Chemical Calculations

Percents

• Percent means “parts of 100” or “parts per 100 parts”

• The formula:

Part

Wholex 100Percent =

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

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 %.

accepted valueX 100 percent

• Density- the ratio of the mass of a substance to the volume of the substance.

- Expressed as:Liquids & solids= grams/cubic centimetersGasses= grams/liters

- Density = Mass/Volume = g/cm3

- Mass = Density X Volume- Volume – Mass / Density

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.

• 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

0.92 g/mL = 227 g / X X = 227 g / 0.92 g/mL

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?

• 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 %

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:

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

74.55 g KCl= 1.14 mol KCl

1 L1000 mL

= 0.4000 L

1.14 mol KCl

0.4000 L= 2.85 mol / L = 2.85 M KCl 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

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

1000 mg= 2.2 X 10-3 g

1,000,000 parts

500 g H2O1 million

= 4.4 ppm Pb

ie: 4.4 parts Pb per million parts H2O

Pg. 461

Specific HeatSpecific 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 material2. The mass of the material3. 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 FormulaSpecific Heat FormulaCp = specific heat at a given pressure (J/g•K)

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 ΔT)Mass = (Cp)(ΔT) / Q

Cp = q___ m X ∆T

Specific Heat Example (pg.61)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 InfoA. Mass (m) of sample = 4.0gB. Initial Temp = 274KC. Final Temp = 314KD. Amt. of Energy absorbed (q) = 32J

2. Plan Work Cp = q___ m X ∆T

3. Calculate- Fill in formula

Cp = _______ = _____ = X

2 SD

2 SD32 J

4.0 g 40 K 160 g•K32 J 0.20 J/g•K

Practice ProblemsPractice Problems

Page 61 1-4

Specific Heat #1Specific 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 #2Specific 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 #3Specific 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 #4Specific 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 FormulaMolar Enthalpy Formula

∆H = molar enthalpy (J/mol)

C = molar heat capacity (J/K•mol)

∆T = change in temperature (K)

C = ΔH / ΔT

Note: A mole is the amount of a substance

∆H = C∆T

Molar Enthalpy HeatingMolar 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

Final Temp = -0.2°C

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

= 267.6 K

= 272.8 K

Pg. 346

Molar Enthalpy CoolingMolar 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 InfoInitial Temp = 19.2°C Final Temp = 4.00°CC = 24.2 J/K•mol

2. Plan Work ∆H = C∆T

3. Calculate ∆H =

= 292 K= 277 K

(24.2 J/K•mol)(277 K – 292 K)

(24.2 J/K•mol)(-15 K) = -363 J/molPg. 347

Enthalpy PracticeEnthalpy 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 gops1 pop = 3 gops 3 mops = 6 jips 7 dips = 2 nips3 kops = 1 fop

6.0 mops

3 mops

6 jips

6 jips1 dip

7 dips2 nips

4 nips2 kops

3 kops1 fop

1 fop3 gops

3 gops1 pop

= 0.0952 pops = 9.5 X 10-2 pops