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Lecture 3. Stoichiometry. Contents. Elements Atomic number, mass number, atomic mass Mole concepts, Avogadro’s number Conversions between mole and number of atoms Relationship between mole and mass Molar mass Percent composition Empirical formula, molecular formula. Stoichiometry is:.

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### Lecture 3

Stoichiometry

Contents
• Elements
• Atomic number, mass number, atomic mass
• Mole concepts, Avogadro’s number
• Conversions between mole and number of atoms
• Relationship between mole and mass
• Molar mass
• Percent composition
• Empirical formula, molecular formula
Stoichiometry is:
• Study of the quantities of materials consumed and produced in chemical reactions
• Mole/mass relationships between reactants and products
Atomic number, Z
• the number of protons in the nucleus
• the number of electrons in a neutral atom
• the integer on the periodic table for each element
Mass Number, A
• integer representing the approximate mass of an atom
• equal to the sum of the number of protons and neutrons in the nucleus
Masses of Atoms

Carbon-12 Scale

Masses of the atoms are compared to the mass of C-12 isotope having a mass of

12.0000 amu

The masses of all other atoms are given relative to this standard

Atomic Masses andIsotopic Abundances

natural atomic masses =

sum[(atomic mass of isotope)

 (fractional isotopic abundance)]

C-12 98.89% mass – 12 amu

C-13 1.11% mass – 13.003355 amu

The average atomic mass for natural carbon is:

12amu x 0.9889+13.0034amu x 0.0111= 12.01amu

The Mole
• a unit of measurement, quantity of matter present
• The number equal to the number of carbon atoms in exactly 12 grams of pure C-12 (6.022  1023)
• Avogadro’s Number

6.022  1023 particles

Chemical Packages—Moles
• Mole = Number of things equal to the number of atoms in 12 g of C-12.
• 1 atom of C-12 weighs exactly 12 amu.
• 1 mole of C-12 weighs exactly 12 g.
• In 12 g of C-12 there are 6.022 x1023 C-12 atoms.

Molecular Mass - Molar Mass ( M )

The Molecular mass of a compound expressed in amuis

numerically the same as the mass of one mole of the

compound expressed in grams, called itsmolar mass.

For water: H2O

Molecular mass = (2 x atomic mass of H ) + atomic mass of O

= 2 ( amu) + amu = amu

Mass of one molecules of water = amu

Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)

= 2 ( g ) + g = g

g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

Molecular Mass - Molar Mass ( M )

The Molecular mass of a compound expressed in amuis

numerically the same as the mass of one mole of the

compound expressed in grams, called itsmolar mass.

For water: H2O

Molecular mass = (2 x atomic mass of H ) + atomic mass of O

= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu

Mass of one molecules of water = 18.02 amu

Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)

= 2 ( 1.008 g ) + 16.00 g = 18.02 g per mole H2O

18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

Molar Mass

The mass in grams of one mole of molecules or formula units of a substance, also called molar mass or molecular weight

Molar mass of:

C - 12.01 g/mole

O - 16.00 g/mole

Fe – 55,85 g/mole

atoms Ag

mol Ag

Example 1. A Silver Ring Contains 1.1 x 1022 Silver Atoms. How Many Moles of Silver Are in the Ring?

Given:

Find:

1.1 x 1022 atoms Ag

moles Ag

Solution Map:

Relationships:

1 mol = 6.022 x 1023 atoms

Solution:

Check:

Since the number of atoms given is less than Avogadro’s number, the answer makes sense.

Write down the quantity to find and/or its units.

Find: ? moles

Information:

Given: 1.1 x 1022 Ag atoms

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Collect needed conversion factors:

1 mole Ag atoms = 6.022 x 1023 Ag atoms.

Information:

Given: 1.1 x 1022 Ag atoms

Find: ? moles

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Write a solution map for converting the units:

Information:

Given: 1.1 x 1022 Ag atoms

Find: ? moles

Conversion Factor:

1 mole = 6.022 x 1023

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

atoms Ag

moles Ag

Apply the solution map:

Information:

Given: 1.1 x 1022 Ag atoms

Find: ? moles

Conversion Factor:

1 mole = 6.022 x 1023

Solution Map: atoms  mole

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

= 1.8266 x 1022 moles Ag

• Significant figures and round:

= 1.8 x 1022 moles Ag

mol Cu

atoms Cu

Practice 1 —Calculate the Number of Atoms in 2.45 Mol of Copper, Continued.

