Chapter 02 Chemical Formulas & Composition Stoichiometry

CHAPTER 2 CHEMICAL FORMULAS & COMPOSITION STOICHIOMETRY

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Chapter Goals 1. 2. 3. 4.

Atoms and Molecules Chemical Formulas Ions and Ionic Compounds Names and Formulas of Some Ionic Compounds 5. Atomic Weights 6. The Mole 2

Chapter Goals 7. Formula Weights, Molecular Weights,

and Moles 8. Percent Composition and Formulas of Compounds 9. Derivation of Formulas from Elemental Composition 10.Determination of Molecular Formulas 11.Some Other Interpretations of Chemical Formulas 12.Purity of Samples 3

Atoms and Molecules Dalton’s Atomic Theory - 1808 Five postulates 1.

2.

3.

4.

5.

An element is composed of extremely small, indivisible particles called atoms. All atoms of a given element have identical properties that differ from those of other elements. Atoms cannot be created, destroyed, or transformed into atoms of another element. Compounds are formed when atoms of different elements combine with one another in small wholenumber ratios. The relative numbers and kinds of atoms are constant in a given compound.

Which of these postulates are correct today? 4

Atoms and Molecules A molecule is the smallest particle of an element that can have a stable independent existence. 

Usually have 2 or more atoms bonded together

Examples of molecules      

H2 O2 S8 H2O CH4 C2H5OH

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Chemical Formulas Chemical formula shows the chemical composition of the substance. 

ratio of the elements present in the molecule or compound

He, Au, Na – monatomic elements O2, H2, Cl2 – diatomic elements O3, S8, P4 - more complex elements H2O, C12H22O11 – compounds  Substance consists of two or more elements

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Chemical Formulas Compound

1 Molecule Contains

HCl H2O

1 H atom & 1 Cl atom 2 H atoms & 1 O atom

NH3

1 N atom & 3 H atoms

C3H8

3 C atoms & 8 H atoms

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Ions and Ionic Compounds Ions are atoms or groups of atoms that possess an electric charge. Two basic types of ions: Positive ions or cations   

one or more electrons less than neutral Na+, Ca2+ , Al3+ NH4+ - polyatomic cation

Negative ions or anions   

one or more electrons more than neutral F-, O2- , N3SO42- , PO43- - polyatomic anions 8

Ions and Ionic Compounds Sodium chloride 

table salt is an ionic compound

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Names and Formulas of Some Ionic Compounds Table 2-3 displays the formulas, charges, and names of some common ions 

You must know the names, formulas, and charges of the common ions in table 2-3.

Some examples are: 

Anions - Cl1- , OH1- , SO42- , PO43-



Cations - Na1+ , NH41+ , Ca2+ , Al3+

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Names and Formulas of Some Ionic Compounds Formulas of ionic compounds are determined by the charges of the ions. 



Charge on the cations must equal the charge on the anions. The compound must be neutral.

NaCl KOH CaSO4 Al(OH)3

sodium chloride (Na1+ & Cl1- ) potassium hydroxide(K1+ & OH1- ) calcium sulfate (Ca2+ & SO42- ) aluminum hydroxide (Al3+ & 3 OH1- )

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Names and Formulas of Some Ionic Compounds Table 2-2 gives names of several molecular compounds. 

You must know all of the molecular compounds from Table 2-2.

Some examples are:   

H2SO4 - sulfuric acid FeBr2 - iron(II) bromide C2H5OH - ethanol 12

Names and Formulas of Some Ionic Compounds You do it! (# 1) What is the formula of nitric acid? HNO3 What is the formula of sulfur trioxide? SO3 What is the name of FeBr3? iron(III) bromide

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Names and Formulas of Some Ionic Compounds You do it! (# 2) What is the name of K2SO3? potassium sulfite What is charge on sulfite ion? SO32- is sulfite ion What is the formula of ammonium sulfide? (NH4)2S 14

Names and Formulas of Some Ionic Compounds You do it! What is charge on ammonium ion? NH41+ What is the formula of aluminum sulfate? Al2(SO4)3 What is charge on both ions? Al3+ and SO4215

Atomic Weights Weighted average of the masses of the constituent isotopes if an element. 



Tells us the atomic masses of every known element. Lower number on periodic chart.

