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Reaction Stoichiometry: Mole-to-Mole and Mass-to-Mass Calculations, Slides of Stoichiometry

An introduction to reaction stoichiometry, focusing on mole-to-mole and mass-to-mass calculations. It covers stoichiometric relations in chemical equations, conversion methods using mole ratios, and mass calculations using balanced equations and gravimetric analysis. Examples are included to illustrate the concepts.

What you will learn

  • What is the role of mole ratios in stoichiometric calculations?
  • How are mass-to-mass calculations performed using balanced equations?
  • What is gravimetric analysis and how is it used in stoichiometry?

Typology: Slides

2021/2022

Uploaded on 09/12/2022

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Reaction Stoichiometry
quantitative relationships between reactants and
products
4.1 Mole-to-Mole Calculations
Stoichiometric relations in chemical equations
2H2+ O2 2H2O
2 mol H2 1 mol O2
1 mol O2 2 mol H2O
Conversion method
mole ratios (conversion factors)
[2 mol H2O/1 mol O2]
(mol given)××(mole ratio)= (mol required)
Example: Determine the number of moles of
water produced from 3.4 mol O2.
OH mol 8.6
O mol 1
OH mol 2
O mol 3.4 2
2
2
2=
×
Stoichiometric conversion factors are reaction
specific
Example: Calculate the amount of O2needed
to produce 3.5 mol H2O by combustion of
methane (CH4).
balanced equation:
CH4+ 2O2 CO2+ 2H2O
mole ratio (conversion factor):
2 mol O2 2 mol H2O
[2 mol O2/2 mol H2O]
2
2
2
2O mol 5.3
OH mol 2
O mol 2
OH mol 3.5 =
×
4.2 Mass-to-Mass Calculations
Conversion method
Fig. 4.4
Example:
Calculate the mass of oxygen needed to
completely burn 5.4 kg of butane (C4H10).
balanced equation:
2C4H10 + 13O2 8CO2+ 10H2O
mole ratio (conversion factor):
13 mol O2 2 mol C4H10
[13 mol O2/2 mol C4H10]
molar masses:
C4H10 58.1 g/mol O2 32.0 g/mol
pf2

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Reaction Stoichiometry

  • quantitative relationships between reactants and products

4.1 Mole-to-Mole Calculations

  • Stoichiometric relations in chemical equations 2H 2 + O 2 →→ 2H 2 O 2 mol H 2 ⇔⇔ 1 mol O 2 1 mol O 2 ⇔⇔ 2 mol H 2 O
  • Conversion method
  • mole ratios (conversion factors) [2 mol H 2 O/1 mol O 2 ] (mol given) ×× (mole ratio)= (mol required)

Example: Determine the number of moles of

water produced from 3.4 mol O 2.

6. 8 molH O

1 molO

2 molHO

3.4 molO 2

2

2

×

  • Stoichiometric conversion factors are reaction specific

Example: Calculate the amount of O 2 needed

to produce 3.5 mol H 2 O by combustion of methane ( CH 4 ). ⇒ balanced equation: CH 4 + 2O 2 →→ CO 2 + 2H 2 Omole ratio (conversion factor): 2 mol O 2 ⇔⇔ 2 mol H 2 O [2 mol O 2 /2 mol H 2 O]

2 2

2

2 3.^5 molO

2 molHO

2 molO

3.5 molHO =

×

4.2 Mass-to-Mass Calculations

  • Conversion method

Fig. 4.

Example:

  • Calculate the mass of oxygen needed to completely burn 5.4 kg of butane ( C 4 H 10 ). ⇒ balanced equation: 2C 4 H 10 + 13O 2 →→ 8CO 2 + 10H 2 Omole ratio (conversion factor): 13 mol O 2 ⇔⇔ 2 mol C 4 H 10 [13 mol O 2 /2 mol C 4 H 10 ]molar masses: C 4 H 10 →→ 58.1 g/mol O 2 →→ 32.0 g/mol
  • Gravimetric analysis - uses measurements of mass to obtain the amount of analyte - precipitation reactions - gas formation reactions

2

2

4 2

2 4 10

2

4 10

4 10 4 10

4 10

3 4 10

19 kg O

  1. 9 10 gO 1 molO

  2. 0 gO 2 molCH

13 molO

  1. 1 gCH

1 molCH 1 kgCH

10 gCH

  1. 4 kgCH

= × =

×

×

×

×

× Example:

  • A sample of ore of mass 5.324 g was analyzed for Ba by dissolving the sample and then precipitating the Ba2+^ ions with sulfuric acid. After drying, the mass of the precipitate was found to be 3.752 g. What is the mass % of Ba in the sample? ⇒ net ionic equation: Ba2+^ + SO 4 2-^ →→ BaSO 4 (s)

mole ratio: [1 mol Ba2+/1 mol BaSO 4 ]molar masses: BaSO 4 →→ 233.40 g/mol Ba →→ 137.34 g/mol

2. 208 g Ba

1 molBa

137. 34 gBa

1 molBaSO

1 molBa

233. 40 gBaSO

1 molBaSO

3. 752 gBaSO

2 4

2

4

4 4

^ =

×

×

×

×

mass %:

100 % 41. 47 % Ba

5. 324 gsample

2. 208 gBa

× =

Fig. 4.