Given:

Find:

2.45 mol Cu

atoms Cu

Solution Map:

Relationships:

1 mol = 6.022 x 1023 atoms

Solution:

Check:

Since atoms are small, the large number of atoms makes sense.

Relationship Between Moles and Mass
• The mass of one mole of atoms is called the molar mass.
• The molar mass of an element, in grams, is numerically equal to the element’s atomic mass,

in amu.

• The lighter the atom, the less a mole weighs.
• The lighter the atom, the more atoms there are

in 1 g.

Mole and Mass Relationships

1 mole

carbon

12.01 g

1 mole

sulfur

32.06 g

g S

mol S

Given:

Find:

57.8 g S

mol S

Solution Map:

Relationships:

1 mol S = 32.07 g

Solution:

Check:

Since the given amount is much less than 1 mol S, the number makes sense.

Example 2:

Calculate the number of moles of sulfur in 57.8 g of sulfur.

Write down the quantity to find and/or its units.

Find: ? moles S

Information:

Given: 57.8 g S

Collect needed conversion factors:

1 mole S atoms = 32.07 g

Information:

Given: 57.8 g S

Find: ? moles S

Write a solution map for converting the units:

Information:

Given: 57.8 g S

Find: ? moles S

Conversion Factor:

1 mole S = 32.07 g

g S

moles S

Apply the solution map:

Information:

Given: 57.8 g S

Find: ? moles S

Conversion Factor:

1 mole S = 32.07 g

Solution Map: g  moles

= 1.802307 moles S

• Significant figures and round:

= 1.80 moles S

g Cu

mol Cu

atoms Cu

Practice 2 —How Many Copper Atoms Are in a Penny Weighing 3.10 g?, Continued

Given:

Find:

3.10 g Cu

atoms Cu

Solution Map:

Relationships:

1 mol Cu = 63.55 g, 1 mol = 6.022 x 1023

Solution:

Check:

Since the given amount is much less than 1 mol Cu, the number makes sense.

Molar Mass of Compounds
• The relative weights of molecules can be calculated from atomic weights.

Formula mass = 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu.

• Since 1 mole of H2O contains 2 moles of H and 1 mole of O.

Molar mass = 1 mole H2O

= 2(1.01 g H) + 16.00 g O = 18.02 g.

mol H2O

g H2O

Example 3—Calculate the Mass of 1.75 Mol of H2O.

Given:

Find:

1.75 mol H2O

g H2O

Solution Map:

Relationships:

1 mol H2O = 18.02 g

Solution:

Check:

Since the given amount is more than 1 mol, the mass being > 18 g makes sense.

g PbO2

mol PbO2

units PbO2

Practice 3 —How Many Formula Units Are in 50.0 g of PbO2? (PbO2 = 239.2), Continued

Given:

Find:

50.0 g PbO2

formula units PbO2

Solution Map:

Relationships:

1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023

Solution:

Check:

Since the given amount is much less than 1 mol PbO2, the number makes sense.

Wednesday, October14.
• Percent composition
• Empirical formula
• Molecular formula
• Chemical equation
• Balancing chemical equation
• Mass - to - mass conversion
Percent Composition
• Percentage of each element in a compound.
• By mass.
• Can be determined from:
• The formula of the compound.
• The experimental mass analysis of the compound.
Example 4 —Find the Mass Percent of Cl in C2Cl4F2.

Given:

Find:

C2Cl4F2

% Cl by mass

Solution Map:

Relationships:

Solution:

Check:

Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

100g MMA

% A mass A (g) moles A

moles A

moles B

100g MMB

% B mass B (g) moles B

Empirical Formulas
• The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.
• Can be determined from percent composition or combining masses.
• The molecular formula is a multiple of the empirical formula.
Empirical Formulas, Continued

Hydrogen Peroxide

Molecular formula = H2O2

Empirical formula = HO

Benzene

Molecular formula = C6H6

Empirical formula = CH

Glucose

Molecular formula = C6H12O6

Empirical formula = CH2O

Practice 5 —Determine the Empirical Formula of Benzopyrene, C20H12, Continued
• Find the greatest common factor (GCF) of the subscripts.

20 factors = (10 x 2), (5 x 4)

12 factors = (6 x 2), (4 x 3)

GCF = 4

• Divide each subscript by the GCF to get the empirical formula.

