Chapter 3 Mass Relations: Stoichiometry

Atomic number # of p + in nucleus

Mass number # of p + and n 0 in nucleus

IsotopeMass #Atomic mass Oxygen Copper Mass # v. Atomic mass

Avg. Atomic Mass (atomic weight) Weighted avg. of atomic masses of naturally occurring isotopes of an element

Isotopes Different forms of the same element with different mass Same # p + Diff. # n 0

e.g. : naturally occurring Cu 69.17% Cu-63 (atomic mass ) 30.83% Cu-65 (atomic mass ) (0.6917)(62.939)+(0.3083)(64.927) =  Found on periodic table

Masses of individual atoms

The Mole (mol)

Mole (mol) Amt. of a substance that contains same # of particles as # of atoms in exactly 12 g of carbon-12

Avogadro’s number Number of particles in exactly one mole of a pure substance (6.02 x ) Named for Amedeo Avogadro [ Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e di Cerreto ( ) ]

Molar mass Mass in grams of one mole of a pure substance Used as a conversion factor, number is taken from the periodic table

What is the mass (in g) of 2.5 mol of cobalt Molar mass Co = g 2.5 mol Co x _________ 2.5 mol Co x g Co / 1 mol Co = g Co = 150 g Co

How many atoms are in g of platinum? g Pt x ________ g Pt x I mol Pt/ g Pt = mol Pt mol Pt x 6.02 x atoms Pt / 1 mol Pt = x atoms 2.53 x atoms

Percent composition ( Mass of element / molar mass cmpd. X 100%) e.g. find % composition of Cu 2 S 2 mol Cu and 1 mol S 2 mol Cu x g Cu / 1 mol Cu = g Cu 1 mol S x 32.06g S / 1 mol S = g S molar mass Cu 2 S = g g Cu/ g Cu 2 S x 100% = % Cu 32.06g S/ g Cu 2 S x 100% = 20.14% S

Determining simplest formula ( Formula showing smallest whole number ratio of atoms) e.g. find the simplest formula for a cmpd. containing 26.56% K, 35.41% Cr, and 38.03% O If 100g of cmpd., then: K = g, Cr = g, O = g

(cont.) 26.56g K x 1 mol K / g K = mol K 35.41g Cr x 1 mol Cr / g Cr = mol Cr 38.03g O x 1 mol O / g O = mol O divide by smallest number  mol K / = mol K mol Cr / = mol Cr mol O / = mol O :1.003 :  2 : 2 : 7 K 2 Cr 2 O 7

Chemical Equations and Chemical Reactions

Chemical equation Represents (w/ symbols and formulas) the reactions and products in a chemical reaction The same # of atoms of each element must appear on each side of the equation Use coefficients to balance equation

Word equations e.g. methane + oxygen  carbon dioxide + water

Formula equations CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) (unbalanced) CH 4 (g) + O 2 (g)  CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)

Rules H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 (ClIF H BrON)

Writing equations e.g., Write a formula equation for the reaction between hydrogen gas and fluorine gas to produce hydrogen fluoride gas H 2 (g) + F 2 (g)  HF(g) H 2 (g) + F 2 (g)  2HF(g)

Balancing equations Balance: Al + Fe 2 O 3  Al 2 O 3 + Fe 2Al + Fe 2 O 3  Al 2 O 3 + Fe 2Al + Fe 2 O 3  Al 2 O 3 + 2Fe

Balance: NH 3 (g) + O 2 (g)  N 2 (g) + H 2 O(g) 2NH 3 (g) + O 2 (g)  N 2 (g) + H 2 O(g) 2NH 3 (g) + O 2 (g)  N 2 (g) + 3H 2 O(g) 2NH 3 (g) + 2O 2 (g)  N 2 (g) + 4H 2 O(g)  wrong 4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O(g)

Mole relationships H 2 (g) + Cl 2 (g)  2HCl(g) 1 molecule of hydrogen reacts with 1 molecule of chlorine to yield 2 molecules of hydrogen chloride or 1 mol H 2 reacts with 1 mol Cl 2 to yield 2 mol HCl or

(cont.) 2g H 2 (1 x molar mass) reacts with 71g Cl 2 (1 x molar mass) to yield 73g HCl (2 x molar mass) I mol H 2 : 1 mol Cl 2 : 2 mol HCl 2g H2 : 71g Cl 2 : 73g HCl

Types of chemical reactions 1.Synthesis, 2Mg(s) + O 2 (g)  2MgO(s) 2.Decomposition, 2H 2 O(l)  2H 2 (g) + O 2 (g) 3.Single replacement, Mg(s) + 2HCl(aq)  H 2 (g) + MgCl 2 4.Double replacement, Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2 KNO 3 (aq)

