chemical accounting. the chemical formula information that can be derived from a chemical formula: -...

Chemical Accounting

The Chemical Formula

Information that can be derived from a chemical formula:- Elements present- Proportions of elements contained (number of atoms contained in the molecule)- Molecular Weight- Review: Mass number vs. Atomic Mass- Carbon – 12 amu or 12.011 amu(C-12 98.9%, C-13 1.10% in nature)

We measure ordinary objects either by counting or weighing them,

depending on which method is more convenient

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.http://weyume.com/wp-content/uploads/2011/04/rice.jpg

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Dozen = 12

Pair = 2

Certain nouns can be used to define a collection of objects

The mole

The mole (n or mol) is the amount of matter that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in

exactly 12 g of the carbon-12 isotope (12C)

• The actual number of atoms in 12 g of carbon-12 was determined experimentally

• Avogadro’s number (NA)

NA = 6.02 x 1023

Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Just as 1 dozen of oranges contains 12 oranges, 1 mole of matter contains 6.02 x 1023 entities

Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

1 mole 12C atoms = 6.02 x 1023 12C atoms

1 mole H2O molecules = 6.02 x 1023 H2O molecules

1 mole NO3- ions = 6.02 x 1023 NO3

- ions

Each of these contains one mole of the substance

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

copper iron

carbon sulfur

mercury

One mole (or an Avogadro’s number)is an extremely big number

• One mole of softdrink cans would cover the surface of the earth to a depth of over 300 kilometers

• If we were able to count the number of atoms at a rate of 10 million per second, it would take about 2 billion years to count a mole of atoms

Molar mass

The molar mass (M) of a substance is the mass of one mole of its entities (atoms, molecules,

ions, or other particles) in units of g/mol

MH2O = 18.0 g/mol

(one mole of H2O molecule weighs 18.0 g)

MNO3- = 62.0 g/mol

(one mole of NO3- ion weighs 62.0 g)

MC = 12.01 g/mol

(one mole of C atom weighs 12.01 g)

Brown, T., E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-

Hill.

The periodic table is indispensable for calculating the molar mass of a substance

• Elements

– M is the numerical value from the periodic table

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MH = 1.008 g/mol

MO = 16.00 g/mol

The periodic table is indispensable for calculating the molar mass of a substance

• Compounds

– M is the sum of the molar masses of the atoms of the elements in the formula

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MSO2 = MS + (2 x MO)

= 32.07 g/mol + (2 x 16.00 g/mol)

= 64.07 g/mol

The periodic table is indispensable for calculating the molar mass of a substance

• Compounds

– M is the sum of the molar masses of the atoms of the elements in the formula

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MK2S = (2 x MK) + MS

= (2 x 39.10 g/mol) + 32.07 g/mol

= 110.27 g/mol

Interconverting moles, mass,and chemical entities

(atoms, molecules, ions, or other particles)

The factor-label method is used to convert from one unit to another

1 peso = 4 25-centavos

1 peso

4 25-centavos= 1

1 peso

4 25-centavos= 1

unit factor

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Alexa bought 18 fresh chicken’s eggs. How many dozens of egg did she buy?

dozens of egg = x18 eggs 1 dozen egg

12 eggs

= 1.5 dozens of egg

unit factor

In order to convert between moles, mass, and chemical entities, the factor label method is used

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

Methane (CH4) is the principal component of natural gas. How many moles of methane

are present in 6.07 g of CH4?

MCH4 = MC + (4 x MH)

= 12.01 g/mol + (4 x 1.01 g/mol)

= 16.05 g/mol

Report final answer with the correct number of significant figures!

nCH4 = x6.07 g CH4 1 mol CH4

16.05 g CH4

= 0.378 mol CH4

How many molecules of methane are present 6.07 g of CH4?

molecules CH4 = x 6.07 g CH4 1 mol CH4

16.05 g CH4

= 2.28 x 1023 molecules CH4

x6.02 x 1023 molecules CH4

mol CH4

Glucose (C6H12O6), also known as blood sugar, is used by the body as energy source.