C20H12 = (C5H3)4

Empirical formula = C5H3

Finding an Empirical Formula

1. Convert the percentages to grams.

a. Skip if already grams.

2. Convert grams to moles.

a. Use molar mass of each element.

3. Write a pseudoformula using moles as subscripts.

4. Divide all by smallest number of moles.

5. Multiply all mole ratios by number to make all whole numbers, if necessary.

a. If ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc.

b. Skip if already whole numbers after Step 4.

### Example 6 — Finding an Empirical Formulafrom Experimental Data

Example:

A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.

C = 60.00%

H = 4.48%

O = 35.53%

Write down the given quantity and its units.

Given: C = 60.00%

H = 4.48%

O = 35.53%

Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O.

Example:Find the empirical formula of aspirin with the given mass percent composition.
Write down the quantity to find and/or its units.

Find: empirical formula, CxHyOz

Information:

Given: 60.00 g C, 4.48 g H, 35.53 g O

Example:Find the empirical formula of aspirin with the given mass percent composition.
Collect needed conversion factors:

1 mole C = 12.01 g C

1 mole H = 1.01 g H

1 mole O = 16.00 g O

Information:

Given: 60.00 g C, 4.48 g H, 35.53 g O

Find: empirical formula, CxHyOz

Example:Find the empirical formula of aspirin with the given mass percent composition.
Write a solution map:

Information:

Given: 60.00 g C, 4.48 g H, 35.53 g O

Find: empirical formula, CxHyOz

Conversion Factors:

1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g

Example:Find the empirical formula of aspirin with the given mass percent composition.

whole

number

ratio

g C

mol C

mole

ratio

empirical

formula

pseudo-

formula

g H

mol H

g O

mol O

Apply the solution map:

Calculate the moles of each element.

Information:

Given: 60.00 g C, 4.48 g H, 35.53 g O

Find: empirical formula, CxHyOz

Conversion Factors:

1 mol C = 12.01 g;

1 mol H = 1.01 g; 1 mol O = 16.00 g

Solution Map: g C,H,O  mol C,H,O 

mol ratio  empirical formula

Example:Find the empirical formula of aspirin with the given mass percent composition.
Apply the solution map:

Write a pseudoformula.

Information:

Given: 4.996 mol C, 4.44 mol H,

2.221 mol O

Find: empirical formula, CxHyOz

Conversion Factors:

1 mol C = 12.01 g;

1 mol H = 1.01 g; 1 mol O = 16.00 g

Solution Map: g C,H,O  mol C,H,O 

mol ratio  empirical formula

Example:Find the empirical formula of aspirin with the given mass percent composition.

C4.996H4.44O2.221

Apply the solution map:

Find the mole ratio by dividing by the smallest number of moles.

Information:

Given: C4.996H4.44O2.221

Find: empirical formula, CxHyOz

Conversion Factors:

1 mol C = 12.01 g;

1 mol H = 1.01 g; 1 mol O = 16.00 g

Solution Map: g C,H,O  mol C,H,O 

mol ratio  empirical formula

Example:Find the empirical formula of aspirin with the given mass percent composition.
Apply the solution map:

Multiply subscripts by factor to give whole number.

Information:

Given: C2.25H2O1

Find: empirical formula, CxHyOz

Conversion Factors:

1 mol C = 12.01 g;

1 mol H = 1.01 g; 1 mol O = 16.00 g

Solution Map: g C,H,O  mol C,H,O 

mol ratio  empirical formula

Example:Find the empirical formula of aspirin with the given mass percent composition.

{ } x 4

C9H8O4

### Example 7 —Finding an Empirical Formulafrom Experimental Data

Write a solution map:

Information:

Given: 3.24 g Ti, 5.40 g compound

Find: empirical formula, TixOy

Conversion Factors:

1 mol Ti = 47.88g;1 mol O = 16.00g

Example:Find the empirical formula of oxide of titanium with the given elemental analysis.

whole

number

ratio

g Ti

mol Ti

mole

ratio

empirical

formula

pseudo-

formula

g O

mol O

Molar massreal formula

Molar massempirical formula

= Factor used to multiply subscripts

Molecular Formulas
• The molecular formula is a multiple of the empirical formula.
• To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound.

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8.

• Determine the empirical formula.
• May need to calculate it as previous.

C5H8

• Determine the molar mass of the empirical formula.

5 C = 60.05, 8 H = 8.064

C5H8 = 68.11 g/mol

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.

