%Ca? %P? %O? 2) Menthol is a flavoring agent extracted from peppermint oil. It c

Question

%Ca?

%P?

%O?

2) Menthol is a flavoring agent extracted from peppermint oil. It contains C,H, and O. In one combustion analysis, 10.0mg of the substance yields 11.53mg of H2O and 28.16mg of CO2. what is the empirical formula of menthol?

CHO (add the subcritps)

3) Determine the number of atoms of each element in the empirical formula of a compound with the following composition.

62.40 percent C, 9.892 percent H, 27.70 percent O

c=atoms? h=atoms? O=atoms?           

show work for all posible points

Explanation / Answer

1. % Ca = ( 100 x 40.078x3)/(310.18) = 38.7627 %

% P = ( 100x30.9738x2)/(310.18) = 19.9715 %

% O = 100-38.7627 -19.9715 =41.2658 %

2. mass of CO2 : 0.02816 g

moles of CO2 : 0.00064 moles : / formula mass of CO2 (44)
moles of carbon : 0.00064 moles : same as the moles of C in the CO2
mass of carbon : 0.00768 g : X the atomic weight of C to get the mass

mass of H2O : 0.01153 g
moles of H2O : 0.00064 moles : / formula mass of H2O (18)
moles of H : 0.00128 moles x 2 the moles of H2O
mass of H : 0.0013 g X the atomic mass of H

Mass of compound : 0.010 g

mass of O : 0.00102 g : mass of the compound - mass of (C+H)
moles of O : 0.00006375 moles : / atomic weight of O to get the number of moles
check mass of C+H+O : 0.010 g

in summary : moles
C : 0.00064
H : 0.00128
O : 0.00006375

divide by smallest to get the ratio
C : 10.10
H : 20.35
O : 1.00

Empirical formula is C10H20O

3. 62.40 percent C, 9.892 percent H, 27.70 percent O

C = 62.40/12 = 5.2

H = 9.892/1 = 9.982

O = 27.70/16 = 1.73125

Divide by smallest

C = 3

H = 6

O = 1

Thus, atoms =

C = 18.069 x 10^23

H = 36.138 x 10^23

O = 6.023 x 19^23

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