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People Window Help the f C For The F takeAssignment/takeCovalentActivity.do?locatoraassignment-take&takeAssignmentSessionLocator..; ge Inbox-nicpathere... (50,424 unread)- n.. GroupMe My Print Center YouTube TV - Watc. so4 Use the References to access important values if needed for this question. Furnace 02 H20 absorber CO2ab Sample When 4.750 grams of a hydrocarbon, CH, were burned in a combustion analysis apparatus, 16.05 grams of CO2 and 3.287 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula- molecular formula Next

Explanation / Answer

Becasue all of the carbon in the sample is converted to CO2, we can use dimensional analysis and the following steps to calculate the mass of C in the sample.

Grams of C = (16.05 g CO2) x ( 1 mol CO2 /44 g CO2) x (1 mol C /1 mol CO2) x (12 g C /1 mol C) =4.377 g C

Becasue all of the hydrogen in the sample is converted to H2O, we can use dimensional analysis and the following steps to calculate the mass of H in the sample.

Grams of H = (3.287 g H2O) x ( 1 mol H2O/18 g H2O) x (2 mol H /1 mol H2O) x (1 g H /1 mol H) =0.365 g H

Now, moles C = 4.377 g C x ( 1 mol C/ 12 g C) = 0.36475 mol C

moles H = 0.365 g H x (1 mol H /1 g H) = 0.365 mol H

Molar ratio of C:H = 0.36475/0.36475 : 0.365 /0.36475 = 1 :1

Thus, the empirical formula of the compound is CH.

Empirical formula weight = 1 x 12 g/mol + 1 x 1 g/mol = 13 g/mol

Now, molecular formula weight / empirical formula weight = 26.04 / 13 = 2

Therefore we can say that the molecular formula is twice that of the empirical formula.

Hence, the molecular formula of the compound is C2H2.

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