• Divide the given molar mass of the compound by the molar mass of the empirical formula.
• Round to the nearest whole number.

Example—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued.

• Multiply the empirical formula by the factor above to give the molecular formula.

(C5H8)3 = C15H24

Emission of Light

Color Change

Release or Absorption of Heat

Formation of Solid Precipitate

Formation of a Gas

Evidence of Chemical Change
Chemical Equations
• Short-hand way of describing a reaction.
• Provides information about the reaction.
• Formulas of reactants and products.
• States of reactants and products.
• Relative numbers of reactant and product molecules that are required.
• Can be used to determine masses of reactants used and products that can be made.
Conservation of Mass
• Matter cannot be created or destroyed.
• Therefore, the total mass cannot change.
• And the total mass of the reactants will be the same as the total mass of the products.
• In a chemical reaction, all the atoms present at the beginning are still present at the end.
• If all the atoms are still there, then the mass will not change.
Chemical Equations

CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)

• CH4 and O2 are the reactants, and CO2 and H2O are the products.
• The (g) after the formulas tells us the state of the chemical.
• The number in front of each substance tells us the numbers of those molecules in the reaction.

Called the coefficients.

Chemical Equations, Continued

CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)

• This equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides.

To obtain the number of atoms of an element, multiply the subscript by the coefficient.

1  C  1

4  H  4

4  O  2 + 2

Symbols Used in Equations
• Symbols used to indicate state after chemical.
• (g) = gas; (l) = liquid; (s) = solid.
• (aq) = aqueous = dissolved in water.
• Energy symbols used above the arrow for decomposition reactions.
• D = heat.
• hn = light.
• shock = mechanical.
• elec = electrical.
Writing Balanced Chemical Equations
• Write a skeletal equation by writing the formula of each reactant and product.
• Count the number of atoms of each element on each side of the equation.
• Polyatomic ions may often be counted as if they are one “element”.
• Pick an element to balance.
• If an element is found in only one compound on both sides, balance it first.
• Metals before nonmetals.
• Leave elements that are free elements somewhere in the equation until last.
• Balance free elements by adjusting the coefficient where it is a free element.
Writing Balanced Chemical Equations, Continued
• Find the least common multiple (LCM) of the number of atoms on each side.
• The LCM of 3 and 2 is 6.
• Multiply each count by a factor to make it equal to the LCM.
• Use this factor as a coefficient in the equation.
• If there is already a coefficient there, multiply it by the factor.
• It must go in front of entire molecules, not between atoms within a molecule.
• Recount and repeat until balanced.
Example 1
• When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide.
• Write a skeletal equation

Mg(s) + O2(g) ® MgO(s)

• Count the number of atoms on each side.

Mg(s) + O2(g) ® MgO(s)

1 ¬ Mg ® 1

2 ¬ O ® 1

Example 1, Continued

Mg(s) + O2(g) ® MgO(s)

• Pick an element to balance.
• Do free elements last.
• Since Mg already balanced, pick O.
• Find the LCM of both sides
• and multiply each side by factor so it equals LCM.

LCM of 2 and 1 is 2.

Mg(s) + O2(g) ® MgO(s)

1 ¬ Mg ® 1

2 ¬ O ® 1

1 x

x 2

Example 1, Continued

Mg(s) + O2(g) ® MgO(s)

• Use factors as coefficients in front of the compound containing the element.
• We do not write 1 as a coefficient, its understood.

Mg(s) + O2(g) ® MgO(s)

1 ¬ Mg ® 1

1 x 2 ¬ O ® 1 x 2

2

Example 1, Continued
• When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide.

Mg(s) + O2(g) ® MgO(s)

• Recount—Mg not balanced now—That’s OK!

Mg(s) + O2(g) ® 2MgO(s)

1 ¬ Mg ® 2

2 ¬ O ® 2

• and Repeat—attacking an unbalanced element.

