Ch 3 Stoichiometry 161 Student.pptx

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3.1 Counting by Weighing 3.2 Atomic Masses

3.3 The Mole 3.4 Molar Mass

3.5 Learning to Solve Problems

3.6 Percent Composition of Compounds

3.7 Determining the Formula of a Compound 3.8 Chemical Equations

3.9 Balancing Chemical Equations

3.10 Stoichiometric Calculations: Amounts of Reactants and Products

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Chemical Stoichiometry

• _{Stoichiometry – The study of quantities of materials}

consumed and produced in chemical reactions.

• _{Branch of chemistry that deals with the relative quantities}

of reactants and products in chemical reactions.

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•

When we count by weighing the average mass

of the object allows us to behave as though

they were all identical.

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Averaging the Mass of Similar Objects

• _{Example: What is the mass of 1000 jelly beans?}

1. Not all jelly beans have the same mass. 2. Suppose we weigh 10 jelly beans and find:

3. Now we can find the average mass of a bean.

4. Finally we can multiply to find the mass of 1000 beans!

Bean 1 2 3 4 5 6 7 8 9 10

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Exercise

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•

Elements occur in nature as mixtures of

isotopes.

•

Carbon = 98.89%

12

C

1.11%

13

C

< 0.01%

14

C

•

A single atom of carbon does not have uniform

mass

•

When we count atoms by weighing we use an

average mass

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• _{What is the unit of mass for an atom?}

• _{It doesn’t make sense to refer to the mass of an atom}

in grams

• _{So chemists defined the Atomic Mass Unit} • _{Defined value}

• _{Originally corresponded the the mass of}1H

– _{Makes sense – smallest atom}

• _{Then for a while is was 1/16 the atomic weight of a}

single atom of oxygen

• _{Both of these roughly corresponded to a nucleon}

(particle that makes up the nucleus)

– _{Either a proton of neutron}

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•

Current definition of an atomic mass unit

•

One atom of

12

C is assigned the mass of

exactly 12 atomic mass units

•

So 1 amu = 1/12 the mass of an atom of

12

C

•

Masses of all other atoms are assigned relative

to this standard.

•

Most accurate method to determine atomic

mass is mass spectrometry.

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1) Vaporize sample

2) Beam of electrons – create positive ions 3) Ions are deflected in magnetic field

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How does deflection of ions help us determine atomic mass?

1. Large ions deflected the least and small ions deflected the most 2. Samples are always compared to a standard that is mixed into

the sample

When 12C and 13C are analyzed 13C is deflected slightly less

(more massive)

The degree of deflection of ions if translated into a ratio of masses

Mass Spectrometry

Mass 13_C

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• _{When we look at the periodic table, the atomic}

weight for carbon is not 12. It is 12.01.

• _{That is because carbon naturally exists as a}

mixture of isotopes.

• _{The mass of carbon is an average of the masses}

of the different isotopes.

• 12C, 13C and 14C

– _{All have 6 protons – that is what makes them}

carbon

– _{Have 6, 7 and 8 neutrons – that is what makes}

them isotopes

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• _{Depict the three isotopes of carbon,}12C, 13C and 14C, using isotopic notation

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• _{98.89% of}12C and 1.11% of 13C (14C is negligible)

• _{98.89% of 12 amu + 1.11% of 13.0034 amu =}

• _{(0.9889)(12 amu) + (0.0111)(13.0034 amu) =}

• _{12.01 amu}

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•

Even though natural carbon does not

contain a single atom with mass 12.01, for

stoichiometric

purposes, we can consider

carbon to be composed of only one type of

atom with an average mass of 12.01.

•

This enables us to count atoms of natural

carbon by weighing a sample of carbon.

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Learning Check

An element consists of 62.60% of an isotope

with mass 186.956 amu and 37.40% of an

isotope with mass 184.953 amu.

