Transcript Moles and Stoichiometry Slides

CH 3: Stoichiometry
Moles
Mole Defined
• Mole = number equal to the # of carbon
atoms in exactly 12 grams of pure C-12.
– Mole = 6.022 x 1023
– Called Avogadro’s number
• One mole of protons = 1g
• One mole of neutrons = 1g
• One mole of an element = atomic mass in
grams
Molar Mass and related terms
• Molar mass – mass in grams of one mole
of a substance
– molecular weight, mass of 1 molecule in amu
– Formula weight, mass of 1 formula unit for an
ionic compound in amu
• Calculate by summing the masses of the
component atoms, units:
– molar mass – grams or grams/mole
– molecular wt & formula wt - amu
Moles, mass and # particles
• Molar mass links the mass of a substance
to the number of particles present
1 mole Mg = 24.31g = 6.022 x 1023 Mg atoms
Molar Mass Related Calculations:
1. Molar mass of a substance.
2. Moles present in a given mass of
substance.
3. Mass of a given number of moles of a
substance.
4. Number of particles in a given mass or a
given number of moles of a substance.
1. What is the molar mass of glucose. The
formula for glucose is C6H12O6.
2. What is the mass of 0.023 moles of
glucose?
3. How many glucose molecules are
present in 0.023 moles of the
compound?
4. How many moles of glucose are present
in 3.5 x 10-3 grams of the substance?
Mass Percent
• Compounds are typically described by
either their chemical formula or their
percent by mass of the component
elements.
Mass % X = n (molar mass X)
x 100%
molar mass of compound
Mass Percent Calculations
• Calculate the mass percent of each
element in C6H12O6
– Assume one mole of the substance.
Empirical and Molecular Formulas
• Molecular formula – ratio of atoms in a
molecule
• Empirical formula – simplest ratio of
elements in a molecule
• Glucose:
– Molecular formula: C6H12O6
– Empirical Formula:
• Molecular formula = n (empirical formula)
Mass %  Empirical Formula
Given mass percent data:
1. “Calculate” mass in grams of each element in
100 g of compound.
2. Convert each mass into moles of the element.
3. Divide each molar answer by the smallest of
the values.
4. If the numbers obtained in step 3 are not whole
numbers, multiply all by an integer so the
results are whole numbers
5. Whole numbers obtained in step ¾ are the
subscripts in the empirical formula.
Empirical Formula  Molecular
Formula
Molar mass needed
• Calculate mass of one mole of the
empirical formula.
Molar Mass
= integer (n)
empirical formula weight
(n)(empirical formula) = molecular formula
Caffeine
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•
•
•
Molar mass = 194 g/mol
49.49 % C
5.19 % H
28.85 % N
16.48 % O
What is the empirical and molecular formula
of caffeine?
Acetaminophen (Tylenol)
Molar mass = 151 g/mol
• 63.56 % C
• 6.00 % H
• 9.27 % N
• 21.17 % O
What is the empirical and molecular formula
of acetaminophen?
CH 3: #86
Urea: Molar Mass 60 g/mol (I looked this up – not in
question)
1.121 g N
0.161 g H
0.480 g C
0.640 g O
What is the empirical and molecular formula
of urea?
3.7/3.8 Chemical Reactions
• Writing chemical reactions
– Write the correct chemical formula for each
reactant and product.
– Use a subscript after each formula to indicate
the state of each substance
• (s) – solid
• (l) – liquid
• (g) – gas
• (aq) – aqueous (dissolved in water)
• Chemical equations must obey the law of
conservation of matter – balancing does
this.
• Balance the equation by adding
coefficients in front of reactants and
products as needed
– DO NOT CHANGE THE FORMULAS FOR
THE COMPOUNDS
• One of my favorite chemical reactions:
Mg(s) +
HCl(aq) 
MgCl2 (aq) + H2 (g)
• As written the reaction does not obey the
law of conservation of matter – reaction
needs to be balanced
Mg(s) + 2 HCl(aq) 
MgCl2 (aq) + H2 (g)
• Now the reaction is balanced!
Balance the Reactions
N2 (g)
+
Na(s)
+
H2 (g) 
Cl2 (g)

NH3 (g)
NaCl(s)
Pb(NO3)2(aq) + NaCl(aq)  PbCl2(s) +
NaNO3(aq)
Meaning of the Balanced Reaction
• Atomic level
• Molar level
• Stoichiometry - relationship between
moles, the balanced reaction and mass
– Page 10? outlines the needed steps
2 Na
+
Cl2
 2 NaCl
2 mol Na + 1 mol Cl2  2 mol NaCl
Cookie example!
2 eggs +
1 bag chips  50 cookies
How many cookies could you make if you
had only one egg?
How many cookies could you make if you
had 2 bags of chips?
Molar Ratios
• Balanced reactions lead to molar ratios
N2
+
3 H2
 2 NH3
N2 (g)
+
3 H2 (g) 
2 NH3 (g)
1. If 3.5 grams of N2 to react, how many
moles and how many grams of NH3
would be made?
N2 (g)
+
3 H2 (g) 
2 NH3 (g)
1. How many grams of H2 are needed to
make 37.8 grams of NH3?
2 LiOH + CO2  Li2CO3 + H2O
1. How many grams of LiOH are needed to
react 750. grams of CO2?
2. How many grams of Li2CO3 will be made
if 750. grams of CO2 react?
Limiting Reagent
• Reactants are not always combined in
stoichiometric quantities.
• Often one reactant is used up before
others.
• This reactant is said to be the limiting
reagent (or reactant.)
• this reactant limits how much product can
be made.
Limiting Reagent Calculations
• When you are given masses for each
reactant you must determine which
reactant is limiting.
1. Convert grams of each reactant to moles.
2. Use the molar ratios to determine which
reactant is limiting.
3. Base the amount of product that can be
made on the limiting reagent.
Another Approach…
1. Convert grams of each reactant to moles.
2. Calculate the moles (or grams) of a
product that could be made from each
reactant.
3. Whichever reactant results in the smaller
quantity of product is the limiting reagent.
4. Base the amount of product made on the
limiting reagent.
N2 (g) + 3 H2 (g)  2 NH3 (g)
• 12.5 grams of nitrogen is combined with
2.30 grams of hydrogen.
• Which reactant is in excess and which is
limiting?
• How many grams of ammonia will be
made?
Percent Yield
% Yield =
actual yield
theoretical yield
x 100 %