Matter and Measurement / Atoms, Molecules, and Ions/Stoichiometry: Calculations with

Chemical Formulas and Equations

H Advanced ChemistryUnit 1

Objective #6 Isotopes / Calculating Average Atomic Mass / Mass Spectrometer

*average atomic masses can be determined from the masses of the various isotopes that naturally occur for an element and their percent of abundance in nature

*examples:*operation of mass spectrometer:*instrument that provides the atomic mass

and percent of abundance of the isotopes that make up a sample of an element

Operation of Mass Spectrometer

*gaseous sample is introduced into instrument and the particles are bombarded with high energy electrons

*positive ions are produced and these are passed through a magnetic field

*the amount of deflection is measured; the more massive the particle the less the deflection

*a graph of the intensity of the detector signal vs. the particle mass is called a mass spectrum

*it is from the data provided from this graph that one obtains the atomic mass and percent abundance necessary to calculate the average atomic mass for an element

Example of Mass Spectrograph for Hg

Most Common Isotopes of Mercury Based on Mass Spectrograph

Isotope Percent Abundance*mercury-196 .15 %*mercury-198 9.97 %*mercury-199 16.87 %*mercury-200 23.1 %*mercury-201 13.18 %*mercury-202 29.86 %*mercury-204 6.87 %

Objective #12 Description of Combustion Analysis

Objective #12 Combustion Analysis

*Example I: An unknown compound is composed of carbon, nitrogen, and hydrogen. When .1156 grams of this compound is reacted with oxygen and burned, the following components are produced: .1638 g of carbon dioxide and .1676 g of water. Calculate the percent composition and empirical formula for this unknown compound.

Step I All of the carbon that was in the original sample is now contained in the carbon dioxide. Convert from grams of carbon dioxide to grams of carbon.

.1638 g CO2 X 12.0 g C/44.0 g CO2 = .04467 g C

Step II Divide the resulting mass of carbon by the original mass of the compound and multiply by 100% to obtain the percentage of carbon in the sample

% C = .04467 g/.1156 g X 100 = 38.64% C

Step III All of the hydrogen that was in the original sample is now contained in the water. Convert from grams of water to grams of hydrogen.

.1676 g H2O X 2.0 g H / 18.0 g H2O = .01862 g

Step IV Divide the resulting mass of hydrogen by the original mass of the compound and multiply by 100% to obtain the percentage of the hydrogen in the sample

%H = .01862 g/.1156 g X 100 = 16.11% HStep V The mass and percentage of the final

component can be determined by mass and percentage difference.

.1156 g - .04467 g - .01862 g = .05231 g N100% - 38.64% - 16.11% = 45.25% N

Step VI Use the percentages or masses of components to compute the empirical formula of the compound.

.04467 g C X 1 mole C/12.0 g C = .003723 moles C

.01862 g H X 1 mole H/1.0 g H = .01862 moles H

.05231 g N X 1 mole N/14.0 g N = .003736 moles N

.003723 moles C/.003723 = 1

.01862 moles H/.003723 = 5

.003736 moles N/.003723 = 1CH5N

Example II Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion analysis of a .225 g sample of this compound produces .512 g of carbon dioxide and .209 g of water. What is the emprical formula of caproic acid? Determine the molecular formula if the molecular mass is 116 g.

.512 g CO2 X 12.0 g C/44.0 g CO2 = .140 g C

.209 g H2O X 2.0 g H/18.0 g H2O = .0232 g HBy difference .0618 g O.140 g C X 1 mole C/12.0 g C = .0117 g C.0232 g H X 1 mole H/1.0 g H = .0232 mole H.0618 g O X 1 mole O/16.0 g O =.00386mole O

.0117 mole C/.00386 = 1 mole C

.0232 mole H/.00386 = 6 mole H

.00386 mole O / .00386 = 1 mole OC3H6O empirical formula116 g/58 g = 2C6H12O2 molecular formula

Objectives #13-14 Stochiometry

1. Calculate the mass of water produced from the decomposition of 5.00 g of ammonium dichromate.

5.00 g X 1mole A.D. / 252.0 g A.D. X 4 mole H2O / 1 mole A.D. X 18.0 g H2O = 14.4 g

2. Calculate the number of water molecules produced from 10. g of A.D.

10. g A.D. X 1 mole A.D. / 252.0 g A.D. X

4 mole H2O / 1 mole A.D.X 6.02 X 1023 molecules / 1 mole H2O = 9.56 X 1022 molecules

3. Calculate the amount of A.D. required to produce 5500 J of energy if the change in energy of the reaction is 2750 J.

5500 J X 1 mole A.D. / 2750 J X 252.0 g A.D. / 1 mole A.D. = 5.0 X 102

g

4. Calculate the number of chromium atoms produced from the decomposition of 10.0 g of A.D.

10.0 g A.D. X 1 mole A.D. / 252.0 g A.D. X 1 mole Cr2O3 / 1 mole A.D. X

6.02 X 1023 f. units Cr2O3 / 1 mole Cr2O3 X 2 Cr atoms / 1 f. unit Cr2O3 =

4.78 X 1022 atoms

Example II: The reaction below produced 4.50 g of aluminum sulfate in the laboratory. Calculate the theoretical yield and the percent yield given 5.00 g of sulfuric acid reacted with 5.00 g of aluminum hydroxide.