•

Calculate the

average atomic mass

and

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17

Avogadro’s Number and the Mole

• _{Samples of matter contain enormous numbers of atoms} • _{So just like we needed a special unit for the mass of an}

atom (amu), we need a unit for a collection of atoms

• _{Can’t use a dozen or a hundred} • _{The unit we use is the Mole}

• _{The number of carbon atoms in exactly 12 grams of}12C

– _{This number was determined to be 6.022 x 10}23 C

atoms (Again from mass spectrometry)

– _{Avogadro’s number}

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One-Mole Quantities

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19

Avogadro’s Number and the Mole

• _{This definition just seems to make things worse until you}

realize the definition has two parts

• _{The mole is both a number of particles and a mass} • _{The number of carbon atoms in exactly 12 grams of}12C

– _{6.022 x 10}23 12C atoms

– _{12 g of}12C

• _{There is a relationship here between the mass of a single}

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20

Avogadro’s Number and the Mole

• _{This means that a mole of any element (or compound) has a}

mass that is equal to its atomic weight expressed in grams.

• _{6.022 x 10}23 C atoms in a mole of natural carbon = 12.01 g

• _{6.022 x 10}23 He atoms in a mole of helium = 4.003 g

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Avogadro’s Number and the Mole

• _{So A mole is both a Number of particles and a Mass}

• _{Mole is a Number of particles} • _{A mole is a unit just like a dozen} • _{1 dozen = 12 of something}

• _{1 mole = 6.022 x 10}23 of something

• _{Mole is also a Mass}

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Avogadro’s Number and the Mole

• _{Avogadro’s number: The number of molecules or}

formula units in a mole. N_A = 6.022 x 1023

• _{Mole: One mole of any substance is the amount whose}

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Avogadro’s Number and the Mole

• _{This equation also gives us a relationship between the}

amu and the gram

• _{This relationship can be used as a conversion factor}

between amu and grams. The 1 is an exact value here because neither the 12 amu nor the 12 g were measured.

• _{The inverse gives the gram value of an amu.}

– _{We will use this on the next slide}

– _{Very close to the mass of a proton/neutron (nucleon)}

€

6.022 x 1023_{atoms 12 amu}

atom ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟

⎟= 12 g

€

6.022 x 1023_{amu = 1 g}

€

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Why is it that atomic mass

(amu/atom) = molar mass (g/mole)?

• _{Atomic mass (amu/atom)} _{= Molar mass( g/mole)}

• _{Has to do with the definition of}_amu_and_mole_{(which are related)}

• _{1. Amu = mass of 1/12 atom of C – 12 1.660578 x 10}– 24_g/amu

• _{2. Mole = # of atoms in 12 g of C – 12 6.022 x 10}23 atoms/mole

• _{Check out Avagadro’s Number and the Mole link on the website.}

24

€

amu

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Problem Solving

• _{So far in this chapter we have the following definitions that we can}

use to solve problems.

• _{6.022 x 10}23 atoms/mole Avogadro’s number

• _{6.022 x 10}23_amu/g

• _{Grams/mole = amu/atom} _{From periodic table}

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Learning Check

• _{Determine the mass of 10 atoms of sodium}

.

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Learning Check

• _{Determine both the number of moles and the number of atoms in}

25.0 g of calcium.

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Learning Check

• _{Determine the number of atoms in 5.50}m_{g of lead}

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Learning Check

• _{Determine the number of moles and the mass of 10.0 x 10}21_atoms

of silicon.

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Why do we need the mole?

► Reactions occur between individual particles or formula units

► Take this reaction for example

– 4FeO(s) + O₂(g) → 2Fe₂O₃(s)

– 4 FeO react with 1 O₂ to make 2 Fe₂O₃

– _{This is all fine and dandy but we cant weigh out}

individual molecules – they are way to small.

– _{We weigh out grams.}

– _{The mole is a way that we can weigh out quantities}

with equivalent numbers of particles

– 4 moles of FeO react with 1 mole of O₂ to make 2

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Why do we need the mole

► Reactions occur between individual particles or formula units

► Take this reaction for example

– 4FeO(s) + O₂(g) → 2Fe₂O₃(s)

– 4 particles FeO 1 particle of O₂ 2 particles of Fe₂O₃ – 4 moles of FeO 1 mole of O₂ 2 moles of Fe₂O₃ – _{4 x 6.022x10}23 1 x 6.022x1023 2 x 6.022x1023

– _{Same ratio of particles}

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•

Mass in grams of one mole of a substance:

Molar Mass of N = 14.01 g/mol atomic weight

Molar Mass of H₂O = 18.02 g/mol molecular weight

(2 × 1.008 g) + 16.00 g

Molar Mass of Ba(NO₃)₂ = 261.35 g/mol formula weight

137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

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Learning Check

• _{Glucose is one of the end products of photosynthesis, the process}

that converts CO₂ and H₂O to complex carbohydrates. The formula for glucose is C₆H₁₂O₆.

