Ch 3.3 Counting (p78) One dozen = 12 things We use a dozen to make it easier to count the amount of substances.
Moles the SI base unit that describes the amount of particles in a substance. Mole is abbreviated mol
So what does a Mole equal? Avogodro s constant is the amount of particles in one mole of a substance. He found that 6.022 x 10 23 is the amount of particles that are found in a pure substance when 1.0 moles are massed out.
Moles Used by chemist when counting large numbers of tiny particles such as atoms It allows chemist to relate the formula mass of an atom to a substance.
Molar Mass The mass in grams of 1 mol of any substance. The molar mass is numerically equal to the formula mass. Most be rounded to a hundredth so to include any isotopes
g/ 1 mole
6.02 x 10 23 particles
Ch3.3 and 7.3 Molar Mass Calculations Chemists must use mole calculations in order to prepare reactions. They act as recipes for chemists.
EXAMPLE PROBLEM #1 How many grams of Calcium fluoride must a chemist prepare if the reaction requires 4.0 moles of calcium fluoride?
Solution The molar mass of CaF 2 is 78.08 g/1 mol. 4.0 moles CaF 2 x 78.08 grams CaF 2 = 1 mole CaF 2 310 g CaF 2
Example Problem #2 How many moles were produced if your reaction yielded 92.0 grams of sodium sulfate?
Solution MM of Na 2 (SO 4 ) is 142.05 g/1 mole 92.0 g Na 2 (SO 4 ) x 1 mole Na 2 (SO 4 ) = 142.05 g Na2(SO4) 0.648 moles Na2(SO4)
Example Problem #3 How many molecules were produced if your reaction yielded 12.0 grams of Carbon monoxide?
Solution 12.0 g CO x 6.02x10 23 molecules CO 28.01 g CO = 2.58x10 23 molecules CO
Ch. 7.3 Percent Composition A measurement that measures the amount of each element in the total compound. Can be used to determine the amount recovered from a compound breaking down
2% HNO 3 22% 76% Hydrogen Nitrogen Oxygen
Ch 7.4 Calculating Empirical Formula A compound is made from 30.5% nitrogen and 69.5% oxygen. What is the empirical formula? Assume you have a 100 gram sample 30.5 g N/ 14 gram per mole = 2.2 moles N 69.5 g O/ 16 grams per mole = 4.3 mole O 2.2 mole N/ 2.2 mole= 1 N 4.3 mole O/2.2 mole = 2 O Formula is NO 2
Determining Molecular Formula The empirical formula is nitrogen dioxide. Its actual molar mass is 92 g/mole. What is the molecular formula? Molar mass of NO 2 is 46 g/ mole. 92g / 46g = 2 NO 2 x 2 = N 2 O 4
Ch 8.1 Chemical Reactions When substances undergo chemical changes, they form new substances. Atoms are rearranged, because bonds are broken and reformed
Chemical Reactions Signs are: 1. evolution of heat and light 2. color change 3. gases emitted 4. a precipitate forms (solid residue)
Equation Make-up Reactants substances that will undergo a chemical change. (Left side)
Reaction Make-up Products substances that are formed as a result of a chemical change. (Right side)
Recipes: 1 cup batter +1/2 cup water + 2 eggs 4 pancake Reactants Products 6CO 2 +6H 2 O C 6 H 12 O 6 + 6O 2
Balancing All equations must have the same type and number of atoms on each side of the equation. Law of Conservation of Mass Tells you the amounts
Beaker Violating Law of Conservation of Mass
Mole Ratios Coefficients show the amount of moles of each substance. Mole ratio is the smallest relative number of moles of the substance involved in a reaction. 2H 2 + O 2 2H 2 O = 2:1:2
Rule to Balancing Can only change coefficients and never subscripts. Always balance H and O last if water is in the equation. In a double replacement RXN, balance the ions (OH) - = H 2 O
Writing Equations Phase identification S, l, g, v, aq All ionic compounds are solids Solutions and acids are mixed with water, so are considered aqueous. Diatomic elements: H 2, N 2, O 2, F 2, Cl 2, Br 2, and I 2
