Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the substance yields 11.53 mg H2O and 28.16 mg CO2. What is the empirical formula of menthol? Add subscripts to complete the empirical formula.

C10H20O

Explanation:

Firstly, for convenience sake, we are converting all mg to g.

This means 10mg = 10/1000 = 0.01g

11.53mg = 11.53/1000 = 0.01153g

28.16mg = 28.16/1000 = 0.02816g

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 0.02816/44 = 0.00064 mole

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.00064= 0.00768

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.01153/18 = 0.00064 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.00064= 0.00128 mole

The mass of hydrogen is thus 0.00128 * 1 = 0.00128g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.01 - 0.00128 - 0.00768 = 0.00104g

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.00104/16 = 0.000065

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.000065

H = 0.00128/0.000065 = 20

O = 0.000065/0.000065 = 1

C = 0.00064/0.000065 = 10

The empirical formula of menthol is thus C10H20O

your answers are a,b,,e and f

i hope this

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.