Chemical Calculations

1
Chemical Calculations
2
Percents

• Percent means parts of 100 or parts per 100
  parts
• The formula
• Part
• Whole

x 100
Percent
3
Percents

• If you get 24 questions correct on a 30 question
  exam, what is your percent?
• A percent can also be used as a RATIO
• A friend tells you she got a grade of 95 on a 40
  question exam. How many questions did she answer
  correctly?

24/30 x 100 80
40 x 95/100 38 correct
4
Percent Error

• Percent error accepted value experimental
  value
• Percent error is used to find out the degree of
  error you have in an experiment.
• There will always be some error, scientists like
  to keep error below 2 .

X 100 percent
accepted value
5
Density

• Density- the ratio of the mass of a substance to
  the volume of the substance.
• - Expressed as
• Liquids solids grams/cubic centimeters
• Gasses grams/liters
• Density Mass/Volume g/cm3
• Mass Density X Volume
• Volume Mass / Density

6
Density D M / V

• Calculate the density of a piece of metal with a
  volume of 18.9 cm3 and a mass of 201.0 g.
• The density of CCl4 is 1.58 g/mL. What is the
  mass of 95.7 mL of CCl4?
• What is the volume of 227 g of olive oil if its
  density is 0.92 g/mL?

D 201.0 g / 18.9 cm3 10.6 g/cm3
1.58 g/mL X / 95.7 mL
X 1.58 g/mL X 95.7 mL
X 151 g
X 227 g / 0.92 g/mL
0.92 g/mL 227 g / X
X 247 or 2.5 X 102 mL
7
Density and Error Practice

• If you were given an object that had a length of
  5.0 cm, a width or 10.0 cm and a height of 2.0
  cm, what would the density of this object be if
  you weighed it and found that it had a mass of
  800.0 g?
• What would the error be for your measurments if
  I told you the accepted value for the density of
  this object is 8.50 g/cm3? Is this acceptable?
  Explain.

5.0 cm X 10.0 cm X 2.0 cm 100 cm3
D 800.0 g / 100 cm3 8.0 g/cm3
8.0 8.5 / 8.5 X 100 5.9
No, the error is greater than 2
8
Concentration Measurements

• Molarity M
• Molarity mol solute / L solution
• Use in solution stoichiometry calculations
• Mole solute Molarity X Liters solution
• Liters solution moles solution / Molarity
• Molality m
• mol solute / kg solvent
• Used with calculation properties such as boiling
  point elevation and freezing point depression
• Parts per Million ppm
• g solute / 1 000 000 g solution
• Used to express small concentrations

Pg. 460
9
Molarity

• What is the molarity of a potassium chloride
  solution that has a volume of 400.0 mL and
  contains 85.0 g KCl?
• Gather Info
• Volume of solution 400.0 mL
• Mass of solute 85.0 g KCl
• Molarity of KCl solution ?
• Plan Work
• Calculate the mass of KCl into moles using molar
  mass
• 85.0 g KCl
• Convert the volume in milliliters into volume in
  liters
• 400.0 mL
• Calculate
• Molarity is moles of solute divided by volume of
  solution

1 mol
1.14 mol KCl
74.55 g KCl
1 L
0.4000 L
1000 mL
1.14 mol KCl
2.85 mol / L 2.85 M KCl
Pg. 465
0.4000 L
10
Parts Per Million

• A chemical analysis shows that there are 2.2 mg
  of lead in exactly 500 g of water. Convert this
  measurement to parts per million.
• Gather Info
• Mass of Solute 2.2 mg
• Mass of Solvent 500 g
• Parts per Million ?
• Plan Work
• First change 2.2 mg to grams
• 2.2 mg
• - Divide this by 500 g to get the amount of lead
  in 1 g water, then multiple by 1,000,000 to get
  the amount of lead in 1,000,000 g water.
• Calculate
• 0.0022 g Pb

1 g
2.2 X 10-3 g
1000 mg
1,000,000 parts
4.4 ppm Pb
1 million
500 g H2O
ie 4.4 parts Pb per million parts H2O
Pg. 461
11
Specific Heat

• Specific Heat the quantity of energy that must
  be transferred as heat to raise the temperature
  of 1g of a substance by 1K.
• The quantity of energy transferred as heat
  depends on
• The nature of the material
• The mass of the material
• The size of temperature change
• Ex 1g of Fe 100C to 50C transfers 22.5J of
  energy. 1g of Ag 100C to 50C transfers
  11.8J of energy.
• Fe has a larger specific heat than Ag
• Meaning that more energy as heat can be
  transferred to the iron than to the silver

12
Explain Specific Heat in My Terms

• Metals Low Specific Heat little energy must
  be transferred as heat to increase temperature.
• Water High Specific Heat (Highest of most
  common substances) can absorb a large quantity
  of energy before temperature increases.

