Stoichiometry: Calculations with Chemical Formulas and Equations
Formula and Molecular WeightsFormula weights (FW): sum of AW for atoms in formula.
FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O)= 2(1.0 amu) + (32.0 amu) + 4(16.0)
= 98.0 amu
Molecular weight (MW) is the weight of the molecular formula.
MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
Atomic and Molecular Weights
Atomic and Molecular Weights
What is the molecular weight of the following molecules?
CH4
H2SO4
Ca(NO3)2
CuSO4 •5H2O
Atomic and Molecular Weights
What is the molecular weight of the following molecules?
CH4 12.011 + 4(1.008) = 16.043 amu
H2SO4
Ca(NO3)2
CuSO4 •5H2O
Atomic and Molecular Weights
What is the molecular weight of the following molecules?
CH4 12.011 + 4(1.008) = 16.043 amu
H2SO4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2
CuSO4 •5H2O
Atomic and Molecular Weights
What is the molecular weight of the following molecules?
CH4 12.011 + 4(1.008) = 16.043 amu
H2SO4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2
40.08 + 2(14.007) + 6(16.00) = 164.09 amu CuSO4 •5H2O
Atomic and Molecular Weights
What is the molecular weight of the following molecules?
CH4 12.011 + 4(1.008) = 16.043 amu
H2SO4 2(1.008) + 32.06 + 4(16.00) = 96.08 amu
Ca(NO3)2
40.08 + 2(14.007) + 6(16.00) = 164.09 amu CuSO4 •5H2O
63.546+32.06+9(16.00)+10(1.008) = 249.68 amu
Percentage Composition from FormulasPercent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:
Atomic and Molecular Weights
100
Compound of FWAWElement of Atoms
Element %
Atomic and Molecular Weights
Determine the Mass Percent Composition of nitrogen in NH4NO3?
Atomic and Molecular Weights
Determine the Mass Percent Composition of nitrogen in NH4NO3?
Formula weight 80.043 amu
Atomic and Molecular Weights
Determine the Mass Percent Composition of nitrogen in NH4NO3?
Formula weight 80.043 amu
%999.34100043.80
007.142% x
amu
amuxN
The Mole strictly a convenience unit
amount that is large enough to see and handle in lab mole = number of things
dozen = 12 things mole = 6.022 x 1023 things
Avogadro’s number = 6.022 x 1023 = NA
The Mole how do you know when you have a mole?
count it out weigh it out
molar mass - mass in grams equal to the atomic weight H is 1.00794 g of H atoms = 6.022 x 1023 atoms Mg is 24.3050 g of Mg atoms = 6.022 x 1023 atoms
Mole: convenient measure chemical quantities.1 mole of something = 6.0221367 x 1023 of that thing.Experimentally, 1 mole of 12C has a mass of 12 g.
Molar MassMolar mass: mass in grams of 1 mole of substance (units g/mol, g.mol-1).Mass of 1 mole of 12C = 12 g.
The Mole
Molar MassMolar mass: sum of the molar masses of the atoms:
molar mass of N2 = 2 x (molar mass of N).Molar masses for elements are found on the periodic table.Formula weights are numerically equal to the molar mass.
The Mole
Molar Masses of Compounds add atomic weights of each atom The molar mass of propane, C3H8, is:
One mole of propane is 44.11 g of propane.
amu 44.11 mass Molar
amu 8.08 amu 1.018H8
amu 36.03amu 12.013C3
Interconverting Masses, Moles, and Numbers of Particles
The Mole
The Mole
Calculate the mass of a single Mg atom, in grams, to 3 significant figures.
The Mole
Calculate the mass of a single Mg atom, in grams, to 3 significant figures.
Mg g 104.04atoms Mg mol 1
Mgg24.30
atoms Mg 106.022
atoms Mg mol 1atom Mg 1Mg g ?
23
23
The Mole
How many atoms are contained in 1.67 moles of Mg?