Ch. 8.2 Reaction Types 5 types of reactions
1. Synthesis Reaction A reaction when two substances form at least one more complex compound.
Synthesis Reaction EXAMPLE: A + B A B 2Fe + O 2 2FeO
Zn + S Al + Br 2
CaO + H2O Ca(OH)2
2. Decomposition Reaction A reaction in which one compound breaks into at least two products.
Decomposition Reaction EXAMPLE: AB A + B 2 NI 3 N 2 + 3I 2
Decomposition of Sugar
Decomposition Products Carbonate turns into CO 2 Hydroxides turn into water Ammonium turns into ammonia
3. Single displacement Reaction A reaction in which atoms of one element take the place of ions of another compound. Cationic or anionic
Single displacement Reaction Example X A + B B A + X 3CuCl 2 + 2Al 2AlCl 3 + 3Cu
Fe 2 O 3 + Al
Ch. 8.3 SR Reaction Activity Series Used by chemists to determine what substances can displace another
4. Double Replacement Reaction A reaction in which the apparent exchange of ions between two compounds or solutions. Most of the reaction will yield an insoluble precipitate If it is a neutralization reaction, water will form.
Double replacement reaction AX + BY AY + BX Pb(NO ) +K (CrO ) 3 2 2 4 Pb(CrO 4 ) + 2K(NO 3 )
5. Combustion reaction Reaction in which an organic compound and oxygen burn. Oxygen always is a reactant
Combustion reaction CO or CO 2 will always be a product H 2 O will always be a product
Combustion reaction C 2 H 5 0H + 3O 2 2CO 2 + 3H 2 O
MUPPET LABS
Stoichiometry Ch. 9.1 Calculations dealing with the THEORICAL amounts of substances needed or produced in a chemical reaction. Mole-mole problems Mass-mole problems Mass-mass problems Percent Yield
Stoichiometry Ch. 9.2 Steps to solving: 1. Balance the chemical equation 2. Determine the mole ratio needed for problem 3. Use Dimensional Analysis to solve
Stoichiometry Mole-mole Problem How many moles of sodium nitrate can be produced from the reaction of 0.67 moles of calcium nitrate with excess sodium chloride? 2NaCl + Ca(NO 3 ) 2 CaCl 2 + 2NaNO 3 0.67 moles Ca(NO 3 ) 2 x 2 mole Na(NO 3 ) 1mole Ca(NO 3 ) 2 = 1.3 moles of Na(NO 3 )
Stoichiometry Mass-mole Problem How many moles of Calcium chloride can be produced from the reaction of 17.0 grams of calcium nitrate with excess sodium chloride? 2NaCl + Ca(NO 3 ) 2 CaCl 2 + 2NaNO 3 17.0 grams Ca(NO 3 ) 2 x 1 mole Ca(NO 3 ) 2 x 1 mole CaCl 2 164 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2 = 0.104 moles of CaCl 2
Stoichiometry Mass-mass Problem How many grams of Copper I sulfide could be produced from 9.90 g of CuCl reacting with an excess of Hydrogen sulfide gas? 2CuCl + H 2 S Cu 2 S + 2HCl 9.90 grams CuCl x 1 mole CuCl x 1 mole Cu 2 S x 160 g Cu 2 S 99 g CuCl 2 mole CuCl 1 mole Cu 2 S = 8.00 grams of Cu 2 S
Ch 9.3 Limiting Reactions Limiting Reactant The reactant that limits how much product can form. It is completely used up. Excess Reactant Substance that is not totally consumed.
Ch 9.3 Stoichiometry Percent Yield The actual amount of product expressed as a percentage of the calculated theoretical yield of that product. The theoretical amount is what is calculated using Stoichiometry Percent Yield = actual amount of product x 100% theoretical amount of product
Stoichiometry Percent Yield Problem What is the percent yield in the following reaction if 5.50 grams of hydrogen react with nitrogen to form 20.4 grams of ammonia? N 2 + 3H 2 2NH 3 5.50 g H 2 x 1 mole H 2 x 2 mole NH 3 x 17 g NH 3 2 g H 2 3 mole H 2 1 mole NH 3 = 31.2 grams of NH 3 20.4 g /31.2g x 100% = 65.4% NH 3