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. C10H20O

2. CH2O

3. CH2

Explanation:

1. Menthol

We can use the information supplied to calculate the number of moles of carbon and hydrogen.

We first divide the mass of carbon iv oxide by its molar mass I. e 44g/mol

28.16:44 = 0.64 moles

Now, we know that 1 mole of carbon iv oxide contains 1 mole carbon. To find the actual mass of carbon, we multiply this number of moles by the atomic mass unit of carbon I. e 12. That equals 0.64 * 12 = 7.68mg

For the hydrogen, we use water.

We first divide by molar mass of water i. e 18

11.53 : 18 = 0.641moles

Since 1 mole water contains 2 moles hydrogen, we multiply the number of moles of water by 2. That equals 0.641 * 2 = 1.282 moles

We multiply this by a. m. u of oxygen = 1.282 * 1 = 1.282mg

To get the mass of oxygen, we subtract the masses of hydrogen and carbon from the total = 10 - 1.282 - 7.68 = 1.038 mg

We now get number of moles of oxygen by dividing by the amu of oxygen = 1.038/16 = 0.065

Thus, we have the following moles

0.065 oxygen, 1.282 hydrogen and 0.64 carbon.

We divide by the smallest

Carbon = 0.64 : 0.065 = 10

Hydrogen = 1.282 : 0.024875 = 20

Oxygen = 1

Possible empirical formula is C10H20O

2. We can use the information supplied to calculate the number of moles of carbon and hydrogen.

Following procedures in 1 above, we get the following:

C = 14.7/44 = 0.3341 moles

This multiplied by its amu = 0.3341 * 12 = 4g

H = 6 / 18 = 0.33

Multiplied by 2 = 0.67 moles

Multiplied by amu of hydrogen = 0.67g

MASS of oxygen = 10 - 0.67 - 3 = 5.33g

Dividing by a. m. u of oxygen = 5.33/16 = 0.33

We divide by smallest number of moles.

C = 0.33/0.33 = 1

H = 0.67/2 = 2

O = 0.33/0.33 = 1

Empirical formula is CH2O

3. We can use the information supplied to calculate the number of moles of carbon and hydrogen.

Following procedures in 1 above, we get the following:

C = 4.71/44 = 0.107g

Multiplied by a. m. u = 0.107 * 12 = 1.285g

H = 1.93g / 18 = 0.107

Multiplied by 2 moles = 0.214

Mass = 0.214 * 1 = 0.214g

Dividing by smallest number:

H = 0.214/0.107 = 2

C = 0.107/0.107 = 1

Empirical formula = CH2