The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a...
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Transcript of The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a...
The electrolysis of water is used to generate 0.500 g of H2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours, calculate Avogadro’s number (the number of atoms of H in 1 gram)
1 amp is defined as the number of coulombs per second.There are 6.24 x 1018 electrons in a coulomb.It takes 1 electron to make 1 H atom
= 3.01 x 1023 at H
x 1 at Hx 6.24 x 1018 el1 el1 coul
x 3.000 coul
s
4.46625 h
x 60 min
1 h
x 60 s
1 min
0.5000 g = 6.02 x 1023 at H
C N OSH
Need to determine the empirical formula
We have a chemical compound
We burn the compound in O2 and measure the amount of CO2, H20, N2, and SO2 produced.
From the mass of CO2 we can calculate moles of C and grams
C N
O
SH
From the mass of H2O we can calculate moles of H and grams
From the mass of N2 we can calculate moles of N and grams
From the mass of SO2 we can calculate moles of S and grams
How do we get moles of O ?
moles
g gg g
moles moles moles
5.43 g
= g
=Total g
minus
= moles
Now for the hard part
Lets get ready to rumble!
A compound was known to contain C, H, N, O, and S. When a 5.43 g sample was burned the products were 8.43 g CO2, 1.15 g H2O, 0.450 g N2, and 3.07 g of SO2. Determine the empirical formula of the compound.
Mass of O = 1.0151 g
Mass of O = 5.43 g – 4.4149
Mass of O = Total C H N S O – Mass of C H N S
Mass of C H N S = 4.4149 g
8.43 g CO2 x 1 mole x 1 mole C = 0.1916 mole C x 12.0 g = 2.299 g C
44.0 g 1 mole CO2 1 mole
1.15 g H2O x 1 mole x 2 mole H = 0.1276 mole H x 1.01 g = 0.1289 g H
18.02 g 1 mole H2O 1 mole
3.07 g SO2 x 1 mole x 1 mole S = 0.04789 mole S x 32.1 g = 1.537 g S
64.1 g 1 mole SO2 1 mole
0.450 g N2 x 1 mole x 2 mole N = 0.03214 mole N x 14.0 g = 0.4500 g N
28.0 g 1 mole N2 1 mole
x 1 mole 16.0 g
= 0.06344 moles O
= 0.06344 moles O
Empirical Formula or Mole Ratio
0.1916 mole C
0.1276 mole H
0.03214 mole N
0.06344 moles O
0.04779 mole S
0.03214 moles
Empirical Formula C12H8N2O4S3
= 6
= 4
= 1
= 2
= 1.5
x 2 = 12
x 2 = 8
x 2 = 2
x 2 = 4
x 2 = 3
0.03214 moles
0.03214 moles
0.03214 moles
0.03214 moles