Chapter 3 Chemical Reactions and Earth’s Composition

Chapter 3

Chemical Reactions and Earth’s Composition

The Development of Earth

• Matter in the Universe condensed into planets.

• The planets closer to the sun have different chemical compositions than the rest of the universe. Many of the volatile chemicals were lost

from these planets.

The Composition of Compounds

• The law of definite proportions states that a specific chemical compound obtained from any source always contains the same proportion by mass of its elements

• H+ + OH- ---> H2O

• 2 H2 + O2 ---> 2 H2O

The Composition of Compounds

• The law of multiple proportions states that the masses of element Y that combine with a fixed mass of elements X to form two or more different compounds are in the ratios of small whole numbers.

• Examples: NO, NO2, N2O, N2O5, etc.

Chemical Equations

• Reactants ↔ Products

Chapter 4 Definitions

Chemical Reaction - A process in which substances are changed into other substances through rearrangement, combination, or separation of atoms.Chemical Equation - A written representation of a chemical reaction, showing the reactants and products, their physical states, and the direction of the reaction.Reactants - The starting material in a chemical reaction or equation.Products - The substances formed in a chemical reaction or equation.Physical States - solids (s), liquids (l), gases (g) and aqueous (aq).Balanced Chemical Equation - A written representation of a chemical reaction that gives the relative amounts of the reactants and products, their physical states, and the direction of the reaction.

Balanced Chemical Equations

1. The reactants appear on the left and the products appear on the right. The adjoining arrow shows the direction of the reaction.

2. The phase of the reactant or product is written after the chemical symbol and is in parentheses.

3. An integer precedes the chemical formula of each substance. This number, known as the stoichiometric coefficient, is the smallest integer that allows the equation to be balanced.

4. Matter and charge are conserved in balanced chemical equations.

Chemical Reactions

Avogadro number and the mole.

NA = 6.0221367 E23 mol-1

A mole is the amount of a substance that contains as many elementary particles (atoms, molecules, or whatever) as there are in exactly 12 g of the carbon-12 isotope.

1 mol = 6.0221367 E23 particles

Problem

Express the following estimates for the year 2010 in nanomoles.

a. USA. 298. million

b. China. 1.34 billion

c: US Debt (to China): 3 Trillion

Mass ↔ moles ↔ molecules ↔ atoms

Problem 39

Aluminum, silicon, and oxygen form minerals known as aluminosilicates. How many moles of aluminum are in 1.50 moles of:

a. pyrophyllite, Al2Si4O10(OH)2

b. Mica, KAl3Si3O10(OH)2

Problem 44. How many moles of O2- ions are in 0.55 mol of Aluminum oxide? What is their mass in grams?

Answers: 1.65 mol O2-, 26.4 grams

Conversions

• Converting between a number of particles and an equivalent number of moles (or vice versa) is a matter of dividing (or multiplying) by Avogadro’s number.

Molar Mass

• The average mass of an atom of helium is 4.003 amu, and the mass of a mole of helium (6.022 x 1023 atoms of He) is 4.003 g.

• The molar mass (M) of helium is 4.003 g/mol.

Molar Mass

• A substance’s molar mass is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per mole

C + 2O

12.01 + 2(16.00) = 44.01g/mol

Problem. Calculate the molar masses of the following:

a. sucrose, C12H22O11

Answer: 342.31

d. fructose, C6H12O6

Answer: 180.158

Mole Calculations

• How many moles of Ca atoms are present in 20.0 g of calcium?

• How many Cu atoms are present in 15.0 g of copper?

Mole Calculations

• How many grams are present in 3.40 moles of nitrogen gas (N2)?

• How many molecules are present in 5.32 moles of chalk (CaCO3)?

A. How many oxygen atoms are present in this sample?

Mole Calculations

The uranium used nuclear fuel exists in nature in several minerals. Calculate how many moles of uranium are found in 100.0 grams of carnotite, K2(UO2)2(VO4)2•3H2O.

ans: 902.176 g/mol; 0.4434 moles of U.