13
Specific Heat Formula

• Cp specific heat at a given pressure (J/gK)
• q energy transferred as heat (J)
• m mass of the substance (g)
• ?T difference btwn. initial and final
  temperatures (K)
• (Final Temp Initial Temp)
• Q Cp (m X ?T)
• Mass (Cp)(?T) / Q

Cp q___ m X ?T
14
Specific Heat Example (pg.61)

• A 4.0g sample of glass was heated from 274K to
  314K and was found to absorb 32J of energy as
  heat. Calculate the specific heat of this glass.
• Gather Info
• A. Mass (m) of sample 4.0g
• B. Initial Temp 274K
• C. Final Temp 314K
• D. Amt. of Energy absorbed (q) 32J
• 2. Plan Work
• Cp q___
• m X ?T
• 3. Calculate
• Fill in formula
• Cp _______ _____
  
• X

2 SD
32 J
32 J
0.20 J/gK
4.0 g
40 K
160 gK
2 SD
15
Practice Problems

• Page 61 1-4

16
Specific Heat 1

• Calculate the specific heat of a substance if a
  35g sample absorbs 48J as the temperature is
  raised from 293K to 313K. Be sure to use the
  correct number of sig. figs. in your answer.

17
Specific Heat 2

• The temperature of a piece of copper with a mass
  of 95.4g increases from 298.0K to 321.1K when the
  metal absorbs 849J of energy as heat. What is the
  specific heat of copper? Use Sig Figs.

18
Specific Heat 3

• If 980kJ of energy as heat are transferred to
  6.2L of H2O at 291K, what will the final temp of
  H2O be? The specific heat of water is 4.18J/gK.
  Assume that 1.0mL of H2O equals 1.0g or H2O. Use
  Sig Figs.

19
Specific Heat 4

• How much energy as heat must be transferred to
  raise them temperature of a 55g sample of Al from
  22.4C to 94.6C? The specific heat of Al is
  0.897J/gK. Note that a temperature change of 1C
  is the same as a temperature change of 1K because
  the sizes of the degree divisions on both scales
  are equal. Use Sig Figs.

20
Enthalpy

• Enthalpy- the sum of the internal energy of a
  system plus the product of the systems volume
  multiplied by the pressure that the system exerts
  on its surroundings. (heat content, total energy
  of the system)
• When calculating enthalpy if the change in
  enthalpy is positive, it means that heating the
  sample requires energy making it an endothermic
  process.(run up the hill)
• When the change is negative, the sample has been
  cooled, meaning that the sample has released
  energy making it an exothermic process.(fall down
  the hill)

21
Molar Enthalpy Formula

• ?H molar enthalpy (J/mol)
• C molar heat capacity (J/Kmol)
• ?T change in temperature (K)
• C ?H / ?T
• Note A mole is the amount of a substance

?H C?T
22
Molar Enthalpy Heating

• How much does the molar enthalpy change when ice
  warms from -5.4C to -0.2C? The molar heat
  capacity of H2O(s) is 37.4J/Kmol
• Gather Info
• Initial Temp -5.4C
• Final Temp -0.2C
• C 37.4J/Kmol
• Plan Work
• ?H C?T
• Calculate
• ?H 37.4J/Kmol (272.8K 267.6K)
• (37.4J/Kmol)(5.2K)
• 194.48 J/mol 194 J/mol

267.6 K
272.8 K
Pg. 346
23
Molar Enthalpy Cooling

• Calculate the molar enthalpy change with an
  aluminum can that as a temperature of 19.2C is
  cooled to a temperature of 4.00C. The molar heat
  capacity for Al is 24.2 J/Kmol.
• Gather Info
• Initial Temp 19.2C
• Final Temp 4.00C
• C 24.2 J/Kmol
• Plan Work
• ?H C?T
• Calculate
• ?H

292 K
277 K
(24.2 J/Kmol)
(277 K 292 K)
(24.2 J/Kmol)
(-15 K)
-363 J/mol
Pg. 347
24
Enthalpy Practice

• Page 346 1 2
• Page 347 1 2

25
Page 346 1

• Calculate the molar enthalpy change of H2O(l)
  when liquid water is heated from 41.7C to 76.2C.

26
Page 346 2

• Calculate the ?H of NaCl when it is heated from
  0.0C to 100.0C.

27
Page 347 1

• The molar heat capacity of Al(s) is 24.2 J/Kmol.
  Calculate the molar enthalpy change when Al(s) is
  cooled from 128.5C to 22.6C.

28
Page 347 2

• Lead has a molar heat capacity of 26.4J/Kmol.
  What molar enthalpy change occurs when lead is
  cooled from 302C to 275C.

29
Simple Conversions

• If you had 6.0 mops, how many pops would you have?

2 kops 4 nips 1 dip 6 jips 1 fop 3 gops 1
pop 3 gops 3 mops 6 jips 7 dips 2 nips 3
kops 1 fop
6 jips
6.0 mops
1 dip
2 nips
2 kops
1 fop
3 gops
1 pop
4 nips
3 gops
6 jips
7 dips
3 kops
1 fop
3 mops
0.0952 pops 9.5 X 10-2 pops