Stoichiometry AP Chemistry Mr. Martin

Topics Law of Conservation of Matter Balancing Chem Eq Mass Relationships in rxn’s Limiting Reagents Theoretical, Act and % yld Empirical 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

Balancing Equations

Steps to Balance

Practice Problems Cu + O > Cu 2 O CaCl 2 + AgNO > AgCl + Ca(NO 3 ) 2 H 2 + O > H 2 O Mg + P 4 ---> Mg 3 P 2 H 2 SO 4 + NaOH ----> H 2 O + Na 2 SO 4

Solutions 4Cu + O > 2Cu 2 O CaCl 2 + 2AgNO > 2AgCl + Ca(NO 3 ) 2 2H 2 + O > 2H 2 O 6Mg + P 4 ---> 2Mg 3 P 2 H 2 SO 4 + 2NaOH ----> 2H 2 O + Na 2 SO 4

Additional Practice Fe 2 O 3 + CO -> Fe 3 O 4 + CO 2 Fe 3 O 4 + CO -> FeO + CO 2 C 12 H 22 O 11 (s) + O 2 (g) -> CO 2 (g) + H 2 O Fe(s) + O 2 (g) -> Fe 2 O 3 Ca(s) + H 2 O(l) -> Ca(OH) 2( aq) + H 2 (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 amu 90% abund Ne amu 10 % abund (0.9 x ) + (0.1 x ) = amu’s atomic mass

Formula Weight Sum of the average atomic masses of each atom in it’s chem ormula EX C 12 H 22 O 11 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 CH 4?

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 H 2 O is equal to 6.02 x 10^23 H 2 O molecules. One mole of HCl is equal to 6.02 x 10^23 formula units of HCl. Molar Mass = g of Carbon Molecular Weight = H 2 O H (2) 1 = 2 0 (1) 16 = 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 H 2 SO 4 )

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 CO 2, the other H 2 O.

Combustion Analysis cont. The change of mass in the tubes gives the mass of CO 2, and H 2 O. A 4.24mg Acedic Acid sample is completely burned yielding 6.24mg of CO 2 and 2.54mg of H 2 O. 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 H 2 O will be produced if 17.5 moles of H 2 combine with excess O 2 ? (17.5) 

Example Problems Cont. How many grams of KClO 3 must decompose to produce moles of O 2 ? (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 CH 3 CHO and 10.0g of O 2 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%)