StoichiometryStoichiometryHow much is consumed and produced in chemical

reactions

Labs#6 Formula of a Hydrate#7 Empirical Formulas#8 Limiting Reactant

Chemical EquationsChapter 7

Atomic Mass UnitAtomic Mass Unit• Standard/unit needed to measure any mass

– SI-standard is platinum-iridium cylinder– Unit is kilogram

• Masses of individual atoms cannot be measured with balance but relative masses of atoms of different elements can be measured– Atomic mass scale used

• Standard is 1 atom C-12 (assigned mass of exactly 12 amu)

• 1 u = 1.661 x 10-27 kg• Unit is atomic mass unit (u)

• Masses of other atoms, molecules, ions, and subatomic particles measured by mass spectrometer and values are reported relative to mass of carbon-12– Proton-1.6726 x 10-27 kg = 1.0073 u– Neutron-1.6749 x 10-27 kg = 1.0087 u– Electron-9.1094 x 10-31 kg = 0.00054858 u

• Use 1 for protons/neutrons while electron’s mass is small enough (0.0005 u) to be neglected for most purposes

Formula weightsFormula weights• Sum of atomic weights of each atom

in chemical formula

• FW of H2SO4 = 2(AW H) + (AW S) + 4(AW O)– = 2(1.0 amu) + 32.1 amu + 4(16.0 amu)– = 98.1 amu

• Of atom = atomic weight of element• Of molecule = molecular weight

Modern Atomic MassesModern Atomic Masses• Most elements mixtures of isotopes-we need to

know– Which isotopes are present naturally– Masses of each isotopes– Abundance of each isotope in element

• Atomic mass of element is weighted average of masses of naturally occurring isotopes forming element– Calculate weighted average by multiplying mass

of each isotope by decimal equivalent of its abundance and then we add each of these products together

• For example, the element lithium has two isotopes that occur in nature: with 7.5 percent abundance, and with 92.5 percent abundance. The atomic mass of lithium-6 is 6.01513 amu, and that of lithium-7 is 7.01601 amu. The average mass of such a mixture of Li atoms is given by:

– average atomic mass = (fraction of isotope X)(mass of isotope X) + (fraction of isotope Y)(mass of isotope Y)

– = (0.075)(6.01513 amu) + (0.925)(7.0161 amu)= 0.45 amu + 6.49 amu= 6.94 amu

• Note that neither lithium-6 nor lithium-7 has an atomic mass of 6.94 amu. This is the average value for the mixture of the two Li isotopes.

• A sample of metal “M” is vaporized and injected into a mass spectrometer. The mass spectrum tells us that 60.10% of the metal is present as 69M and 39.90% is present as 71M. The mass value for 69M and 71M are 68.93 amu and 70.92 amu, respectively. – What is the average atomic mass of

the element? 69.22 amu– What is the element? Ga

• The element indium exists naturally as two isotopes. 113In has a mass of 112.9043 amu, and 115In has a mass of 114.9041 amu. The average atomic mass of indium is 114.82. Calculate the percent relative abundance of the two isotopes of indium.– 114.82 = 112.9043 (X) + 114.9041 (Y)– X + Y = 1 so Y = 1 –X– Substitute and get 113In = 4.2% and 115In

= 95.8%

The Mole (mol)The Mole (mol)• Mole used because atoms, ions, and simple

molecules very small• Must work with very large numbers of them to

obtain quantities measurable in lab• = #C atoms in exactly 12 grams of pure 12C

– SI unit defined in relation to mass of C-12 isotope– Amount of substance that contains as many elementary

entities as there are atoms in 0.012 kg of carbon-12– In 0.012 kg of carbon-12 there are 6.0221421 x 1023

carbon-12 atoms (experimentally determined )

Molar mass Molar mass of element is mass equal to its atomic mass expressed in grams

• Any element will contain same number of atoms as molar mass of any other element

– 28.09 g Si contains same # of atoms as 12.01 g C

• # atoms in a molar mass, called Avogadro's number, is equal to 6.022 x 1023 atoms

• Quantity of a substance that contains Avogadro's number of atoms or other entities is called mole

– 1 mole Na+ ions = 6.022 x 1023 Na+ ions = 23.00 g Na + – 1 mole O2 molecules = 6.022 x 1023 O2 molecules = 32.00

g O2

– 1 mole O3 molecules = 6.022 x 1023 O3 molecules = 48.00 g O3

• How many grams does a sample containing 34 atoms of neon weigh?

34 atoms Ne 20.18 amu Ne 1 g Ne = 1.139 x 10-21 g Ne

1 atom Ne 6.022 x 10-23 amu Ne

• A sample of elemental silver (Ag) has a mass of 21.46 g. – How many moles of silver are in the sample?– How many atoms of silver are in the sample?

