NAME _____________________________ WCSU CHEM 100 UNIT 6 1 THE MOLE & STOICHIOMETRY

An Introduction to the term: MOLE

The word, mole is from the Greek for "pile or heap". It was first proposed by Amedeo Avogadro, around 1811 ... and it

was immediately, pretty well ignored. It is believed that August Hortsmann coined the term, gram-molecular weight

(which is a synonym for a mole mass) in the 1880's and Wilhelm Ostwald was first to use the term, mole in 1900. The

first numerical value for the number of entities represented by 1 mole was first calculated in 1908 by Jean Baptiste

Perrin. http://rd.springer.com/article/10.1007/s10698-011-9128-7 and http://www.iupac.org/publications/ci/2010/3201/2_lorimer.html

The term mole refers to the "quantity of substance". It has the symbol of n , and the term mole, is abbreviated as mol.(No, really, I kid you not!)

For example: You will see the symbol of “n” used in

arithmetic equations such as the Ideal Gas Law: PV = nRT

polymerization reaction equations, where “n” stands for moles of reactant as well as moles of repeating subunits such as in the polymerization of ethene: n CH2 =CH2 [-CH2-CH2-]n

the hydrocarbon general formula: CnH2n+2. (Yep! The “n” represents moles of carbon atoms in 1 mol of molecules of the compound….)

The term mole is analogous to terms such as: dozen, couple, few, pair, baker's dozen ...etc. It is just a number, but a big number.

I) Mole: Definition: the unit of amount of substance of a specified elementary entity, which may be; an atom,

molecule, ion, electron, any other particle or a specified group of such particles, such that the

Avogadro

constant is equal to exactly 6.022 141 79 x 1023 per mole. http://www.iupac.org/publications/ci/2010/3201/2_lorimer.html

A) Thus, we now come to the concept that 1 mol of a substance has a number of entities (or species) that is equivalent to Avogadro's constant (NA). For our course, the value of a mol is simplified as, 6.022 x1023 elementary species (atoms, molecules, ions, electrons...)

1) The mole at its most fundamental level is simply a number of species, which can be converted to a mass, or into a specific number of species, or into a volume of gas

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a) but the most important of these, for the first-year student, is mass.

2) The value 6.022 x1023, comes from the relationship between 1 gram and 1 atomic mass unit (µ).

a) Recall from our work on atomic structure than 1 µ is equivalent to 1.6605402 x 10-24 grams. When you ask: How many atomic mass units are equal

to 1.00 gram ….you get:

1.00 gram & this equals: 6.02 x 1023

1.66 x 10-24 gram

http://www.orbromart.com/moleday/

WHAT ANALOGIES CAN BE USED TO IMAGINE 1 MOLE OF SOMETHING? Think about Avogadro's number [NA ≅ 6.022 x 1023 ] Now … in your opinion…estimation…guess-timation….

Would 1 mol of rice grains be contained, completely in a 2000 mL beaker?

Would 1 mol of rice grains be contained, completely in our classroom?

Would 1 mol of rice grains be contained in the space of this floor of classrooms?

Would 1 mol of rice grains be contained by the entire building, assuming the use of closets, rooms…etc?

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Count Lorenzo Romano Amedeo Carlo Avogadro, di Quaregna e di Cerreto(1776 – 1856)

A PILLAR OF THE CHEMISTRY COMMUNITY

http://fr.wikipedia.org/wiki/Amedeo_Avogadro

WELL … LET’S THINK OF THE “SIZE OF 1 MOLE LIKE THIS… 1.00 mole of rice grains would fill a cube 120. miles on each edge 1.00 mole of rice grains equals more grains than all grain grown since the beginning of time 1.00 mole of rice grains would cover all the land area in the world to a depth of 75 meters OR…..

IBM announced in 1989 that its fastest computer chip could process 6,000 mips (million instructions per sec) and could count the entire population of the US in 4/100 second (0.04 seconds) but the same computer would require over 3 million years to count to Avogadro's number (6.022 x 1023 = 602,200,000,000,000,000,000,000).

In order to fit 1.0 mole of raindrops into a 30. meter diameter tank, the sides of the tank would have to be 280 times higher than the distance from the earth to the sun. (The distance from the earth to the sun is approx. 93 million miles)

If 1.0 mole of pennies were divided equally between every person on earth, each person would receive 1.2 x 1014

pennies. Personal spending, at the rate of one million dollars/day would use up each person's wealth in just under

3,000 years. However, the surface of the earth would be buried in pennies to a depth of about 420 meters.

Pretend you get a counting job, the day you learn how to count. Assume you counted by ones, eight hours per day,

five days per week for 50 weeks a year. (We'll give you 2 weeks of vacation.) If you reached five billion by the

time you retired at the age of 65, you would be considered to be a "good counter". Now, if every living person

were to count by ones from the time they learned to count, until they reached the age of 65 years old, this army

could count all of the leaves on all the trees, shrubs and bushes in the whole world in about two months.