5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 1 mole Al2(SO4)3 / 3 mole H2SO4 X 342.3 g Al2(SO4)3 / 1 mole = 5.82 g

5.00 g Al(OH)3 X 1 mole Al(OH)3 / 78.0 g X 1 mole Al2(SO4)3 / 2 mole Al(OH)3 X 342.0 g Al2(SO4)3 =

11.0 g %Yield = 4.50 g / 5.82 g X 100 = 77%

Use the above information to determine the amount of the excess reactant leftover.

5.00 g H2SO4 X 1 mole H2SO4 / 98.1 g X 2 mole Al(OH)3 / 3 mole H2SO4 X 78.0 g Al(OH)3 = 2.65 g (amount used)

5.00 g – 2.65 g = 2.35 g leftover

Objectives #15-16 Molarity, Molality, Solution Stoichiometry and Titrations

Example I: Determine the molarity of a solution containing 2.37 moles of potassium nitrate in enough water to give 650 ml of solution.

M = 2.37 moles / .650 L = 3.65 M

Example II: Determine the molarity of a solution containing 25.0 g of sodium hydroxide dissolved in enough water to give 2.50 L of solution.

M= 25.0 g NaOH / 40.0 g / 2.50 L = .250 M

Example III: How many grams of sucrose are present in 125 ml of a 1.07 M sucrose solution?

Moles = .125 ml X 1.07 M = .134 moles

.134 moles sucrose X 342.0 g / 1 mole= 45.8 g

*MolalityExample I: Calculate the molality of a

solution containing 5.0 g of sodium chloride dissolved in 250 g of water

5.0 g NaCl X 1mole NaCl / 58.5 g / .250 kg H2O

= .34 m

Example II: Calculate the molality of a solution containing 5.0 moles of NaCl dissolved in 250 g of water.

5.0 moles NaCl / .250 kg H2O = 20. m

*Dilution*moles before dilution = moles after

dilution*formula: M1V1 = M2V2

Example I: What is the molarity of a solution prepared by diluting 12.0 ml of .405 M NaCl to a final volume of

80.0 ml?12.O ml X .405 M / 80.0 ml = .0608 M

Example II: What volume of .25 M HCl solution must be diluted to prepare

1.00 L of .040 M HCl?.040 M X 1.00 L / .25 M = .16 L

*Precipitation Reactions and Solution Stoichiometry*key concept: to find moles of substance in

solution multiply moles by volumeExample I: How many grams of precipitate can be

produced from the reaction of 1.00 L of .375 M silver nitrate solution with an excess of sodium chloride solution?

.375 M X 1.00 L silver nitrate = .375 moles

.375 moles AgNO3 X 1 mole AgCl / 1 mole AgNO3 X 143.4 g AgCl / 1 mole AgCl = 53.8 g

Example II: How many grams of precipitate can be produced from the reaction of 2.50 L of a .200 M calcium chloride solution with an excess of phosphoric acid?

.200 M calcium chloride X 2.50 L = .500 moles CaCl2

.500 moles CaCl2 X 1 mole Ca3(PO4)2 / 3 moles CaCl2 X 310.3 g Ca3(PO4)2 X 1 mole = 51.7 g

.

Example III: What volume in ml of a .300 M silver nitrate solution is needed to react with 40.0 ml of a .200 M potassium phosphate?

.200 M potassium phosphate X .0400 L = .00800 moles

.00800 moles K3PO4 X 3 mole AgNO3 / 1 mole K3PO4 = .0240 moles AgNO3

moles / M = L .0240 moles AgNO3 / .300 M = 80.0 ml

*Titration Problems:Example I: What is the molarity of a 37.5 ml

sample of a sulfuric acid solution that will completely react with 23.7 ml of a .100 M sodium hdyroxide solution?

.100 M NaOH X .0237 L = .00237 moles NaOH

.00237 moles NaOH X 1 mole H2SO4 / 2 mole NaOH = .00119 moles H2SO4

M = .00119 moles H2SO4 / .0375 L = .0317 M

Example II: What volume in liters of a 1.00 M hydrogen sulfide solution is needed to react completely with .500 L of a 4.00 M nitric acid solution?

4.00 moles HNO3 X .500 L = 2.00 moles HNO3

2.00 moles HNO3 X 3 moles H2S / 2 moles HNO3 = 3.00 moles H2S

L = moles / M 3.00 moles H2S / 1.00 L = 3.00 L