100 gram of K2(UO2)2(VO4)2•3H2O

Law of Conservation of Mass

• The law of conservation of mass states that the sum of the masses of the reactants of a chemical equation is equal to the sum of the masses of the products.

Chemical Change

• Chemical reactions follow the law of conservation of mass.

Balanced Chemical Equations

• Chemical equations should be balanced to follow the law of conservation of mass. Total mass of each element on the

reactant side must equal the total mass of each element on the product side.

Total charge of reactant side must equal the total charge of product side.

Combustion Reactions

• Reactions that occur between oxygen (O2) and another element in a compound. When the other compound is a

hydrocarbon, the products of complete combustion are carbon dioxide and water vapor.

• Hydrocarbons are molecular compounds composed of only hydrogen and carbon and are a class of organic compounds.

Practice

Balance the following equations for the following combustions reactions.

C3H8 + O2 CO2 + H2O

C5H10 + O2 CO2 + H2O

Stoichiometric Calculations

• Calculating the mass of a product from the mass of a reactant requires: A balanced chemical reaction Molar mass of the reactant Molar mass of the product

Example

How much carbon dioxide would be formed if 10.0 grams of C5H12 were completely burned in oxygen?

C5H12 + 8 O2 ---> 5 CO2 + 6 H2O

Percent Composition

• Mass percent of an element in a compound

Mass % = mass of element in compound x 100% mass of compound

• Practice: Calculate the percent of iron in iron(III) oxide, (Fe2O3).

Mass Percent and Empirical Formulas

Problem . A sample of an iron compound is 22.0% Fe, 50.2% oxygen, and 27.8% chlorine by mass. What is the empirical formula of this compound.

Answer: FeO8Cl2

Simplified Carbon Cycle

Problem

Sodium carbonate (105.988 g/mol) reacts with hydrochloric acid (36.461) to produce sodium chloride, water, and carbon dioxide. How much hydrochloric acid is required to produce 10.0 g of carbon dioxide (44.01g/mol)?

Combustion Analysis

CaHb + excess O2 ---> a CO2(g) + b/2 H2O

The percent of carbon and hydrogen in CaHb can be determined from the mass of H2O and CO2 produced.

Percent Composition and Empirical Formulas

1. Assume there is 100 g of the sample, so the percent composition will equal the number of grams of each element.

2. Convert the grams of each element into the moles of each element with their molar mass.

3. Divide the smallest number of moles of an element into the moles of each element present.

4. Convert the fractional ratios for each element into whole numbers by multiplying all the ratios by the same number.

5. The resulting numbers are the subscripts for the each element in the empirical formula.

Example

Asbestos was used for years as an insulating material in buildings until prolonged exposure to asbestos was demonstrated to cause lung cancer. Asbestos is a mineral containing magnesium, silicon, oxygen, and hydrogen. One form of asbestos, chrysotile (520.27 g/mol), has the composition 28.03% magnesium, 21.60% silicon, 1.16% hydrogen. Determine the empirical formula of chrysotile.

Mass Spectrometry and Molecular Mass

• All Mass spectrometers separate atoms and molecules by first converting them into ions and then separating those ions based on the ratio of their masses to their electric charges.

• Mass spectrometers are instruments used to determine the mass of substances.

Mass Spectrometer

Mass Spectra

Determining the Molecular Formula

• The molecular formula can be determined from the percent composition and mass spectral data.

Example

A combustion analysis of an unknown compound indicated that it is 92. 23% C and 7.82% H. The mass spectrum indicated the molar mass is 78 g/mol. What is the molecular formula of this unknown compound?

Limiting Reactants

During photosynthesis a reaction mixture of carbon dioxide and water is converted to a molecule of glucose.

Ham and Cheese as “limiting” reagents…

Limiting Reagents

• The limiting reactant is completely consumed in the chemical reaction.

The amount of product formed

depends on the amount of the limiting reagent available.

Example

10.0 g of methane (CH10.0 g of methane (CH44) is burned in 20.0 ) is burned in 20.0 g of oxygen (Og of oxygen (O22) to produce carbon ) to produce carbon dioxide (COdioxide (CO22) and water (H) and water (H22O). O).

a.a. What is the limiting reactant?What is the limiting reactant?

b.b. How many grams of water will be How many grams of water will be produced?produced?

