Chemical Formulas and Composition Stoichiometry 化學計量法

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22Chemical Formulas and

Composition Stoichiometry化學計量法

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Chapter GoalsChapter Goals Chemical Formulas 化學式 Ions and Ionic Compounds 離子及離子化合物 Names and Formulas of Some Ionic Compounds Atomic Weights 原子重 The Mole 莫耳數 Formula Weights, Molecular Weights, and Moles Percent Composition and Formulas of Compounds Derivation of Formulas from Elemental Composition Determination of Molecular Formulas 分子式 Some Other Interpretations of Chemical Formulas Purity of Samples

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Chemical Formulas Chemical Formulas 化學式化學式• Chemical formula shows the chemical

composition of the substance.– ratio of the elements present in the molecule or

compound• He, Ne, Na – monatomic elements• O2, H2, Cl2 – diatomic elements• O3, S4, P8 - more complex elements• H2O, C12H22O11 – compounds

• Substance consists of two or more elements

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Chemical FormulasChemical FormulasAllotropic modifications (allotropes) 同素異形物 :different forms of the same element in the same physical state (chapter 13)

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Chemical FormulasChemical Formulas

2 H atoms 1 O atom1 H atom 1 Cl atom

1 N atom 3 H atoms

3 C atoms 8 H atoms

4 C atoms 10 H atoms 1 O atom

Organic compounds: contain C―C or C―H bonds or both, often in combination with nitrogen, oxygen, sulfur and other elements

Inorganic compounds: do not contain C―H bonds

丙烷乙醚

• Compound contain two or more elements in chemical combination in fixed proportions

-Law of Definite Proportions 定組成定律

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Fig 2-1 Formula and models for some molecules

( 比例模型 ) ( 球棍模型 )

四氯化碳

Chemical formula the number of atoms of each

type in the moleculeStructural Formula the order in which atoms are

connected(chemical bonds between atoms) Ball-and-Stick Model three-dimensional shape of

molecules Space-Filling Model Relative size of atoms and the

shapes of molecules

Example 2-1 Chemical FormulaLook at each of following molecular models. For each one, write the structural formula and the chemical formula. (color code: black=carbon; white=hydrogen; red=oxygen; blue=nitrogen; light green=fluorine; dark green=chloride.)a)1-butanol 丁醇

b)Freon-12 二氯二氟代甲烷 ( 氟氯烷 )

c)Nitrogen mustard HN1

Occurs in some fruits, dried beans, cheese, and nuts; used

as an additive in certain plastics, detergents, and some

medicinal formulations

Formerly used as a refrigerant; implicated in

atmospheric ozone depletion

A highly toxic substance, used as a chemotherapy drug in the treatment of Hodgkin’s disease and of some forms of chronic

leukemia

C O HHH

HCH

HCH

HCH

HC4H10O

ClF CCl

FCF2Cl2

C O ClClH

HCH

HNH

HCH

H

HC4H9NCl2

C H O N F Cl

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Ions and Ionic CompoundsIons and Ionic Compounds

Fig 2-2 The arrangement of ions in NaCl

Formula Unit: the small repeat of a substance- for non-ionic substances, the molecule NaCl, CaCl2

•Ions are atoms or groups of atoms that possess an electric charge•Two basic types of ions:

Positive ions or cations 正離子one or more electrons less than neutralNa+, Ca2+, Al3+

NH4+ - polyatomic cation

Negative ions or anions 負離子one or more electrons more than neutralF-, O2-, N3-

SO42-, PO4

3- - polyatomic anions

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Ions and Ionic CompoundsIons and Ionic Compounds

(NH4)2SO4

Polyatomic compound

Polyatomic ion

Metals form more than one kind of ion

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Names and Formulas of Names and Formulas of Some Ionic CompoundsSome Ionic Compounds

• Formulas of ionic compounds are determined by the charges of the ionsCharge on the cations = the charge on the anions

(add to zero)– The compound must be neutral

NaCl sodium chloride (Na1+ & Cl1-) KOH potassium hydroxide (K1+ & OH1-) CaSO4 calcium sulfate (Ca2+ & SO4

2-) Al(OH)3 aluminum hydroxide (Al3+ & 3 OH1-)