How do we know what the values of these numbers are? 16

The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of things  

Just like a dozen = 12 things One mole = 6.022 x 1023 things

Avogadro’s number = 6.022 x 1023 

Symbol for Avogadro’s number is NA. 17

The Mole How do we know when we have a mole?  

count it out weigh it out

Molar mass - mass in grams numerically equal to the atomic weight of the element in grams. H has an atomic weight of 1.00794 g 

1.00794 g of H atoms = 6.022 x 1023 H atoms

Mg has an atomic weight of 24.3050 g 

24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

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The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. ? gM g =

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The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures. ? g Mg = 1 Mg atom

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The Mole Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.  1 mol Mg atoms ? g Mg = 1 Mg atom  23 6.022 × 10 Mg atoms 

  × 

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The Mole Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

 1 mol Mg atoms ? g Mg = 1 Mg atom  23 6.022 × 10 Mg atoms   24.30 gMg   1 mol Mg atoms

  × 

  = 4.04 ×10 −23 g Mg 

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The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

? Mg atoms =

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The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.  1 mol Mg   ? Mg atoms = 1.00 × 10 g Mg   24.30 g Mg  −6

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The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. ? Mg atoms =1.00 ×10  6.022 ×10 23 Mg atoms   1 mol Mg atoms 

−6

 1 mol Mg g Mg   24.30 g Mg 

   

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   

The Mole Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures. ? Mg atoms = 1.00 ×10

−6

 1 mol Mg   g Mg   24.30 g Mg 

 6.022 ×10 23 Mg atoms    = 2.48 ×1016 Mg atoms  1 mol Mg atoms 

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The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

? Mg atoms =

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The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

? Mg atoms = 1.67 mol Mg

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The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

 6.022 ×10 23 Mg atoms   ? Mg atoms = 1.67 mol Mg  1 mol Mg  

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The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

 6.022 ×10 23 Mg atoms   ? Mg atoms = 1.67 mol Mg  1 mol Mg   24 = 1.00 ×10 Mg atoms

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The Mole Example 2-3. How many atoms are contained in 1.67 moles of Mg?

 6.022 × 1023 Mg atoms   ? Mg atoms = 1.67 mol Mg 1 mol Mg   = 1.00 × 1024 Mg atoms

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The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg? You do it!

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The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

? mol M g = 73.4 g M g

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The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

 1 mol Mg atoms   ? mol Mg = 73.4 g Mg   24.30 g Mg 

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The Mole Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

 1 mol Mg atoms  ? mol Mg = 73.4 g Mg   24.30 g Mg  = 3.02 mol Mg IT IS IMPERATIVE THAT YOU KNOW HOW TO DO THESE PROBLEMS 35

Molecular Weights, and Moles How do we calculate the molar mass of a compound? 

add atomic weights of each atom

The molar mass of propane, C3H8, is:3 × C=3 × 1 .20 1 a m =3 u 6.0 3 a m u

8 ×H=8× .10 1 a m u=  .80 8 a m u M o l ma r a s s =4 4.1 1 a m u 36

Molecular Weights, and Moles The molar mass of calcium nitrate, Ca(NO3)2 , is: You do it!

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Molecular Weights, and Moles 1× Ca = 1× 40.08 amu = 40.08 amu 2 × N = 2 ×14.01 amu = 28.02 amu 6 × O = 6 ×16.00 amu = 96.00 amu Molar mass

= 164.10 amu 38

Molecular Weights, and Moles One Mole of 

Cl2 or 70.90g

Contains 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms



C3H8

You do it!

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Molecular Weights, and Moles One Mole of Contains 



Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms

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Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. ? C H molecules = 74.6 g C H × 3

8

3

8

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Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of ?propane. C 3 H 8 molecules = 74.6 g C 3 H 8 ×

 1 mole C 3 H 8     44.11 g C H  3 8  

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Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of ? C3 H8 molecules= 74.6 g C3 H8 × propane.

 1 mole C3 H8  6.022× 1023 C3 H8 molecules  =    1 mol C3 H8  44.11 g C3 H8   

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Molecular Weights, and Moles Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. ? C3H 8 molecules = 74.6 g C3 H 8 ×

 1 mole C3 H 8  6.022 ×10 23 C3H 8 molecules     = 44.11 g C3 H 8  44.11 g C3H 8   24 1.02 × 10 molecules 44

Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. You do it!

