Answer:
Question 1 is 21.0g
Explanation:
50 x 4.18 x 3.1 / 0.444 / 69.6= 21.0
Use a ratio to find out that x/190.2 = 74.5/100
hope this helps
Manipulation of the independent variable should change the dependent variable, the value of which "depends" on the independent variable's change in value.
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ + H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
O moles - 0.64 mmol
Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C H O
0.64 1.28 0.064
x1000 x1000 x1000 to get whole numbers
640 1280 64
10 20 1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O
The molecule for ethane is C2H6. or CH3-CH3.
Carbon LOVES hydrogen. If possible, Carbon would have four bonds attach to hydrogens. IN this molecule, one of the bond is used to attach a carbon to another carbon, so instead of 4 hydrogens, each carbon would have 3 hydrogens..