Determine the molar mass of glucose.

Determine the number of moles in 50.0 g of glucose.

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Learning Check

• Magnesium hydroxide (Mg(OH)₂) is the active ingredient in many

over the counter antacids.

Determine the molar mass of Mg(OH)₂.

Determine the number of moles in 6.25 g of Mg(OH)₂.

Determine the mass of OH –_{ions present in this sample.}

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•

Mass percent of an element:

€

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•

Mass percent the elements in glucose C

₆

H

₁₂

O

₆

:

€

mass % C = _{6 (12.01) +12 (1.008)+ 6(16.00) =}6 (12.01) _{180.16 x 100 = 40.00 %}72.06

€

mass % H = _{6 (12.01) +12 (1.008)+ 6(16.00) =}12 (1.008) _{180.16 x 100 = 6.716 %}12.10

€

mass % O = _{6 (12.01) +12 (1.008)+ 6(16.00) =}6 (16.00) _{180.16 x 100 = 53.29 %}96.00

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Exercise

•

Determine the mass percent of iron and oxygen

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•

Empirical formula = CH



Simplest whole-number ratio

•

Molecular formula = (empirical formula)

_n

[

n

= integer]

•

Molecular formula = C

₆

H

₆

= (CH)

₆



Actual formula of the compound

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•

When new compounds are discovered or

isolated the first thing that needs to be

determined is the chemical composition

•

Weigh the sample

•

Decompose the determine constituent

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•

One way to determine hydrogen and carbon

content of a compound is to react it with

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• _{How can this information be used to determine the formula of a}

compound?

• _{If a substance has been isolated that contains only C, H and N.} • _{If 0.1156 g of this sample is reacted with oxygen and 0.1638 g of}

CO₂ and 0.1676 g of H₂O are collected.

• _{The nitrogen is what is left over}

• _{0.1156 – (0.4470 + 0.01875) = 0.5215 g N}

• _{To determine formulas though we need numbers of atoms and}

grams don’t really help us with that

€

0.1638 g CO2 x 12.01 g C

44.01 g CO2 = 0.04470 g C

€

0.1676 g H2O x 2.016 g H

18.02 g H2O = 0.01875 g H

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• _{Ratio of moles is the same as ratio of atoms (Section 3.3 The}

Mole)

• _{Divide by the smallest value}

€

0.04470 g C x 1 mole C

12.01 g C = 0.00372 mole C

€

0.01875 g H x 1 mole

1.008 g H = 0.0186 mole H

€

0.05215 g N x 1 mole

14.00 g N = 0.00373 mole N

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• _{So we need to convert grams to moles}

• Empirical Formula = C₁H₅N₁

€

0.04470 g C x 1 mole C

12.01 g C = 0.00372 mole C/0.00372 = 1

€

0.01875 g H x 1 mole

1.008 g H = 0.0186 mole H/0.00372 = 5

€

0.05215 g N x 1 mole

14.00 g N = 0.00373 mole N/0.00372 = 1

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• _{The book has you convert the grams of each compound to a}

percentage

• _{We start by determining the percentages of each type of element}

• _{Since the compound contains only C, O and N we can figure out}

the percentage of nitrogen by subtraction

• _{100.00 % - (38.67% + 16.22%) = 45.11% N} 0.04470 g C

0.1156 g Compound x 100 = 38.67 % C

€

0.01875 g H

0.1156 g Compound x 100 = 16.22 % H

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• _{So we have 38.67 % C, 16.22 % H and 45.11 % N} • _{So if we had 100 g of the compound we would have} • _{38.67 g carbon}

• _{16.22 g hydrogen} • _{45.11 g nitrogen}

• _{To figure out the formula we convert grams to moles.}

• _{The ratio of moles in a compound is the same as the ratio of}

atoms in a compound

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Determine the smallest whole number ratio of atoms in the compound by dividing each of the values by the smallest value.