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H+ 1+ hydrogen

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

Some examples are:H2SO4 - sulfuric acidFeBr2 - iron(II) bromideC2H5OH - ethanol

Example 2-2 Formulas for Ionic CompoundsWrite the formula of the following ionic compounds: (a) sodium

fluoride, (b) calcium fluoride, (c) iron(II) sulfate, (d) zinc phosphate

(b) Ca2+ F-

CaF2

(a) Na+ F-

NaF(c) Fe2+ SO4

2-

FeSO4

(d) Zn2+ PO43-

Zn3(PO4)2

Example 2-3 Name for Ionic CompoundsName the following ionic compounds: (a) (NH4)2S, (b)

Cu(NO3)2, (c) ZnCl2, (d) Fe2(CO3)3

(b) Cu2+ NO3-

copper(II) nitrate

(a) NH4+ S2-

ammonium sulfide(c) Zn2+ Cl-

Znic Chloride

(d) Fe3+ CO32-

iron(III) carbonate

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You do it!• What is the formula of nitric acid? HNO3• What is the name of FeBr3? iron(III) bromide• What is the name of K2SO3? potassium sulfite • What is charge on sulfite ion? SO3

2- is sulfite ion

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You do it!• What is the formula of ammonium sulfide? (NH4)2S • 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 SO4

2-

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Atomic Weights Atomic Weights 原子量原子量Atomic weights (AW)– an early observation was that

carbon and hydrogen have relative atomic masses, of approximately 12 and 1

Atomic mass unit (amu)– exactly 1/12 of the mass of a

particular kind of carbon atom, called carbon-12

H Hydrogen 1.00794 amu Na Sodium 22.989768 amu Mg Magnesium 24.305 amu

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The MoleThe Mole 莫耳數莫耳數

Molar mass (g/mol) 莫耳質量 the mass of 1 mole of atoms of a pure element in grams =

the atomic weight of the element in atomic mass unit 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

1 mole = 6.022 x 1023 particlesAvogadro’s number (NA) = 6.022 x 1023 亞佛加厥常數Helium exists as discrete He atom 1 mole of He consist of 6.022x1023 He atoms Hydrogen commonly exists as diatomic (two-atom) H2 1 mole of Hydrogen is 6.022x1023 H2 molecule (2x(6.022x1023) H atoms)

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The Mole The Mole

C 12Ti 47.9

Au 197H 1

S 32

Table 2-3, p. 57

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One mole of atomsThe Mole The Mole

Fig. 2-4, p. 57

Bromine (79.9g)溴 Aluminum (27.0g) 鋁Mercury (200.6g)水銀

Copper (63.5g)銅

Sulfur (32.1g)硫Znic (65.4g)鋅 iron (55.8g) 鐵

Example 2-4; 2-5 Moles of atoms; Numbers of atomsHow many moles of atoms does 136.9g of iron metal

contain? And How many atoms?

6.022x1023 atoms 1 mole atoms and

6.022x1023 atoms 1 mole atoms

? Fe atoms= 2.451 mole atoms x 6.022x1023 atoms 1 mole atoms

=1.476x1024 Fe atoms

55.85g Fe 1 mole atoms55.85g Fe

1 mole atoms

? Mole Fe atoms=136.9g Fe x 55.85g Fe 1mole Fe atoms

=2.451 mol Fe atoms

Exercise 32, 40 and 42

Example 2-6 Masses of AtomsCalculate the average mass of one iron atom in

grams.? g Fe Fe atom =

55.85g Fe 1 mole atomsx 6.022x1023 Fe atoms

1 mole Fe atoms

=9.274x10-23 g (the average mass of 1 Fe atom)

Example 2-7 Avogadro’s NumberA stack of 500 sheets of typing paper is 1.9 inches thick.

Calculate the thickness, in inches and in miles, of a stack of typing paper that contains one mole of sheets.

? In.= 1 mole atoms sheets x 6.022x1023 atoms 1 mole sheets

=2.3x1021 in.x 500 sheets

1.9 in.

? mi.= 2.3x1021 in. x 1ft 12 in.=3.6x1016 mi.

x 5,280ft 1 mi.

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The MoleThe MoleExample 2-1 Calculate the mass of a single Mg

atom in grams to 3 significant figures.

? g Mg = x6.022x1023 Mg atoms 1 mole Mg atoms1 Mg atom x 1mole Mg atoms

24.3g Mg

= 4.04 x10-23 g Mg

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The MoleThe MoleExample 2-2 Calculate the number of atoms in

one-millionth of a gram of Mg to 3 significant figures.

