stoichiometric

BY :

AGUNG RAHMADANIDEPARTMENT OF PHARMACY

UNIVERSITY OF MULAWARMANSAMARINDA

2014

1

STOICHIOMETRIC

General Chemistry

Stoichiometric2

Atomic Masses, Mole concept, and Molar Mass (Average atomic mass).

Number of atoms per amount of element.

Percent composition and Empirical formula of molecules.

Chemical equations, Balancing equations, and Stoichiometric calculations including limiting reagents.

Chemical Stoichiometry3

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

4

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

Atomic Masses5

Elements occur in nature as mixtures of isotopes

Carbon = 98.89% 12C 1.11% 13C <0.01% 14C

Carbon, average atomic mass = 12.01 amu

6

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016100

= 6.941 amu

Average atomic mass of lithium:

The Mole7

The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

1 mole of anything = 6.022 1023 units

AVOGADRO’S NUMBER EQUALS 6.022 1023 UNITS

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Molar Mass9

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

C=12.01 O=16

CO2 = (12.01+16+16) = 44.01 grams per mole

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Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)

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1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

Calculating Atoms, Moles, and Mass

We use the following conversion factors:

• Density converts grams – milliliters

• Atomic mass unit converts amu – grams

• Avogadro’s number converts moles – number of atoms

• Molar mass converts grams – moles

Strategy for Calculations

Map out a pattern for the required conversion

Given a number of grams and asked for number of atoms

Two conversions are required • Convert grams to moles

1 mol S/32.06 g S OR 32.06 g S/1 mol S• Convert moles to atoms

mol S x (6.022 x 1023 atoms S) / 1 mol S

Practice Calculations

1. Calculate the number of atoms in 1.7 moles of boron.

2. Find the mass in grams of 2.5 mol Na (sodium).

3. Calculate the number of atoms in 5.0 g aluminum.

4. Calculate the mass of 5,000,000 atoms of Au (gold)

Interconversion Between Moles, Particles, and Grams

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Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

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Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Proportional Relationships

StoichiometryStoichiometry mass relationships between substances in a

chemical reaction based on the mole ratio

Mole RatioMole Ratio indicated by coefficients in a balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

Stoichiometry Steps

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

Mole ratio - moles moles Molar mass - moles grams Molarity - moles liters soln Molar volume - moles liters gas

1 mol of a gas=22.4 Lat STP

Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

Stoichiometry Problems

How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?


3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

O2 KClO3

x mol KClO3 = 9 mol O2 = 6 mol KClO3

2 mol KClO3

3 mol O26 mol


How many grams of KClO3 are required

to produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L2KClO3 2KCl + 3O2




? g 9.00 L2KClO3 2KCl + 3O2

O2 KClO3

x g KClO3 = 9.00 L O222.4 L O2

= 32.8 g KClO3


3 mol O2

122.55 g KClO3

1 mol KClO3

32.8 g



9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L2KClO3 2KCl + 3O2

O2 KClO3

x g KClO3 = 9.00 L O222.4 L O2

= 32.8 g KClO3


3 mol O2

122.55 g KClO3

1 mol KClO3

32.8 g

How many grams of silver will be formed from 12.0 g copper?

12.0g Cu

1 molCu

63.55g Cu

= 40.7 g Ag

Cu + 2 AgNO3 2 Ag +

Cu(NO3)2 2 mol

Ag

1 molCu

107.87g Ag

1 molAg

12.0 g ? g

Cu Ag

x g Ag = 12.0 g Cu63.55 g Cu

= 40.7 g Ag1 mol Cu 2 mol Ag

1 mol Cu107.87 g Ag

1 mol Ag40.7 g

useAvogadro’s

number

use molar mass

Mole Calculations

Volume, cm3 Mass, g Moles Atomsuse density

gcm3x

mol gx

atoms molx

A graduated cylinder holds 25.4 cm3 of mercury. If the density of mercuryat 25 oC is 13.534 g / cm3, how many moles of mercury are in the cylinder?

HINT: Volume of solids/liquids and moles are not directly connected. You must first use the density to convert the volume to a mass, and then derive the quantity of mercury, in moles, from the mass. Finally, the number of atoms is obtained from the number of moles.

How many atoms of mercury are there?

Therefore, the mass of mercury is found to be equivalent to 344 g of mercury.

