Part A Combustion analysis of toluene, a common organic solvent, gives 8.79 mg o

Question

Part A

Combustion analysis of toluene, a common organic solvent, gives 8.79 mg of CO2 and 2.06 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Express your answer as a chemical formula.

Part B

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1307 mg sample of menthol is combusted, producing 0.3678 mg of CO2 and 0.1507 mg of H2O. What is the empirical formula for menthol?

Express your answer as a chemical formula.

Part C

If the compound has a molar mass of 156 g/mol, what is its molecular formula?

Express your answer as a chemical formula.

Explanation / Answer

mass of CO2 = 0.00879 g or since the molar mass of CO2 is 44g / mole we have
0.0002 moles of CO2 since there is 1 mole so C in CO2
we have 0.00020 moles of C or 0.0024 g of C
mass of H2O = 0.00206 g or since the molar mass of water is 18 g / mol we have
0.000114 moles of H2O since there are2 moles of H in 1 moles of H2O =
0.000229 moles of H or 0.000231 g of H
molar ratio of C : H = 0.000200 : 0.000229

or after dividing by the smallest 0.000200
molar ratio of C : H = 1.000 : 1.145
multiply by 7 to get whole numbers 7.000 8.015 :

empirical formula is C7H8

we use the mass of carbon dioxide to work out the mass of carbon
molar mass of CO2 = 44 g/ mole
0.0003678 g of CO2 has 0.0003678 /44 = 0.000008359 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2
so moles of C in the compound = 0.000008359 moles
mass of C = 0.000008359 x 12 = 0.0001003 g

we use the mass of water to find out the mass of hydrogen
molar mass of H2O = 18.0152 g/ mole
0.0001507 g of H2O has 0.0001507 / 18.0152 = 0.000008365 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.00001673 moles
mass of H = 0.00001673 x 1.0079 = 0.00001686 g

we calculate the O by difference from the mass of sample once we know the mass of C and H
mass of H + C = 0.0001172 g
mass of sample = 0.0001307 g
mass of O by difference = 0.00001354 g
moles of O = 0.00001354 /16 = 0.0000008461 moles

we now work out the molar ratio of all the elements in the compound
molar ratio of C : H : O = 0.000008359 : 0.00001686 : 0.0000008461
smallest number 0.0000008461
divide the ratio by the smallest number we get
molar ratio of C : H : O = 10.0 : 20.0 : 1.0

empirical formula is C10H20O
empirical formula mass of C10H20O is 156 g which is also the molar mass

so the empirical formula is also the molecular formula

C10H20O

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