However, all of the efforts of this same group would just about count all of the atoms of iron in the head of

a straight pin…which is nowhere near the mass of 1 mol of iron.

In order to think of 1 mole you could just go get * 18.0 grams of water. (Do you now have an idea of how small a molecule of water is?)

Really Nice: Check out: https://www.wimp.com/how-big-is-a-mole/

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II) The Mole and the Chemical Formula

A) The subscripts of a chemical formula are really used to indicate the molar ratio between the species that comprise a compound. (TAKE HOME MESSAGE: The subscripts are related to n)

1) The empirical formula indicates the simplest whole number ratio between elements

a) It may or may not express how the chemical exists in natureb) Ionic compounds are written only in their empirical formula

i) the ionic compound is not discrete, and we do not know with surety which cation is bonded to which anion … but we are aware of the ratio.

ii) Hence ionic compounds are written in the empirical formula and are considered to exist as formula units.

http://www.personal.kent.edu/~cearley/ChemWrld/compounds/ions.htm

c) 1 mol of NaCl formula units = 1 mol of Na1+ ions and 1 mol Cl1- ions

1 mol of CaBr2 formula units = 1 mol of Ca2+ ions and 2 mol of Br-1 ions

1 mol of Fe2(SO4)3 formula units = 2 mol Fe3+ ions and 3 mol (SO4)2- ions or

2 mol Fe3+ ions 3 mol S atoms and 12 mol O atoms

3) The molecular formula represents a discrete species, and expresses the actual molecule But the subscripts continue to represent the moles of species in 1 mol of substance.

a) 1 mol CH4 molecules = 1 mol of C atoms and 4 mol of H atoms

1 mol C8H14 molecules = 8 mol of C atoms and 14 mol of H atoms

1 mol SO2 molecules = 1 mol of S atoms and 2 mol of O atoms

1 mol CH3OH molecules= 1 mol of C atoms, 4 mol of H atoms & 1 mol of O atoms

1 mol CH3CH2CH2CH2COOH = 5 mol C atoms, 10 mol H atoms & 2 mol of O atoms

Take Home Message: The subscript of a formula is essentially the number of moles of each species in 1 mol of the substance.

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Making the connection:

Be very aware of what you are being asked. You may be asked for the number of moles of species or total number of species.

When asked for the total number of moles of species in 1 mol of substance, just add up the subscripts.

When asked for the total number of species in 1 mol of a substance, multiply the sum of the subscripts, by 6.022 x 1023

When stuck interpreting this: ask yourself 2 very different questions:

Given 1 dozen CO2 … How many dozen O atom do you have? How many O atoms do you have?

Formula Total Moles ofAtoms or Ions per 1 mol

Total Number of Species

e.g.) In 1 mole of CO2 there are 3 mol of atoms or 1.806(6) x 1024 atoms

CH3CH2CH2CH2COOH 17 1.023(4) x 1025

Na2O 3 1.806(6) x 1024

Li2CO3 6 3.612 x 1024

C4H10 14 8.428 x 1024

B) The coefficient of a substance applies to the entire substance … hence ….

In 5 moles of CO2 molecules [or 5 CO2] …. 5 mol of C atoms and 10 mol of O atoms

Given 8 moles of Al2O3 …. or 8 Al2O3 ….. 16 mol Al3+ ions and 24 mol O2- ions

Given 3 mol of O2 molecules or 6 O2 …. 12 mol of O atoms

1) This is particularly important when reading a balanced chemical reaction.

TRY THIS….each question is essentially the same … Each asks for the number of molecules given a number of moles. Remember, 1 mol of anything equals 6.022 x 1023

1. How many molecules are in 1 mol of C12H22O11 ?

1) 6.022 x 1023 3) 9.033 x 1023

2) 3.011 x 1023 4) 1.204 x 1024

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2. How many molecules are in 2 mol of C12H22O11?

1) 6.022 x 1023 3) 9.033 x 1023

2) 3.011 x 1023 4) 1.204 x 1024

3. How many molecules of C12H22O11 are in 0.5000 mol of the compound?

1) 6.022 x 1023 3) 9.033 x 1023

2) 3.011 x 1023 4) 1.204 x 1024

4. 9.033 x 1023 molecules are the equivalent of what number of moles?

1) 1.00 3) 1.502) 0.500 4) 2

TRY THIS….each question is a bit different. Read very critically to ascertain the “guts” of each question.