Volume, cm3 Mass, g Moles Atomsuse density use

molar massuse

Avogadro’snumber

gcm3x

mol gx

. 13.534 g Hg 1 cm3 Hg

25.4 cm3 Hg =344 g Hg

Knowing the mass, you can now find the quantity in moles.

. 1 mol Hg . 200.6 g Hg

344 g Hg =1.71 mol Hg

Finally, because you know the relation between atoms and moles (Avogodro’snumber), you can now find the number of atoms in the sample.

. 6.02 x 1023 atoms Hg . 1 mol Hg

1.71 mol Hg =1.03 x 1024 atoms Hg

A

A B C

B

CA graduated cylinder holds 25.4 cm3 of mercury. If the density of mercuryat 25 oC is 13.534 g / cm3, how many moles of mercury are in the cylinder?How many atoms of mercury are there?

344 g Hg

1.71 mol Hg

atoms molx

2 Na + Cl2 2 NaCl2 grams 1 gram 2 grams W R O N G

Violates Law of Conservation of Matter

2 atoms 1 molecule 2 molecules*

*Better name would be “formula unit”

2 moles 1 mole 2 moles

100 g x L x g

Na Cl2

x L Cl2 = 100 g Na23 g Na

= 49 L Cl21 mol Na 1 mol Cl2

2 mol Na22.4 L Cl21 mol Cl2

48.69 L

Right side of room…calculate how many grams of NaCl will be produced from 100 g of Na.

Left side of room…calculate how many grams of NaCl will be produced from 48.69 L of Cl2.

Na NaCl

x g NaCl = 100 g Na23 g Na

= 254 g NaCl1 mol Na 2 mol NaCl

2 mol Na58.5 g NaCl1 mol NaCl

Cl2 NaCl

x g NaCl = 48.69 L Cl222.4 L Cl2= 254 g NaCl

1 mol Cl2 2 mol NaCl1 mol Cl2

58.5 g NaCl1 mol NaCl

Stoichiometry

KClO3 KCl + O232 2500 g x g x L

KClO3 O2

x L O2 = 500 g KClO3122.5 g KClO3

= 137 L O2

1 mol KClO3 3 mol O2

2 mol KClO3

22.4 L O2

1 mol O2

KClO3 KCl

x g KCl = 500 g KClO3122.5 g KClO3

= 304 g KCl1 mol KClO3 2 mol KCl

2 mol KClO3

74.5 g KCl1 mol KCl

(304 g)

(196 g)

x g O2 = 137 L O2 22.4 L O2

= 196 g O2

1 mol O2 32 g O2

1 mol O2

137 L

Stoichiometry

2 TiO2 + 4 Cl2 + 3 C CO2 + 2 CO + 2 TiCl4

4.55 molx mol

C Cl2

x mol C = 4.55 mol Cl24 mol Cl2= 6.07 mol Cl2

3 mol C

C TiO2

x molecules TiCl4 = 115 g TiO280 g TiO2

= 8.66x1023 molecules TiCl4

1 mol TiO2 2 mol TiCl42 mol TiO2

6.02x1023 molecules TiCl41 mol TiCl4

x g TiO2 = 4.55 mol C3 mol C

= 243 g TiO2

1 mol TiO2 80 g TiO2

2 mol TiO2

How many moles of chlorine will react with 4.55 moles of carbon?

How many grams of titanium (IV) oxide will react with 4.55 moles of carbon?

x g

How many molecules of TiCl4 will react with 115 g TiO2?

TiO2 TiCl4

x molecules115 g

6.02x1023 atoms H2O6.02x1023 molecules H2O

Which has more atoms: 30 g aluminum metal or 18 mL distilled water?

x atoms Al = 30 g Al27 g Al

= 6.69x1023 atoms Al1 mol Al 6.02x1023 atoms Al

1 mol Al

How many atoms of aluminum are in 30 g of aluminum?

Al

x atoms Al = 30 g Al27 g Al

= 6.69x1023 atoms Al6.02x1023 atoms Al

Al

x atoms = 18 mL H2O1000 mL H2O

= 1.45x1021 atoms

1 L H2O 1 mol H2O22.4 L H2O

How many atoms are in 18 mL of water? H2O

x atoms = 18 mL H2O1 mL H2O

= 1.81x1024 atoms

1 g H2O 1 mol H2O 18 g H2O

How many atoms are in 18 mL of water?