The Mole
How many atoms are contained in 1.67 moles of Mg?
atoms Mg 101.00
Mg mol 1
atoms Mg 106.022Mg mol 1.67atoms Mg ?
24
23
The Mole
How many moles of Mg atoms are present in 73.4 g of Mg?
The Mole
How many moles of Mg atoms are present in 73.4 g of Mg?
MgmolMgg
atomsMgmolMggMgmol 02.3
30.24
14.73?
Mole Problems with Compounds
Calculate the number of C3H8 molecules in 74.6 g of propane.
Mole Problems with Compounds
Calculate the number of C3H8 molecules in 74.6 g of propane.
molecules 101.02HC g 44.11
molecules HC 106.022
HC g 74.6molecules HC ?
24
83
8323
8383
Mole Problems with Compounds
Calculate the number of O atoms in 26.5 g of Li2CO3.
Mole Problems with Compounds
Calculate the number of O atoms in 26.5 g of Li2CO3.
atoms O 106.49
CO Liunitform. 1
atoms O 3
CO Limol 1
CO Liunitsform.106.022
CO Lig 73.8
COLi mol 1COLi g 26.5atoms O ?
23
3232
3223
32
3232
1 Mole of Some Common Molecular Substances
1 Mole Br2 159.81g
C4H10
Contains 6.022 x 1023 Br2 molecules 2(6.022 x 1023 ) Br atoms
1 Mole of Some Common Molecular Substances
1 Mole Br2 159.81g
C4H10 58.04 g
Contains 6.022 x 1023 Br2 molecules 2(6.022 x 1023 ) Br atoms
6.022 x 1023 C4H10 molecules 4 (6.022 x 1023 ) C atoms 10 (6.022 x 1023 ) H atoms
Formulas from Elemental Composition empirical formula - simplest molecular formula,
shows ratios of elements but not actual numbers of elements
molecular formula - actual numbers of atoms of each element in the compound
determine empirical & molecular formulas of a compound from percent composition percent composition is determined experimentally
Empirical Formulas
Simplest whole number ratio of atoms or ions present in a compound
Empirical Formula H2O
C6H12O6
H2O2
Empirical Formulas
Simplest whole number ratio of atoms or ions present in a compound
Empirical Formula H2O H2O
C6H12O6
H2O2
Empirical Formulas
Simplest whole number ratio of atoms or ions present in a compound
Empirical Formula H2O H2O
C6H12O6 CH2O
H2O2
Empirical Formulas
Simplest whole number ratio of atoms or ions present in a compound
Empirical Formula H2O H2O
C6H12O6 CH2O
H2O2 HO
Once we know the empirical formula, we need the MW to find the molecular formula.Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula.
Empirical Formulas and Molecular Formula
Combustion Analysis
Empirical Formulas from Analyses
Percent Composition mass of that element divided by the mass of
the compound x 100% percent composition of C in C3H8
81.68%100%g44.11
g12.013
100%HC mass
C mass C %
83
Percent Composition
percent composition of H in C3H8
Percent Composition
percent composition of H in C3H8
18.32%81.68%100%
or18.32%100%g 44.11
g 1.018
100%HC
H8100%
HC mass
H mass H%
8383
Percent Composition
Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig.
Percent Composition Calculate the percent composition of Fe2(SO4)3 to
3 sig. fig.
100%Total
O48.0%100%399.9g
16.0g12100%
)(SOFe
O12O %
S24.1%100%399.9g
32.1g3100%
)(SOFe
S3S %
Fe27.9%100%399.9g
55.8g2100%
)(SOFe
Fe2%Fe
342
342
342
Start with mass % of elements (i.e. empirical data) and calculate a formula, orStart with the formula and calculate the mass % elements.
