Stoichiometry: Calculations with Chemical Formulas and Equations

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


What is the molecular weight of the following molecules?

CH4

H2SO4

Ca(NO3)2

CuSO4 •5H2O



CH4 12.011 + 4(1.008) = 16.043 amu

H2SO4

Ca(NO3)2

CuSO4 •5H2O



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



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



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:


100

Compound of FWAWElement of Atoms

Element %


Determine the Mass Percent Composition of nitrogen in NH4NO3?



Formula weight 80.043 amu



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.


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


Calculate the number of O atoms in 26.5 g of Li2CO3.


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


Empirical Formula H2O H2O

C6H12O6

H2O2

Empirical Formulas



C6H12O6 CH2O

H2O2

Empirical Formulas



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


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


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


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


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 ?


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



for Co Co for O O 01110

011101

01480

011101333

.

.

.

..



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

.

.

.

..

.

:


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


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


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)


CH4(g) + 2O2(g) CO2(g) + 2H2O(g)


CH4(g) + 2O2(g) CO2(g) + 2H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g)




CH4(g) + 2O2(g) CO2(g) + 2H2O(g)



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


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?


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


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


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?



Fe O + 3 CO 2 Fe + 3 CO2 3 2



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.

.


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?


? 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

..


How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?


?.

. 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



434434

434

PO)(NH mol 0.357N mol 3

PO)(NH mol 1N mol 1.07

N mol 1.07N g 14.0


g/mol 149.0PO)(NH of massmolar



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


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


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.


Combination and Decomposition ReactionsCombination reactions: fewer reactants than products.Decomposition reactions: more products than reactants.


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


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


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? ..


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 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%


CH COOH + C H OH CH COOC H H O

Calculate the theoretical yield3 2 5 3 2 5 2


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?

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Stoichiometry: Calculations with Chemical Formulas and Equations