5. How many moles of atoms are in 1 mol of C12H22O11 molecules?

1) 45 3) 2.709 x 1025

2) 22.5 4) 1.355 x 1025

6. Calculate the total number of atoms equal to 1 mol of C12H22O11 molecules

1) 45 3) 2.709 x 1025

2) 22.5 4) 1.355 x 1025

7. How many moles of atoms are in 0.500 mol of C12H22O11 molecules?

1) 45 3) 2.709 x 1025

2) 22.5 4) 1.355 x 1025

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ans : 1) 1 2) 4 3) 2 4) 3

ans : 5) 1 2) 3 3) 2

III) Balancing Equations by Trial and Error (a.k.a. Balancing by Inspection)

A) Predicated upon the Law of the Conservation of Matter

Take Home Message: The number of moles of each element (or species) on the

reactant side

must equal the number of moles of the same element (species) on the product side.

1) During a chemical reaction, atoms can't just appear or disappear. We must account for every atom or ion.

a) Balancing an equation gives you the * mole ratio . You may change the

coefficients BUT, you may NEVER change * a subscript, in order to balance

2) Coefficients (as a rule) will be written in the simplest, whole number ratio

3) Please note: Fractional coefficients are sometimes used in more advanced work … We will stick to the whole number coefficient)

For instance: Some advanced course work will see the balanced chemical written as:

CH3OH + 3 O3 → CO2 + 2 H2O + 726 kJ 2

We will balance and work with equations of whole-number coefficients. So let’s double those coefficients…

(This approach impacts your Gen Chem work in an area called Kinetics)

2 CH3OH + 3 O3 → 2 CO2 + 4 H2O + 1452 kJ

B) PROCESS BALANCING BY INSPECTION: For each and every species....

On the reactant side: multiply: (coefficient)(by all appropriate subscripts for the species)

Assume a coefficient of 1 when there is none written Assume a subscript of 1 when there is none written

On the product side: multiply: (coefficient )(by all appropriate subscripts species)

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Notice how the change in enthalpy doubles when the coefficients were doubled.

Compare /Adjust the coefficients so that the results of the two multiplications are equal.

1) First year students often have 1 major misconception … they believe that the coefficients of the reactant side and the coefficients of the product side need to add up to each other. They are WRONG!! The number of moles of each element on the reactant side must be the same as the number of moles of that element on the product side … but their coefficients may be different.

2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 CO2(g)

for example: note that 2 +3 = 5 and that 4 + 3 = 7 …The sum of the coefficients is NOT the same but matter is conserved…Analyze the

moles of atom & ions on both sides. They are the same number on both sides

2) Helpful Hints for balancing by trial and error: These are just hints ☺

These first three suggestions are

Balance metal atoms or metal ionsinterchangeable as a starting point.

Balance nonmetal atoms or ions (but leave O until the very end)

You can balance polyatomic ions which are INTACT on both sides of the reaction arrow against each other (as 1 entity), if you choose.

Use the H2O MOH hint when possible

With combustion reactions, or those with O2 as a reactant, think about using the fraction rule

Sometimes a reaction is balanced already, so be prepared, to leave every coefficient as "1".

C) Straight-Up Balancing By Inspection Examples:

1) _____ Al(s) + _____ Cl2(g) _____ AlCl3(s)

2) _____ PbCl2(aq) + _____ KI(aq) _____ KCl(aq) + _____ PbI2(s)

3) _____ C2H4(g) + _____ O2(g) _____ CO2(g) + _____ H2O(l)

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4) _____ Na(s) + _____ Cl2(g) _____ NaCl(s)

5) _____ HgO(s) _____ Hg(l) + _____ O2(g)

6) _____ O2(g) ______ O3(g)

7) _____Al(s) + _____ O2(g) _____ Al2O3(s)

8) _____ KClO3(s) _____ O2(g) + _____ KCl(s)

9) _____ C6H12(l) + _____ O2(g) _____ CO2(g) + _____ H2O(g)

D) Simple Examples with Polyatomic Ions ....If it helps, put ( ) around the polyatomic ion if none exist.

REVIEW: Polyatomic ion: A chemically unstable group of bonded atoms, which end up with an unequal number of protons and electrons.

Since most PAI are made of bonded nonmetal atoms, the bonding that holds the PAI together, is often covalent bonds. The charge is shared by the entire polyatomic ion (sort of like a like a group of passengers sharing a taxi … the taxi is like the charge)

A polyatomic ion tends to act chemically like a single species .

1) Emphasize that last statement … a PAI is a set of atoms which acts as a single species.

2) Lastly, most inorganic compounds that contain a polyatomic ion are ALSO classified as ionic compounds. The whole compound exists due to ionic bonding, yet it also contains some covalent bonds because of the polyatomic ion’s structure.

Bonding is due to opposite charge attraction

Thus: NaNO3(s) Is an ionic compound. Na+1 NO3-1

But this is put together with covalent bonds

Remember: Just because something is an ION does NOT MEAN it is IONIC

Hence: NO3-1 is an ion... but made via covalent bonds and is not ionic, itself. NO3

-1 is found though in many ionic compounds ... such as NaNO3, Pb(NO3)2, Fe(NO3)3

http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/Ions/ions.html

Cyanide PAI Bicarbonate PAI

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e.g. On a test, you are given: ___Cu + ___AgNO3 ___Cu(NO3)2 + ___ Ag

you might find it helpful to put ( ): ___Cu + ___ Ag(NO3) ___Cu(NO3)2 + ___ Ag How many are represented here? Note, how many of the polyatomic ion are represented?

