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.
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Welcome back Yvette — been a while! Let's pick up where you left off. Here are the solutions to the examples: Problem 1: A hydrocarbon containing 86% of carbon and 14% of hydrogen has a molecular mass of 70. a) Determine its Empirical formula. Step 1: Calculate the moles of each element in 100 g of the compound. Moles of C = 86 g12 g/mol = 7.167 mol Moles of H = 14 g1 g/mol = 14 mol Step 2: Find the simplest whole number ratio of moles. Divide by the smallest number of moles (7.167 mol): Ratio C:H = (7.167)/(7.167) : (14)/(7.167) ≈ 1 : 1.95 ≈ 1:2 Step 3: Write the empirical formula. The empirical formula is CH_2. a) The empirical formula is CH_2. b) Determine its Molecular formula. Step 1: Calculate the empirical formula mass. Empirical formula mass of CH_2 = (1 × 12) + (2 × 1) = 14 g/mol. Step 2: Determine the factor n by which the empirical formula must be multiplied. n = Molecular massEmpirical formula mass = 70 g/mol14 g/mol = 5 Step 3: Write the molecular formula. Molecular formula = (CH_2)_5 = C_5H_10. b) The molecular formula is C_5H_10. --- Problem 2: On complete combustion in oxygen, 0.42g of a gaseous hydrocarbon (X) gave 1.32g of carbon dioxide and 0.54g of water. a) What is the empirical formula of X? Step 1: Calculate the mass of carbon and hydrogen in the products. Mass of C in CO_2 = 1.32 g CO_2 × 12 g C44 g CO_2 = 0.36 g C Mass of H in H_2O = 0.54 g H_2O × 2 × 1 g H18 g H_2O = 0.06 g H Step 2: Verify that the total mass of C and H equals the initial mass of X. Total mass of C and H = 0.36 g + 0.06 g = 0.42 g. This matches the given mass of hydrocarbon X. Step 3: Calculate the moles of each element. Moles of C = 0.36 g12 g/mol = 0.03 mol Moles of H = 0.06 g1 g/mol = 0.06 mol Step 4: Find the simplest whole number ratio of moles. Divide by the smallest number of moles (0.03 mol): Ratio C:H = (0.03)/(0.03) : (0.06)/(0.03) = 1 : 2 Step 5: Write the empirical formula. The empirical formula is CH_2. a) The empirical formula of X is CH_2. b) 0.21g of X occupied a volume of 120cm³ at r.t.p, what is the relative molecular mass of X? Step 1: Convert the volume to dm³ and use the molar volume at r.t.p. (24 dm³/mol) to find the moles of X. Volume of X = 120 cm^3 = 0.120 dm^3 Moles of X = 0.120 dm^324 dm^3/mol = 0.005 mol Step 2: Calculate the relative molecular mass (M) of X. M = Mass of XMoles of X = 0.21 g0.005 mol = 42 g/mol b) The relative molecular mass of X is 42. c) Hence, determine the molecular formula of X. Step 1: Calculate the empirical formula mass from part (a). Empirical formula mass of CH_2 = (1 × 12) + (2 × 1) = 14 g/mol. Step 2: Determine the factor n by which the empirical formula must be multiplied. n = Molecular massEmpirical formula mass = 42 g/mol14 g/mol = 3 Step 3: Write the molecular formula. Molecular formula = (CH_2)_3 = C_3H_6. c) The molecular formula of X is C_3H_6. Drop