A compound containing only carbon, hydrogen, and oxygen was found to contain \(70.58 \%\) carbon and \(5.92 \%\) hydrogen by weight. Calculate the empirical formula for this compound. If the compound has a molecular weight of approximately \(135 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula?

To determine the empirical formula of a compound, we start by knowing the percentage composition of each element. Here, the compound contains 70.58% carbon, 5.92% hydrogen, and the rest is oxygen. Assume you have a 100g sample of the compound:

• 70.58g of Carbon
• 5.92g of Hydrogen
• 23.50g of Oxygen (100 - 70.58 - 5.92)

Next, convert these masses to moles by dividing by the atomic masses of each element. For carbon (12 g/mol), hydrogen (1 g/mol), and oxygen (16 g/mol):
\[ \text{Moles of C} = \frac{70.58g}{12g/mol} = 5.882 \text{ moles} \ \text{Moles of H} = \frac{5.92g}{1g/mol} = 5.92 \text{ moles} \ \text{Moles of O} = \frac{23.50g}{16g/mol} = 1.469 \text{ moles} \] Identify the simplest whole number ratio by dividing each by the smallest number of moles found:
\[ \text{Ratio of C} = \frac{5.882}{1.469} \thickapprox 4 \ \text{Ratio of H} = \frac{5.92}{1.469} \thickapprox 4 \ \text{Ratio of O} = \frac{1.469}{1.469} = 1 \] This gives the simplest whole number ratio of 4:4:1. Thus, the empirical formula is \( C_4H_4O\).

To find the molecular formula, we need the molecular weight. The compound's molecular weight is given as approximately 135 g/mol. First, calculate the empirical formula weight using atomic masses:
\[ \text{Empirical formula weight} = (4 \times 12) + (4 \times 1) + (1 \times 16) = 48 + 4 + 16 = 68 \text{ g/mol} \] Now, divide the molecular weight by the empirical formula weight to find the multiplication factor:
\[ \frac{135 g/mol}{68 g/mol} \thickapprox 2 \] Therefore, the molecular formula is the empirical formula multiplied by this factor: \( (C_4H_4O)_2 = C_8H_8O_2 \).

Converting mass to moles is a crucial step in finding the empirical formula. Use the molar mass of each element, which is the atomic weight expressed in g/mol. For example, given the masses: 70.58g of Carbon, 5.92g of Hydrogen, and 23.50g of Oxygen. Convert each mass to moles:

\[ \text{Moles of C} = \frac{70.58g}{12g/mol} = 5.882 \text{ moles} \]
\[ \text{Moles of H} = \frac{5.92g}{1g/mol} = 5.92 \text{ moles} \]
\[ \text{Moles of O} = \frac{23.50g}{16g/mol} = 1.469 \text{ moles} \]
Eleven these values help determine the ratio of atoms in the compound's simplest form.

Elemental composition analysis involves determining the percentage by weight of each element in a compound. This is done through experimental methods such as combustion analysis. In our example, the compound contains 70.58% carbon and 5.92% hydrogen. The remaining 23.50% is oxygen. These percentages are crucial for calculating the empirical formula.

Stoichiometry is the calculation of reactants and products in chemical reactions. Here, it's used to establish the ratios of atoms in the empirical formula. After converting masses to moles, the key step is to find the simplest ratio. This ratio is determined by dividing each amount by the smallest number of moles:
\[ \text{Ratio of C} = \frac{5.88}{1.47} \thickapprox 4, \text{ Ratio of H} = \frac{5.92}{1.47} \thickapprox 4, \text{ Ratio of O} = \frac{1.47}{1.47} = 1 \]
This 4:4:1 ratio accurately represents the empirical formula, \( C_4H_4O\), demonstrating the application of stoichiometry in molecular chemistry.