1) _____Al(s) + _____ Cu(NO3)2(aq) _____ Al(NO3)3(aq) + _____ Cu(s)

2) ____Al2(SO4)3(aq) + ____ Ca(OH)2(aq) ____ CaSO4(aq) + ____Al(OH)3(s)

3) _____Al(s) + _____H2SO4(aq) _____H2(g) + _____ Al2(SO4)3(aq)

4) ____Ba(NO3)2(aq) + ____K2CrO4(aq) ____BaCrO4(s) + ____ KNO3(aq)

5) _____ Ca(s) + _____ HNO3(aq) _____ Ca(NO3)2(aq) + _____H2(g)

6) _____ Ca(s) + _____ Zn(OH)2(aq) _____ Ca(OH)2(s) + _____ Zn(s)

7) _____ Cu(s) + _____ AgNO3(aq) _____ Ag(s) + _____ Cu(NO3)2(aq)

8) _____ Pb(NO3)2(aq) + _____ K3PO4(aq) _____ Pb3(PO4)2(s) + _____ KNO3(aq)

9) _____BaBr2(aq) + _____Al2(S2O3)3(aq) _____BaS2O3(s) + _____AlBr3(aq)

10) ____K3PO4(aq)  +  ____MgCl2(aq) ____Mg3(PO4)2(s) +  ____ KCl(aq)

11) _____Pb(NO3)2(aq) + _____H3AsO4(aq) _____PbHAsO4(s) + _____ HNO3(aq)

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E) "Fraction Rule" Use a decimalized value (or fraction) to balance the reactant O2 in an equation.

LOOK FOR: O2 as a reactant & use a decimalized coefficient (e.g. 1.5 , 4.5) to balance the O2

You don't need to use this EVERY time there is O2… It's just a strategy for your bag of tricks...

1) _____ C4H6(g) + _____ O2(g) _____ CO2(g) + _____ H2O(l)

2) _____ C5H10(g) + _____ O2(g) _____ CO2(g) + _____ H2O(l)

3) _____ C6H6(g) + _____ O2(g) _____ CO2(g) + _____ H2O(l)

4) _____ C6H14(g) + _____ O2(g) _____ CO2(g) + _____ H2O(l)

5) _____ C7H14 + _____ O2 _____ CO2 + _____ H2O

6) _____ C8H14 + _____ O2 _____ CO2 + _____ H2O

7) _____ C8H18 + _____ O2 _____ CO2 + _____ H2O

8) _____ FeS2(s) + _____ O2(g) _____ Fe2O3(s) + _____ SO2(g)

9) _____ NH3(g) + _____ O2(g) _____ NO (g) + _____ H2O(l)

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F) H2O MOH Hint: When balancing an equation with H2O on one side of the reaction arrow and a metal hydroxide on the other, re-write the water as H(OH) and balance the lone hydrogen (H) separately from the (OH)

metal hydroxide : * metal cation (hydroxide polyatomic ion) e.g. NaOH, Fe(OH)2

LOOK FOR : * a metal hydroxide on one side and water on the other side

1) _____ Na(s) + _____ H2O(l) _____ NaOH(aq) + _____ H2(g)

2) _____Li(s) + _____ H2O(l) _____ LiOH(aq) + _____ H2(g)

3) _____ HNO3(aq) + _____ Mg(OH)2(aq) _____ H2O(l) + _____Mg(NO3)2(aq)

4) ____ NH4Cl(s) + ____ Ca(OH)2(s) ____CaCl2(s) + ____H2O(l) + ____ NH3(g)

5) ____NH4Br(aq) + ____KOH(aq) ____H2O(l) + ____ NH3(g) + _____KBr(aq)

6) _____ Al(s) + _____ H2O(l) _____ Al(OH)3(aq) + ______ H2(g)

7) _____ LiOH (aq) + _____ H2S(g) _____ Li2S(s) + _____H2O(g)

TRY THIS: BALANCING EQUATIONS 274

Directions: Begin to push yourself to remember the rules, procedures, processes, reference tables, facts…etc.