How many moles of glucose are present in 1.75 x 1022 molecules of glucose?

nC6H12O6 = x 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6

6.02 x 1023 molecules C6H12O6

= 0.0291 mol C6H12O6

How many grams of glucose are present in 1.75 x 1022 molecules of glucose?

MC6H12O6 = (6 x MC) + (12 x MH) + (6 x MO)

= (6 x 12.01 g/mol) + (12 x 1.01 g/mol)

+ (6 x 16.00 g/mol)

= 180.18 g/mol

How many grams of glucose are present in 1.75 x 1022 molecules of glucose?

nC6H12O6 = x 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6

6.02 x 1023 molecules C6H12O6

180.18 g C6H12O6x

1 mol C6H12O6

= 5.24 g C6H12O6

1. Urea [(NH2)2CO] is used in animal feeds, as fertilizer and in the manufacture of polymers.

a. Draw the Lewis structure of area. C is surrounded by O and the N’s. (Where are the H’s connected?)

b. Calculate its molar mass.

c. Consider 25.6 g of urea. How many moles of urea present?

d. How many moles of N are present?

e. How many moles of C are present?

f. How many molecules of urea are present?

g. How many atoms of N are present?

2. Vitamin C, ascorbic acid, is often sold as sodium ascorbate.

a. Calculate its molar mass.

b. Consider a 500.-mg tablet. How many moles of sodium ascorbate are present?

c. How many moles of C are present?

d. How many moles of Na are present?

e. How many formula units of sodium ascorbate are present?

f. How many atoms of Na are present?

OO

OH

-O

HOOH

Na+

1. Monosodium glutamate (MSG), a food flavor enhancer, has been blamed for “Chinese Restaurant Syndrome”. Its molecular formula is C5H8NO4Na.

a. Calculate the molar mass of MSG.

b. Consider a 2.00-g sachet. How many moles of MSG are present?

c. How many moles of C are present?

d. How many moles of N are present?

e. How many formula units of MSG are present?

f. How many atoms of O are present?

HO O-

O O

NH2

Na+

Chemical reactions and chemical equations

A chemical reaction shows the process in which a substance (or substances) is

changed into one or more new substances

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants product

(g) (g) (l)

“Two molecules of hydrogen react with one molecule of oxygen to yield two moles of water”

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

The Law of Conservation of Mass states that matter is neither created nor destroyed

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

To conform with the Law of Conservation of Mass, there must be the same number of each type of atom on both sides of the arrow. Hence, we balance the equation by

adding coefficients before each chemical symbol

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Calculating the amounts of reactant and product

Stoichiometry a double cheeseburger

2 bun slices + 2 cheese slices + 2 burger patties =

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 1 mol O2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

1 mol O2

= 11 mol O2

2 mol CO= 1

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

2 mol CO2

= 12 mol CO2

2 mol CO= 1

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

1 mol O2 = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

1 mol O2

2 mol CO2

= 12 mol CO2

1 mol O2

= 1

The amount of one substance in a reaction is related to that of any other

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

All alkali metals react with water to produce hydrogen gas and the corresponding

alkali metal hydroxide

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water?

nH2 = x6.23 mol Li 1 mol H2

2 mol Li

= 3.12 mol H2

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

How many grams of H2 will be formed by the complete reaction of 80.57 g of Li with water?

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

mH2 = x x80.57 g Li 1 mol Li 1 mol H2

6.941 g Li 2 mol Li

2.016 g H2x

1 mol H2

= 11.70 g H2

In a lifetime, the average American uses about 794 kg of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores (such as Cu2S) by a multistep process. After an initial

grinding, the first step is to “roast” the ore (heat it strongly with O2) to form Cu2O and SO2

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

How many moles of oxygen are required to roast 10.0 mol of Cu2S?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

nO2 = x10.0 mol Cu2S 3 mol O2

2 mol Cu2S

= 15.0 mol O2

How many grams of SO2 are formed when 10.0 mol of Cu2S is roasted?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mSO2 = x x 10.0 mol Cu2S 2 mol SO2 64.07 g SO2

2 mol Cu2S 1 mol SO2

= 641 g SO2

How many grams of O2 are required to form 2.86 kg of Cu2O?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mO2 = x x2.86 kg Cu2O 1000 g Cu2O 1 mol Cu2O

1 kg Cu2O 143.10 g Cu2O

3 mol O2 32.00 g O2x x

2 mol Cu2O 1 mol O2

= 960 g O2

Within the cylinders of a car’s engine, the hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor.

a.How much carbon dioxide ( in kg) is produced when 2.00 kg of octane is burned ?

b.How much oxygen(in kg) is required to burn the same amount of octane?

