Stoichiometry

AP Chemistry

Mr. Martin

Topics

Law of Conservation of MatterBalancing Chem EqMass Relationships in rxn’sLimiting ReagentsTheoretical, Act and % yldEmpirical and Molecular formulas

A. Balancing Equations

Law of Conservation of Mass – matter is neither created or destroyed – during chem rxn’s some matter is converted to energy – why we must bal equations

Stoichiometry – the relationship among the substances in chem rxn’s.

Chemical Equations

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Balancing Equations

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Steps to Balance

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Practice Problems

Cu + O2 ----> Cu2O CaCl2 + AgNO3 ----> AgCl + Ca(NO3)2 H2 + O2 ----> H2O Mg + P4 ---> Mg3P2

H2SO4 + NaOH ----> H2O + Na2SO4

Solutions

4Cu + O2 ----> 2Cu2OCaCl2 + 2AgNO3 ----> 2AgCl + Ca(NO3)2

2H2 + O2 ----> 2H2O6Mg + P4 ---> 2Mg3P2

H2SO4 + 2NaOH ----> 2H2O + Na2SO4

Additional Practice

Fe2O3 + CO -> Fe3O4 + CO2 Fe3O4 + CO -> FeO + CO2 C12H22O11(s) + O2(g) -> CO2(g) + H2O Fe(s) + O2(g) -> Fe2O3

Ca(s) + H2O(l) -> Ca(OH)2(aq) + H2(g)

B. Atomic and Molecular Weights

Atomic Mass Scale uses atomic mass units (AMU’S) relative mass scale originally based on hydrogen = 1 amu

Now 1 amu = 1/12 of the carbon-12 isotope Average atomic masses (atomic weight)

elements are a mixture of isotopes and therefore the mass of an element is a weighted average of the naturally occurring isotopes.

Calculation of Atomic Mass

AM = sum (abundance) x (isotope mass)

of all the naturally occurring

isotopes

Ex: Ne-20 19.992 amu 90% abund

Ne-22 21.991 amu 10 % abund

(0.9 x 19.992) + (0.1 x 21.991) =

20.192 amu’s atomic mass

Formula Weight

Sum of the average atomic masses of each atom in it’s chem ormula

EX C12H22O11

C (12) 12 = 144 amu’s

H (22) 1 = 22 amu’s

O (11) 16 = 176 amu’s

342 amu’s formula weight

Percent Composition by Weight

What is the % comp of Methane CH4?

The Mole

A number 6.02 x 10^23 – like a dozen = 12 , a pair = 2, or a gross = 144

A mol is the amount of matter that contains as many objects (atoms, molecules or formula units) as the # of atoms in exactly 12g of the carbon-12 isotope.

The Mole and Molar Mass

One mole of Carbon = 6.02x10^23 atoms of carbon

One mole of H2O is equal to 6.02 x 10^23 H2O molecules.

One mole of HCl is equal to 6.02 x 10^23 formula units of HCl.

Molar Mass = 12.01 g of Carbon

Molecular Weight = H2O

H (2) 1 = 2

0 (1) 16 = 16

18 g/mol Formula Weight = HCl

36.5g/mol

Conversion Problems

Gram moles atoms, fu’s,

molecules

How many moles are in 200g of sulfuric acid. ( 2.04 mole of H2SO4)

Conversion Problems

How many formula units are in 10g of sulfuric acid? ( 6.14 x 10^22 fu’s)

Empirical and Molecular Formulas

Empirical Formulas – gives the lowest whole # ratio of atoms (moles) in a compound.

EF’s are derived from exp info most notably % comp.

Example Problems

An analysis of sodium dichromate gives the following mass percentages, 17.5% Na, 39.7 % Cr, and 42.8% O. Calc the emp. formula?

Example Problem Mol. Formula

Molecular formula – is a multiple of the emp. formula. n = exp mm/ef weight

The % composition of acetaldehyde is 54% C, 9.2% H, and 36.3% O. Its molar mass is 88 amu’s. What is its molecular formula.

Combustion Analysis

A method of analysis where a sample cont. C,H, O undergoes complete combustion. The gas stream goes through two tubes one absorbs CO2, the other H2O.

Combustion Analysis cont.

The change of mass in the tubes gives the mass of CO2, and H2O.

A 4.24mg Acedic Acid sample is completely burned yielding 6.24mg of CO2 and 2.54mg of H2O. What is the mass % of each element?

Quantitive Information from Bal. Eq

grms mol mol grms

Ex. A (moles to moles) How many moles of H2O will be produced if 17.5 moles of H2 combine with excess O2? (17.5)

Example Problems Cont.

How many grams of KClO3 must decompose to produce 185.0 moles of O2? (1.51 x 10^4 g)

Example Prob Cont.

How many grams of silver chloride can be produced from the reaction of 17.5 g of silver nitrate with excess sodium chloride solution? (14.4g)

Example Problems

How many mg of aluminum nitrate would be required to completely react with 1.75 kg of barium hydroxide? (1.45 x 10^6mg)

Limiting Reagent & Percent Yld

The reactant that is entirely consumed when a reaction goes to completion is the limiting reagent.

The reactant that is not completely consumed is said to be in excess.

The amount of product that is produced is determined by the limiting reactant.

Ex. Limiting Reagent Prob.

In a lab test reaction 20.0g of CH3CHO and 10.0g of O2 were put in a reaction vessel. (a) How many grams of acedic acid can be produced (b) How many grams of excess reactant remain after the reaction is complete.

Theoretical and Percent Yld.

The max amount of product that can be obtained is the theoretical yield. ( based on stoichiometric calc.)

The actual yield is the mass obtained by the actual reaction.

% yield = actual yld/theorical yld x 100

Problem

The amount of acedic acid produced from the previous reaction was 23.8g. What is the percent yield of the reaction. (87.2%)