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Molecular Weights, and Moles Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3. 1 mol Li 2 CO 3 ? O atoms = 26.5 g Li 2 CO 3 × × 73.8 g Li 2 CO 3 6.022 ×10 23 form.units Li 2 CO 3 3 O atoms × = 1 mol Li 2 CO 3 1 formula unit Li 2 CO 3 6.49 ×10 O atoms 23

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Molecular Weights, and Moles Occasionally, we will use millimoles.  

Symbol - mmol 1000 mmol = 1 mol

For example: oxalic acid (COOH)2  

1 mol = 90.04 g 1 mmol = 0.09004 g or 90.04 mg 47

Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2. You do it!

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Molecular Weights, and Moles Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.

? mmol (COOH)2 = 0.234 g (COOH)2 ×  1 mmol (COOH)2    = 2.60 mmol (COOH)2  0.09004 g (COOH)2  49

Percent Composition and Formulas of Compounds % composition = mass of an individual element in a compound divided by the total mass of the compound x 100% Determine the percent composition of C in C 3H 8.

mass C %C= ×100% mass C 3 H 8 3 ×12.01 g = ×100% 44.11 g = 81.68%

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Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? You do it!

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Percent Composition and Formulas of Compounds What is the percent composition of H in C3H8? mass H %H= ×100% mass C3 H 8 8×H = ×100% C3H 8 8 ×1.01 g = ×100% = 18.32 % 44.11 g or 18.32% = 100% − 81.68%

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Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures. You do it!

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Percent Composition and Formulas of Compounds Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 2×F e fig. 2×5 5.8 g 3 sig. % F=e ×1 0 0 %= ×1 0 0 % =2 7.9 % F e

F e2 S O4 3 3 9 9.9 g 3×S 3×3 2.1 g % S= ×1 0 0 %= ×1 0 0 %   =2 4.1 % S F e2 S O4 3 3 9 9.9 g 1 2×O 1 2×1 6.0 g % O= ×1 0 0 %= ×1 0 0 %=4 8.0 % O F e2 S O4  3 3 9 9.9 g T o ta l =1 0 0 % 54

from Elemental Composition Empirical Formula - smallest whole-number ratio of atoms present in a compound  

CH2 is the empirical formula for alkenes No alkene exists that has 1 C and 2 H’s

Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound  

Ethene – C2H4 Pentene – C5H10

We determine the empirical and molecular formulas of a compound from the percent composition of the compound. 

percent composition is determined experimentally

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from Elemental Composition Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula? Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are:   

24.74 g of K 34.76 g of Mn 40.50 g of O

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from Elemental Composition 1 mol K ? mol K = 24.74 g K × = 0.6327 mol K 39.10 g K

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from Elemental Composition 1 mol K ? mol K = 24.74 g K × = 0.6327 mol K 39.10 g K 1 mol Mn ? mol Mn = 34.76 g Mn × = 0.6327 mol Mn 54.94 g Mn

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from Elemental Composition 1 mol K = 24.74 g K × = 0.6327 mol K 39.10 g K

? mol K

1 mol Mn ? mol Mn = 34.76 g Mn × = 0.6327 mol Mn 54.94 g Mn ? mol O

1mol O = 40.50 g O × = 2.531 mol O 16.00 g O

obtain smallest whole number ratio

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from Elemental Composition ? mol K

1 mol K = 24.74 g K × = 0.6327 mol K 39.10 g K

1 mol Mn ? mol Mn = 34.76 g Mn × = 0.6327 mol Mn 54.94 g Mn 1mol O ? mol O = 40.50 g O × = 2.531 mol O 16.00 g O obtain smallest whole number ratio 0.6327 for K ⇒ =1K 0.6327

0.6327 for Mn ⇒ = 1 Mn 0.6327

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from Elemental Composition 1 mol K = 0.6327 mol K 39.10 g K 1 mol Mn ? mol Mn = 34.76 g Mn × = 0.6327 mol Mn 54.94 g Mn 1mol O ? mol O = 40.50 g O × = 2.531 mol O 16.00 g O ? mol K

= 24.74 g K ×

obtain smallest whole number ratio 0.6327 for K ⇒ =1 K 0.6327

0.6327 for Mn ⇒ = 1 Mn 0.6327

2.531 =4 O 0.6327 thus the chemical formula is KMnO 4 for O ⇒

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from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? You do it!

