Title: Stoichiometry and the Mole
1Chapter 8
- Stoichiometry and the Mole
2Objectives
- Explain what a mole is.
- Perform conversions with Avogadros Number and
Molar Mass
3Chemistry and Cooking
- Chemistry is not so different from cooking
- Amount to be made must be considered
- Ingredients (reactants) must be measured
- Cooking conditions must be correct
- Techniques must be learned
- These are the basics of stoichiometry
- I wonder why there arent more chemistry
channels? (Hmmmmmm?)
4Stoichiometry
- Definition Relationship between quantities of
reactant and product in a chemical reaction. - Literally means Measure the elements
- Answers the question of how much
- Always based on a chemical reaction
- The chemists recipe
- You already know how to do stoichiometry!
5Consider the FollowingDorothys Sour Cream
Raisin Bars
1.75 C Oatmeal 1.00 C Brown Sugar 1.00 C Butter 1.75 C Flour
4.00 Egg Yolks 3.00 Tbs. Corn Starch 2.00 C Sour Cream 1.50 C White Sugar
2.00 C Raisins Makes 24 Yummy Bars
Combine the Oatmeal, Brown Sugar, Butter, and
Flour with pastry cutter. Hold out 2.00C. Place
the rest in the bottom of a 9x13 greased pan.
Bake 15 min at 350 F. Combine yolks,
starch, sour cream, sugar and raisins in a pan.
Bring to a boil and pour over crust. Crumble hold
out over top. Bake 20 min. at 350 F
6What Does This Tell Us?
- A number of things!
- The amount and type of ingredients
- The conditions required
- The number of bars made
- Also
- How to change the recipe
- How to scale it up or down
7How Many Bars Could You Make If
1.75 C Oatmeal 1.00 C Brown Sugar 1.00 C Butter 1.75 C Flour
4.00 Egg Yolks 3.00 Tbs. Corn Starch 2.00 C Sour Cream 1.50 C White Sugar
2.00 C Raisins Makes 24.0 Yummy Bars
Ans. 6.00 Bars
8How Many Bars Could You Make If
1.75 C Oatmeal 1.00 C Brown Sugar 1.00 C Butter 1.75 C Flour
4.00 Egg Yolks 3.00 Tbs. Corn Starch 2.00 C Sour Cream 1.50 C White Sugar
2.00 C Raisins Makes 24.0 Yummy Bars
Ans. 41.1 Bars
9How Many Bars Could You Make If
1.75 C Oatmeal 1.00 C Brown Sugar 1.00 C Butter 1.75 C Flour
4.00 Egg Yolks 3.00 Tbs. Corn Starch 2.00 C Sour Cream 1.50 C White Sugar
2.00 C Raisins Makes 24.0 Yummy Bars
- You have 4.33 pints sour cream
Calc. 103.92 Bars
Ans. 104 Bars
10Chemical Reactions
- 4.04 g of hydrogen gas and 32.00 grams of oxygen
gas react to form 36.04 grams of water. - How many grams of water can be made from
- 8.08 g hydrogen gas
Ans. 72.1 g water
11Its Just That Easy
- When Doing Stoichiometry
- Keep the recipe (balanced equation) in mind
- Set up ratios that are true
- Check to see your answer makes sense
- Check those significant figures
12The Mole
From www.wildcru.org/research/farming/moles/mole1
.jpg
13The Mole
- One mole of anything is equal to
6.022x1023particles - Also called Avogadros Number
- Inconvenient for large things like hamburgers and
hammers - Works well for small things like atoms, ions, and
molecules - Mole Song (Michael Offutt Chemistry Songbag 1)
14Conversions With Moles
- Easy to convert from particles to moles and back
- Use conversion factors
- Set up a ratio that cancels units
15Convert the Following
- 6.0x1023 atoms gold to moles
- Ans. 1.0 moles Au
16Convert the Following
- 9.12x1024 molecules hydrogen to moles
- Calc. 15.144470
- Ans. 15.1 mol H2
17Convert the Following
- 3.12 moles strontium to atoms
- Calc.1.8788x1024
- Ans.1.88x1024 atoms Sr
18Moles in Equations
- Coefficients in balanced equations represent
ratios - 2H2 O2 ? 2H2O
- This means 2 molecules of hydrogen react with 1
molecule of oxygen to produce 2 molecules of
water - Or
- 2 moles of hydrogen react with 1 mole of oxygen
to produce 2 moles of water
19How Much Is A Mole?
- 1 mole of any atom has a mass equal to its atomic
mass in grams - This is an atoms Molar Mass (MM)
- Mass of a Mole
- Units of grams/mole (g/mol)
- What is the Molar Mass of sodium?
- 22.99 g/mol
- What is the Molar Mass of aluminum?
- 26.98 g/mol
20About 1 Mole of Aluminum
27.405g
From http//www.tufts.edu/as/wright_center/person
al_pages/george_l/Al.jpg
21Converting Grams and Moles
- We often need to convert between grams and moles.