€

38.67 g C x 1 mol C_{12.01 g C = 3.220 mol C}

€

16.22 g H x 1 mol H_{1.008 g H = 16.09 mol H}

€

45.11 g N x 1 mol N_{14.01 g N = 3.220 mol N}

€

C: 3.220_{3.220 = 1.000 =1} H: 16.09_{3.220 = 4.997 = 5} N: 3.220_{3.220 = 1.000 =1}

CH

5N = Empirical Formula

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Empirical formula = CH₅N. Simplest ratio of atoms in a compound What is the actual molecular formula?

C₂H₁₀N₂? C₃H₁₅N₃? (CN₅N) _n

To determine the molecular formula you need the molar mass of the

compound in question. Can determine molar mass using any number of chemical methods (boiling point elevation, freezing point depression, etc.

Molar mass/empirical formula mass = n

So if the molar mass of our unknown compound determined to be 31.06 then it is the same as the empirical formula mass and n = 1

Molecular formula = CH₅N

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A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the

empirical formula of the compound.

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A compound contains only C, H and N. Combustion of 35.0 mg of the compound produces 33.5 mg of CO₂ and 41.1 mg of H₂O. What is the empirical formula of this compound?

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•

A representation of a chemical reaction:

•

Liquid ethanol reacts with gaseous oxygen to

produce carbon dioxide gas and water vapor.

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

State Symbol

Solid (s)

Liquid (l)

Gas (g)

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C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g) • _{Reactants on the left, products on the right.} • _{States of reactants and products are included} • _{The equation is balanced.}

– _{1 molecule of ethanol reacts with 3 molecule of oxygen}

to produce 2 molecule of carbon dioxide and 3 molecule of water.

– _{1 mole of ethanol reacts with 3 moles of oxygen to}

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1. Determine what reaction is occurring.

What are the reactants, the products, and

the physical states involved?

2. Write the

unbalanced

equation that

summarizes the reaction described in

step 1.

3. Balance the equation by inspection,

starting with the most complicated

molecule(s). The same number of each

type of atom needs to appear on both

reactant and product sides.

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4 NH

3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)

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Concept Check

Which of the following correctly balances the chemical equation given below?

CaO + C

→

CaC

₂

+ CO

₂

I. CaO

₂

+ 3C

→

CaC

₂

+ CO

₂

II. 2CaO + 5C

→

2CaC

₂

+ CO

₂

III. CaO + (2.5)C

→

CaC

₂

+ (0.5)CO

₂

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•

The number of atoms of each type of

element must be the same on both sides

of a balanced equation.

•

Subscripts must not be changed to

balance an equation.

•

A balanced equation tells us the ratio of

the number of molecules which react and

are produced in a chemical reaction.

•

Coefficients can be fractions, although

they are usually given as lowest integer

multiples.

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Learning Check

Elemental boron is produced by heating solid diboron trioxide with magnesium metal. Solid magnesium

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Learning Check

Balance the following equations.

FeO (s) + O₂(g) → Fe₂O₃(s)

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Learning Check

Balance the following equations.

Pb(NO₃)₂(aq) + Na₃PO₄(aq) → Pb₃(PO₄)₂(s) + NaNO₃(aq)

(59)

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•

Chemical equations can be used to relate the

masses of reacting chemicals.

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For the following equation: 4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)

How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas?

We can’t solve this by going from grams to grams because reactions take place between molecules or moles

We need moles as a conversion factor in between

grams to moles to moles to grams - GMMG

15.0 g of Cr to moles of Cr to moles of Cr₂O₃ to g of Cr₂O₃

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1. Balance the equation for the reaction.

2. Convert the known mass of the reactant

or product to moles of that substance.

3. Use the balanced equation to set up the

appropriate mole ratios.

4. Use the appropriate mole ratios to

calculate the number of moles of desired

reactant or product.

5. Convert from moles back to grams if

required by the problem.

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(63)

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4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)

15.0 g of Cr to moles of Cr to moles of Cr₂O₃ to g of Cr₂O₃

Grams of Cr to moles of Cr.

Moles of Cr to moles of Cr₂O₃

Moles of Cr₂O₃to grams of Cr₂O₃.