2Mg(s) + O2(g) ® 2MgO(s)

2 x 1 ¬ Mg ® 2

2 ¬ O ® 2

Example 2
• Under appropriate conditions at 1000°C, ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and steam
• write the skeletal equation
• first in words
• identify the state of each chemical

ammonia(g) + oxygen(g) ® nitrogen monoxide(g) + water(g)

• then write the equation in formulas
• identify diatomic elements
• identify polyatomic ions
• determine formulas

NH3(g) + O2(g) ® NO(g) + H2O(g)

Example 2, Continued

NH3(g) + O2(g) ® NO(g) + H2O(g)

• count the number of atoms of on each side

NH3(g) + O2(g) ® NO(g) + H2O(g)

1 ¬ N ® 1

3 ¬ H ® 2

2 ¬ O ® 1 + 1

Example 2, Continued
• Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous steam

NH3(g) + O2(g) ® NO(g) + H2O(g)

• pick an element to balance - H
• avoid element in multiple compounds on same side - O
• find least common multiple of both sides (6)
• multiply each side by factor so it equals LCM

NH3(g) + O2(g) ® NO(g) + H2O(g)

1 ¬ N ® 1

3 ¬ H ® 2

2 ¬ O ® 1 + 1

2 x

x 3

Example 2, Continued

3

• use factors as coefficients in front of compound containing the element

2 NH3(g) + O2(g) ® NO(g) + H2O(g)

1 ¬ N ® 1

2 x 3 ¬ H ® 2 x 3

2 ¬ O ® 1 + 1

Example 2, Continued
• Recount – N & O not balanced

2 NH3(g) + O2(g) ®NO(g) + 3 H2O(g)

2 ¬ N ® 1

6 ¬ H ® 6

2 ¬ O ® 1 + 3

• and Repeat – attack the N

2 NH3(g) + O2(g) ® 2NO(g) + 3 H2O(g)

2 ¬ N ® 1 x 2

6 ¬ H ® 6

2 ¬ O ® 1 + 3

Example 2, Continued
• Recount Again – Still not balanced and the only element left is O!

2 NH3(g) + O2(g) ® 2NO(g) + 3 H2O(g)

2 ¬ N ® 2

6 ¬ H ® 6

2 ¬ O ® 2 + 3

Example 2, Continued
• and Repeat Again
• A trick of the trade - when you are forced to attack an element that is in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction!

2 NH3(g) + ? O2(g) ® 2NO(g) + 3 H2O(g)

2 ¬ N ® 2

6 ¬ H ® 6

2 ¬ O ® 2 + 3

• We want to make the O on the left equal 5, therefore we will multiply it by 2.5

2 NH3(g) + 2.5 O2(g) ® 2NO(g) + 3 H2O(g)

2 ¬ N ® 2

6 ¬ H ® 6

2.5 x 2 ¬ O ® 2 + 3

Example 2, Continued
• You can’t have a coefficient that isn’t a whole number. Multiply all the coefficients by a number to eliminate fractions
• If ?.5, then multiply by 2; if ?.33, then 3; if ?.25, then 4

{2 NH3(g) + 2.5 O2(g) ® 2NO(g) + 3 H2O(g)} x 2

4 NH3(g) + 5 O2(g) ® 4NO(g) + 6 H2O(g)

4 ¬ N ® 4

12 ¬ H ® 12

10 ¬ O ® 4 + 6

Example 3

When aluminum metal reacts with air, it produces a white, powdery compound called aluminum oxide.

• Reacting with air means reacting with O2:

Aluminum(s) + oxygen(g) ® aluminum oxide(s)

Al(s) + O2(g) ® Al2O3(s)

Example 3, Continued

When aluminum metal reacts with air, it produces a white, powdery compound called aluminum oxide.

• Reacting with air means reacting with O2:

Aluminum(s) + oxygen(g) ® aluminum oxide(s)

Al(s) + O2(g) ® Al2O3(s)

2Al(s) + O2(g) ® Al2O3(s)

2Al(s) + 1.5O2(g) ® Al2O3(s)

{2Al(s) + 1.5O2(g) ® Al2O3(s)}x2

4 Al(s) + 3 O2(g) ® 2 Al2O3(s)

Example 4

Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen.

• Acids are always aqueous.
• Metals are solid except for mercury.

Al(s) + HC2H3O2(aq) ® Al(C2H3O2)3(aq) + H2(g)

Al(s) +3 HC2H3O2(aq) ® Al(C2H3O2)3(aq) +1.5 H2(g)

{Al(s) +3 HC2H3O2(aq) ® Al(C2H3O2)3(aq) +1.5 H2(g)}x2

2 Al(s) + 6 HC2H3O2(aq) ®2 Al(C2H3O2)3(aq) + 3 H2(g)

Example 5

Combustion of ethyl alcohol (C2H5OH) in flambé (a brandied flaming dessert).