Empirical Formulas from Analyses
Empirical Formulas from Analysis
Determine the empirical formula (CrxOy) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g 2.) Determine mol of each element 3.) Divide through by smallest subscript 4.) Convert all subscripts to integers
Empirical Formulas from Analysis
Determine the empirical formula (CrxOy) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g
68.42g Cr and 31.58 g O
Empirical Formulas from Analysis
Determine the empirical formula (CrxOy) for a compound with the percent composition: 68.42% Cr, 31.58% O. 1.) Assume 100g
68.42g Cr and 31.58 g O
2.) Determine mol of each element
molOgO
molOgOmolO
molCrgCr
molCrgCrmolCr
974.100.16
158.31
316.1996.51
142.68
Empirical Formulas from Analysis
Divide through by smallest whole number1.316/1.316 = 1 Cr
1.974/1.316 = 1.50 O
Convert to whole numbers
Cr1O1.5 x 2 = Cr2O3
Empirical Formulas from Analysis
A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?
Make the simplifying assumption that you have 100.0 g of compound. in 100.0 g of compound there is
24.74 g of K 34.76 g of Mn 40.50 g of O
4 KMnO O 40.6327
2.531Ofor
Mn10.6327
0.6327Mnfor K 1
0.6327
0.6327Kfor
rationumber wholesmallest obtain
O mol 2.531Og16.00
Omol1Og 40.50 O mol ?
Mn mol 0.6327 Mng 54.94
Mnmol 1 Mng 34.76 Mnmol ?
Kmol 0.6327K g 39.10
K mol 1 Kg 24.74 K mol ?
Empirical Formulas from Analysis
A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
Empirical Formulas from Analysis
A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
? ..
.
? ..
.
mol Co g Comol Co
g Comol Co
mol O g Omol O
g Omol O
find smallest whole number ratio
6 5411
58 9301110
2 3681
16 0001480
Empirical Formulas from Analysis
A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
for Co Co for O O 01110
011101
01480
011101333
.
.
.
..
Empirical Formulas from Analysis
A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?
for Co Co for O O
multipy both by to turn fraction to whole number
Co Co O O
thus our formula is
Co O
01110
011101
01480
011101333
3
1 3 3 1333 3 4
3 4
.
.
.
..
.
:
Empirical Formulas from Analysis
Chemical Equations
Lavoisier: mass is conserved in a chemical reaction. (Law of Conservation of Mass)
Chemical equations: symbolic representation of a chemical reaction.
Two parts to an equation: reactants and products:
2H2 + O2 2H2O reactants on left side of reaction products on right side of equation relative amounts of each with stoichiometric
coefficients
Chemical Equations
attempt to show on paper what is happening at the molecular level
Chemical Equations
look at what an equation tells us
reactants yields products
1 form. unit 3 mol. 2 atoms 3 mol.
1 mole 3 moles 2 moles 3 moles
159.7g 84g 111.7g 132g
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products.
Chemical Equations
Chemical Equations
Law of Conservation of Matter
no detectable change in quantity of matter in an ordinary chemical reaction
discovered by Lavoisier balance chemical reactions to obey this law propane,C3H8, burns in oxygen to give carbon
dioxide and waterC3H8 + 5 O2 3 CO2 + 4 H2Onote that there are equal numbers of atoms of each
element on both sides of equation
Law of Conservation of Matter
NH3 burns in oxygen to form NO & water
OH 6 + NO 4 O 5 + NH 4
correctlyor
OH 3 + NO 2 O + NH 2
223
2225
3
Law of Conservation of Matter
C5H12 burns in oxygen to form carbon dioxide & water
balancing equations is a skill acquired only with lots of practice work many problems
C H + 8 O 5 CO + 6 H O5 12 2 2 2
Balance the following Equations
CH4(g) + O2(g) CO2(g) + H2O(g)
CaCO3(s) CaO(s) + CO2(g)
AgNO3(aq) + NaI(aq) NaNO3(aq) + AgI(s)
H2SO4(aq) + KOH K2SO4(aq) + H2O(l)
Balance the following Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Balance the following Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
CaCO3(s) CaO(s) + CO2(g)
Balance the following Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
CaCO3(s) CaO(s) + CO2(g)
AgNO3(aq) + NaI(aq) NaNO3(aq) + AgI(s)
Balance the following Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
CaCO3(s) CaO(s) + CO2(g)
AgNO3(aq) + NaI(aq) NaNO3(aq) + AgI(s)
H2SO4(aq) + 2KOH K2SO4(aq) + 2H2O(l)
Balanced chemical equation gives number of molecules that react to form products.Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product.These ratios are called stoichiometric ratios.
