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Atomic vs. Molecular Weight Calculate Molecular Mass ..., Study notes of Chemistry

First find molecular masses of substances. 2. Convert 122.45 g of KClO3 to moles. 3 Find molar ratios from balanced equation.

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Chapter 17
Atomic vs. Molecular Weight
Atomic weight on periodic table is
average of natural abundance of
isotopes
Atomic mass is the number of nucleons
in a particular atom—specified by
isotope
Molecular mass is the mass of one mole
of molecules
One atomic mass number of grams
6.0221367x1023 molecules
Calculate Molecular Mass
O atomic weight 15.9996 (round to 16
for this class)
O-16 atomic mass 16 u
Molecular oxygen
O
atomic mass 32 u
Molecular
oxygen
O
2
atomic
mass
32
u
Molecular O2molecular mass 32 g/mole
CO2molecular mass
C=12 g/mole, O2=32 g/mole
CO2=12+32=44 g/mole
Na2CO3molecular mass
Na = 23 g/mole
C = 12 g/mole
O = 16 g/mole
2N 3O lti l
2
N
a,
3
O
, so mu
lti
p
l
y
2(23) + 12 + 3(16) = 106 g/mole
Write these down
C = 12
H = 1
O = 16
N14
N
=
14
K = 39
Cl = 35.5
Use for some questions
pf3
pf4
pf5

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Chapter 17

Atomic vs. Molecular Weight

 Atomic weight on periodic table is average of natural abundance of isotopes  Atomic mass is the number of nucleons in a particular atom—specified by isotope  Molecular mass is the mass of one mole of molecules  One atomic mass number of grams  6.0221367x10 23 molecules

Calculate Molecular Mass

 O atomic weight 15.9996 (round to 16 for this class)  O-16 atomic mass 16 u  Molecular oxygen O atomic mass 32 uMolecular oxygen O 2 atomic mass 32 u  Molecular O 2 molecular mass 32 g/mole

CO 2 molecular mass

 C=12 g/mole, O 2 =32 g/mole  CO 2 =12+32=44 g/mole

Na 2 CO 3 molecular mass

 Na = 23 g/mole  C = 12 g/mole  O = 16 g/mole  2 N 2 Na, 3 O, so multiply 3 O lti l  2(23) + 12 + 3(16) = 106 g/mole

Write these down

 C = 12
 H = 1
 O = 16
 NN = 14 14
 K = 39

 Cl = 35.  Use for some questions

2 KClO 3  2 KCl + 3 O 2

How many grams of oxygen can be produced from 122.45 g of KClO 3?

  1. First find molecular masses of substances
  2. Convert 122.45 g of KClO 3 to moles 3 3. Find molar ratios from balanced equationFind molar ratios from balanced equation
  3. Calculate moles of O 2
  4. Convert that to grams of O (^2)

2 KClO 3  2 KCl + 3 O 2

How many grams of oxygen can be produced from 122.45 g of KClO 3?

  1. 32.00 g
  2. 48.00 g 3 3. 61 22 g61.22 g
  3. 122.45 g

1 2 3 4

0% 0% 0% 0%

1 CH 4 + 2 O 2  2 H 2 O + 1 CO 2

How many grams of water are produced from 1. moles of methane?

  1. 11.25 grams
  2. 22.5 grams 3 3. 45 grams45 grams
  3. 90 grams

1 2 3 4

0% 0% 0% 0%

Moles calculated from Grams

 176 g of CO 2 = Number of moles?  Molar mass of CO 2 = 44  If you multiply,  176 g x 44 gl^ results in units of g^

2

mole

g

176 g x 44 results in units of  you get a unit mess  UNITS alert you that you made an error  KEEP UNITS WITH NUMBERS!!

mole (^) mole

Mass to moles relationship To convert between grams and moles

 Find molar mass  Divide grams by molar mass

Problem

H 2 S  O 2  SO 2  H 2 O

  • 32 grams SO (^2)
  • How many grams OHow many grams O 2 2 used?used?

Problem

 Balance first

H (^) 2 SO 2  SO 2  H 2 O

2 H (^) 2 S  3 O 2  2 SO 2  2 H 2 O

  • 32 grams SO (^2)
  • How many grams O 2 used?

 Then determine molar ratios  2 SO 2 to 3 O 2

2 H (^) 2 S  3 O 2  2 SO 2  2 H 2 O

Problem

 Find molar masses

2 H (^) 2 S  3 O 2  2 SO 2  2 H 2 O

  • 32 grams SO (^2)
  • How many grams O 2 used?

 SO 2 = 32 + 32 = 64 g/mol SO 2

 O 2 = (2x1) +16 = 32 g/mol O 2

 H 2 O = (2x1) +16 = 18 g/mol H 2 O

 H 2 S = (2x1) +32 = 34 g/mol H 2 S

Problem

2 H (^) 2 S  3 O 2  2 SO 2  2 H 2 O

  • 32 g SO 2 needs how many grams O 2?
  • How many moles is 32 g SO 2?
    • 32 grams SO (^2)
    • How many grams O 2 used?

2 0.^5 SO 2

64 g

32 gSO moles

mole

Problem

2 H (^) 2 S  3 O 2  2 SO 2  2 H 2 O

  • How many moles O 2 is needed?
  • 0.5 moles SO 2 in 2:3 ratio with O 2
    • 32 grams SO (^2)
    • How many grams O 2 used?

2 0.^5 SO 2

64 g

32 gSO moles

mole

  • 0.75 moles O (^2)

Set up

proportion

? O  

with the unknown on top (the O 2 )

  • 32 grams SO (^2)
  • How many grams O 2 used?

 

 

 

? g O 2

Problem 6

Is it balanced? Molar ratio 1:1: 4 grams oxygen  Grams carbon consumed?

CO 2  CO 2

 Grams carbon dioxide produced?

Problem 6

Molar ratio 1:1: 4 grams oxygen  1 mole O 2 = 32 g

CO 2  CO 2

2 0.^1252 32

1 4 molesO g

mole g O  

Problem 6

Molar ratio 1:1:

0.125 moles O 2 in 1:1 ratio

CO 2  CO 2

0.125 moles C 0.125 moles CO (^2)

Problem 6

0.125 moles C Grams carbon consumed?

g C

g

moles C 15

CO 2  CO 2

g C

mole

moles C 1. 5

2 5.^52

0. 125 mole gCO

mole

g

CO  

Reaction Speed

Collision of molecules required for it to occur  Increase concentration  Increase temperatureIncrease temperature  Catalyst can facilitate reaction

Chlorine catalyst