1. When balanced using the simplest whole number ratio, what is the molar ratio between carbon and silicon?

_____SiO2(s) + _____C(s) _____Si(s) + _____CO(g)

a) 4:1 b) 1:2 c) 2:1 d) 5:3

2. When balanced, using the simplest whole number ratio, what is the coefficient of zinc metal?

_____Zn(s) + _____HCl(aq) _____ZnCl2(aq) + _____H2(g)

a) 1 b) 2 c) 3 d) 4

3. When balanced, using the simplest whole number ratio what is the coefficient in front of oxygen gas?

_____PH3(g) + _____O2(g) _____P4O10(g) + ____ H2O(l)

a) 8 b) 6 c) 16 d) 4

4. When balanced using the simplest whole number ratio, what is the coefficient of NaCl?

_____Na2Cr2O7(aq) + _____AlCl3(aq) _____NaCl(aq) + _____Al2(Cr2O7)3(aq)

a) 1 b) 12 c) 3 d) 6

5. When balanced using the simplest whole number ratio, what are the molar ratios between the substances?

____C2H3Cl(g) + ____O2(g) ____CO2(g) + ____H2O(l) + ____HCl(g)

a) 2:5:4:2:2 b) 1:3:4:2:6 c) 4:10:3:4:1 d) 1:1:3:4:8

6. When balanced using the simplest whole number ratio what is the coefficient in front of H2O? (Think about one of the hints..)

_____Al4C3(s) + _____H2O(l) _____CH4(g) + _____Al(OH)3(g)

a) 24 b) 2 c) 3 d) 12

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7. When balanced using the simplest whole number ratio, what is the coefficient in front of water?

_____HNO3(aq) + _____Mg(OH)2(aq) _____H2O(l) + _____Mg(NO3)2(aq)

a) 1 b) 2 c) 3 d) 4

8. When balanced using the simplest whole number ratio, what is the coefficient for barium bromide?

_____BaBr2(aq) + _____Al2(S2O3)3(aq) _____BaS2O3(aq) + ____AlBr3(aq)

a) 1 b) 7 c) 3 d) 6

9. When balanced using the simplest whole number ratio, what is the number of moles of KNO3 produced?

_____Pb(NO3)2(aq) + _____K3PO4(aq) _____Pb3(PO4)2(aq) + _____KNO3(aq)

a) 18 b) 2 c) 6 d) 7

10 When balanced using the simplest whole number ratio, what is the coefficient for carbon dioxide?

_____C5H10(g) + _____O2(g) ______CO2(g) + ______H2O(l)

a) 10 b) 2 c) 15 d) 5

11 When balanced using the simplest whole number ratio, what is the coefficient of oxygen gas?

_____C5H12(g) + _____O2(g) ______CO2(g) + ______H2O(l)

a) 6 b) 5 c) 10 d) 8

ANSWERS:

1. c 2. a remember, you can assume a coefficient of 1, if there is no other coefficient written in front of a symbol3. a4. d5. a The molar ratios are represented by the coefficients of the balanced equation : this requires the use of the "fraction" rule and then the doubling of the coefficients.6. d Did you use the “Water-Hydroxide Hint“ ? It would help7. b Did you use the “Water-Hydroxide Hint “8. c 9. c10 a11 d

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MORE BALANCING EQUATIONS (only if you wish for more practice…)

1) C6H6(g) + O2(g) CO2(g) + H2O(l)

2) C2H6(g) + O2(g) CO2(g) + H2O(l)

3) C3H8(g) + O2(g) CO2(g) + H2O(l)

CHALLENGE (a minor one)

4) CH3OH(g) + O2(g) CO2(g) + H2O(l)

5) FeS2(s) + O2(g) Fe2O3(s) + SO2(g)

CHALLENGE

6) Ca3(PO4)2(s) + SiO2(s) + C(s) P4(g) + CaSiO3(l) + CO(g)

7) IBr(g) + NH3(g) NI3(s) + NH4Br(s)

8) BCl3(s) + P4(s) + H2(g) BP(s) + HCl(g)

9) Al(s) + H2SO4(aq) Al2(SO4)3(aq) + H2(g)

10) Al(s) + Cu(NO3)2(aq) Al(NO3)3(aq) + Cu(s)

11) PbCl2(aq) + KI(aq) PbI2(s) + KCl(aq)

12) Cu + HNO3 Cu(NO3)2 + H2O + NO2

13) Fe3(PO4)2(aq) + NaOH(aq) Na3PO4(aq) + Fe(OH)2(s)

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CHALLENGE (a minor one)

14) PbO(s) + SO2(g) PbS(s) + O2(g)

15) Fe2O3(s) + C(s) Fe(s) + CO2(g)

16) Mg(s) + O2(g) MgO(s)

17) Na(s) + I2(g) NaI(s)

18) Al(s) + S(s) Al2S3(s)

19) N2(g) + H2(g) NH3(g)

20) C2H3SNH2 + H2O C2H3O2NH4 + H2S (not nearly as tough as it looks)

21) CO + H2 CH3OH

22) Re(s) + Br2(l) ReBr3(s)

23) F2 + KCl KF + Cl2

24) Fe + CuSO4 Cu + FeSO4

25) NaClO3 NaCl + O2

26) PF3(g) + H2O H3PO3(aq) + HF(aq)