• Combustion is burning, and therefore, reacts with O2.
• Combustion of compounds containing C and H always make CO2(g) and H2O(g) as products.

C2H5OH(l) + O2(g) ® CO2(g) + H2O(g)

Example 5, Continued

Combustion of ethyl alcohol (C2H5OH) in flambé (a brandied flaming dessert).

• Combustion is burning, and therefore, reacts with O2.
• Combustion of compounds containing C and H always make CO2(g) and H2O(g) as products.

C2H5OH(l) + O2(g) ® CO2(g) + H2O(g)

C2H5OH(l) + O2(g) ®2CO2(g) + H2O(g)

C2H5OH(l) + O2(g) ® 2CO2(g) +3H2O(g)

C2H5OH(l) +3O2(g) ®2 CO2(g) 3H2O(g)

C2H5OH(l) + 3 O2(g) ® 2 CO2(g) + 3 H2O(g)

Example 6

Combustion of liquid butane (C4H10) in a lighter.

C4H10(l) + O2(g) ® CO2(g) + H2O(g)

Example 6 Continued

Combustion of liquid butane (C4H10) in a lighter.

C4H10(l) + O2(g) ® CO2(g) + H2O(g)

C4H10(l) + O2(g) ®4CO2(g) + 5H2O(g)

C4H10(l) +6.5 O2(g) ®4CO2(g) + 5H2O(g)

{C4H10(l) +6.5 O2(g) ®4CO2(g) + 5H2O(g)}x2

2 C4H10(l) + 13 O2(g) ®8 CO2(g) + 10 H2O(g)

Mole Relationships inChemical Formulas
• Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound.

mol CaCO3

mol O

Given:

Find:

1.7 mol CaCO3

mol O

Solution Map:

Relationships:

1 mol CaCO3 = 3 mol O

Solution:

Check:

Since the given amount is much less than 1 mol S, the number makes sense.

g C10H14O

mol C10H14O

mol C

g C

Example —Find the Mass of Carbon in 55.4 g C10H14O.

Given:

Find:

55.4 g C10H14O

g C

Solution Map:

Relationships:

1 mol C10H14O = 150.2 g, 1 mol C = 12.01 g,

10 mol C : 1 mol C10H14O

Solution:

Check:

Since the amount of C is less than the amount of C10H14O, the answer makes sense.

Making MoleculesMole-to-Mole Conversions
• The balanced equation is the “recipe” for a chemical reaction.
• The equation 3 H2(g) + N2(g)  2 NH3(g) tells us that 3 molecules of H2 react with exactly 1 molecule of N2 and make exactly 2 molecules of NH3 or:

3 molecules H2 1 molecule N2  2 molecules NH3

• Since we count molecules by moles:

3 moles H2 1 mole N2  2 moles NH3

mol Cl2

mol NaCl

Example—How Many Moles of NaCl Result from the Complete Reaction of 3.4 Mol of Cl2? 2 Na(s) + Cl2(g) → 2 NaCl

Given:

Find:

3.4 mol Cl2

mol NaCl

Solution Map:

Relationships:

1 mol Cl2 2 NaCl

Solution:

.

Mass-to-Mass Conversions
• We know there is a relationship between the mass and number of moles of a chemical.

1 mole = Molar Mass in grams.

• The molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other.

mol C6H12O6

g C6H12O6

g CO2

mol CO2

Example —How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO2 in Photosynthesis?
• Photosynthesis:

6 CO2(g) + 6 H2O(g)  C6H12O6(s) + 6 O2(g)

• The equation for the reaction gives the mole relationship between amount of C6H12O6 and CO2, but we need to know the mass relationship, so the solution map will be:
Write a solution map:

Information:

Given: 58.5 g CO2

Find: g C6H12O6

Conversion Factors:

1 mol C6H12O6 = 180.2 g

1 mol CO2 = 44.01 g

1 mol C6H12O6 6 mol CO2

mol

CO2

mol

C6H12O6

g

C6H12O6

Example:How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction?6 CO2(g) + 6 H2O(l)  6 O2(g) + C6H12O6(aq)

g

CO2

Apply the solution map:

Information:

Given: 58.5 g CO2

Find: g C6H12O6

Conversion Factors:

1 mol C6H12O6 = 180.2 g

1 mol CO2 = 44.01 g

1 mol C6H12O6 6 mol CO2

Solution Map: g CO2 mol CO2

mol C6H12O6  g C6H12O6

Example:How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction?6 CO2(g) + 6 H2O(l)  6 O2(g) + C6H12O6(aq)

= 39.9216 g C6H12O6

• Significant figures and round:

= 39.9 g C6H12O6

Carvone (C10H14O) is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liqueurs, soaps, and perfumes. Find the mass of carbon in 55.4 g of carvone.

