reflect the actual number of atoms, but it simplifies the structure.
In our exercise, after finding the mole ratios:

• C: approximately 3
• H: approximately 5
• O: 1
• S: 2

We form the empirical formula: C₃H₅OS₂. This means, there are 3 Carbon, 5 Hydrogen, 1 Oxygen, and 2 Sulfur atoms in the simplest form of the compound.

molecular formula

The molecular formula shows the exact number of atoms of each element in a molecule and is a multiple of the empirical formula.
To find it, calculate the empirical formula mass and compare it with the molar mass. For instance, in our example:

• Molar mass of the empirical formula (C₃H₅OS₂) = 145.25 g/mol
• Given molar mass = 242.4 g/mol
• Factor: 242.4 / 145.25 ≈ 1.67

Multiplying the empirical formula by this factor:
C₃H₅OS₂ * 1.67 = C₅H₈O₂S₃.
This is the molecular formula.

percent composition

Percent composition indicates the percentage by mass of each element in a compound. It helps in determining the formula of a compound by providing the ratio of each element present.
For example, in the original exercise, the compound's percent composition is:

• Carbon (C): 29.73%
• Hydrogen (H): 4.16%
• Oxygen (O): 13.20%
• Sulfur (S): 52.91%

Assuming a 100 g sample, these percentages convert directly to grams (e.g., 29.73% C = 29.73 g of C).
From here, we can find the moles of each element which helps in determining the empirical and molecular formulas.