The number of hydrogen atoms that are in 4.40 mol of ammonium sulfide is 2.12 x10^25 atoms
calculation
find the number of moles of Hydrogen in ammonium sulfide (NH4)2S
that is 4.40 x number of hydrogen atoms in (NH4)2S ( 4x2= 8 atoms)
moles is therefore= 4.40 x8= 35.2 moles
by use of Avogadro's law constant
that is 1mole = 6.02 x10^23 atoms
35.2 moles=?
by cross multiplication
{35.2 moles x 6.02 x10^23} /1 mole = 2.12 x10^25 atoms
<span>Ethoxyethane; trifluoroborane; BF3.Et2O; Boron trifluoride ethyl ether; Boron trifluoride diethyl ether; Boron trifluoride-diethyl ether; Boron
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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
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