Write a solution map for converting the units:

Information:

Given: 55.4 g C10H14O

Find: g C

Conversion Factors:

1 mol C10H14O = 150.2 g

1 mol C10H14O  10 mol C

1 mol C = 12.01 g

g

C10H14O

mol

C10H14O

mol

C

g

C

Apply the solution map:

Information:

Given: 55.4 g C10H14O

Find: g C

Conversion Factors:

1 mol C10H14O = 150.2 g

1 mol C10H14O  10 mol C

1 mol C = 12.01 g

Solution Map:

g C10H14O  mol C10H14O 

mol C  g C

= 44.2979 g C

• Significant figures and round:

= 44.3 g C

THE LIMITING REACTANT

The limiting reactant present in a mixture of reactants is the reactant that will run out first, and thus, it determines the amount of product that can be produced.

A useful approach to solving limiting reactant problems is to calculate the amount of product that could be produced by each of the quantities of reactant that are available. The reactant that gives the least amount of product is then the limiting reactant.

Example —What Is the Limiting Reactant and Theoretical Yield When 0.552 Mol of Al React with 0.887 Mol of Cl2? 2 Al(s) + 3 Cl2(g) → 2 AlCl3

mol Cl2

mol Al

Pick least

amount

mol AlCl3

mol AlCl3

Limiting

reactant and

theoretical

yield

Given:

Find:

0.552 mol Al, 0.887 mol Cl2

mol AlCl3

Solution Map:

Relationships:

3 mol Cl2 2 AlCl3; 2 mol Al  2 mol AlCl3

Solution:

Limiting

Reactant

Theoretical

Yield

Write a solution map:

Example:What is the limiting reactant and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl2? 2 Al(s) + 3 Cl2(g) → 2 AlCl3

Information:

Given: 0.552 mol Al, 0.877 mol Cl2

Find: limiting reactant, theor. yield

Conversion Factors:

2 mol AlCl3  2 mol Al,

2 mol AlCl3  3 mol Cl2

}

mol

Al

mol

AlCl3

Smallest

amount is

theoretical

yield

mol

Cl2

mol

AlCl3

Information:

Given: 0.552 mol Al, 0.877 mol Cl2

Find: limiting reactant, theor. yield

Conversion Factors:

2 mol AlCl3  2 mol Al,

2 mol AlCl3  3 mol Cl2

Solution Map:

mol each reactant  mol AlCl3

Apply the solution map:

Example:What is the limiting reactant and theoretical yield when 0.552 mol of Al react with 0.887 mol of Cl2? 2 Al(s) + 3 Cl2(g) → 2 AlCl3

Limiting reactant = Al

Smallest

amount

Theoretical yield = 0.552 mol AlCl3

g NH3

g CuO

mol NH3

mol CuO

mol N2

mol N2

g N2

g N2

Practice—How Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)If 4.61 g of N2 Are Made, What Is the Percent Yield?, Continued

Given:

Find:

9.05 g NH3, 45.2 g CuO

g N2

Solution Map:

Relationships:

1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g

2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2

Choose

smallest

Practice—How Many Grams of N2(g) Can Be Made from 9.05 g of NH3 Reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)If 4.61 g of N2 Are Made, What Is the Percent Yield?, Continued

Solution:

Theoretical

yield

Check:

Since the percent yield is less than 100, the answer makes sense.

REACTION YIELDS

The amount of product calculated in the last three examples are not the amounts that would be produced if the reactions were actually done in the laboratory. In each case, less product would be obtained than was calculated. There are numerous causes. Some materials are lost during transfers from one container to another and side reactions take place that are different from the one that is intended to take place.

The amount of product calculated in the examples is called the theoretical yield. The amount of product actually produced is called the actual yield. These two quantities are used to calculate the percentage yield using the following equation:

Example: Suppose the mixture of reactants calculated earlier to give 73.7 g SO2 was done in the laboratory and only 42.7 g of SO2 was collected. What is the percentage yield of the reaction?

Solution: