8 - 1 stoichiometry molar mass is the mass of 1.00 mole of particles of that substance. to find the...

8 - 1

StoichiometryStoichiometry

Molar mass is the mass of 1.00 mole ofparticles of that substance.

To find the molar mass of an element, simplytake the atomic mass from the periodic table,round it off to 2 decimal places, and add theunit g (grams).

1 mol Al = 26.98 g Al

1 mol H2 = 2.02 g H2

8 - 2

StoichiometryStoichiometry

Remember your diatomics!

Those elements ending in -ine and -gen arediatomic molecules.

H2, F2, Cl2, I2, Br2, O2, N2

8 - 3

Converting Grams to MolesConverting Grams to Moles

Calculate the number of moles in 250 mg ofaspirin, C9H8O4.

m = 250 mg C9H8O4

n = 250 mg C9H8O4

n = 1.4 × 10-3 mol C9H8O4

1 g103 mg

× ×1 mol C9H8O4

180.17 g C9H8O4

8 - 4

Converting Grams to MolesConverting Grams to Moles

Calculate the number of moles in 25.0 g ofammonium chloride.

m = 25.0 g NH4Cl

n = 25.0 g NH4Cl

n = 0.467 mol NH4Cl

× 1 mol NH4Cl 180.17 g NH4Cl

8 - 5

Converting Moles to GramsConverting Moles to Grams

Calculate the mass in grams of 5.75 mol ofnitrogen atoms.

m = 5.75 mol N

n = 5.75 mol N

n = 80.6 g N

×14.01 g N1 mol N

8 - 6

Converting Moles to GramsConverting Moles to Grams

Calculate the mass in grams of 5.75 mol ofnitrogen molecules.

m = 5.75 mol N2

n = 5.75 mol N2

n = 161 g N2

× 28.02 g N2

1 mol N2

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Mass Percent ProblemsMass Percent Problems

The mass percent of an element found in a compound is expressed:

m% =m% = ×100%masscpd

massele

8 - 8

Mass Percent ProblemsMass Percent Problems

Calculate the percent composition ofammonium found in ammonium carbonate?

m% =m% =

m% = m% =

×

100%

36.10 g 96.11 g ×

100%

= 37.56%

masscpd

massele

8 - 9

Magnesium hydroxide is 54.87% oxygen bymass.

(a) Determine the mass of oxygen in 165 g of this compound.

(b) Determine the number of moles of oxygen.

(a) m% = 54.87% O

m% = m% =

m = 165 g × 0.5487 = 90.54 g O

× 100%

massele

masscpd

massele

8 - 10

(b) nO = 90.54 g O ×

nO = 5.659 mol O

1 mol O16.00 g O

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More Mass Percent ProblemsMore Mass Percent Problems

How many grams of phosphorus arecontained in 0.853 g of magnesium

phosphate?

m = 0.853 g Mg3(PO4)2

m = 0.853 g Mg3(PO4)2 ×

1 mol Mg3(PO4)2

262.84 g Mg3(PO4)2

2 mol P1 mol

Mg3(PO4)2

× 30.97 g P1 mol P

=0.201 g P

×

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How many grams of carbon are found in0.764 g CO2?

m = 0.764 g CO2

m = 0.764 g CO2 × 1 mol CO2

44.01 g CO2

× 1 mol C 1 mol CO2

×12.01 g C 1 mol C

m = 0.208 g C

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Mass Percent WrapupMass Percent Wrapup

Remember:

%’s are based on 100. All %’s when added must equal 100.

(NH4)2CO3 is an ionic compound.

In one formula unit of (NH4)2CO3 there are two ammonium ions and one carbonate ion.

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Mass Percent WrapupMass Percent Wrapup

Remember:

In one mole of (NH4)2CO3 there are two moles of ammonium ions and one

mole of carbonate.

8 - 15

Mass Percent WrapupMass Percent Wrapup

Remember:

CO2 is a covalent compound.

In one molecule of CO2 there is one atom

of carbon and two atoms of oxygen.

In one mole of CO2 there is one mole of carbon and two moles of oxygen.

8 - 16

Mass-Mass ProblemsMass-Mass Problems

(a)How many grams of oxygen are required when 1.50 g of ethane (C2H6) undergo complete combustion?

(b) What mass of carbon dioxide is produced when 18.5 g of oxygen are consumed?

Before solving this problem, the many facets

of a mass-mass problem will be discussed.

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When given the word equation, the chemical equation must be correctly written.

To have a correctly written chemical equation means that all chemical formulas must be correct, all atoms

are balanced, and the correct physical states are used.

Remember, single and double replacement reactions require certain conditions.

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Remember, the complete combustion of a hydrocarbon produces carbon dioxide and water.

A mass-mass problem consists of being given the mass of one of the reactants or products and determining the mass of one of the other reactants or products.

A mass-mass problem consists of three distinct steps.

Step 1: The given mass in grams goes to the right of the equal sign.

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Step 2: The second term will be a mole ratio obtained from the balance equation.

Step 3: The third term will be a conversion converting moles

to grams.

Always use the problem solving strategysteps when solving a stoichiometry problem.

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

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m = 1.50 g C2H6

m = 1.50 g C2H6

× 1 mol C2H6

30.08 g C2H6

×

7 mol O2

2 mol C2H6

× 32.00 g O2

1 mol O2

= 5.59 g O2

given info in problem

grams to moles

mole ratio frombalanced equation

moles to grams

8 - 21

m = 18.5 g O2

m = 18.5 g O2

× 1 mol O2

32.00 g O2

×

4 mol CO2

7 mol O2

× 44.01 g CO2

1 mol CO2

= 14.5 g CO2

8 - 22

Mass-Volume ProblemsMass-Volume Problems

A piece of magnesium is ignited in a jarcontaining 0.500 dm3 of oxygen at STP.

Howmany grams of magnesium oxide is

formed?

Before solving this problem, the facets of amass-volume problem will be discussed.

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Avogadro’s hypothesis states that when samples of different ideal gases at the same temperature and pressure contain the same number of molecules, then

the volumes of all the samples are equal.

At STP, the molar volume of any ideal gas is 22.4 dm3 = 22.4 L.

The term molar means /mol (per one mole).

Ideal gases will be discussed at a later time.

8 - 24

Remember that the coefficients in a balanced equation indicate the

relative number of moles or the relative

volumes.

The setup of a mass-volume problem is the same as that of a mass-mass problem except for one of the conversions.

That conversion converts moles to volume or volume to moles.

8 - 25

V = 0.500 dm3 O2

2Mg(s) + O2(g) → 2MgO(s)

m = 0.500 dm3 O2

×

1 mol O2

22.4 dm3 O2

×

2 mol MgO 1 mol O2

×

40.30 g MgO 1 mol MgO = 1.80 g MgO

Converts volume to moles.

8 - 26

Limiting Reactant ProblemsLimiting Reactant Problems

Limiting reactant problems are what thename implies. One of the reactants is

presentin a smaller amount than is required by thebalanced chemical equation.

When the limiting reactant is consumed, the

reaction stops.

What is present after the reaction stops?

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When the reaction stops:

The non-limiting reactant (reactant or reagent in excess) is present but in a smaller amount.

Some product(s) is/are formed.

An example of a limiting reactant scenario is

the following.

Suppose you have 50 bolts, 75 nuts, and

75 washers.

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An assembled set of nuts/bolts/washers consists of 2 washers, 1 bolt, and 1 nut.

(a) How many complete assembled sets can you make?

(b) What acts as the limiting component?

You can always identify a limiting reactantproblem because the starting amounts(grams or moles) of all the reactants is given.

8 - 29

Limiting Reactant ProblemLimiting Reactant Problem

A solution containing 5.00 g of calciumchloride is mixed with another solution of8.00 g of potassium phosphate.

(a)Determine the theoretical yield of the precipitate.(b) Which of the reactants is limiting?(c) How much of the reactant in excess is

left over?(d) What mass of chloride is produced?

8 - 30

m = 5.00 g CaCl2

m = 8.00 g K3PO4

3CaCl2(aq) + 2K3PO4(aq) → Ca3(PO4)2(s) + 6KCl(aq)

Note the mass of both reactants

given!

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m 5.00 g CaCl21 mol CaCl2

110.98 g CaCl2××

1 mol Ca3(PO4)2

3 mol CaCl2×

310.18 g Ca3(PO4)2

1 mol Ca3(PO4)2

=

=

4.66 g Ca3(PO4)2

8 - 32

m 8.00 g K3PO4

The theoretical yield is 4.66 g Ca3(PO4)2

because it is the smallest of the two yields.

1 mol K3PO4

212.27 g K3PO4××

1 mol Ca3(PO4)2

2 mol K3PO4×

310.18 g Ca3(PO4)2

1 mol Ca3(PO4)2

=

=

5.85 g Ca3(PO4)2

8 - 33

What to know about theoretical yields:

The theoretical yield is always determined from the stoichiometry of

the balanced equation.

Theoretical yields represent the maximum amount of product(s)

provided there are no experimental errors.

Actual yields are the amount of productactually formed, usually much less than thetheoretical yield but never greater.

8 - 34

The percent yield is defined as:

%yield

(b) CaCl2 is the limiting reactant because it produces the smallest amount of Ca3(PO4)2.

actual yieldtheoretical yield= ×

100%

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(c)

m 5.00 g CaCl2

The 6.38 g K3PO4 is the amount that isrequired to produce 4.66 g Ca3(PO4)2.

1 mol CaCl2110.98 g CaCl2

××

2 mol K3PO4

3 mol CaCl2×

212.27 g K3PO4

1 mol K3PO4

=

=

6.38 g K3PO4

8 - 36

If 6.38 g K3PO4 is the required amount, then

8.00 g – 6.38 g = 1.62 g K3PO4 is left over.

(d) Using the conservation of mass-energy,

mr = mp

mCaCl2 + mK3PO4 = mCa3(PO4)2 + mKCl

5.00 g + 6.38 g = 4.66 g + mKCl

mKCl = 6.72 g KCl

8 - 37

Avogadro’s NumberAvogadro’s Number

Avogadro’s number is an enormous countingnumber and is analogous to a dozen, a pair, atrio, or a gross.

Avogadro’s number, N, is an unfathomablenumber.

N = 6.02 × 1023 atoms/molecules/ions

This number would be written as 602 followedby 21 zeroes!

8 - 38

Avogadro’s number is an alternative way toexpress the quantity of 1 mol and does notreplace the previous definition.

For example,

1 mol H2O = 18.02 g = 6.02 × 1023 H2O

molecules

The type of problem given dictates whichdefinition is used as shown in the followingslides.

8 - 39

Avogadro’s Number ProblemsAvogadro’s Number Problems

How many water molecules are present in173 g H2O?

n = 173 g H2O N = 6.02 × 1023 H2O molecules

n = 173 g H2O

n = 5.78 × 1024 H2O molecules

×

× 6.02 × 1023 H2O molecules1 mol H2O

1 mol H2O18.02 g H2O

8 - 40

What is the mass of a H2O molecule?

n = 1 H2O moleculeN = 6.02 × 1023 H2O molecules

m = 1 H2O molecule×1 mol H2O

18.02 g H2O×

6.02 × 1023 H2O molecules

1 mol H2O=2.99 × 10-23 g H2O

8 - 41

(a)How many molecules are there in 2.53 × 1022 NH3 molecules?(b) Determine the mass of 2.53 × 1023 NH3

molecules?(c) How many atoms are there in 2.53 ×

1023 NH3 molecules?

(a)n = 2.53 × 1022 NH3 molecules

2.53 × 1022 NH3 molecules ×

1 mol NH36.02 × 1023 NH3 mole…=0.0420 mol NH3

8 - 42

(b)

m = 2.53 × 1022 NH3 molecules

m = 0.716 g NH3

×

1 mol NH3

6.02 × 1023 NH3 molecules×

17.04 g NH3

1 mol NH3

8 - 43

(c)

n = 2.53 × 1022 NH3 molecules

n = 1.01 × 1023 atoms

×

4 atoms

1 NH3 molecules

8 - 44

Empirical FormulasEmpirical Formulas

Empirical formulas show the type and relative

number of atoms.

An empirical formula shows the simplestwhole number ratio of bonded atoms.

It is easy to recognize an empirical formulabecause the subscripts can not be reduced

toany simpler whole numbers.

8 - 45

Empirical Formula ProblemEmpirical Formula Problem

A 100.0 g sample of an oxide of chromium is

found to contain 68.4 g of chromium and31.6 g of oxygen. Determine the empiricalformula of this compound.

m = 68.4 g Cr m = 31.6 g O

n= 68.4 g Cr × 1 mol Cr52.00 g Cr

=1.32 mol Cr

8 - 46

31.6 g O × 1 mol O16.00 g O

= 1.98 mol O

Cr1.32O1.98 => CrO1.5 => Cr2O3

The subscript 1.98 is close enough to round but not the 1.32. In this case, you divide by the smaller of the subscripts. A sanity check is to compare your value to what it should be using oxidation (valence numbers).

8 - 47

A compound contains 2.62 g of nitrogen,0.750 g of hydrogen, and 6.63 g of chlorine.What is the empirical formula of thecompound?

m = 2.62 g N m = 0.750 g H m = 6.63 g Cl

n =2.62 g N ×1 mol N

14.01 g N=0.187 mol N

n =0.750 g H× 1 mol H1.01 g H

=0.740 mol H

8 - 48

N0.187H0.740Cl0.187 NH3.96Cl NH4Cl

n =6.63 g Cl × 1 mol Cl35.45 g Cl

= 0.187 mol Cl

=> =>

8 - 49

Empirical Formula WrapupEmpirical Formula Wrapup

Note that N, H, and Cl are not written asdiatomic molecules.

