9701 ChemistryAtom, Molecules & Stoichiometry

Chemical ReactionsThe actual phenomenon that occurs when chemical interact with each other.Balanced Chemical Equations represent events that occur at the atomic level which we cannot perceive; but they explain the mass ratio of substances involved in a chemical equation (stoichiometry)

CHEMICAL EQUATIONSa A + b B c C + d Da, b, c, d are stoichiometric coefficientsAlonsos Rules for BE:Easy element 1st hard elements last.One element at a time.Use fractions when necessary.

The moleA term for a certain number of something.A mole of something is 6.02 x 1023 of something.602, 000, 000, 000, 000, 000, 000, 000

Molecular WeightM.W. = the weight (in grams) of a mole of substanceRound to the nearest tenthHydrogen is 1.00797 1.0 g/mol

Mole Weight is an INTENSIVE property, it doesnt depend on amount

Try these MWsCa40.1

H22 (1.0) = 2.0

BaF2[137.3 + 2(19.0)] = 175.3

MW of 2BaF2?? is still 175.3 1 mole of Ca weighs 40.1 grams1 mole of H2 weighs 2.0 grams

Avogadros NumberNA = 6.02 x 1023 of anythingAvogadro's law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

Percent Composition

A. Determined from Formulas (Accepted Value)Is NaCl 50.0% Na by weight?No, Na is 23.0 g/mole and Cl is 35.5 g/mole To Prove, % Na =

Formula Weight (FW) Sum of the atomic weights for the atoms in a chemical formula unit (ionic compound)Molecular Weight (MW) Sum of the atomic weights of the atoms in a molecule (covalent compound)For the molecule ethane, C2H6, the molecular weight would be

Empirical FormulaDefinition: The simplest formula indicating the mole ratio of elements in a compoundExamples:H2O2 HOC6H6CHN2O4?CO2 ?NO2CO2

Empirical Formula STEPSChange grams to molesDivide by the least # moles for a RATIOApply ratio to the formula

Try This0.556 g Carbon and 0.0933 g Hydrogen

Why is the Empirical Formula a ratio of small WHOLE numbers?Cant have half of an atomAtoms combine as whole unitsShows the simplest way that atoms can pair

Try This: 70.5 % Fe and 29.5 % O

___ moles Fe, ___ moles O Assume 100g of substance: Fe2O31.26 moles1.26 molesFeO1.5

Try This: 40.0% C6.7% H 53.3% O

Why doesnt the ratio of the % give the empirical formula? Must account for differing masses of elements. Why does the ratio of the moles give the empirical formula?The ratio of the # of atoms normalizes for mass differences.

Molecular Formulas

Definiton: Formula of an actual compound as it exists in molecules.Benzene exists as C6H6 not CH. Hydrogen Peroxide exists as H2O2 not HO.

Stoichiometry The Big LeaguesA. Define: Problem Solving involving mass-mass relationships in chemical changesEx. How many grams of rust are formed when 12.00 g of Fe reacts with oxygen.

B. Must use balanced equations for the correct mole ratios

C. Coefficients yield the mole ratio!!! 2 H2 + O2 2 H2O 2 : 1 : 2

0.200mol of a hydrocarbon undergo complete combustion to give 35.2g of carbon dioxide and 14.4g of water as the only products.What is the molecular formula of the hydrocarbon?

A. C2H4 B. C2H6 C. C4H4 D. C4H8

[AS Nov 2009 Paper 12 Q1]

CxHy (g) + O2 (g) CO2 (g) + H2O (g)

Calculating Empirical FormulasA 5.00 g sample of an unknown hydrocarbon was burned and produced 14.6 g of CO2. What is the empirical formula of the unknown compound?5.00 g14.6 g

2006 A?g N =

A pure hydrocarbon is used in bottled gas for cooking and heating. When 10cm3 of the hydrocarbon is burned in 70cm3 of oxygen (an excess), the final gaseous mixture contains 30cm3 of carbon dioxide and 20cm3 unreacted oxygen. All gaseous volumes were measured under identical conditions. What is the formula of the hydrocarbon?