- We measure in grams but,
- Reactions are in moles
- Use molar mass for grams and moles
- Use Avogadros number for moles and particles
22Convert the Following
- 8.00 grams of helium to moles
- Ans. 2.00 mol He
23Convert the Following
- 6.25 moles of chlorine to grams
- Calc.221.5625
- Ans. 222 g Cl
24Convert the Following
- And this is a tough one!
- 49.00 grams of calcium to atoms
- Calc.7.36222x1023
- Ans. 7.362x1023 atoms Ca
25Moles and Compounds
- Formulas tell us mole information too
- The formula H2O means
- 2 moles of hydrogen and 1 mole of oxygen makes up
1 mole of water - The formula Ca(NO3)2 mean
- 1 mole of calcium, 2 moles of nitrogen, and 6
moles of oxygen make 1 mole of calcium nitrate
26Molar Mass and Compounds
- Compounds have molar masses too.
- Determine the moles of each element
- Add the molar masses of each
- Ex. Find the molar mass of water (H2O)
- 2 moles of H and 1 mole of O
- 2(1.01g/mol) 1(16.00 g/mol) 18.02g/mol
- Always put 2 digits after the decimal pt.
27Find The Molar Mass Of
- Sodium Carbonate
- Na22.99, C12.01, O16.00
- Na2CO3
- MM2(22.99)(12.01)3(16.00)
- Ans.105.99 g/mol
- Calcium Phosphate
- Ca40.08, P30.97, O16.00
- Ca3(PO4)2
- MM3(40.08)2(30.97)8(16.00)
- Ans.310.18 g/mol
28Converting Moles and Mass
- Conversions with compounds are just like elements
- Use molar mass of the compound
- Convert 2.33 grams of water to moles
29Convert the Following
- 2.34 g Sodium Carbonate to moles
- 105.99 g/mol
- Calc. 0.022077
- Ans.0.0221 mol Na2CO3
30Convert the Following
- 1.87 moles of Calcium Phosphate to grams
- 310.18 g/mol
- Calc.580.0366
- Ans.580. g Ca3(PO4)2
31Calculate the Following
- Grams of oxygen in 16.11 grams of Calcium
Phosphate (MM 310.18 g/mol) - Calc. 6.64801
- Ans. 6.648 g O
32The Conversion Works!
33Balanced Equations and Mass
- What masses are required to complete the reaction
below - 2H2 O2 ? 2H2O
- Hydrogen need 2 moles
- So 4.04 g are needed
- Oxygen need 1 mole
- So 32.00 g are needed
- Water 2 moles produced
- So 36.04 grams are produced
34Homework
- p309 's 37,41,47,103,107,109
35Objectives
- Calcualte Mass Percent
- Describe Empirical and Molecular Formulas
- Calculate Empirical Formulas
36Mass Percent
- Definition The percentage, by mass, that each
element contributes to the compound - How to
- Find the Molar Mass of the compound
- Divide the mass of each element by the MM
- Multiply by 100.
- Gives you percent
37Example
- Find the mass percent of Glucose
- C6H12O6 C-12.01, H-1.01, O-16.00
- C 39.99 H 6.73 O 53.28
38Example
- Find the mass percent of Sodium Carbonate
- Na2CO3 Na-22.99, C-12.01, O-16.00
- Na 43.38 C 11.33 O 45.29
39Empirical Formulas
- The formula with the simplest ratio of elements
present - May or may not be the actual formula
- Often determined from percent composition
- Percent found by lab analysis
- Combustion analysis or other means
40Molecular Formulas
- Actual formula of the compound
- Formula is a whole number multiple of the
empirical formula - Molar mass is a whole number multiple of
empirical formula molar mass - CH2 Empirical Formula, 14.04 g/mol
- C2H4- Molecular Formula, 28.08 g/mol
41How To Find Empirical Formulas
- If percent composition is given
- Assume 100.00 grams
- All percents become grams
- Find the moles of each element
- Divide each by the smallest number
- If you have whole numbers write the formula OR
- Multiply all number to get a whole number
42Cont.
- If masses are given
- Start with the masses
- Everything else is the same
- When you write the formula
- Ionic Compounds
- Metals are usually written first
- Covalent Compounds
- Usually written Carbon, Hydrogen, Oxygen
43Example
- Find the empirical formula of a compound with the
following percent composition - K 26.56, Cr 35.41, O 38.03
44Example
- A compound is composed of Chlorine and Oxygen.
The percent of Chlorine is 38.77. What is the
empirical formula?
45Example
- CH2 is the empirical formula of a compound that
has a molar mass of 70.15 g/mol. What is the
molecular formula?
46Example
- A Compound is made of only phosphorus and oxygen
and has a molar mass of approximately 284 g/mol.
The compound is made of 56.36 O. What is the
molecular formula
47Smores (Yum!)