27) PCl5(l) + H2O(l) H3PO4(aq) + HCl(aq)ANSWERS :

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1) 2 C6H6(g) + 15 O2(g) 12 CO2(g) + 6 H2O(l)

2) 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)

3) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

4) 2 CH3OH(g) + 3 O2(g) 2 CO2(g) + 4 H2O(l)

5) 4 FeS2(s) + 11 O2(g) 2 Fe2O3(s) + 8 SO2(g)

6) 2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s) P4(g) + 6 CaSiO3(l) + 10 CO(g)

7) 3 IBr(g) + 4 NH3(g) NI3(s) + 3 NH4Br(s)

8) 4 BCl3 + P4 + 6 H2 4 BP + 12 HCl

9) 2 Al(s) + 3 H2SO4(aq) Al2(SO4))3(aq) + 3 H2(g)

10) 2 Al(s) + 3 Cu(NO3)2(aq) 2 Al(NO3)3(aq) + 3 Cu(s)

11) PbCl2(aq) + 2 KI(aq) PbI2(s) + 2 KCl(aq)

12) Cu + 4HNO3 Cu(NO3)2 + 2 H2O + 2 NO2

13) Fe3(PO4)2(aq) + 6 NaOH(aq) 2 Na3PO4(aq) + 3 Fe(OH)2(s)

14) 2 PbO(s) + 2 SO2(g) 2 PbS(s) + 3O2(g)

15) 2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 CO2(g)

16) 2 Mg(s) + O2(g) 2MgO(s)

17) 2 Na(s) + I2(g) 2 NaI(s)

18) 2 Al + 3 S Al2S3

19) N2(g) + 3H2(g) 2NH3(g)

20) C2H3SNH2 + 2 H2O C2H3O2NH4 + H2S

21) CO + 2 H2 CH3OH

22) 2 Re(s) + 3 Br2(l) 2 ReBr3(s)

23) F2 + 2KCl 2KF + Cl2

24) Fe + CuSO4 Cu + FeSO4

25) 2 NaClO3 2 NaCl + 3 O2

26) PF3(g) + 3 H2O(l) H3PO3(aq) + 3 HF(aq)

27) PCl5(l) + 4 H2O(l) H3PO4(aq) + 5 HCl(aq)279

IV) MOLE (abbreviation = mol) [Greek = heap] We may speak of the mole in at least three ways:

1 mol equals

a # of grams a # of particles the volume of 1 mol of gas (a # of liters [gaseous volume])

called called called

mole mass Avogadro's constant molar volume

A different calculation 6.022 x 1023 22.4 L for each substance.

Calculated using the a constant used only for gases @ STP chemical formula and according to KMT gases exist as molecules values from the periodic table table a constant number for ideal gases

A) Definition: Technically, 1 mol is the SI base unit for the quantity of substance. Remember, when the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles. (http://goldbook.iupac.org/M03980.html)

1) Species:

molecules = descriptor for substances made from atoms chemically united via covalent bonds(this applies to covalent compounds and a few elements e.g.) diatomic elements)

formula units = species for ionic compounds

atoms = species for most elements

ions = species for the products of dissolved / dissociated ionic compounds

2) mass in grams totaling 6.02 x 1023 molecules or formula units or atoms

3) volume in Liters

http://www.gluegrant.org/burnresearch.htm

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... up to 6.02 x 1023

22.4 Liters containing6.02 x 1023

molecules @STP

IV) The MOLE as a MASS

Mole Mass (BEST TERM: can't go wrong using this phrase although there are other names...)

the number of grams in 1 mol of a substance (element or compound)

calculated using values off the PT and the formula of the substance

one name another name

Gram Atomic Mass Gram Formula Mass (used for most elements) (used when speaking of compounds) EXCEPT the 7-H club and members of the 7-H club abbrev. abbrev.

GAM GFM (mole mass for an element) (mole mass for a compound or diatomic elements)

A) Definition: Gram Atomic Mass (GAM) * The mass, in grams of 1 mol of atoms of an element

or rather it is 1 mol mass, in grams of an element

1) to determine: Look up the relative atomic mass and give that number a unit of “grams”.

2) 1 GAM is a mass that represents 6.022 x 1023 atoms of most elements 1 GAM (1 mole mass) of an element represents 1 mol of the element

AND(!) That mass is listed on your periodic table.

Try This: … All answers are in whole number-ratios

1) What is the mole ratio between the species in 1 mol of H2O molecules? * 2: 1

…. now use the idea of mole mass and gram atomic mass to answer a very different question….

2) What is the ratio, by mass between hydrogen and oxygen in 1 mol of H2O? * 1: 8

3) What is the mole ratio between the species in 1 mol of C4H10 molecules? * 2: 5

4) What is the ratio, by mass between carbon and hydrogen in 1 mole of C4H10? * 24 : 5

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22.98977 +1

Na 112-8-1

B) Gram Formula Mass (GFM) * The mass of 1 mol of a diatomic element or any compound …

Essentially it is 1 mol mass, in grams of a compound.

