A gas of molecular weight 58 was found to contain \(82.8 \%\) \(\mathrm{C}\) and \(17.2 \% \mathrm{H}\). What is the molecular formula of this compound?

Let's assume that we have 100 grams of this compound, so that the percentages of each element directly correspond to their mass in grams. Therefore, we have: - 82.8 g of Carbon (C) - 17.2 g of Hydrogen (H) Now, we will convert these masses into moles by dividing by the respective atomic weights of Carbon (12.01) and Hydrogen (1.008): Moles of C: \( \frac{82.8}{12.01} = 6.89 \) Moles of H: \( \frac{17.2}{1.008} = 17.06 \) To find the ratio of Carbon to Hydrogen, we will divide each value by the smaller of the two: Ratio of C: \( \frac{6.89}{6.89} = 1 \) Ratio of H: \( \frac{17.06}{6.89} = 2.47 \) Since the ratio of Hydrogen is close to 2.5 and we need whole numbers in our empirical formula, we will multiply both ratios by 2 to get a simpler whole number ratio: Ratio of C: \( 1 * 2 = 2 \) Ratio of H: \( 2.47 * 2 = 4.94 \approx 5 \) Now, the empirical formula is C₂H₅.

First, we need to find the empirical weight of the empirical formula, which is the sum of the atomic weights of all elements in the empirical formula: Empirical weight of C₂H₅: \( (2 * 12.01) + (5 * 1.008) = 30.05 \) Now we will find the ratio between the given molecular weight (58) and the empirical weight (30.05): Molecular weight ratio: \( \frac{58}{30.05} = 1.93 \approx 2 \) Since the molecular weight ratio is close to 2, it means that the molecular formula is twice the empirical formula: Molecular formula: (C₂H₅) * 2 = C₄H₁₀ The molecular formula of this compound is C₄H₁₀.