? Mg atoms = 6.022x1023 Mg 1 mole Mg1.00x10-6g Mg x x1mole Mg

24.3g Mg

= 2.48 x10-16 Mg atoms

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The MoleThe MoleExample 2-3 How many atoms are contained in 1.67

moles of Mg?

? Mg atoms = 6.022x1023 Mg atoms 1 mol Mg1.67 mol Mg x

= 1.0 x1024 Mg atoms

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The MoleThe MoleExample 2-4 How many moles of Mg atoms are

present in 73.4 g of Mg?? mol Mg = 73.4g Mg x 1mole Mg

24.3g Mg

= 3.02 mol Mg

IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS

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Formula Weights, Molecular Formula Weights, Molecular Weights, and Moles Weights, and Moles 式量式量 ,, 分子分子量及莫耳數量及莫耳數

• How do we calculate the molar mass of a compound?– add atomic weights of each atomFormula Weight (FW)

• The molar mass of propane ( 丙烷 ), C3H8, is:3 x C = 3 x 12.01 amu = 36.03 amu8 x H = 8 x 1.01 amu = 8.08 amuMolar Mass = 44.11 amu

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Formula Weights, Molecular Formula Weights, Molecular Weights, and MolesWeights, and Moles

• The molar mass of calcium nitrate, Ca(NO3)2 , is:You do it!You do it!

1 x Ca = 1 x 40.08 amu = 40.08 amu 2 x N = 2 x 14.01 amu = 28.02 amu6 x O = 6 x 16.00 amu = 96.00 amuMolar Mass = 164.10 amu

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Molar mass of the substance = the formula weight of the substance

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• One Mole of Contains– Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms– C3H8 You do it!You do it!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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Example 2-9 Masses of MoleculesWhat is the mass in grams of 10.0 million SO2 molecules?

? g SO2 = 1.0x106 SO2 molecules x 6.022x1023 SO2 molecule 64.1g SO2

=1.06x10-15 g SO2 Exercise 44

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Example 2-10 MolesHow many (a) moles of O2 (b) O2 molecules, and (c) O atoms are contained

in 40g of oxygen gas (dioxygen) at 25oC?

? Mol O2 = 40.0g O2 x 32.0g O2

1 mol O2 = 1.25 mol O2(a)

One mole of O2 contains 6x1023 O2 molecules, and its mass is 32.0g

? O atoms = 40.0g O2 x 6.02x1023 O2 molecules 32.0g O2

=1.5x1024 O atomsExercise 36

(c)

? O2 molecules =1.25 mol O2 x 6.02x1023 O2 molecules 1mol O2

(b) = 7.52x1023 O2 molecules

x 1 O2 molecule 2 O atoms


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Example 2-11 Numbers of atomsCalculate the number of hydrogen atoms in 39.6g of

ammonium sulfate, (NH4)2SO4

One mole of (NH4)2SO4 is 6x1023 formula units and has a mass of 132.1g

? H atoms = 39.6g (NH4)2SO4 x

=1.44x1024 H atoms Exercise 34

x 1 formula units (NH4)2SO4

8 H atoms

g of (NH4)2SO4

mol of (NH4)2SO4

Formula units of (NH4)2SO4

H atoms

1 mol (NH4)2SO4

132.1g (NH4)2SO4

6.02x1023 formula units (NH4)2SO4

1mol (NH4)2SO4

x


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Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane

? C3H8 molecules =

x 6.022x1023 C3H8 molecules 1 mole C3H8

74.6g C3H8 x1mole C3H844.11g C3H8

= 1.02 x1024 molecules

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Example 2-6. What is the mass of 10.0 billion propane molecules?

?g C3H8 molecules =6.022x1023 C3H8 molecules

1 mole C3H81.00x1010 molecules x

= 7.32 x10-13 g of C3H8

1mole C3H8x 44.11g C3H8

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Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

? moles O3 = 60.0g O3 x = 1.25 moles48.0 gO3

1mole(a)

? molecules O3 = 1.25 moles x= 7.53 x1023 molecules O3

1 mole6.022x1023 molecules(b)

? O atoms = 7.53x1023 molecules O3 x= 2.26 x1024 atoms O

1 O3 molecule3O atoms(c)

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Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.

? O atoms =

x

= 6.49 x1023 O atoms1 form. unit Li2CO3

3 O atoms

26.5g Li2CO3 x 73.8g Li2CO3 1 mol Li2CO3

1 mole Li2CO3

6.022x1023 form. units Li2CO3 x

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• Occasionally, we will use millimoles – Symbol - mmol– 1000 mmol = 1 mol 1 mmol = 10-3 mole

• For example: oxalic acid 草酸 (COOH)2 – 1 mol = 90.04 g – 1 mmol = 0.09004 g or 90.04 mg

milli10-3

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Example 2-9 Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.