1) to determine: Multiply each element’s mol mass by its subscript from the formula and add the products of each element together.

2) Calculate 1 mole mass of C2H5Cl

Calculate the GFM of Al(NO3)3 Note: This helps you to understand what the ( ) around a polyatomic ion imply. Learn this!!!

Calculate 1 mole mass of Mg3(PO4)2

Directions: This is just a "drill" exercise regarding the calculation of a mole mass. Use your Periodic Tables to calculate the mole mass of each substance. For this exercise, use the GAM rounded to a whole-number

1. Calculate the GFM of SF6

a) 51 grams b) 146 gramsc) 231 gramsd) 25 grams

2) Calculate the mole mass of N2. a) 14 gramsb) 56 gramsc) 28 gramsd) 43 grams

3) Calculate the GFM of Al2O3

a) 102 gramsb) 114 gramsc) 73 gramsd) 79 grams

4) What is 1 mole mass (the GAM) of chromium metal?

a) can not be determined b) 24 gramsc) 52 gramsd) 18 grams

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12.0111

C 1.00794

H 35.453

Cl

26.98154

Al14.0067

N15.9994

O

5) Calculate the mole mass of SiO2.a) 60 gb) 91 gc) 26 gd) 32 g

6) Calculate 1 mol mass of Na2Cr2O7.a) 262 gramsb) 507 gramsc) 82.0 gramsd) 117 grams

7) Calculate the GFM of Ca(NO3)2a) 180 gb) 164 gc) 147 gd) 23 g

8) Calculate 1 mole mass of Ca3(PO4)2.a) 133 gb) 49 gc) 151 gd) 310 g

9) What is the GFM of Al2(CrO4)3 ?a) 402 gb) 210 gc) 197 gd) 445 g

10) Calculate a mole mass of C8H18a) 66 gramsb) 1,728 gramsc) 114 gramsd) 37 grams

11) Calculate the GFM of Fe2(S2O3)3a) 448 gramsb) 932 gramsc) 506 gramsd) 774 grams

12) Calculate 1 mol mass of NH4Cla) 53 gramsb) 3 grams

c) 104 gramsd) 92 grams

13) Calculate the GFM of F2.a) 9 gramsb) 18 gramsc) 19 gramsd) 38 grams

14) Calculate 1 mole mass of MgOa) 20 gramsb) 40 gramsc) 384 gramsd) 77 grams

15) Calculate the GFM of C6H12O6.a) 342 gb) 201 gc) 299 gd) 180 g

16) Calculate the mass of 3.00 moles of H2O.a) 18 gramsb) 36 gramsc) 54 gramsd) 90 grams

17) Calculate the mass of 0.50 mole of H2O.a) 18 gramsb) 5.0 gramsc) 72 gramsd) 9.0 grams

18) Calculate the mass of 2.5 moles of NaCl mass.a) 145 gramsb) 58 gramsc) 309 gramsd) 75 grams

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SELECTED ANS. 1) b 2) c 3) a 4) c 6) a

7) b 8) d 12) a 13) d 15) d 18) a

VI) Percent Composition by Mass (An application of Mole Mass)

In general, to determine any percent, you must be concerned with the Part you want x 100 Whole

A) In order to determine the percent composition, by mass, we are particularly interested in

Mass of the element you want x 100 1 mol mass of the compound

B) Process for determing percent composition using Mole Mass: (use the E.S.A. Method)

Step 1) Determine 1 mole mass of the compound. (any mass will work ...but 1 mol mass is such a nice beginning if a total mass of a sample is not given)

Step 2) Multiply the GAM of the element in question by its subscript to find the element's total mass

Step 3) Divide the element's total mass of the element by the mole mass (GFM) of the compound

from step 1. (This fulfills the equation of part/whole)

Step 4) Multiply this quotient (from step 3) by 100 so that you may obtain a percent or percentage.

Most answers will be recorded in two or three sig figs…

Use the following as an example of a percent composition problem.

Calculate the percent composition, by mass of potassium ion in potassium sulfide (K2S)? (You assume 1 mol)

a) Equation: % K = Total Mass of K x 100 1 mole mass of K2S Taken from the periodic table

by looking up potassium (#19).b) Substitution: (2) (39.1) g x 100

110.3 g In lab, we should use the entire value

which equals: 78.2 g x 100 of the relative atomic mass. 110.3 g

c) Answer : 70.89 % or 70.9 % (sig figs are a bit iffy …because of the rounding. Thus, for this topic, I focus on process & have been limiting answers to 3 sig figs …based upon the rounded atomic mass value.)

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1) Calculate the percent composition of chloride ion, by mass in 1 mol mass of lithium chloride (LiCl). Use E.S.A.

