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Prep Chem

04/25/09

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The Mole Concept


�Before You Begin:

To master this material you will need to be able to perform unit conversions and round answers to the correct number of significant figures. These topics are found in Measurement. Note that the atomic weights can be found on any periodic table. Isotope abundances and masses are more difficult to find.


A mole is a quantity of matter that has the same mass in grams as its atomic or formula weight.

Atomic Weight

Atomic weight—the average mass of the atoms of an element taking into account the masses of the various isotopes and their relative abundance. For example, oxygen has an atomic weight of 15.9994. This means that the average of the atoms’ masses is 15.9994 atomic mass units (amu).  

Calculating atomic weights—the atomic weight is a weighted average of all of the atomic masses of the naturally occurring isotopes. This means that, instead of each mass counting equally toward the average, the mass is scaled according to each isotope’s abundance. To find the atomic weight multiply each isotope’s atomic mass by its percent abundance in decimal form then find the sum of these weighted masses.

 


4Concept Check: Chromium has four naturally occurring isotopes: 50Cr, 52Cr, 53Cr, and 54Cr. The atomic masses and abundance data are:

Chromium Isotopic Abundances
Isotope Mass Abundance

50Cr

49.9460464 amu

4.345 %

52Cr

51.9405098 amu

83.789 %

53Cr

52.9406513 amu

9.501 %

54Cr

53.9388825 amu

2.365 %

What is chromium's atomic weight?

Answer: Find the weighted average after converting the percentages to decimals.

Atomic mass Cr = 0.04345(49.9460464 amu) + 0.83789(51.9405098 amu)

 + 0.09501(52.9406513 amu) + 0.02365(53.9388825 amu) = 51.996 amu

This answer is rounded to the third decimal place because we apply the multiplication rule and then the addition rule for finding significant figures.



4Concept Check: There are two naturally occurring isotopes of boron, 10B and 11B. The atomic weight of boron is 10.811 amu, the atomic mass of 11B is 11.0093054 and its abundance is 80.1%. What is the atomic mass of 10B? 

Answer: Set the atomic weight equal to the weighted average, with 19.9% of the 10B isotope at an unknown mass. Solve for this unknown.

10.811 amu = 0.199(10B) + 0.801(11.0093054 amu)

10B = [10.811 amu - 0.801(11.0093054 amu)] ÷ 0.199 =  10.0 amu

The answer is rounded to three significant figures because of the degree of uncertainty of the abundance data. The experimental value for the atomic mass of 10B is 10.0129369 amu.

 

Atomic mass and abundance data were found at Web Elements (September 16, 2004).


 

Formula and molecular weights

The formula weight is the sum of the atomic weights of all of the ions in one formula unit of an ionic compound. The molecular weight is the sum of the atomic weights of all of the atoms in a molecule of a molecular compound. Remember that the subscripts in a formula indicate the number of ions in an ionic compound and the number of atoms in a molecule or polyatomic ion.

 


4Concept Check: What is the formula weight of ammonium sulfate?

Answer: The formula of ammonium sulfate is (NH4)2SO4.

FW = 2(14.0067 amu) + 8(1.00794 amu) + 32.065 amu + 4(15.9994 amu) = 132.140 amu

Note that the answer is rounded to the third decimal place due to application of the addition rule for significant figures.


 

Avogadro’s Number and the Mole

A mole is a collection of atoms with a mass equal to the atomic weight in grams. The number of atoms in a mole is 6.02 x 1023. This works because one amu is 1.66 x 10-24 gram, so it takes 6.02 x 1023 of them to make one gram. This is one of the most useful concepts in chemistry. The atomic weight of lithium is 6.941amu. This means that a sample with a mass of 6.941 grams has 6.02 x 1023 atoms of lithium in it. Note that this works for molecules and ions, too. The number 6.02 x 1023 is known as Avogadro’s number in honor of the chemist who first proposed a law relating the volume of a gas to the number of gaseous particles it contains. Note that the law works for mass, too, but it was easier for chemists in the 17th century to measure the volume of a gas than to weigh it.

 


4Concept Check: How many atoms of tin are in a sample with a mass of 9.74 grams?

Answer: Use unit analysis and treat the atomic weight and Avogadro’s number as conversion factors.

           9.74gSn(1molSn/118.710gSn)(6.022EE23atomsSn/1molSn)=4.94EE22atomsSn

 


 

Percent composition by mass

Percent composition by mass—the percent composition by mass is the ratio of the masses of one element in a compound to the mass of the entire compound expressed as a percent. Due to the law of constant composition, the relative amount of an element in a compound is independent of the source of the compound. That means that we can use either experimentally measured masses or atomic weights from the periodic table; either source of data would five the same result.

 

From experimental results—to find the percent composition by mass from experimental data, divide the mass of the element by the mass of the compound and express the result as a percent.


4Concept Check: 0.5627 grams of elemental silver are recovered from 0.8861 grams of an unknown compound. What is the percent silver by mass of this compound?

Answer: divide the mass of the elemental silver by the mass of the unknown compound to find the percent silver:

                                       %Ag=100(0.5627g/0.8861g)=63.50%

This answer has four significant figures by the division rule and the 100 is exact.


 

From a formula—to find the percent composition by mass from atomic weights divide the atomic weight of the total amount of the element in the compound by the formula weight (or molecular weight) of the compound and express the result as a percent.


4Concept Check: What is the percent silver in silver carbonate? Is the unknown compound from the previous problem silver carbonate?

Answer: The formula of silver carbonate is Ag2CO3. Assume one mole.

%Ag = 100{2(107.8682 g)/[2(107.8682 g) + 12.0107 g + 3(15.9994 g)]} = 78.2376%

This is not the same compound because 63.50 ≠ 78.2376. The answer has seven significant figures. The small whole numbers and the 100 are exact numbers. The numerator has seven significant figures. The denominator has seven significant figures. Note that the number 12.0107 does not limit the significant figures because, by the addition rule, the result of the addition in the denominator is rounded to the fourth decimal place.

 

 

 

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