NB: Stoichiometric ratios are ideal proportionsReal ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles).
Quantitative Information from Balanced Equations
The ratio of grams of reactant cannot be directly related to the grams of product.
Quantitative Information from Balanced Equations
Calculations Based on Chemical Equations
can do this in moles, formula units, etc. most often work in grams (or kg or pounds or tons)
How many CO molecules are required to react with 25 formula units of Fe2O3?
Fe O + 3 CO 2 Fe + 3 CO2 3 2
? CO molecules = 25 FU Fe O3 CO molecules
1 FU Fe O
= 75 CO molecules
2 32 3
Calculations Based on Chemical Equations
How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?
Calculations Based on Chemical Equations
How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?
? Fe atoms = 2.50 10 FU Fe O2 Fe atoms
1 FU Fe O
= 5.00 10 Fe atoms
52 3
2 3
5
Calculations Based on Chemical Equations
What mass of CO is required to react with 146 g of iron (III) oxide?
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Calculations Based on Chemical Equations
What mass of CO is required to react with 146 g of iron (III) oxide?
Fe O + 3 CO 2 Fe + 3 CO2 3 2
? g CO = 146 g Fe O1 mol Fe O
g Fe O
3 mol CO
1 mol Fe O
28.0 g CO
1 mol CO g CO
2 32 3
2 3 2 3
159 6
76 8
.
.
Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Calculations Based on Chemical Equations What mass of carbon dioxide can be produced by
the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?
Fe O + 3 CO 2 Fe + 3 CO2 3 2
? g CO mol Fe O3 mol CO
1 mol Fe O
g CO
mol CO
= 71.3 g CO
2 2 32
2 3
2
2
2
0 54044 0
1.
.
Calculations Based on Chemical Equations
What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?
Calculations Based on Chemical Equations
? g Fe O 8.65 g CO1 molCO
44.0 g CO
mol Fe O
3 mol CO
g Fe O
mol Fe O g Fe O
2 3 22
2
2 3
2
2 3
2 32 3
1
159 6
110 5
..
Calculations Based on Chemical Equations
How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?
Calculations Based on Chemical Equations
?.
. lb Co = 125 lb Fe O lb CO
159.6 lb Fe O lb CO2 3
2 3
84 0658
Fe O + 3 CO 2 Fe + 3 CO2 3 2
159.6 g 84.0 g 111.6 g 132.0 g
or
159.6 lb 84.0 lb 111.6 lb 132.0 lb
Other Interpretations of Chemical Formulas
What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar 434
Other Interpretations of Chemical Formulas
What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
434434
434
PO)(NH mol 0.357N mol 3
PO)(NH mol 1N mol 1.07
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar
Other Interpretations of Chemical Formulas
What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?
434434
434434
434434
434
PO)(NH g 53.2PO)(NH mol 1
PO)(NH g 149.0PO)(NH mol 0.357
PO)(NH mol 0.357N mol 3
PO)(NH mol 1N mol 1.07
N mol 1.07N g 14.0
N mol 1N of g 15.0N mol ?
g/mol 149.0PO)(NH of massmolar
Using the Periodic Table Properties of compounds vary systematically
because of good ordering in the periodic table.