1 mol H2O3 atoms

1 molecule H2O

1 moL H2O6.02x1023 molecules H2O 3 atoms

1 molecule H2O

W R O N GW R O N G

Recall, density of water

LITERS is ONLY used for GASES @ STPLITERS is ONLY used for GASES @ STP

Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen.

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Determining Elemental Composition(Formula)

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The masses obtained (mostly CO2 and H2O and sometimes N2)) will be used to determine:

1. % composition in compound2. Empirical formula3. Chemical or molecular formula if the

Molar mass of the compound is known or given.

Writing Chemical Reactions

Consider the following reaction:

hydrogen reacts with oxygen to produce

water Write the above reaction as a

chemical equationH2 + O2 H2O

Don’t forget the diatomic elements

Law of Conservation of Mass

Law of conservation of mass - matter cannot be either gained or lost in the process of a chemical reaction The total mass of the products must

equal the total mass of the reactants

A Visual Example of the Law of Conservation of Mass

4.4 Balancing Chemical Equations

A chemical equation shows the molar quantity of reactants needed to produce a particular molar quantity of products

The relative number of moles of each product and reactant is indicated by placing a whole-number coefficient before the formula of each substance in the chemical equation

)(O )2Hg( )2HgO( 2 gls

Coefficient - how many of that substance are in the reaction

Balancing

The equation must be balanced All the atoms of every reactant must

also appear in the products Number of Hg on left? 2

on right 2 Number of O on left? 2

on right 2

Examine the Equation

H2 + O2 H2O

Is the law of conservation of mass obeyed as written? NO

Balancing chemical equations uses coefficients to ensure that the law of conservation of mass is obeyed

You may never change subscripts!

WRONG: H2 + O2 H2O2

Step 1. Count the number of moles of atoms of each element on both product and reactant sides

Reactants Products

2 mol H 2 mol H 2 mol O 1 mol O

Steps in Equation Balancing

The steps to balancing:H2 + O2 H2O

H2 + O2 H2O

H2 + O2 2H2O

This balances oxygen, but is hydrogen still balanced?


Step 2. Determine which elements are not balanced – do not have same number on both sides of the equation Oxygen is not balanced

Step 3. Balance one element at a time by changing the coefficients

Reactants Products

4 mol H 4 mol H

2 mol O 2 mol O


H2 + O2 2H2O

How will we balance hydrogen?

2H2 + O2 2H2O

Step 4. Check! Make sure the law of conservation of mass is obeyed

Balancing an Equation

Practice Equation Balancing

Balance the following equations:

1. C2H2 + O2 CO2 + H2O

2. AgNO3 + FeCl3 Fe(NO3)3 + AgCl

3. C2H6 + O2 CO2 + H2O

4. N2 + H2 NH3

Calculations Using the Chemical Equation

Calculation quantities of reactants and products in a chemical reaction has many applications

Need a balanced chemical equation for the reaction of interest

The coefficients represent the number of moles of each substance in the equation

General Principles

1. Chemical formulas of all reactants and products must be known

2. Equation must be balanced to obey the law of conservation of mass

• Calculations of an unbalanced equation are meaningless

3. Calculations are performed in terms of moles

• Coefficients in the balanced equation represent the relative number of moles of products and reactants

Using the Chemical Equation

Examine the reaction:

2H2 + O2 2H2O

Coefficients tell us?

2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O

What if 4 moles of H2 reacts with 2 moles of O2?

It yields 4 moles of H2O

2H2 + O2 2H2O

Using the Chemical Equation

The coefficients of the balanced equation are used to convert between moles of substances

How many moles of O2 are needed to react with 4.26 moles of H2?