Percent Yield

• Theoretical Yield:Theoretical Yield: the calculated the calculated amount of product formed from a given amount of product formed from a given amount of reactant(s)amount of reactant(s)

• Actual Yield:Actual Yield: the measured amount of the measured amount of product formedproduct formed

Percent YieldPercent Yield = Actual Yield x 100%= Actual Yield x 100% Theoretical Yield Theoretical Yield

Percent Yield Problem

Aluminum burns in bromine liquid producing aluminum bromide. In a certain experiment, 6.0 g of aluminum was reacted with an excess of bromine to yield 50.3 g aluminum bromide. Calculate the theoretical and percent yields for this experiment.

The Empirical* molecular formula can be directly calculated from percent composition data

*based on experiment(s)

ChemTour: Avogadro’s Number

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This ChemTour provides a step-by-step explanation of how to use Avogadro’s number to convert from moles to molecules or atoms and vice versa.

ChemTour: Balancing Equations


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This ChemTour reviews the procedure for balancing the mass of reactants on the left side of the equation against the mass of products on the right side.

ChemTour: Carbon Cycle


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This ChemTour shows how carbon is cycled and stored in Earth’s environment.

ChemTour: Percent Composition


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This ChemTour provides a step-by-step explanation for finding the percent composition of elements in a substance from the molecular formula and for finding the empirical formula from the percent composition data.

ChemTour: Limiting Reactant


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Learn to identify the limiting reactant for a reaction and to determine the maximum yield of products.

Function of Water: Moles

A balloon is filled with 0.2 mole of hydrogen (H2) gas and 0.1 mole of oxygen (O2) gas. If the mixture in the balloon is ignited, how many moles of water will form?

A) 0.1 mole B) 0.2 mole C) 0.3 mole

Function of Water: Moles

Consider the following arguments for each answer and

vote again:

A. Since each water molecule (H2O) contains 1 oxygen atom, 0.1 mole of oxygen should form 0.1 mole of water.

B. One mole of H2O has 2 moles of hydrogen atoms and 1 mole of oxygen atoms, so 0.4 mole of hydrogen atoms and 0.2 mole of oxygen atoms will form 0.2 mole of H2O.

C. In a chemical reaction the number of moles must always be conserved, so 0.3 mole of water (0.1 + 0.2) will form.

Function of Water: Masses

If the mixture in the balloon is ignited, approximately how many grams of water will form? Note: A gas is about 1000 times less dense than its corresponding liquid.

A 1-liter balloon is filled with hydrogen (H2) and oxygen (O2) such that the molar ratio of hydrogen to oxygen is 2:1.

A) ~1 milligram B) ~1 gram C) ~1 kilogram

Function of Water: Masses

Consider the following arguments for each answer and vote again:

A. Gases are very light, so even if the reaction goes to completion, only a milligram or so of water will form.

B. A 1-liter gas mixture of hydrogen and oxygen will react to form ~1 liter of water vapor, which has a mass of ~1 gram.

C. The balloon contains 1 liter of gas, so the reaction will form ~1 liter of liquid water, which has a mass of ~1 kilogram.

Combustion of Phosphorus

Elemental analysis shows the product to contain 43.6% phosphorus and 56.4% oxygen by mass. What is the empirical formula of the phosphorus oxide?

A piece of elemental phosphorus is combusted in an atmosphere of excess oxygen until it is completely converted to an oxide of phosphorus, PxOY.

A) PO B) P2O3 C) P2O5

Combustion of Phosphorus


A. Atoms of phosphorus and oxygen react one by one to form a compound whose empirical formula is therefore PO.

B. Since the product is ~40% phosphorus and ~60% oxygen, the ratio of phosphorus to oxygen is ~2:3, suggesting that the formula is P2O3.

