This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Hey m, good to see you again. Here's how to solve problem 3.44: a) Step 1: Assume a 100 g sample of the oxide. The mass of nitrogen (N) is 30.45 g. The mass of oxygen (O) is 100 g - 30.45 g = 69.55 g. Step 2: Convert the mass of each element to moles using their molar masses (N = 14.01 g/mol, O = 16.00 g/mol). Moles of N = 30.45 g14.01 g/mol ≈ 2.173 mol Moles of O = 69.55 g16.00 g/mol ≈ 4.347 mol Step 3: Divide each mole value by the smallest number of moles to find the simplest whole-number ratio. Ratio for N = 2.173 mol2.173 mol = 1 Ratio for O = 4.347 mol2.173 mol ≈ 2.00 The empirical formula is NO_2. b) Step 4: Calculate the empirical formula mass (EFM) for NO_2. EFM = (1 × 14.01 g/mol) + (2 × 16.00 g/mol) = 14.01 + 32.00 = 46.01 g/mol Step 5: Determine the integer n by dividing the given molar mass by the empirical formula mass. We'll use 90 g/mol for the molar mass, noting the ± 5 g/mol range. n = Molar MassEFM = 90 g/mol46.01 g/mol ≈ 1.956 This value is very close to 2. If we use a molar mass of 92 g/mol (which is within the 90 ± 5 g/mol range), n = 92 g/mol46.01 g/mol ≈ 2.00. So, n=2. Step 6: Multiply the subscripts in the empirical formula by n to get the molecular formula. Molecular Formula = (NO_2)_2 = N_2O_4 Send me the next one 📸