? Mmol (COOH)2 = 0.234g (COOH)2 x1mmole (COOH)2 0.09004g (COOH)2

= 2.6mmol (COOH)2

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Percent Composition and Percent Composition and Formulas of CompoundsFormulas 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 C3H8.

% C = x 100%Mass C Mass C3H8

= 3x12.01g 3x12.01+8x1.01g = 81.68%x 100%

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• What is the percent composition of H in C3H8?

% H = x 100%Mass H Mass C3H8

= 8x1.01g 3x12.01+8x1.01g = 18.32%x 100%

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Example 2-12 Percent CompositionCalculate the percent composition by mass of HNO3

Exercise 62

1xH=1x1.0g=1.0g 1xN=1x14.0g=14.0g 3xO=3x16g=48g Mass of 1 mol of HNO3= 1+14.0+48.0 = 63.0g

% H = x 100%Mass C Mass HNO3

= 1.6% H= 1.0g 63.0g x 100%

% N = x 100%Mass N Mass HNO3

= 22.2% N= 14.0g 63.0g x 100%

% O = x 100%Mass O Mass HNO3

= 76.2% O= 48.0g 63.0g x 100%

Total =100.0%

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Example 2-10 Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.

% Fe = x 100%2Fe Fe2(SO4)3

= 2x55.8g 2x55.8+3x32.1+12x16.0 g

= 27.9%

x 100%

% S = x 100%3S Fe2(SO4)3

= 2x32.1g 399.9g

= 24.1%x 100%

= 48.0%% O = x 100%12O Fe2(SO4)3

= 12x16g 399.9g x 100%

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Derivation of Formulas from Derivation of Formulas from Elemental CompositionElemental Composition

• Empirical Formula ( 實驗式 ) - 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 (CH2)2–Pentene – C5H10 (CH2)5

• 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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Determination of Molecular Determination of Molecular FormulasFormulas

• Butane C4H10 Empirical formula C2H5 2x (C2H5) = C4H10 Molecular Formula • Benzene C6H6 Empirical formula CH 6x (CH) = C6H6

Molecular formula = n x simplest formulaMolecular weight = n x simplest formula weight

n= Molecular weightSimplest formula weight

Example 2-13 Simplest FormulasCompounds of containing sulfur and oxygen are serious air

pollutants; they represent the major cause of acid rain. Analysis of a simple of a pure compound reveals that it contains 50.1% sulfur and 49.9% oxygen by mass. What is the simplest formula of the compound?


Exercise 54

? mol S atoms = 50.1g S x 1 mol S atoms 32.1g S = 1.56 mol S atoms

? mol O atoms = 49.9g O x 1 mol O atoms 16.0g O

= 3.12 mol O atoms

S mol : O mol = 1.56 mol S atoms : 3.12 mol O atoms =1:2 SO2

Example 2-14 Simplest FormulasA 20.882g sample of an ionic compound is found to

contain 6.072g of Na, 8.474g of S, and 6.336g of O. What is its simplest formula?


Exercise 56

? mol Na = 6.072g Na x 1 mol Na23.0g Na = 0.264 mol Na

Na mol: S mol : O mol = 0.264 : 0.264 : 0.396 =2 : 2 : 3 Na2S2O3

? mol S = 8.474g S x 1 mol S32.1g S = 0.264 mol S

? mol O = 6.336g O x 1 mol O16.0g O = 0.396 mol O

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Example 2-11 A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?

? mol K = 24.74g K x 1 mol K39.10g K = 0.6327 mol K

K mol: Mn mol : O mol = 0.6327 : 0.6327 : 2.531 =1 : 1 : 4 KMnO4

? mol Mn = 34.76g Mn x 1 mol Mn32.1g Mn = 0.6327 mol Mn


Let the compound weight 100g

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Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?? mol Co = 6.541g Co x

1 mol Co58.93g Co = 0.111 mol Co

Co mol: O mol = 0.111 : 0.148 = 3 : 4 Co3O4


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A combustion train used for carbon-hydrogen analysis 燃燒反應器

Magnesium perchlorate 過氯酸鎂 Mg(ClO4)2 H

Sodium hydroxid C

O??