• _{Conversion string:}

1 mol Cr

15.0 g Cr = 0.288 mol Cr

52.00 g Cr 

2 3

2 mol Cr O

0.288 mol Cr = 0.144 mol Cr O

4 mol Cr 

2 3

2 3 2 3

2 3

152.00 g Cr O

0.144 mol Cr O = 21.9 g Cr O

1 mol Cr O 

2 3 2 3

2 mol Cr O 152.00 g Cr O 1 mol Cr

15.0 g Cr = 21.9 g Cr O

52.00 g Cr 4 mol Cr 1 mol Cr O

(64)

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Learning Check

Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is made according to the following

equation:

Sn(s) + 2HF(aq) → SnF₂(aq) + H₂(g)

(65)

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Learning Check

Consider the following reaction:

PCl₃ + 3H₂O → H₃PO₃ + 3HCl

(66)

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Learning Check

Consider the following reaction:

P

₄

(s) + 5 O

₂

(g)

→

2 P

₂

O

₅

(s)

(67)

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Learning Check (Part I)

Methane (CH

₄

) reacts with the oxygen in the

air to produce carbon dioxide and water.

Ammonia (NH

₃

) reacts with the oxygen in the

air to produce nitrogen monoxide and water.



Write

balanced equations

for each of

(68)

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Exercise (Part II)

What mass of ammonia would produce the

same

amount

of water as 1.00 g of methane reacting

(69)

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•

Where are we going?

 To find the mass of ammonia that would

produce the same amount of water as 1.00 g of methane reacting with excess oxygen.

•

How do we get there?

 We need to know:

• _{How much water is produced from 1.00 g of}

methane and excess oxygen.

– _{Do we need grams or moles?}

• _{How much ammonia is needed to produce}

(70)

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Exercise (Part II)

What mass of ammonia would produce the

same

amount

of water as 1.00 g of methane reacting

(71)

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• _{Contains the relative amounts of reactants that}

matches the numbers in the balanced equation.

N

₂

(

g

) + 3H

₂

(

g

)

→

2NH

₃

(

g

)

(72)

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•

Limiting reactant – the reactant that is

consumed first and therefore limits the

amounts of products that can be formed.

•

Determine which reactant is limiting to

calculate correctly the amounts of

products that will be formed.

(73)

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•

Hydrogen and nitrogen will react to form

ammonia according to the equation:

N₂ (g) + 3H₂(g) → 2NH₃ (g)

(74)

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•

The amount of products that can form is

limited by the hydrogen.

•

Hydrogen is the limiting reactant.

•

Nitrogen is in excess.

•

For this particular set of reaction

conditions

(75)

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•

We cannot simply add the total moles of

all the reactants to decide which reactant

mixture makes the most product. We

must always think about how much

product can be formed by using what we

are given, and the ratio in the balanced

equation.

(76)

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Example

How many grams of NH₃ can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? N₂ + 3H₂ → 2NH₃

€

3.00 g N2 x 1 mol N2

28.02 g N2= 0.107 mol N2 (how much N2 we have)

€

3.00 g H2 x 1 mol H2

2.016 g H2=1.49 mol H2 (how much H2 we have)

€

0.107 mol N2 x 3 mol H2

1 mol N2 = 0.321 mol H2 (H2 needed - have 1.49 so H2 not limiting)

€

1.49 mol H2 x 1 mol N2

3 mol H2= 0.497 mol N2 (N2 needed - only have 0.107 so N2 is limiting reactant)

0.107 mol N2 x 2 mol NH3

1 mol N2 x 17.03 g NH 3

1mol NH3

(77)

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Learning Check

(78)

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•

An important indicator of the efficiency of

a particular laboratory or industrial

reaction.

Percent Yield

Actual yield

(79)

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Example

Find the percent yield of product if 1.50 g of SO₃ is produced from 1.00 g of O₂ and excess sulfur via the reaction:

2S + 3O₂→ 2SO₃

(80)

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Example

Find the percent yield of product if 1.50 g of SO₃ is produced from 1.00 g of O₂ and excess sulfur via the reaction:

• 2S + 3O₂→ 2SO₃

•_{First we find the maximum yield (theoretical yield) of product:}

•_{We then divide the actual yield by the theoretical yield:}

€

Percent Yield =1.50 g_{1.67 g x 100 = 89.8 %}

€

1.00 g O2 x 1 mol O2

32.00 g O2 x 2 mol SO 3

3 mol O2 x 80.06 g SO 3

(81)

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Learning check

Consider the following reaction:

P

₄

(

s

) + 6F

₂

(

g

)

→

4PF

₃

(

g

)



What

mass of P

₄

is needed to produce

85.0 g of PF

₃

if the reaction has a 64.9%