C. Although the mass ratio is 4.36/5.64, the molar ratio is 1.4/3.5=2/5, so the empirical formula can only be P2O5

Product Yield of Phosphorus Oxide

A 1-liter bulb is filled with pure oxygen. Elemental phosphorus is added to the bulb in increments of 2 milligrams, corresponding to the blue points in the plots below.

A) B) C)

Which plot correctly shows the accumulated amount of P4O10 produced (y-axis) as a function of the mass of phosphorus added (x-axis)?

Product Yield of Phosphorus Oxide


A. Until sufficient phosphorus is added to achieve the correct molar ratio of phosphorus to oxygen (2:5), no reaction will occur, after which P4O10 will form as more phosphorus is added.

B. The amount of P4O10 product will accumulate as phosphorus is added until all oxygen is used up, at which point no more P4O10 can be produced.

C. As long as oxygen is present, the amount of P4O10 will be constant. When the oxygen is depleted, the amount of P4O10 will steadily decrease.

Mass Spectrum of Oxygen

A sample of oxygen (O2) gas is isotopically enriched such that 50% of the atoms are 16O and 50% of the atoms are 18O.

A) B) C)

What is the correct mass spectrum for this gas?

Mass Spectrum of Oxygen


A. Three isotopic species, 16O-16O, 16O-18O, and 18O-18O, exist with equal probabilities, so there should be three lines of equal intensity in the mass spectrum.

B. Three isotopic species, 16O-16O, 16O-18O , and 18O-18O, exist, and 16O-18O has twice the probability (16O-18O, 18O-16O) so there should be a 1:2:1 intensity pattern.

C. Two isotopic species, 16O-16O and 18O-18O, exist with equal probabilities because molecules can only be formed from the same isotopes, so 16O-18O cannot exist.

Oxidation of Bromides

Consider the oxidation of 1 gram of a metal bromide, MBr, in excess MnO2/H2SO4

4 MBr + MnO2 + 2 H2SO4 →

2 M2SO4 + MnBr2 + 2 H2O + Br2

A) Potassium B) Rubidium C) Cesium

Which metal bromide will yield the largest quantity of Br2?

Oxidation of Bromides


A: Potassium has the lowest molar mass and so 1.0 gram of KBr will have the most moles and produce the most Br2.

B: Rubidium's molar mass is closer to bromine's than either of the other two metals, so RbBr will produce the most Br2.

C: Cesium has the highest molar mass and so 1.0 gram of CsBr has the most metal and will produce the most Br2.

Empirical Formula of H:O:C:S

A compound contains hydrogen, oxygen, carbon, and sulfur in the mass ratios 1:2:3:4.

A) H8OC2S B) HO16C12S32 C) HO2C3S4

What is the empirical formula of the compound?

Empirical Formula of H:O:C:S


A. Although the mass ratios are 1:2:3:4, the molar ratios are 1:0.125:0.25:0.125, so the empirical formula is H8OC2S.

B. Although the mass ratios are 1:2:3:4, the ratios of the relative atomic masses are 1:16:12:32, so the molecular formula should be HO16C12S32.

C. Since the molecule is 10% hydrogen, 20% oxygen, 30% carbon and 40% sulfur the empirical formula is obviously HO2C3S4.

Molecular Formula of a Hydrocarbon

One mole of an unknown hydrocarbon is combusted with an excess of oxygen to produce 88 grams of carbon dioxide (44 g/mol) and 36 grams of water vapor (18 g/mol).

What is the molecular formula of the hydrocarbon?

A) C2H2 B) CH2 C) C2H4

Molecular Formula of a Hydrocarbon

Consider the following arguments for each answer

and vote again:

A. Two moles of CO2 and 2 moles of H2O are formed from 1 mole of CxHY, so the molecular formula must be C2H2.

B. The products contain 2 moles of carbon atoms and 4 moles of hydrogen atoms, which is a ratio of 1:2, so the molecular formula is simply CH2.

C. The products coming from 1 mole of CxHY molecules contain 2 moles of carbon atoms and 4 moles of hydrogen atoms, so the molecular formula is C2H4.

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Chapter 3 Chemical Reactions and Earth’s Composition