Chemical Calculations
Percents
• Percent means “parts of 100” or
“parts per 100 parts”
• The formula:
Part
x
100
Percent = Whole
Percents
• If you get 24 questions correct on a 30
question exam, what is your percent?
24/30 x 100 = 80%
• 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?
40 x 95/100 = 38 correct
Percent Error
• Percent error = |accepted value – experimental value| X 100 percent
accepted 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 %.
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
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.
D= 201.0 g / 18.9 cm3 = 10.6 g/cm3
• The density of CCl4 is 1.58 g/mL. What is the mass of
95.7 mL of CCl4?
1.58 g/mL = X / 95.7 mL
X = 1.58 g/mL X 95.7 mL
X = 151 g
• What is the volume of 227 g of olive oil if its density is
0.92 g/mL?
X = 227 g / 0.92 g/mL
0.92 g/mL = 227 g / X
X= 247 or 2.5 X 102 mL
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?
5.0 cm X 10.0 cm X 2.0 cm = 100 cm3
D = 800.0 g / 100 cm3 = 8.0 g/cm3
• 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.
8.0 – 8.5 / 8.5 X 100 = 5.9%
No, the % error is greater than 2 %
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
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:
1 mol
85.0 g KCl
= 1.14 mol KCl
74.55 g KCl
– Convert the volume in milliliters into volume in liters
400.0 mL 1 L
= 0.4000 L
1000 mL
• Calculate
– Molarity is moles of solute divided by volume of solution
1.14 mol KCl
= 2.85 mol / L = 2.85 M KCl
0.4000 L
Pg. 465
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
1g
2.2 mg
= 2.2 X 10-3 g
1000 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
500 g H2O
1,000,000 parts
1 million
= 4.4 ppm Pb
ie: 4.4 parts Pb per million parts H2O
Pg. 461
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:
1. The nature of the material
2. The mass of the material
3. The size of temperature change
•
Ex: 1g of Fe 100°C to 50°C transfers 22.5J of energy.
1g of Ag 100°C to 50°C 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
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.
Specific Heat Formula
Cp = specific heat at a given pressure (J/g•K)
q = energy transferred as heat (J)
Cp =
m = mass of the substance (g)
∆T = difference btwn. initial and final temperatures (K)
(Final Temp – Initial Temp)
Q = Cp (m X ΔT)
Mass = (Cp)(ΔT) / Q
q___
m X ∆T
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.
1.
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.
3.
-
Plan Work
Cp = q___
m X ∆T
Calculate
Fill in formula
32 J
Cp =
_______
=
4.0 g X 40 K
2 SD
32 J =
_____
160 g•K
2 SD
0.20 J/g•K
Practice Problems
Page 61 1-4
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.
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.
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/g•K. Assume that
1.0mL of H2O equals 1.0g or H2O. Use Sig Figs.
Specific Heat #4
• How much energy as heat must be transferred to raise them temperature of a
55g sample of Al from 22.4°C to 94.6°C? The specific heat of Al is 0.897J/g•K.
Note that a temperature change of 1°C is the same as a temperature change of
1K because the sizes of the degree divisions on both scales are equal. Use Sig
Figs.
Enthalpy
• Enthalpy- the sum of the internal energy of a system plus
the product of the system’s 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)
Molar Enthalpy Formula
∆H = molar enthalpy (J/mol)
C = molar heat capacity (J/K•mol)
∆T = change in temperature (K)
∆H = C∆T
C = ΔH / ΔT
Note: A mole is the amount of a substance
Molar Enthalpy Heating
•
How much does the molar enthalpy change when ice warms from -5.4°C
to -0.2°C? The molar heat capacity of H2O(s) is 37.4J/K•mol
1.
Gather Info
Initial Temp = -5.4°C = 267.6 K
Final Temp = -0.2°C = 272.8 K
C = 37.4J/K•mol
2.
Plan Work
∆H = C∆T
3.
Calculate
∆H = 37.4J/K•mol (272.8K – 267.6K)
= (37.4J/K•mol)(5.2K)
= 194.48 J/mol = 194 J/mol
Pg. 346
Molar Enthalpy Cooling
•
Calculate the molar enthalpy change with an aluminum can that as a
temperature of 19.2°C is cooled to a temperature of 4.00°C. The molar
heat capacity for Al is 24.2 J/K•mol.
1.
Gather Info
Initial Temp = 19.2°C = 292 K
Final Temp = 4.00°C = 277 K
C = 24.2 J/K•mol
2.
Plan Work
∆H = C∆T
3.
Calculate
∆H = (24.2 J/K•mol)(277 K – 292 K)
(24.2 J/K•mol)(-15 K) = -363 J/mol
Pg. 347
Enthalpy Practice
Page 346 1 & 2
Page 347 1 & 2
Page 346 #1
• Calculate the molar enthalpy change of H2O(l) when liquid water is heated
from 41.7°C to 76.2°C.
Page 346 #2
• Calculate the ∆H of NaCl when it is heated from 0.0°C to
100.0°C.
Page 347 #1
• The molar heat capacity of Al(s) is 24.2 J/K•mol. Calculate the molar
enthalpy change when Al(s) is cooled from 128.5°C to 22.6°C.
Page 347 #2
• Lead has a molar heat capacity of 26.4J/K•mol. What molar enthalpy change
occurs when lead is cooled from 302°C to 275°C.
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
6.0 mops 6 jips
1 dip 2 nips 2 kops 1 fop 3 gops 1 pop
3 mops 6 jips 7 dips 4 nips 3 kops 1 fop 3 gops
= 0.0952 pops = 9.5 X 10-2 pops