- Given the following
- 1 bag 30 mallows, 1 bar16 rectangles
- 2 grahams 4 rec. 1 mallow ? 1 smore
- If you have 24 bars and plenty of grahams and
mallows how many smores can you make?
48Smores (Yum!)
- Given the following
- 1 bag 30 mallows, 1 bar16 rectangles
- 2 grahams 4 rec. 1 mallow ? 1 smore
- If you have 3 bags and plenty of bars and grahams
how many smores can you make? -
49Objectives
- Describe reaction stoichiometry
- Determine mole ratios
- Calculate theoretical, actual, and percent yields
50Reaction Stoichiometry
- Objectives
- Describe reaction stoichiometry
- Calculate theoretical yields for reactions
- Compare actual yields to theoretical yields
- Calculate the percent yield of a reaction
- Determine a limiting reactant
51Reaction Stoichiometry
- A method for determining how much in chemical
reactions - How much product can be made from X grams of
reactant - How much reactant is necessary to make X grams of
product
52Mole Ratios
- Calculations ALWAYS use mole ratios
- Come from balanced equation
2H2 O2 ? 2H2O - Mole Ratios could be
53Process for Calculations
- Write balanced equation
- Find necessary molar masses
- Set up calculation
- Basic Setup
- Grams A ? Moles A ? Moles B ? Grams B
Mole Ratio
MM A
MM B
54Assumptions
- Base your calculations off of the given reactant
- Assume that there is plenty of the other reactant
- We say that other reactant is in excess
- Dont worry about the reactant that is not stated
- You wont use it in your calculations
55Theoretical Yield
- Maximum mass of product you can obtain in a
reaction - This is what you are calculating for
- When the reaction is run you may not get exactly
what you calculated - Could be more or less
- This is called the actual yield
56Actual Yield
- Amount of product obtained when a reaction is
carried out - Reasons for differences in theoretical and actual
yield - Loss of product
- Residue in reaction vessel
- Equilibrium
- Moisture in product
57Percent Yield
- Percent of theoretical yield obtained when a
reaction is carried out. - Represents the efficiency of a reaction
- Goal is to have a high percent yield
- Equation
58Example
- 84.12g of O2 is reacted with excess NO2 according
to the following equation.
2NO O2 ? 2 NO2 - What mass of NO2 is formed? O232.00
g/mol, NO246.01g/mol
59Example
- What mass of NO is required to react with 84.12
grams of O2 according to the equ. 2NO O2 ? 2
NO2 - O2 32.00 g/mol, NO 30.01 g/mol
60Example
- What mass of NO is required to produce 39.12
grams NO2 if excess O2 is present 2NO O2 ? 2
NO2 - NO 30.01 g/mol, NO2 46.01 g/mol
61Example
- What mass of Silver Bromide is produced when 2.33
grams of Silver Nitrate is mixed with excess
Sodium Bromide? - AgNO3 NaBr ? AgBr NaNO3
- AgBr187.77g/mol, AgNO3169.88g/mol
62Example
- The previous reaction was run and only 0.88 grams
of AgBr were produced. Find the percent yield.
63Homework
64Who Bought the Smores Stuff?
- Given the following
- 1 bag 30 mallows, 1 bar16 rectangles
- 2 grahams 4 rec. 1 mallow ? 1 smore
- If you have 24 bars, 3 bags and plenty of grahams
how many smores can you make?
65Objectives
- Describe limiting and excess reactants
- Determine the limiting reactant for a reaction
- Calculate theoretical yields from limiting
reactants
66Limiting Reactant
- Reactant that is consumed first in a chemical
reaction - Left over reactant is considered to be excess
reactant - Limiting reactant is not necessarily the one with
the smallest mass - Depends on mole ratio
67Limiting Reactant
- Limiting reactant problems will give you the mass
of both reactants - Both masses need to be considered
- You will need to figure out which reactant is
limiting - The reactant that gives the smallest product mass
is the limiting reactant - The other is in excess
- The reactants could run out simultaneously
68Process
- Consider the reaction A B ? C
- Convert grams of A to grams of C
- Convert grams of B to grams of C
- The reactant that gives the smallest mass of C is
the limiting reactant AND - The smallest mass of C is the theoretical yield
69Example
- 152.50 g of CO reacts with 24.50 g of H2 What
mass of CH3OH can be produced? CO 2H2 ? CH3OH - CO 28.01, H2 2.02, CH3OH 32.05
70Cont.
- What mass of the excess reactant is left? CO
2H2 ? CH3OH - CO 28.01, H2 2.02, CH3OH 32.05
71Example
- 92.7 g of N2 react with 265.8 g of NH3. What
mass of Ammonia is produced? N2 3H2 ?
2NH3 - N2 28.02, H2 2.02, NH3 17.04
72Cont.
- What mass of the excess reactant is left? N2
3H2 ? 2NH3 - N2 28.02, H2 2.02, NH3 17.04
73Cont.
- Determine the percent yield if only 82.0 grams of
ammonia are produced
74Homework