2K(s) + 2H2O(l) 2KOH(aq) + H2(g)
2M(s) + 2H2O(l) 2MOH(aq) + H2(g)
Combustion in Air Combustion is the burning of a substance in
oxygen from air:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Patterns of Chemical Reactivity
Patterns of Chemical Reactivity
Combination and Decomposition Reactions2Mg(s) + O2(g) 2MgO(s)
There are fewer products than reactants; the Mg has combined with O2 to form MgO.
2NaN3(s) 2Na(s) + 3N2(g) (the reaction that occurs in an air bag)
There are more products than reactants; the sodium azide has decomposed into Na and nitrogen gas.
Patterns of Chemical Reactivity
Combination and Decomposition ReactionsCombination reactions: fewer reactants than products.Decomposition reactions: more products than reactants.
Patterns of Chemical Reactivity
If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess).Limiting Reactant: one reactant that is consumed.
Limiting Reactant Concept
Limiting Reactant Concept
What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110 g of oxygen?
CS O CO 2 SO
1 mol 3 mol 1 mol 2 mol
76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
2 2 2 2 3
Limiting Reactant Concept
What do we do next?
CS O CO 2 SO
mol SO g CS1 mol CS
76.2 g
mol SO
1 mol CS
g SO
1 mol SO g SO
2 2 2 2
2 22 2
2
2
22
3
95 62 641
161? ..
Limiting Reactant Concept
What can we conclude from our data now? Which is limiting reactant?
What is maximum mass of sulfur dioxide?
CS O CO 2 SO
mol SO g CS1 mol CS
76.2 g
mol SO
1 mol CS
g SO
1 mol SO g SO
mol SO 110 g O1 mol O
32.0 g O
mol SO
3 mol O
g SO
1 mol SO g SO
2 2 2 2
2 22 2
2
2
22
2 22
2
2
2
2
22
3
95 62 641
161
2 641147
? ..
?.
Limiting Reactant Concept
limiting reactant is O2
maximum amount of SO2 is 147 g
CS O CO 2 SO
mol SO g CS1 mol CS
76.2 g
mol SO
1 mol CS
g SO
1 mol SO g SO
mol SO 110 g O1 mol O
32.0 g O
mol SO
3 mol O
g SO
1 mol SO g SO
2 2 2 2
2 22 2
2
2
22
2 22
2
2
2
2
22
3
95 62 641
161
2 641147
? ..
?.
The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield.The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:
Theoretical yield
100yield lTheoretica
yield ActualYield %
Percent Yields from Reactions
theoretical yield is what we have been calculating actual yield is what you have in your flask
A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?
% yield = actual yield
theoretical yield100%
Percent Yields from Reactions
CH COOH + C H OH CH COOC H H O
Calculate the theoretical yield3 2 5 3 2 5 2
Percent Yields from Reactions
CH COOH + C H OH CH COOC H H O
Calculate the theoretical yield
? g CH COOC H = 10.0 g C H OH 88.0 g CH COOC H
g C H OH
g CH COOC H
% yield =14.8 g CH COOC H
19.1 g CH COOC H
3 2 5 3 2 5 2
3 2 5 2 53 2 5
2 5
3 2 5
3 2 5
3 2 5
46 0
191
100% 77 5%
.
.
.
Synthesis Problem In 1986, Bednorz and Muller succeeded in
making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La2CuO4 which superconducts at 35K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3.56 mg of La2CuO4?
molar mass of La CuO = 405.3 g / mol
3.56 mg La CuO g
1000 mg
mol La CuO
405.3 g La CuO mol La CuO
mol La CuO molecules La CuO
mol La CuO
2 La atoms
molecule La CuO La atoms
2 4
2 4
2 4
2 42 4
2 42 4
2 4
2 4
1
18 78 10
8 78 106 022 10
1
106 10
6
623
19
.
( . ).
.
Synthesis Problem
Group Activity Within a year after Bednorz and Muller’s initial
discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa2Cu3O7, that began to superconduct at 100 K. If we wished to make 1.00 pound of YBa2Cu3O7, how many grams of yttrium must we buy?