Equation: % = Part x 100Whole

Substitution:

Answer: __________% chloride ion

2) Assuming 1 mol of substance, determine the percent composition of iodide ion, by mass in BaI2

Equation:

Substitution:

Answer: ___________% iodide ion

3) Find the percent composition of chloride ion in CrCl3

4) Find the percent composition of cadmium ion in CdCO3

5) Find the percent composition of barium ion in BaF2

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6) What is the percent composition of hydrogen in CH3COOH (Hint: There are multiple C's, H's,& O's. What do you think you need to do to find the mole mass?....Yep, group the C’s with each other & group the hydrogen ... and oxygen….)

7) Find the percent composition of carbon in CH3CH2CH2CH2OH

8) Find the percent composition of nitrogen in (NH4)3PO4

ANSWERS: 1) 83.6 % Cl ion 2) 64.9 % I ion 3) 67.2 % Cl ion 4) 65.2 % Cd ion 5) 78.3 % Ba ion 6) 6.67 % H 7) 64.9 % C 8) 28.1 % N

Now try something a bit more complex...Percent Composition and Hydrates

Hydrates are compounds that have a specific number of water molecules trapped in them. This translates nicely to mole theory. For every 1 mole of formula units of an ionic compound, a specific number of moles of water are trapped inside the crystals.

Many ionic compounds when crystallized from aqueous solutions retain a definite proportion of water as part of their crystalline structure. This water is referred to as the water of hydration. These ionic crystals appear to be perfectly dry. When an ionic crystal is in possession of this "trapped water", the crystal is called a HYDRATE. Upon heating a sample of the ionic crystal, this water may be driven off. As the water is driven off, the crystalline structure is shattered and all that remains is a dry powder. This dry powder is said to be the ANHYDROUS salt. (The prefix "AN-" simply means "WITHOUT" and "HYDROUS" refers to water, so, an "anhydrous salt" refers to an ionic crystal WITHOUT WATER.)

For instance, MgSO4•7H2O (called magnesium sulfate heptahydrate) has 7 moles of water molecules per 1 mole of magnesium sulfate formula units. The prefix hepta means 7, and hydrate refers to the water.

In the example of BaCl2•10 H2O (barium chloride decahydrate), there are 10 moles of water trapped within 1 mole of the barium chloride. This trapped water is insufficient to dissolve away the crystal. The prefix deca means 10.

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The calculation of the mole mass is pursued along the same lines as any other compound. However, the coefficient for the moles of water matters tremendously.

Can you see how you can calculate the gram formula mass of MgSO4•7H2O?

mass of 1 mol Mg + mass of 1 mol S + mass of 4 mols O + mass of 14 mols H + mass of 7 mols O 24 g + 32 g + 4(16 g) + 14(1 g) + 7 (16 g) = 246 g/mol

To determine the percent composition, follow the same procedure for any other compound.

1) Calculate the percent composition of water, by mass in one mole of MgSO4•7H2O

2) Calculate the percent composition of water, by mass, in one mole of BaCl2•10 H2O

3) Calculate the percent composition of water, by mass, in one mole of CuSO4•5 H2O

ans: 1) 51.2% 2) 46.5% 3) 36%

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NAME ____________________________________ PREQUIZ 1: PERCENT COMPOSITION DIRECTIONS: Complete each question by selecting the most correct answer. You do not need to show your work - but following the ESA process limits the "human" error dramatically.

___ 1) What is the percent composition of iron, ion, by mass in the compound, Fe2O3(s)?

a) 75 % c) 70 %

b) 23 % d) 67 %

___2) Which of the following compounds is mostly chloride ion, according to its percent composition by mass?

a) MnCl2 c) ICl

b) NaClO4 d) LiCl

___ 3) Which of the following compounds has the greatest percent composition by mass, of oxygen?

a) BaO c) Al2O3

b) CO d) NaClO

4) Calculate the percent composition by mass of water in 1 mol of the hydrate LiCl•H2O

Answers: PQ #1: 1) c 112/160 x 100 2) d 3) b 4) 30% …

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NAME ____________________________________ PREQUIZ 2: PERCENT COMPOSITION

DIRECTIONS: Complete each question by selecting the most correct answer. You do not need to show your work - but following the ESA process limits the "human" error dramatically.

___ 1) What is the percent composition, by mass, of manganese in the compound, MnSO4(s)?

a) 55.7 % c) 10.8 %

b) 36.4 % d) 27.1 %

___ 2) In which of the following compounds is mostly fluoride ion, according to its percent composition by mass?

a) HF c) NaF

b) CCl3F d) CrF2

___ 3) Which of the following compounds has the greatest percent composition by mass, of nickel?

a) NiClO2 c) NiCl3

b) Ni2S3 d) NiO

4) Calculate the percent composition by mass of water in 1mol of the hydrate BaCl2•2 H2O

PQ # 2 1) b 2) a 3) d 4) 14.8%

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