chocolate chip cookies
DESCRIPTION
Chocolate Chip Cookies. 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters 1 bag chocolate chips. How much?. Of what?. What units?. Chocolate Chip Cookies. 2.25 cups flour - PowerPoint PPT PresentationTRANSCRIPT
Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips
Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips
How much?
What units?
Of what?
Chocolate Chip Cookies• 2.25 flour• 8 butter• 0.5 shortening• 0.75 sugar• 0.75 brown sugar• 1 salt• 1 baking soda• 1 vanilla• 0.5 Egg Beaters• 1 chocolate chips
How much?
Of what?
Chocolate Chip Cookies• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters• 1 bag chocolate chips
How much?
What units?
Of what?
Get on with it!
What does this have to do with CHEMISTRY?
2.25 cups flour + 8 Tbsp butter + 0.5 cups shortening +
0.75 cups sugar + 0.75 cups brown sugar + 1 tsp salt +
1 tsp baking soda + 1 tsp vanilla + 0.5 cups Egg Beaters + chips
(a synthesis reaction)
(177ºC)
1 batch of chocolate chip cookies!
coefficient
unitsubstance
Welcome to STOICHIOMETRY
What is stoichiometry?
• Composition stoichiometry deals with the mass relationships of elements in compounds.
• Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.
• All reaction stoichiometry calculations start with a balanced chemical equation.
Let’s Revisit the Cookies…
• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters
For 1 batch: The Egg Beaters I have are close to expiring! I’d like to use the rest of them in this recipe. I have 1.5 cups of Egg Beaters.
How many batches of cookies can I make with that many Egg Beaters?
Let’s Revisit the Cookies…
• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters
For 1 batch: I have 1.5 cups of Egg Beaters.
How many batches of cookies can I make with that many Egg Beaters?
1.5 cups E.B.x
1 batch cookies
0.5 cups E.B.=
3.0 batches of cookies
Let’s Revisit the Cookies…
• 2.25 cups flour• 8 Tbsp butter• 0.5 cups shortening• 0.75 cups sugar• 0.75 cups brown sugar• 1 tsp salt• 1 tsp baking soda• 1 tsp vanilla• 0.5 cups Egg Beaters
For 1 batch: I have 1.5 cups of Egg Beaters.
How much butter do I need to deplete (use up) the Egg Beaters?
1.5 cups E.B.x
8 Tbsp butter
0.5 cups E.B.=
24 Tablespoons of butter
… Back to Chemistry• There are three types of stoichiometry
problems:– Mole-Mole problems (1 conversion)– Mass-Mole problems (2 conversions)– Mass-Mass problems (3 conversions)
given required
Stoichiometry Problems
• Stoichiometric problems are solved by using ratios from balanced chemical equations to convert the given quantity.
• A mole ratio is a conversion factor that relates the amount in moles of any two substances involved in a chemical reaction.
• This information is obtained from the balanced chemical equation.
Mole Ratios
• Example:
• The relationships between product and reactants or reactants can be expressed in the following mole ratios:
2 H2 + O22 H2O
Mole Ratios
2 H2 + O2 2 H2O
2
2
2
2
H mol 2OH mol 2or
OH mol 2H mol 2
2
2
2
2
O mol 1OH mol 2or
OH mol 2O mol 1
2
2
2
2
H mol 2O mol 1or
O mol 1H mol 2
Practice
• For each reaction, write all possible mole ratios.
)g(O3)s(Al4)l(OAl2 232
)g(O)l(Hg2)s(HgO2 2
32
32
OAl mol 2Al mol 4or
Al mol 4OAl mol 2
32
2
2
32
OAl mol 2O mol 3or
O mol 3OAl mol 2
Al mol 4O mol 3or
O mol 3Al mol 4 2
2
HgO mol 2Hg mol 2or
Hg mol 2HgO mol 2
HgO mol 2O mol 1or
O mol 1HgO mol 2 2
2 2
2
O mol 1Hg mol 2or
Hg mol 2O mol 1
Mole-Mole ProblemsExample:
2 H2 + O22 H2O
How many moles of water can be formed from 0.5 mol H2?0.5 mol H2 x
2 mol H2
2 mol H2O=
0.5 mol H2O
Mole-Mole Practice
3CuSO4 + 2 Al Al2(SO4)3 + 3 Cu
1. Convert 0.5 mol CuSO4 to mol Cu
0.5 mol CuSO4 x 3 mol Cu
3 mol CuSO4
= 0.5 mol Cu
2. Convert 0.5 mol Al to mol CuSO4
0.5 mol Al x 3 mol CuSO4
2 mol Al= 0.8 mol CuSO4
Mass – Mole Problems
• Step 1: Write a BALANCED EQUATION.• Step 2: Calculate the molar mass of your
given substance and convert from mass to moles.
• Step 3: Determine the mole ratio from the coefficients in the balanced equation.
• Step 4: Set up the conversion and solve.
Mass-Mole ProblemsExample:
2 H2 + O22 H2O
How many moles of water can be formed from 48.0 g O2?48.0 g O2 x
1 mol O2
2 mol H2O=
3.00 mol H2O
32.00 g O2
1 mol O2x
Mass-Mole PracticeCuSO4 Al Al2(SO4)3 Cu3 2 3+ +
1. a. 13.5 g Al 1 mol Al
26.98 g Al=x 0.751 mol
CuSO4
b.
c.
13.5 g Al 1 mol Al2(SO4)3
2 mol Al=x 0.250 mol
Al2(SO4)3
13.5 g Al 3 mol Cu
2 mol Al=x 0.751 mol
Cu
Mole ratio
x3 mol CuSO4
2 mol Al
1 mol Al
26.98 g Al
x
1 mol Al
26.98 g Alx
Mole – Mass Practice
Ca + AlCl3 CaCl2 + Al3 2 3 2
0.095 mol AlCl3 x 3 mol Ca
2 mol AlCl3
x 40.08 g Ca
1 mol Ca= 5.7 g Ca
Mass-Mass ProblemsExample:
2 H2 + O22 H2O
How many grams of water can be formed from 48.0 g O2?48.0 g O2 x
1 mol O2
2 mol H2O=
32.00 g O2
1 mol O2x x
18.02 g H2O
1 mol H2O
54.1 g H2O
Mass-Mass PracticeCuSO4 Al Al2(SO4)3 Cu3 2 3+ +
1. a. 8.5 g Al 1 mol Al
26.98 g Al
=x
75 g CuSO4
b. 8.5 g Al 1 mol Al2(SO4)3
2 mol Al=x
54 g Al2(SO4)3
Mole ratio
x3 mol CuSO4
2 mol Al
1 mol Al
26.98 g Al
x
1 mol CuSO4
342.14 g Al2(SO4)3
1 mol Al2(SO4)3
x
x
159.61 g CuSO4
Mass-Mass Practicec. 8.5 g Al 3 mol Cu
2 mol Al=x
30. g Cu
1 mol Al
26.98 g Al
x x63.55 g Cu
1 mol Cu