This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to your questions: 2.5.1 Step 1: Calculate the mass of oxygen in one mole of the compound. Given molar mass (M) = 74 g·mol^-1 and percentage of oxygen by mass = 43.24\%. Mass of oxygen = 0.4324 × 74 g·mol^-1 = 31.9976 g·mol^-1. Step 2: Determine the number of oxygen atoms (z). The molar mass of oxygen is 16 g·mol^-1. Number of oxygen atoms (z) = 31.9976 g·mol^-116 g·mol^-1 ≈ 2. So, there are 2 oxygen atoms in the compound. The formula is C_xH_yO_2. Step 3: Calculate the remaining mass for carbon and hydrogen. Mass of 2 oxygen atoms = 2 × 16 g·mol^-1 = 32 g·mol^-1. Remaining mass for carbon and hydrogen = 74 g·mol^-1 - 32 g·mol^-1 = 42 g·mol^-1. Step 4: Determine the number of carbon (x) and hydrogen (y) atoms. Let the formula for the carbon and hydrogen part be C_xH_y. We know 12x + 1y = 42. By trial and error, or by considering common organic structures: If x=1, y=30 (too many hydrogens). If x=2, y=18 (too many hydrogens). If x=3, y=42 - (12 × 3) = 42 - 36 = 6. This gives C_3H_6. This ratio (C_3H_6) is plausible for an organic compound. Step 5: Combine the atoms to form the molecular formula. The molecular formula is C_3H_6O_2. 2.5.2 The molecular formula C_3H_6O_2 can represent several functional isomers. Two common functional groups for this formula are carboxylic acids and esters. 1. Propanoic acid (a carboxylic acid): CH_3CH_2COOH This is a 3-carbon chain with a carboxylic acid group. 2. Methyl ethanoate (an ester): CH_3COOCH_3 This is an ester formed from ethanoic acid and methanol. Last free one today — make it count tomorrow, or type /upgrade for unlimited.