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from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 1 mol Co ? mol Co = 6.541 g Co × = 0.1110 mol Co 58.93 gCo 1mol O ? mol O = 2.368 g O × = 0.1480 mol O 16.00 g O find smallest whole number ratio 63

from Elemental Composition Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound? 0.1110 0.1480 for Co ⇒ = 1 Co for O ⇒ = 1.333 O 0.1110 0.1110 multipy both by 3 to turn fraction to wh ole number 1 Co ×3 = 3 Co

1.333 O × =3 4 O

Thus the compound's formula is: Co 3O 4 64

Determination of Molecular Formulas Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 

short cut method

1 mol contains 56.1 g 85.63% is C and 14.37% is H 56.1 g × 0.8563 = 48.0 g of C 56.1 g × 0.1437 = 8.10 g of H

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Determination of Molecular Formulas convert masses to moles 1 mol C 48.0 g of C × = 4 mol C 12.0 g C 1 mol H 8.10 g of H × = 8 mol H 1.01 g H Thus the formula is: C 4 H8

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Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? molar mass of (NH 4 )3PO 4 = 149.0 g/mol 1 mol N ? mol N = 15.0 g of N × = 1.07 mol N 14.0 g N

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Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? molar mass of (NH 4 ) 3 PO 4 = 149.0 g/mol 1 mol N ? mol N = 15.0 g of N × = 1.07 mol N 14.0 g N 1 mol (NH 4 ) 3 PO 4 1.07 mol N × = 0.357 mol (NH 4 ) 3 PO 4 3 mol N 68

Interpretations of Chemical Formulas Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N? molar mass of (NH4 )3 PO 4 = 149.0 g/mol

1 mol N ? mol N = 15.0 g of N × = 1.07 mol N 14.0 g N 1 mol (NH4 )3 PO 4 1.07 mol N × = 0.357 mol (NH4 )3 PO 4 3 mol N 149.0 g (NH4 )3 PO 4 0.357 mol (NH4 )3 PO 4 × = 53.2 g (NH4 )3 PO4 1 mol (NH4 )3 PO 4 69

Purity of Samples The percent purity of a sample of a substance is always represented as mass of pure substance % purity = ×100% mass of sample mass of sample includes impurities

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Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 98.3 g Na 3 PO 4 unit factor 100.0 g sample

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Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 98.3 g Na 3 PO 4 unit factor 100.0 g sample 98.3 g Na 3 PO 4 ? g Na 3 PO 4 = 250.0 g sample × 100.0 g sample = 246 g Na 3 PO 4

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Purity of Samples Example 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4? 98.3 g Na 3 PO 4 unit factor

100.0 g sample 98.3 g Na 3PO 4 ? g Na 3PO 4 = 250.0 g sample × 100.0 g sample =246 g Na 3 PO 4 ? g impurities = 250.0 g sample - 246 g Na 3 PO 4 = 4.00 g impurities

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Synthesis Problem In 1986, Bednorz and Muller succeeded in making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La2CuO4 which superconducts at 35K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3.56 mg of La2CuO4?

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Synthesis Problem molar mass of La 2 CuO 4 = 405.3 g/mol  1g   ( 3.56 mg La 2CuO 4 ) ×   1000 mg   1 mol La 2 CuO 4  −6   = 8 . 78 × 10 mol La 2 CuO 4  405.3 g La CuO  2 4  

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Synthesis Problem molar mass of La 2 CuO

4

= 405.3 g/mol

 1g   ( 3.56 mg La 2 CuO 4 ) ×  1000 mg   1 mol La 2 CuO 4  −6   = 8 . 78 × 10 mol La 2 CuO  405.3 g La CuO  2 4   (8.78 ×10

−6

4

 6.022 ×10 23 molecules La 2 CuO mol La 2 CuO 4 )  1 mol La 2 CuO 4 

 2 La atoms   molecule La CuO 2 

4

4

 19  = 1 . 06 × 10 La atoms  

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  × 

Group Activity Within a year after Bednorz and Muller’s initial discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa2Cu3O7, that began to superconduct at 100 K. If we wished to make 1.00 pound of YBa2Cu3O7, how many grams of yttrium must we buy?

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End of Chapter 2 The mole concept and basic stoichiometry ideas introduced in this chapter are essential components for the remainder of this course.

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