Answer to Question #175826 in Organic Chemistry for Lawrence akpa

Question

A researcher used a combustion method to analyze a compound use as a anti-knock additive in gasoline a 9.394mg sample of the yielded 31.154 of CO2 and 7.977mg of water in the combustion. calculate the percentage composition of the compound and determine it empirical formula

Answer 1

Step 1: Parameters:

Mass of compound = 9.394 mg

Mass of CO₂ yielded = 31.154 mg

Mass of H₂O yielded = 7.977 mg

Molar mass of CO₂ = 44.01 g/mol

Molar mass of H₂O = 18.02 g/mol

Step 2: Calculate amount of substance

moles of CO₂ = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO₂

moles of C = moles of CO₂ * (1 mol C /1 mol of CO₂) = 7.08 * 10^-4 mol

moles of H₂0 = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H₂O

moles of H moles of H = moles of H20 * (2 mol H/ 1 mol H₂O) = 4.43* 10^-4 *2 = 8.86 * 10^-4

Step 3: Calculate mass of C and H

mass of mass of C = moles C × molar mass C

mass of C = 7.08 × 10^-4 mol × 12.01 g/mol

mass of C = 0.0085 grams

mass of H = moles H × molar mass H

mass of H = 8.86 × 10^-4 mol × 1.01 g/mol

mass of H = 0.000894 grams

Step 4: Calculate total mass of compound Total mass of the compound = 0.0085 grams + 0.000894 grams = 0.009394 grams 9.394 mg

$\therefore$ no other element is missing.

Step 5: Calculate the percent composition

%C = (8.50 mg / 9.394 mg) x 100 = 90.5%

%H = (0.894 mg / 9.394 mg) x 100 = 9.5%

Step 6: Empirical formula

C = 90.5/9.5 = 9.5/12 = 0.8

H = 9.5/9.5 = 1/1 = 10

$\therefore$ The empirical formula is $C_8H_{10}$