Example 2-15 Percent CompositionHydrocarbons are organic compounds composed

entirely of hydrogen and carbon. A 0.1647g sample of pure hydrocarbon was burned in a C-H combustion train to produce 0.4931g of CO2 and 0.2691g of H2O. Determine the masses of C and H in the sample and the percentage of these elements in this hydrocarbon.

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Exercise 66

? g C = 0.4931g C x 12.01g C44.01g CO2

= 0.1346 g C

? g H = 0.2691g H x 1.01g H18.1g H2O

= 0.03010g H

? % C = 0.1346g C0.1647g sample

= 81.72% C x 100%

? % H = 0.03010g H0.1647g sample = 18.28% H x 100%

Example 2-16 Percent CompositionA 0.1014g sample of purified glucose was burned in a C-

H combustion train to produce 0.1486g of CO2 and 0.0609g of H2O. An elemental analysis showed that glucose contains only carbon, hydrogen and oxygen. Determine the masses of C,H, and O in the sample and the percentage of these elements in glucose.

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Exercise 68

? g C = 0.1486g C x 12.01g C44.01g CO2

= 0.04055g C

? g H = 0.0609g H x 1.01g H18.1g H2O

= 0.00681g H

? % C = 0.04055g C0.1014g sample = 39.99% C x 100%

? % H = 0.00681g H0.1014g sample = 6.72% H x 100%

? g O = 0.1014 – [0.04055g C + 0.00681g H] = 0.0540g O

? % O = 0.0540 g O0.1014g sample = 53.2% O x 100%

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Example 2-17 Molecular FormulaIn example 2-16, we found the elemental composition of

glucose. Other experiments show that its molecular weight is approximately 180 amu. Determine the simplest formula and the molecule formula.

Exercise 49 and 50

C 0.04055 0.040512.01 = 0.003376 = 1.00 C0.003376

0.003376

H 0.00681 0.00681.008 = 0.00676 = 2.00 H0.00676

0.003376

O 0.0540 0.054016.00 = 0.00338 = 1.00 O0.00338

0.003376

ElementRelative Mass

Of elementRelative Number of Atoms

(divide mass by AW) Divide by Smallest

CH2O

CH2O formula weight = 30.03 amuThe molecular weight of glucose weight 180 amu

n=180/30.03=6Molecular weight = (CH2O)x6

C6H12O6

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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?

1 mol = 56.1g/mol

? mol C = 85.63g C x 1 mol C12.0g C = 7.1 mol C

? mol H = 14.37g H x 1 mol H1.01g H = 14.2 mol H

C mol: H mol = 7.1 : 14.2 = 1:2Simplest formula CH2

Molecular formula = (Simplest formula) x

56.1 = (C+2xH)xX =(12+2x1.01)xXX = 4 (CH2)4 C4H8

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Example 2-18 Law of Multiple ProportionsWhat is the ratio of the numbers of oxygen atoms

that are combined with a given number of nitrogen atoms in the compound N2O3 and NO?

Oxygen ratio = N2O3

2x(NO) = 3O/2N2O/2N = 3O

2O = 32

Exercise 71 and 72

Law of Multiple Proportions 倍比定律• When two elements, A and B, form more than one compound,

the ratio of the masses of element B that combine with a given mass of element A in each of the compound can be expressed by small whole numbers.

H2O H2O2 (1:2 oxygen ratio) SO2 SO3 (2:3 oxygen ratio)

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Example 2-19 Composition of CompoundsWhat mass of Chromium is contained in 35.8g of (NH4)2Cr2O7?

Some Other Interpretations of Some Other Interpretations of Chemical FormulasChemical Formulas

Mass (NH4)2Cr2O7 Mol (NH4)2Cr2O7 Mass Cr Mol Cr

? mol (NH4)2Cr2O7 = 35.8g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7

252.0g (NH4)2Cr2O7= 0.142 mol (NH4)2Cr2O7

? mol Cr atoms = 0.142mol (NH4)2Cr2O7 x2 mol Cr atoms

1 mol (NH4)2Cr2O7= 0.284 mol Cr atoms

? g Cr = 0.284 mol Cr atoms x 52.0g Cr1 mol Cr atoms

= 14.8g Cr Exercise 76

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Example 2-20 Composition of CompoundsWhat mass of potassium chlorate, KClO3, would contain 40.0g

of oxygen?Mol KClO3 Mass O Mol O

? g KClO3 = 40g O x

= 102g KClO3 Exercise 78

Mass KClO3

1 mol O atoms 16.0g O atoms

1 mol KClO3

3 mol O atomsx 122.6g KClO3 x 1 mol KClO3

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Example 2-21 Composition of Compounds(a) What mass of sulfur dioxide, SO2, would contain the same

mass of oxygen as is contained in 33.7g of arsenic pentoxide, As2O5?

