1.- Menthol is a flavoring agent extracted frompeppermintoil. It contains C, H,

Question

1.- Menthol is a flavoring agent extracted frompeppermintoil. It contains C, H, and O. In one combustion analysis,10.00 mg of the substance yields 11.53 mg H2O and 28.16mg CO2. What is the empirical formula of menthol? 2.- Lactic acid, which consists of C, H, and O, has long beenthought to be responsible for muscle soreness following strenuousexercise. Determine the empirical formula of lactic acid given thatcombustion of a 10.0 g sample produces 14.7 g CO2 and6.00 g H2O. 3.- Determine the empirical formula of an organic compoundcontaining only C and H given that combustion of a 1.50 g sample of the compound produces 4.71 g CO2and 1.93 g H2O 1.- Menthol is a flavoring agent extracted frompeppermintoil. It contains C, H, and O. In one combustion analysis,10.00 mg of the substance yields 11.53 mg H2O and 28.16mg CO2. What is the empirical formula of menthol? 2.- Lactic acid, which consists of C, H, and O, has long beenthought to be responsible for muscle soreness following strenuousexercise. Determine the empirical formula of lactic acid given thatcombustion of a 10.0 g sample produces 14.7 g CO2 and6.00 g H2O. 3.- Determine the empirical formula of an organic compoundcontaining only C and H given that combustion of a 1.50 g sample of the compound produces 4.71 g CO2and 1.93 g H2O

Explanation / Answer

I. Menthol is a flavoring agent extracted frompeppermintoil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the substance yields 11.53 mg H2O and 28.16mg CO2. What is the empirical formula of menthol?

The reaction:

CxHyOz + O2 = xCO2 + yH2O

Information (atomic masses and compound molar masses):

1) Calculate the moles of C present in those 28.16mg CO2 using dimensional analysis:

moles C = 28.16 mg CO2 (1g CO2 / 1000mg CO2) (1mol CO2 / 44g CO2) (1mol C/ 1mol CO2)

moles C = 0.00064 moles

2) Calculate the moles of H present in those 11.53mg H2O using dimensional analysis:

moles H = 11.53 mg H2O (1g H2O / 1000mg H2O) (1mol H2O / 18g H2O) (2mol H/ 1mol H2O)

moles H = 0.00128 moles

3) Convert the above moles of C and H to grams, then add them and then calculate grams of O by substracting the sum from the sample mass (10.00mg = 0.01g). Then convert those grams of O to moles of O:

mass C = 0.00064 moles C (12g C / 1mol C) = 0.00768g

mass H = 0.00128 moles H (1g H / 1mol H) = 0.00128g

mass O = 0.01g - 0.00768g - 0.00128g

mass O = 0.00104g

moles O = 0.00104g (1 mol O / 16g O)

moles O = 0.000065 moles

4) Divide the total moles of each element by the lowest number of moles (in this case the moles of O) to calculate the number of atoms in the empiral formula (round up if needed):

moles C = 0.00064 moles / 0.000065 moles = 9.85 --> 10

moles H = 0.00128 moles / 0.000065 moles = 19.70 --> 20

moles O = 0.000065 moles / 0.000065 moles = 1

So, the empirical formula for menthol is C10H20O (which happens to be its molecular formula as well).

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