They are written as diatomics only when they are by themselves.

The order of the nonmetals N and H areimportant. Because you are determining theempirical formula for an ionic compound, youmust have a cation containing N and H whichis the ammonium polyatomic ion.

8 - 50

Molecular FormulasMolecular Formulas

Determine the molecular formula for acompound with the following composition:40.00% C, 6.72% H, and 53.29% O and amolar mass of 180. g/mol.

mC = 40.00 g mH = 1.01 g mO = 53.29 g

MM = 180. g/mol

Solve this portion of the problem as a typical

empirical formula problem.

8 - 51

.

nC = 40.00 g C 1 mol C

12.01 g C× = 3.331 mol C

nH = 6.72 g H × 1 mol H 1.01 g H

= 6.667 mol H

nO = 53.29 g O ×1 mol O

16.00 g O=3.331 mol O

C3.331H6.667O3.331 => CH2O

8 - 52

MMCH2O = 30.03 g/mol

Dividing the molar mass of the compound by

the molar mass from the empirical formulagives a multiplier of 6. Multiply the

subscriptsof the empirical formula by 6 to give

C6H12O6.

n =180. g/mol30.03 g/mol

= 6

8 - 53

Molecular Formula WrapupMolecular Formula Wrapup

Molecular formulas show the actual numberof each type of atom in a molecule.

When doing either an empirical or amolecular formula problem, when givenpercents, the percents can readily bechanged to grams.

Remember percents are based on 100.

8 - 54

To determine the multiplier, divide the molar

mass of the compound by the molar mass of

the empirical formula.

There will be instances when the empiricaland the molecular formula will be identical.

H2O, NH3

The molecular formula does not tell you how

the atoms are arranged in the molecule.

8 - 55

HydratesHydrates

A hydrate is an ionic compound with watermolecules attached usually in crystal form.

An example of a hydrate is CuSO4•5H2Owhich is called copper(II) sulfatepentahydrate.

The name of the empirical formula follows the

normal convention and a prefix that is usedfor covalent compounds precedes the namehydrate.

8 - 56

The raised dot in the formula, CuSO4•5H2O,does not mean multiplication.

When determining the molar mass of thehydrate, you add 5 times the molar mass tothe molar mass of CuSO4.

Keep in mind that 1 mol CuSO4•5H2O contains

1 mole of CuSO4 and 5 moles of H2O.

8 - 57

Empirical Formulas of HydratesEmpirical Formulas of Hydrates

The formula for a hydrate of sodiumcarbonate is Na2CO3•xH2O. A 2.714 g

sampleis heated to a constant mass of 1.006 g ofNa2CO3. Determine the formula of thehydrate.

m = 2.714 g Na2CO3•xH2Om = 1.006 g Na2CO3

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mH2O = mNa2CO3•xH2O - mNa2CO3

mH2O = 2.714 g – 1.006 g = 1.708 g H2O

1.708 g H2On= ×1 mol H2O

18.02 g H2O

n= 0.09478 mol H2O

n= 1.006 g Na2CO3×1 mol Na2CO3

105.99 g Na2CO3

n=0.009491 mol Na2CO3

8 - 59

0.009491Na2CO3•0.09478H2O

Na2CO3•9.986H2O

Na2CO3•10H2O

sodium carbonate decahydrate

Divide both coefficients by 0.009491.

8 - 60

More HydratesMore Hydrates

A hydrate of potassium aluminum sulfate has

the empirical formula KAl(SO4)2•xH2O. Asample of this hydrate with a mass of 5.459

gis heated to remove all the water and 2.583

gof the hydrate remains.

(a)What is the mass percent of water in the hydrate?

(b)What is x?

8 - 61

a) m = 5.459 g KAl(SO4)2•xH2O m = 2.583 g KAl(SO4)2

m%

=mH2Omcmp ×100%

=5.459 g – 2.583 g

5.459 g×100%

= 52.68%

8 - 62

b) n = 2.876 g H2O× 1 mol H2O18.02 g H2O

n =0.1596 mol H2O

n = 2.583 g KAl(SO4)2 ×1 mol

KAl(SO4)2 258.22 g KAl(SO4)2

n = 0.01000 mol KAl(SO4)2

0.01000 mol KAl(SO4)2•0.1596 mol H2O

KAl(SO4)2•16H2O