(cont.) 5. Combustion, C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g)

Specific examples Decomposition of metal hydroxides, Ca(OH) 2 (s)  CaO(s) + H 2 O(g) Decomposition of metal chlorates 2KClO 3 (s)  2 KCl(s) + 3O 2 (g) Replacement of hydrogen in water by a metal, 2K(s) + 2H 2 O(l)  2 KOH(aq) + H 2 (g)

Stoichiometry

Molar mass Mass in grams of one mole of a pure substance Used as a conversion factor, number is taken from the periodic table

Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactionscalculation quantitativereactantsproductschemical reactions

Stoichiometric air-fuel ratios of common fuels Fuel By weight Percent fuel Gasoline 14.7 : 1 6.8% Natural Gas 17.2 : 1 5.8% Ethanol 9 : % Diesel 14.6 : 1 6.8%

Stoichiometry (mass relationships) Start with a balanced equation  mole ratio 2Al 2 O 3  4Al + 3O 2 2 mol Al 2 O 3 : 4 mol Al : 3 mol O 2 e.g. 2 mol Al 2 O 3 / 4 mol Al

? Mol Al produced f/ 15.0 mol Al 2 O mol Al 2 O 3 x 4 mol Al / 2 mol Al 2 O 3 = 30.0 mol Al

4 types of stoichiometry problems 1.Mole – mole 2.Mole – mass 3.Mass – mole 4.Mass - mass

mole - mole CO 2 + 2LiOH  Li 2 CO 3 + H 2 O How many of LiOH are required to react with 30 mol of CO 2

(cont.) 30 mol CO 2 x 2 mol LiOH / 1 mol CO 2 = 60 mol LiOH

Mole - mass 6 CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2

Given 3.00 mol of water and an excess of carbon dioxide how many grams of glucose will be produced? 3.00 mol H 2 O x 1 mol C 6 H 12 O 6 / 6 mol H 2 O x 180 g C 6 H 12 O 6 / 1 mol C 6 H 12 O 6 = 90.0 g C 6 H 12 O 6

Mass-mole C + SO 2  CS 2 + CO If 8.00 g of SO 2 reacts with an excess of carbon how many moles of CS 2 are formed? 5C + 2SO 2  CS 2 + 4CO 8.00 g SO 2 x 1 mol SO 2 / 64.1 g SO 2 x 1 mol CS 2 / 2 mol SO 2 = mol CS 2

Mass -mass Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many g of SnF 2 is produced from the reaction of 30.00g of HF with an excess of Sn?

30.00g HF x 1 mol HF/ 20.01g HF x 1 mol SnF 2 / 2 mol HF x 156.7g SnF 2 / 1 mol SnF 2 = g SnF 2

Limiting Reactant

Limiting reactant Reactant that limits the amt. of the other reactants that can combine, and the amt. of product formed

Limiting reactant Silicon dioxide (quartz) reacts with hydrogen fluoride according to the following reaction: SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l)

(cont.) If 2.0 mol of HF is combined with 4.5 mol of SiO 2, which is the limiting reactant? 2.0 mol HF x 1 mol SiO 2 / 4 mol HF = 0.50 mol SiO 2 Therefore 2.0 mol HF requires 0.50 mol of SiO 2 to completely react, 4.5 mol SiO 2 is more than 0.50 mol, therefore HF is the limiting reactant

Percent yield Percent yield = actual yield/ theoretical yield x 100% C 6 H 6 + Cl 2  C 6 H 5 Cl + HCl If 36.8 g of C 6 H 6 reacts with an excess of Cl 2 the actual yield is 38.8 g of C 6 H 5 Cl. What is the percent yield?

(cont.) 36.8 g C 6 H 6 x 1 mol C 6 H 6 / 78.1 g C 6 H 6 x 1 mol C 6 H 5 Cl/ 1 mol C 6 H 6 x 113 g C 6 H 5 Cl/ 1 mol C 6 H 5 Cl = 53.2 g C 6 H 5 Cl (theoretical yield)

(cont.) percent yield = 38.8 g C 6 H 5 Cl/ 53.2 g C 6 H 5 Cl x 100% = 72.9%

Calculation of Molecular Formulas The simplest (empirical) formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experiment shows that the formula mass of this compound is g. What is the molecular formula of this compound? 

(cont.) molecular mass P 2 O 5 = g g = g (P 2 O 5 ) 2 = P 4 O 10