2C8H18(l) + 25O2 (g) 16CO2 (g) + 18H2O (g)

Limiting Reactants

The reactant used up first in a chemical reaction is called the limiting reactant. Excess reactants are present in quantities greater than necessary to react with the quantity of the limiting reactant.

A + B --- C + D

Given the amounts of A and B, which is the limiting reactant? How much C and D are produced?

Urea is prepared by reacting ammonia with carbon dioxide:

2NH3(g) + CO2(g) -- (NH2)2CO(aq) + H2O(l)

In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.

• Which is the limiting reactant?

• How much urea (in grams) is produced?

• How much of the excess reactant (in grams) is left at the end of the reaction?

Strategy

• Check if the equation is balanced.

• Convert mass of each reactant to moles.

• Calculate the amount of product formed from the each of the reactants.

• The reactant the produces the less amount is the limiting reactant.

1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000⁰C and is used in welding metals:

Al + Fe2O3 -- Al2O3 + 2Fe

In one process, 124 g of Al are reacted with 601 g of ferric oxide.

1. Which is the limiting reactant?

2. How much Al2O3 (in grams) is produced?

3. How much of the excess reactant (in grams) is left at the end of the reaction?

2. Titanium is a strong & light metal used in rockets & aircrafts. It is prepared by the reaction between titanium (IV) chloride with molten magnesium at around 1000⁰C:

TiCl4 + Mg -- Ti + 2MgCl2

In a certain industrial operation, 3.54 x 107g of TiCl4 are reacted with 1.13 x 107 g of magnesium.•Which is the limiting reactant?•How much Ti (in grams) is produced?•How much of the excess reactant (in grams) is left at the end of the reaction?

COMPOSITION STOICHIOMETRY:

Mass Percentage / % Composition of Elements in a

Compound

What is the percent composition of carbon in the following:a. CO2

b. C6H12O6

Percent Composition

1. Compute for the % composition of the elements in H2SO4.2. Determine the formula for a compound that consists of 40% C, 6.72% H and 53.29% O. Hint: Assume 100g of the compound

Elemental Composition

Empirical formula: Simplest formula for a compound (usually computed through elemental composition)

Molecular formula: Represents the actual number of atoms of each element present in a molecule of the compound

Empirical vs. Molecular Formulae

The elemental composition of a compound was found to be 4.8% H, 57.1% C and 38.0% S. Determine the molecular formula if the molecular weight of the actual compound is 168 g/mol.

Molecular Formula Sample Problem

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The Mole In Gases

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The Mole

Molar volume of a gas: One mole of any gas occupies a volume of 22.4 L at standard temperature and pressure (STP).

STP is defined as 1 atmosphere (atm) of pressure and a temperature of 0 oC.

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The Mole

SOLUTION STOICHIOMETRY

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Solutions and Concentration

The amount of solute in a given amount of solvent is defined as solution concentration.

A dilute solution contains relative small amounts of solute in a given amount of solvent.

A concentrated solution contains relatively large amounts of solute in a given amount of solvent.

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Solutions

Molarity (M) is defined as the moles of solute per liter of solution.

M =liter

mol

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Solutions

Percent Concentration

Percent by volume = x 100

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Solutions

Percent Concentration

Percent by mass = x 100

Sample Problems• How many moles of nitric acid is present in 10.0

mL of a 1.25 M solution?

• Suppose you have 2.00 mole of sucrose (approx. 685 grams) diluted in 1.75 L of water. What is the concentration of your sugar solution? What is its concentration in percent by mass? (Assume density of water is 1.00 g/mL)

• What is the molarity of a solution that consists of 2.00 grams of KMnO4 (MW: 158.0 g/mol) dissolved into 100.0 mL of solution?