Transcript ap chemistry chapter 3 stoichiometry

AP CHEMISTRY
CHAPTER 3
STOICHIOMETRY
Average atomic massweighted average based on isotopic
composition
To calculate :
% Isotope A (mass of A) + % Isotope
B (mass of B) + …= avg. atomic mass
What is the average atomic mass in grams of
lithium if 7.42% exists as 6Li (6.015 g/mol)
and 92.58% exists as 7Li (7.016 g/mol)?
(0.0742)(6.015) + (0.9258)(7.016) = ?
6.94 g/mol
Don’t forget to change percents to decimals!
Don’t divide by the number of different isotopes!
Don’t round until the end.
Your answer must be between the masses of the individual isotopes.
Check!
Mass Spectrometer
A mass spectrometer is an instrument used to determine the relative
masses of atoms by the deflection of their ions in a magnetic field.
One use is to determine the isotopic abundance of a sample of an
element.
Samples are
vaporized and
ionized. The
ions are then
separated by
mass and the
data is
graphed.
Figure 3.2
Neon Gas
These show the Mass spec data for samples of neon. Using this data,
what would the approximate average atomic mass of neon be?
Figure
3.3
Mass
Spectrum
of
Natural
Copper
What element results in this
mass spec data?
Mole
-the # of C atoms in 12g of pure
carbon-12
Avogadro’s Number = 6.022 × 1023
The mass of one mole of an
element is equal to its atomic mass
in grams.
Cody found a gold nugget that had a
mass of 1.250 oz. How many moles was
this? How many atoms?
(1 lb = 16 oz, 453.59g = 1 lb)
1.250 oz Au
1 lb 453.59g 1 mol Au
16 oz 1 lb
196.97g Au
= 0.1799 moles
0.1799 moles Au 6.022 × 1023 atoms
1 mol Au
= 1.083 × 1023 atoms
Molar Mass = mass in grams of
one mole of a substance
Calculate the molecular mass of
cisplantin, Pt(NH3)2Cl2.
Pt
N
H
Cl
1 × 195.08 = 195.08
2 × 14.01 = 28.02
6 × 1.01 = 6.06
2 × 35.45 = 70.90
300.06 g
How many grams are 3.25 moles
of cisplantin?
3.25 mol cisplantin 300.06g
1 mol cisplantin
= 975g
Percent Composition -“mass percent” or “percent by mass”
Total mass of element
× 100 = % comp of element
Total mass of compound
Find the percent composition of all
elements in cisplantin, Pt(NH3)2Cl2.
Pt 195.08 × 100 = 65.01%
300.06
N 28.02 × 100 = 9.34%
300.06
H 6.06 × 100 = 2.02%
300.06
Cl 70.90 × 100 = 23.63%
300.06
Determining the Formula of a
Compound
Empirical formula- simplest whole
number ratio of the various types of
atoms in a compound
Molecular formula
- the exact formula
(empirical formula)x
A sample of a compound contains 11.66g of iron and
5.01g of oxygen. What is the empirical formula of
this compound?
11.66g Fe 1 mol Fe = 0.2088 mol Fe
55.85g Fe
5.01g O 1 mol O = 0.3131 mol O
16.00g O
Fe:O 0.2088 : 0.3131
1:1.5 or 2:3
0.2088 0.2088
Fe2O3
What is the empirical formula of
hydrazine, which contains 87.5% N and
12.5%H?
87.5gN 1 mol N = 6.246 mol N
14.01g N
12.5g H 1 mol H = 12.38 mol H
1.01 g H
N:H = 6.246 : 12.38 1 : 1.98 = 1 : 2
6.246 6.246
NH2
Combustion Analysis
Additional O2 is added to burn the sample of a compound
containing C, H, and sometimes O. All of the H2O is absorbed in
the first chamber and all of the CO2 produced is absorbed in the
second chamber. Increases in the masses of the two chambers are
used to determine the mass of CO2 and H2O produced.
Suppose you isolate an acid from clover leaves and
know that it contains only the elements C, H, and O.
Heating 0.513g of the acid in oxygen produces
0.501g of CO2 and 0.103g of H2O. What is the
empirical formula of the acid? Given that another
experiment has shown that the molar mass of the
acid is 90.04g/mol, what is its molecular formula?