Use the factor-label method to perform this calculation

2H2 + O2 2H2O

2

22 H mol__

O __molH mol 26.4

1

2

2.13 mol O2

Use of Conversion Factors

Digits in the conversion factor come from the balanced equation

Conversion Between Moles and Grams

Requires only the formula weight Convert 1.00 mol O2 to grams

Plan the path Find the molar mass of oxygen

32.0 g O2 = 1 mol O2 Set up the equation Cancel units 1.00 mol O2 x 32.0 g O2

1 mol O2 Solve equation 1.00 x 32.0 g O2 =

32.0 g O2

moles ofOxygen

grams ofOxygen

Conversion of Mole Reactants to Mole Products

Use a balanced equation C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

1 mol C3H8 results in: 5 mol O2 consumed 1 mol C3H8 /5 mol O2

3 mol CO2 formed 1 mol C3H8 /3 mol CO2

4 mol H2O formed 1 mol C3H8 /4 mol H2O

This can be rewritten as conversion factors

Calculating Reacting Quantities

Calculate grams O2 reacting with 1.00 mol C3H8

Use 2 conversion factors Moles C3H8 to moles O2

Moles of O2 to grams O2

Set up the equation and cancel units

1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 = 1 mol C3H8 1 mol O2

1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2

moles Oxygen

grams Oxygen

moles C3H8

Calculating Grams of Product from Moles of Reactant

Calculate grams CO2 from combustion of 1.00 mol C3H8

Use 2 conversion factors Moles C3H8 to moles CO2 Moles of CO2 to grams CO2


1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 = 1 mol C3H8 1 mol CO2

1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2

moles CO2

grams CO2

moles C3H8

Relating Masses of Reactantsand Products

Calculate grams C3H8 required to produce 36.0 grams of H2O

Use 3 conversion factors Grams H2O to moles H2O Moles H2O to moles C3H8 Moles of C3H8 to grams C3H8


36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8 18.0 g H2O 4 mol H2O 1 mol C3H8

36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8

moles H2O

grams C3H8

moles C3H8

grams H2O

Calculating a Quantity of Reactant

Ca(OH)2 neutralizes HCl Calculate grams HCl neutralized by 0.500

mol Ca(OH)2 Write chemical equation and balance

Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l) Plan the path


0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl 1 mol Ca(OH)2 1 mol HCl

Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl

molesCa(OH)2

grams HCl

molesHCl

General Problem-solving Strategy

Na + Cl2 NaCl

Sample Calculation

1. Balance the equation

2. Calculate the moles Cl2 reacting with 5.00 mol Na

3. Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2

4. Calculate the grams Na reacting with 5.00 g Cl2

2Na + Cl2 2NaCl

%100yield ltheoretica

yield actual yield %

Theoretical and Percent Yield

Theoretical yield - the maximum amount of product that can be produced Pencil and paper yield

Actual yield - the amount produced when the reaction is performed Laboratory yield

Percent yield: = 125 g CO2 actual x 100% = 97.4% 132 g CO2 theoretical

Sample Calculation

If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield:

2 Al(s) + Fe2O3(s) Al2O3(aq) + 2Fe(aq)

[25.0 g / 30.0 g] x 100% = 83.3%

Calculate the % yield if 26.8 grams iron was collected in the same reaction

LIMITING REACTANT

The limiting reactant is the reactant that is completely consumed or finished first, controlling the amounts of products formed.

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63

6 green used up6 red left over

Limiting Reagents

Limiting Reactant64

5 cars + 200 drivers Limiting cars or drivers?

50 chairs + 15 students Limiting chairs or students?

Solving a Stoichiometry Problem

65

1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is

limiting. 4. Use moles of limiting reactant

and mole ratios to find moles of desired product.

5. Convert from moles to grams.

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Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted. Its amount isCalculated using the balanced equation.

Actual Yield is the amount of product actually obtainedfrom a reaction. It is always given.

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Percent Yield

Actual yield = quantity of product actually obtained

Theoretical yield = quantity of product predicted by stoichiometry using limiting reactant

68

Percent Yield Example

14.4 g excess

Actual yield = 6.26 g

Copyright©2000 by Houghton Mifflin Company. All rights reserved. 69

For the reaction: (Molar Masses: Cr = 52.00; Cl = 35.45 g/mol)

2 Cr (s) + 3 Cl2 (g) ========> 2 CrCl3 (s)

(a) How many grams of CrCl3 are produced by 1.00 g of Cr ?

(b) How many grams of CrCl3 are produced by 2.00 g of Cl2 ?

(c) If the Actual yield of CrCl3 is 2.65 g, then what is the Percent Yield of the CrCl3 ?

70

Sample Exercise

Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine -

TiO2(s) + 2 Cl2(g) + C(s) TiCl4(l) + CO2(g)

If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced?

Practice Example 171

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

Solution: 1. Determine C and H, the rest from 33.5mg is N. 2. Determine moles from masses. 3. Divide by smallest number of moles.

Practice Example 272

Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula.

Solution:1. Convert mass to moles.2. Determine empirical formula.3. Determine actual formula.

C8H10N4O2

stoichiometric

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