A. C2H6 B. C3H6 C. C3H8 D. C4H10

[AS June 2005 Paper I Q1]

Stoichiometry

(mass relationships within chemical equations)

The coefficients in the balanced equation can also be interpreted as mole ratios of reactants and productsStoichiometric Calculations

Limiting & Excess ReactantsThe limiting reactant is the reactant present in the smallest stoichiometric amountIn other words, its the reactant youll run out of firstLimiting H2; Excess O2

Limiting & Excess Reactants ProblemIf 5.0g of both Mg and O2 are used:Which is the limiting and the excess reactants?How much of the excess will be left unreacted ?How much MgO will be produced ?2 Mg (s) + O2 (g) 2 MgO (s)

Mass spectrometerA highly accurate instrument!

Mass spectrometer consists of 6 parts:1.8 Mass spectrometer (SB p.20)

Mass spectrum of Cl2:1.8 Mass spectrometer (SB p.21)

m/e ratioCorresponding ion3535Cl+3737Cl+7035Cl35Cl+7235Cl 37Cl+7437Cl 37Cl+

Relative atomic massThe relative atomic mass of an element is the weighted average of the relative isotopic masses of its natural isotopes on the carbon-12 scale.1.9 Relative isotopic, atomic and molecular masses (SB p.23)

1.9 Relative isotopic, atomic and molecular masses (SB p.23)What is the relative atomic mass of Cl?The relative abundances of Cl-35 and Cl-37 are 75.77 and 24.23 respectively

Mass Spectrometer 5 StagesOnce the sample of an element has been placed in the mass spectrometer, it undergoes five stages.Vaporisation the sample has to be in gaseous form. If the sample is a solid or liquid, a heater is used to vaporise some of the sample.

X (s) X (g)

or X (l) X (g)

Mass Spectrometer 5 StagesIonisation sample is bombarded by a stream of high-energy electrons from an electron gun, which knock an electron from an atom. This produces a positive ion:

X (g) X + (g) + e- Acceleration an electric field is used to accelerate the positive ions towards the magnetic field. The accelerated ions are focused and passed through a slit: this produces a narrow beam of ions.

Mass Spectrometer 5 StagesDeflection The accelerated ions are deflected into the magnetic field. The amount of deflection is greater when:

the mass of the positive ion is less the charge on the positive ion is greater the velocity of the positive ion is less the strength of the magnetic field is greater

Mass SpectrometerIf all the ions are travelling at the same velocity and carry the same charge, the amount of deflection in a given magnetic field depends upon the mass of the ion.For a given magnetic field, only ions with a particular relative mass (m) to charge (z) ration the m/z value are deflected sufficiently to reach the detector.

Mass SpectrometerDetection ions that reach the detector cause electrons to be released in an ion-current detectorThe number of electrons released, hence the current produced is proportional to the number of ions striking the detector.The detector is linked to an amplifier and then to a recorder: this converts the current into a peak which is shown in the mass spectrum.

The isotopic composition of an element is indicated below.

What is the relative atomic mass of the element?

A. 10.2 B. 10.5 C. 10.8 D. 11.0

[AS June 2007 Paper I Q1]

(c) A sample of iron has the following isotopic composition by mass:

(i) Define the term relative atomic mass. (ii) By using the data above, calculate the relative atomic mass of iron to three significant figures. [5][AS June 2005 Paper II Q1]

Q: 1FC1 is a solution containing 16.75g dm-3 of hydrated sodium carbonate, Na2CO3.xH2O.FC2 is 0.125mol dm-3 hydrochloric acid, HCl.25.0cm3 of FC1 is pipetted into a conical flask and add a few drops of the indicator and titrate with FC2 from the burette. The titre value obtained is 23.10cm3. (a) Calculate how many moles of the acid were run from the burette into the conical flask during the titration of FC1 with FC2. 1(b)Calculate the number of moles of anhydrous sodium carbonate, Na2CO3, in 25.0cm3 of FC1. Na2CO3 + 2HCl 2NaCl + H2O + CO2 1

(c) Calculate the concentration, in mol dm-3, of sodium carbonate, Na2CO3, in FC1. 1(d) Calculate the mass of anhydrous sodium carbonate present in 1.00dm3 of FC1.[Ar: Na, 23.0; C, 12.0; O, 16.0.] 1(e) Calculate the mass of water present in the hydrated sodium carbonate. 1(f) Find the value of x. 1[AS Nov 2001 Paper III Q1]

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