Quantitative Analysis Balanced Equations The Mole Percent Composition The Chemical Formula Solutions Stoichiometry Limiting Reagents Percent Yield
Relative Atomic Mass The Atomic Mass for each element is found on the periodic table Determined by Law of Definite Proportions Atomic mass determined by comparing a standard mass Originally Hydrogen Current Standard is Carbon – 12
Carbon-12 Standard Carbon is a very common substance found everywhere Assigning a value of 12 to C-12 allows all other elements to have value close to a whole number One atomic mass unit = 1/12 the mass of a C-12 atom
Atoms in 12 grams of C-12 A defined amount of substance Value = 6.02 x 1023 Estimated through experimentation as the number of atoms present in 12 grams of Carbon-12 12 g = 6.02 x 1023 atoms = 1 mole C-12
1 Mole = 6.02 x 1023 things The mass of 1 Mole of an element = The Mole 1 Mole = 6.02 x 1023 things The mass of 1 Mole of an element = the atomic mass of that element
Moles to Mass Given the number of moles of a compound or element Find the Molar Mass of the substance on the Periodic Table Moles x Molar Mass = Mass of substance
Mass to Moles Given the Mass of a compound or element Find the Molar mass of the substance on the Periodic Table Mass / Molar Mass = # of Moles
Moles to Particles Given the number of moles of a compound or element Use Avogadro’s Constant 6.02 x 1023 Moles x 6.02 x 1023 = # of particles
Particles to Moles Given the number of particles of a compound or element Use Avogadro’s Constant 6.02 x 1023 number of particles / 6.02 x 1023 = # of Moles
Mass to Particles Given the Mass of a compound or element Find the Molar Mass of the substance Mass / Molar Mass = # of Moles Take the number of moles Use Avogadro’s Constant = 6.02 x 1023 Moles x 6.02 x 1023 = # of particles
Particles to Mass Given the number of particles of a compound or element Use Avogadro’s Constant = 6.02 x 1023 number of particles / Avogadro’s Constant = # of Moles Take the number of moles Find the Molar mass of the substance Moles x Molar Mass = Mass of substance
Moles Mass (g) Particles molar mass 6.02 x 1023 x molar mass
Percent Composition A descriptive method of indicating how much there is of an element in a compound by comparing its mass to the mass of the compound Ex. CO2 molar mass = 44g C = 12g / 44g x 100 = 27% O = (16gx2) / 44g x 100 = 73%
Empirical Molecular Formula Formula vs. Provides the lowest ratio of atoms of the elements in the compound CH3O C2H6O2CH3O C4H12O4CH3O Provides the actual ratio of atoms of each element in the compound CH3O C2H6O2 C4H12O4
Empirical Formulas From Percent Composition Determine relative masses of each element (assume sample=100g) Convert relative masses into moles Divide each number of moles by the smallest number of moles Express number of moles of each element as the smallest ratio
Example: C=85.7%;H=14.3% C = 85.7g H = 14.3g C = 85.7g/12g = 7.14 moles H = 14.3g/1g = 14.3 moles C:H = 7.14:14.3 = 1:2 CH2
Molecular Formulas Using Molar Mass Determine the Empirical Formula Calculate the molar mass of the Empirical Formula Compare with the actual molar mass of the unknown (will be given) Actual molar mass empirical formula molar mass Multiply the empirical formula by (n) = n
Concentrations in Solutions Solutions are homogeneous mixtures Solutions have definite percent compositions Solutions are made up of: Solute = substance in smaller amount Solvent = substance in larger amount Concentration(c) = Amount of Solute Amount of Solvent
Concentrations in Solutions Dilute Solutions have low solute and high solvent ratios Concentrated Solutions have high solute and low solvent ratios
Solution Concentration Percentage by volume Percentage by weight Vsolute (smaller volume of two substances) Vsolution (larger volume of two substances) V/V = X 100 msolute (mass of substance) Vsolution (volume of solution) W/V = X 100
Solution Concentration Molar Concentration: Molarity Moles of solute per litre of solution Grams of solute/molar mass Volume of solution in litres Cmol = Mol L Unit =
Solution Concentration Parts per Million (ppm) Used to describe very low concentrations in solutions 1:1 000 000 1 g : 1 000 000 ml 1 g : 1 000 L 1 mg: 1 L Normally expressed as mg/L
Diluting Solutions Many lab activities include creating and using a series of concentrations of a particular solution. Stock Solution provides a concentrated version of a solution that can quickly be diluted for other purposes. Laboratory solutions are dilutions of the stock solution
civi = cfvf Diluting Solutions Need to know three of the following: concentration of the stock solution (ci) final concentration of the solution (cf) volume of stock solution (vi) volume of the lab solution (vf) Relationship: civi = cfvf
Balanced Chemical Reactions Count the number of atoms of each element on the reactant side and the product side Add coefficients in front of molecular formulas to balance Coefficients affect the entire molecule Don’t alter chemical formulas Don’t add additional compounds or elements to balance
Balance the Reaction Ba(OH)2(aq) + Na2SO4(aq) -> BaSO4(s) + NaOH(aq) Left Side Right Side Ba= 1 Ba= 1 OH= 2 OH = 1 Na= 2 Na = 1 SO4= 1 SO4= 1 The equation is not balanced.
Balancing the Reaction Ba(OH)2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaOH(aq) Left Side Right Side Ba= 1 Ba= 1 OH= 2 OH = 2 Na= 2 Na = 2 SO4= 1 SO4= 1 The equation is now balanced.
Mole Ratios in Balanced Equations 2H2 + O2 2H2O Moles in the equation Two moles of hydrogen One mole of oxygen Two moles of water Mole Ratios present H2:O2 = 2:1 H2:H2O = 2:2 O2:H2O = 1:2
Mole Ratio Applied 2H2 + O2 2H2O If 4 moles of H2 are used you need 2 moles of O2 If 10 moles of O2 are used you get 20 moles of H2O If 40 moles of water was made 40 moles of hydrogen and 20 moles of oxygen are needed
Calculating Masses of reactants and products Given mass of substance Convert mass of substance to moles Find mole ratio of given to required substance Calculate moles of required Convert moles of required to mass of required
Limiting Reagents Goal: To discover which reactant will run out first. The mole ratio is used to determine the required amounts The reactant left over it is labeled the excess reagent The reactant completely used up is labeled the limiting reagent
Yield of a Chemical Reaction Yield is the amount of product produced in a reaction Theoretical yield is the amount of product produced