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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The electrolysis of water is used to generate 0.500 g of H 2 (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 10 18 electrons in a coulomb. It takes 1 electron to make 1 H atom = 3.01 x 10 23 at H x 1 at H x 6.24 x 10 18 el 1 el 1 coul x 3.000 coul s 4.4662 5 h x 60 min 1 h x 60 s 1 min 0.5000 g = 6.02 x 10 23 at H

Transcript of The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a...

Page 1: The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

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

Page 2: The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

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

Page 3: The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

Now for the hard part

Lets get ready to rumble!

Page 4: The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

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

Page 5: The electrolysis of water is used to generate 0.500 g of H 2 (g). If the hydrogen was generated at a rate of 3.000 amps over a period of 4.46625 hours,

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