C + O2  CO2
0.501g CO2 1 mol CO2 1 mol C 12.01gC
44.01g CO2 1 mol CO2 1 mol C
= 0.1367 g C in compound
2H2 + O2  2H2O
0.103g H2O 1 mol H2O 2 mol H 1.01g H
18.02g H2O 1 mol H2O 1 mol H
= 0.01155 g H
0.513g cmpd  (0.137 + 0.012)= 0.364g O
0.364g O 1 mol O = 0.02275 mol O
16.00g O
0.137g C 1 mol C = 0.01141 mol C
12.01g C
0.01155g H 1 mol H = 0.01144 mol H
1.01 g H
C:H:O
0.01141: 0.01144 : 0.02275
0.01141: 0.01144 : 0.02275
0.01141 0.01141 0.01141
1: 1: 1.98
1:1:2
CHO2
efm = 12.01 + 1.01 + 32.00 = 45.02
90.04/45.02 = 2
C2H2O4
Chemical Equations
 Products
Yields
A balanced equation must have the
same # of atoms of each element on each
side.
Reactants
Symbols representing physical states:
(s) (l) (g) (aq)
Balancing Chemical Equations
We balance equations by adding
coefficients, never by changing
formulas.
Most equations can be balanced
by inspection. Some redox
reactions require a different
method.
Ex.
Al(s) + Cl2 (g)  AlCl3(s)
2Al(s) + 3 Cl2 (g)  2AlCl3(s)
N2O5(s) + H2O(l)  HNO3(l)
N2O5(s) + H2O(l)  2 HNO3(l)
Pb(NO3)2(aq) + NaCl(aq) 
PbCl2(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2 NaCl(aq) 
PbCl2(s) + 2 NaNO3(aq)
Phosphine, PH3(g) is combusted in air
to form gaseous water and solid
diphosphorus pentoxide.
2PH3(g) + 4O2(g)  3H2O(g) + P2O5(s)
When ammonia gas is passed over hot liquid sodium
metal, hydrogen is released and sodium amide, NaNH2, is
formed as a solid product.
2NH3(g) + 2Na(l)  H2(g) + 2NaNH2(s)
Stoichiometric Calculations
Mole ratio =
moles required/moles given
What mass of NH3 is formed when 5.38g
of Li3N reacts with water according to
the equation:
Li3N(s) + 3H2O  3LiOH(s) + NH3(g)?
5.38g Li3N 1 mol Li3N 1 mol NH3 17.04g NH3
34.83g Li3N 1 mol Li3N 1 mol NH3
= 2.63g NH3
Limiting reagent- reagent that limits
or determines the amount of product
that can be formed
Animation
If you are given amounts of two or more
reactants in a stoichiometry problem and
asked to determine how much product
forms, the easiest thing to do is to work a
problem with each reactant and take the
smaller of the answers.
You work in a sandwich shop.
You MUST make sandwiches
according to the following
formula: You use two pieces of
bread, 4 pieces of meat, 2 pieces
of cheese, 3 pieces of bacon and 2
ounces of special sauce. If you
have 20 pieces of bread, 32 pieces
of meat, 20 pieces of cheese, 21
pieces of bacon and one quart of
special sauce, how many
sandwiches can you make?
X + 2Y  XY2
X
Before reaction
Y
XY2
After reaction
Assume that the reaction goes to completion. Draw the resulting
particles in the right-hand box. What is the limiting reagent?
X + 2Y  XY2
X
Before reaction
Y
XY2
After reaction
Assume that the reaction goes to completion. Draw the resulting
particles in the right-hand box. What is the limiting reagent?
How many moles of Fe(OH)3(s) can be produced by
allowing 1.0 mol Fe2S3, 2.0 mol H2O and 3.0 mol O2
to react?
2Fe2S3(s) + 6H2O(l) + 3O2(g)  4Fe(OH)3(s) +6S(s)
1.0 mol
2.0 mol 3.0 mol
limiting
2.0 mol H2O 4 mol Fe(OH)3
6 mol H2O
= 1.3 mol Fe(OH)3
If 17.0g of NH3(g) were reacted with 32.0g of
oxygen in the following reaction, how many grams
of NO(g) would be formed?
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
17.0g
32.0g
Xg
17.0g NH3 1 mol NH3 4 mol NO 30.01 g NO = 29.9g NO
17.04g NH3 4 mol NH3 1 mol NO
32.0 g O2 1 mol O2 4 mol NO 30.01 g NO = 24.0g NO
32.00g O2 5 mol O2
1 mol NO
24.0g is the smaller yield, so O2 is limiting and 24.0 g is the
correct answer.
Theoretical yield- amount of product that
should form according to stoichiometric
calculations
Actual yield- experimental yield
Percent yield = actual yield
× 100
Theoretical yield
In the reaction of 1.00 mol of CH4 with an
excess of Cl2, 83.5g of CCl4 is obtained.
What is the theoretical yield, actual yield and
% yield?
Actual yield = 83.5g CCl4
CH4 + 2Cl2  CCl4 + 2H2
1.00 mol CH4 1 mol CCl4 153.81g CCl4
1 mol CH4 1 mol CCl4
= 154g CCl4 (theoretical yield)
83.5 × 100 = 54.2% yield
154