(b) What mass of calcium chloride, CaCl2, would contain the same number of chloride ions as are contained in 48.6g of sodium chloride, NaCl?

Mol O atoms

? g SO2 = 33.7g As2O5 x

= 23.5g SO2

Exercise 78

Mol SO2 Mass SO2 Mass As2O5 Mol As2O5

Mol Cl- ions Mol CaCl2 Mass CaCl2 Mass NaCl Mol NaCl

1 mol As2O5 229.8g As2O5

5 mol O atoms x 1 Mol As2O5 1mol SO2 x2 mol O atoms

64.1g SO2 x 1mol SO2

? g CaCl2 = 48.6g NaCl x

= 46.2g CaCl2

1 mol NaCl58.4g NaCl

1 mol Cl- x 1 Mol NaCl

1mol CaCl2x 2 mol Cl-

111.0g CaCl2 x 1mol CaCl2

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Water H2O 18

Hydrated copper (II)Sulfate, (CuSO4•5H2O)1mol=249.7g

Anhydrous oxalic acid, (COOH)21mol=90g

Mercury (II) oxide1mol=216.6g

Hydrated oxalic acid, (COOH)2•2H2O)1mol=126.1g

Hydrate : with waterAnhydrous: without water

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Hydrated copper (II) Sulfate, (CuSO4•5H2O)1mol=249.7g

copper (II) Sulfate, (CuSO4)1mol=159.6g

Blue

Gray

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Some Other Some Other Interpretations of Interpretations of


Example 2-22 Composition of CompoundsA reaction requires pure anhydrous calcium sulfate, CaSO4,

Only an unidentified hydrate of calcium sulfate,CaSO4•x H2O, is available.

(a) We heat 67.5g of unknown hydrate until all the water has been driven off. The resulting mass of pure CaSO4 is 53.4g. What is the formula of the hydrate, and what is the formula weight?

(b) Suppose we wish to obtain enough of this hydrate to supply 95.5g of CaSO4 after heating. How many grams should we weight out?

x = = 2 ? mol H2O mol CaSO4

14.1g H2O = 53.4g CaSO4 1mol H2Ox 18.0g H2O

136.2 g CaSO4 x 1mol CaSO4

? g CaSO4 x 2H2O =

(a) ? g water driven off =67.5g CaSO4•x H2O - 53.4g CaSO4=14.1g H2O

CaSO4 •2H2O FW=136.2g/mol+2x(18.0g/mol)=172.2g/mol

95.5g CaSO4 desired x 172.2 g CaSO4•2H2 O136.2g CaSO4

=121g CaSO4•2H2 O

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Example 2-16 What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?

? mol N = 15.0g N x 1 mol H14.0g N = 1.07 mol N

Molar mass of (NH4)3PO4 = 149.0 g/mol

1.07 mol N x1 mol (NH4)3PO4

3mol N = 0.357 mole (NH4)3PO4

0.357mol (NH4)3PO4 x149.0g (NH4)3PO4

1 mol (NH4)3PO4= 53.2g (NH4)3PO4

X g (NH4)3PO415g N=149.0g (NH4)3PO4

3x14 N X =53.2g

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Purity of SamplesPurity of Samples• The percent purity of a sample of a substance is

always represented as

% purity = Mass of pure substance Mass of sample x100%

Mass of sample includes impurities

Impurity: 1.8%

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Purity of SamplesPurity of SamplesExample 2-23 Percent PurityCalculate the masses of NaOH and impurities in 45.2g of

98.2% pure NaOH.? g NaOH =

98.2 g NaOH 100g sample =44.4g NaOH45.2g sample x

? g impurities = 1.8 g impurities 100g sample =0.81g impurities45.2g sample x

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Purity of SamplesPurity of SamplesExample 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? ? g Na3PO4 =

98.3 g Na3PO4 100g sample

=246g Na3PO4250.0g sample x

? g impurities =1.8 g impurities 100g sample = 4g impurities250.0g sample x

? g impurities = 250.0g sample - 246g Na3PO4 = 4g impuritiesor

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Chemical Formulas and Composition Stoichiometry 化學計量法