AT Chapter 3 Notes 15.notebook September 29, 2015

Measuring Atomic Masses

Mass Spectrometer ­ used to isolate of an element and determine their mass.

1 AT Chapter 3 Notes 15.notebook September 29, 2015

• An element sample is heated to vaporize it and the gaseous atoms are “zapped” with an electron beam that knocks electrons off the gaseous atoms to produce positively charged gaseous ions.

• The positively charged ions are accelerated by passing through an electric field and slit to produce an accelerated ion beam.

• This positively charged beam is subjected to a magnetic field, which separates the ions in the beam based on differences in mass; the more massive the ion, the less it is deflected.

• The spectrometer automatically detects these differences quantitatively and displays the mass of each and its percent abundance.

• Standard = C­12 isotope; assigned a value of exactly 12.0000 amu; the masses of all other atoms is based on comparison with the mass of C­12.

• One amu = 1/12 the mass of one C­12 atom.

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Calculation of Average Atomic Masses From Isotopic Data

average = Σ(mass of isotope) x (percent abundance)

A sample of metal "M" is vaporized and injected into a mass spectrometer. The mass spectrum tells us that 60.10% of the metal is present as M­69 and 39.90% is present as M­71. The mass values for M­69 and M­71 are 68.93 amu and 70.92 amu, respectively. Calculate the average atomic mass of metal "M".

Calculation of Relative Isotope Abundances

The element exists naturally as two isotopes. In­113 has a mass of 112.9043 amu, and In­115 has a mass of 114.9041 amu. The average atomic mass of indium is 114.82 amu. Calculate the percent relative abundance of the two isotopes of indium.

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THE MOLE/FORMULA STOICHIOMETRY/CONVERSIONS

How do chemists connect the atomic scale (amu) to the macroscopic scale (grams)? The bridge is the “mole”. The mole is the key to many chemical calculations. A mole is defined as the number equal to the number of atoms in exactly 12 grams of pure C­12. One mole of a substance contains Avogadro's number of particles. Avogadro's number = 6.02 X 1023 particles. The mole is often called “the chemist’s dozen”. The key problem­solving relationship is: the average mass of one atom of a substance expressed in amu's is the same number as the mass of one mole of a substance expressed in grams.

1 atom of Ne­20 = 20.18 amu 1 mole of Ne­20 = 20.18 grams 1 mole of Ne­20 = 6.02 X 1023 atoms of Ne 1 gram (exactly) = 6.02 X 1023 amu Recall,

1 mole = 6.02 X 1023 particles = 22.4 L gas at STP = molar mass (gram formula mass)

MASS

MOLE GAS VOLUME AT STP

NUMBER OF PARTICLES

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Problem: Calculate the number of molecules in 76.2 g of .

Problem: Calculate the number of atoms in 1.45 g of chlorate.

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Percent Composition

Problem: Determine the mass percentage of each of the elements in potassium tartrate, NaKC4H4O6.

Determining Empirical/Molecular Formulas

The analysis of a rocket fuel showed that it contained 87.4% and 12.6% by weight. Mass spectral analysis showed the fuel to have a molecular weight of 32.06 grams. Determine the empirical and molecular formulas of the compound.

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Determining Empirical Formula From Combustion Data

Combustion Analysis: When a compound containing carbon and hydrogen is combusted in an apparatus diagrammed below, all of the carbon is converted to CO2 and all of the hydrogen is converted to H2O. The amount of CO2 is measured by the mass increases in the CO2 absorber, and the amount of water is measured by the mass increase in the H2O absorber.

Problem: A compound contains C, H, and O. Combustion of 1.000 g of this compound yields 1.500 g of CO2 and 0.405 g H2O. Determine the empirical formula of the compound. If the molecular mass of this compound was found to be 176 g/mol, determine its molecular formula.

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Problem: When 1.24 grams of a compound containing only carbon, hydrogen, and is burned in oxygen, 1.76 grams of CO2, 1.08 grams of H2O, and 1.28 grams of SO2 are formed. Calculate the empirical formula for the substance.

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CHEMICAL EQUATIONS/BALANCING/EQUATION STOICHIOMETRY

Review Problem: The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate for fuel: à 3 Al(s) + 3 NH4ClO4(s) Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

Show how the mole stoichiometric amounts verify the Law of Mass Conservation.

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Problem: Over the years the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid fueled rocket motors. The reaction is: à Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s)

What masses of (III) oxide and aluminum must be used to produce 15.0 g of iron? What mass of aluminum oxide would be produced?

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Limiting Reactant Calculations

In any stoichiometric calculation, it is essential to determine which reactant is the limiting one in order to calculate correctly the amounts of products that will be formed. To do this, you can compare the mole ratio of substances required by the balanced equation to the mole ratio of reactants actually present. The one that "runs out first" is limiting. I prefer to just do a mass to mass problem. Calculate the mass of one of the products separately from the masses of each of the reactants. The reactant that gives you less product had to run out first and is the limiting reactant, while some of the other reactant will be left unreacted and is in excess.

When you calculate the amount of product from an amount of limiting reactant using stoichiometry, you are assuming the reaction runs perfectly to completion and you are thus calculating the maximum possible amount of product. This is called the theoretical yield. In real life we never get a theoretical yield for a variety of reasons. The amount of product that is obtained is called the actual yield. We are often interested in how close the actual yield is to the theoretical yield and this is measured by calculating a percent yield:

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Problem: A strip of Zn metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of nitrate.

a) Write a balanced equation representing this reaction. What class of reaction is this?

b) Determine the mass of silver that will form. What is the limiting reactant? Which reactant is in excess?

c) If 1.24 g of silver is actually formed, what is the percent yield?

d) Calculate the mass of excess reactant that remains.

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A Challenging Equation Stoichiometry Problem

When a 10.0 g sample of a mixture of methane (CH4) and ethane (C2H6) is burned uin excess oxygen, exactly 525 kJ of heat is produced. What is the percentage by mass of CH4 in the original mixture? à CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆H = ­890.4 kJ (per mole CH4) à C2H6(g) + 3.5O2(g) 2CO2(g) + 3H2O(g) ∆H = ­1560.0 kJ (per mole C2H6)

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A slightly more challenging problem

A 2.000 g sample of a mixture of CaCl2 and RbCl is treated with excess AgNO3(aq), causing AgCl(s) to precipitate from the solution. If the mass of AgCl obtained is 3.45 g, then what is the percentage by mass of RbCl in the original mixture?

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