21.46 g Ag 1 mol Ag = 0.1989 mol Ag 107.9 g Ag0.1989 mol Ag 6.022 x 10-23 atoms Ag = 1.198 x 10-23 atoms

Ag 1 mol Ag

The molar mass of a compound is its molecular mass expressed in grams

• 1 mol NO2 = 6.022 x 1023 NO2 molecules = 46.01 g NO2

1 mol N2O5 = 6.022 x 1023 N2O5 molecules = 108.0 g N2O5

• Also note that 1 mole of NO2 consists of 1 mole of N atoms, and 2 moles of O atoms

• 1 mol NO2 contains 1 mol N atoms = 14.01 g N1 mol NO2 contains 2 mol O atoms = 32.00 g O1 mol NO2 = 46.01 g NO2

• Just as molecular mass is sum of atomic masses, molar mass of compound is sum of molar masses of atoms in molecule

Molecular Mass-Formula Molecular Mass-Formula MassMass

• Formula mass is an extension of atomic mass• Multiply # atoms of each element in formula by atomic

mass of element and add results together• For example, the molecular masses of two nitrogen

oxides, NO2 and N2O5, are as follows:– molecular mass of NO2 = atomic mass of N + 2(atomic

mass of O)• = 14.01 amu + 2(16.00 amu)

= 46.01 amu– molecular mass of N2O5 = 2( atomic mass of N) + 5(atomic

mass of O)• = 2(14.01 amu) + 5(16.00 amu)

= 108.02 amu

Homework:Read 3.1-3.3, pp. 81-93Q pp. 22, 24, 28, 32, 46, 48 b/c, 52

Percent Composition• Relative contribution of mass of element to mass

of formula in which it appears

• Mass composition-when mass of each element in a substance is specified, either as a % or in grams

• 3 steps:– Compute molecular mass of compound– Calculate how much of molecular mass comes from

each element– Divide each element’s mass contribution by total

molecular mass and multiply by 100 to convert to %

massmass of element in compound

mass of compound% 100%

• Sodium chloride (NaCl)-The molar mass, 58.44 g, is the sum of the mass of 1 mole of Na, 22.99 g, and the mass of 1 mole of Cl, 35.45 g. – % Na by mass is 22.99 g

Na/58.44 g NaCl x 100 = 39.33%– % Cl by mass is 35.45 g Cl/58.44

g NaCl x 100 = 60.66%

• Calculate the mass percent of each element in potassium ferricyanide, K3Fe(CN)6.– 35.62% K– 16.96% Fe– 21.88% C– 25.53% N

Formulas

• molecular formula = (empirical formula)n– [n = integer] (actual ratio of atoms in

compound)

• molecular formula = C6H6 = (CH)6

• empirical formula = CH (simplest whole-number ratio of atoms in compound)

Empirical Formula Determination

1. Base calculation on 100 grams of compound (If % given, assume 100 g)

2. Determine moles of each element in 100 grams of compound (divide given mass by atomic mass)

3. Divide each value of moles by smallest of values

4. Multiply each number by integer to obtain all whole numbers

1. Calculated from empirical formula when molar mass is known

2. Divide molar mass of compound by empirical molar mass

3. Multiply empirical formula by quotient obtained from division

Molecular Formula Determination

• Determine empirical/molecular formulas for a deadly nerve gas that gives the following mass percent analysis:

• C=39.10% 3.26 mol C/.0.542• H=7.67% 7.61 mol H/.0.542• O=26.11% 1.63 mol O/.0.542• P=16.82% 0.543 mol P/.0.542• F=10.30% 0.542 mol F/.0.542• Known molar mass = 184.1 g.• C6H14O3PF• Empirical formula = 184.1 g. So

molecular formula also.

When CH and CHO are combusted in an excess of O2, the only products formed are CO2 and H2O.

• By measuring mass of original sample and masses of products, can calculate empirical formula of compound

• Since all of C/H appear as CO2 and H2O respectively, masses of these elements can be determined

• If original compound contains oxygen, its mass is determined by subtracting mass of compound from sum of masses of C/H

• Masses of element are converted to moles and empirical formula is determined.

Combustion of 11.5 grams of ethanol produces 22.0 grams of CO2 and 13. 5 grams of H2O.

Determine the empirical formula of ethanol.

• A 0.6349 g sample of an unknown produced 1.603 g of CO2 and 0.2810 g of H2O. Determine

the empirical formula of compound.• 1.603 g CO2 1 mol CO2 1 mol C 12.01 g C = 0.4374 g C

44.01 g CO2 1 mol CO2 1 mol C

• 0.2810 g H2O 1 mol H2O 2 mol H 1.01 g H = 0.0315 g H

18.016 g H2O 1 mol H2O 1 mol H

• 0.6349 – (0.4374 + 0.0313) = 0.166 g O• 0.4374/12.01 = 0.03642 mol C/0.01038• 0.0315/1.01 = 0.03119 mol H/0.01038• 0.166/16.00 = 0.01038 mol O/0.01038

• C3.5H3O = multiply everything by 2

• C7H6O2

Homework:Read 3.4-3.5, pp. 93-101Q pp. 125-126, #54a, 56, 58 (you’ll

love this one), 65, 68

Chemical Equation-symbolic way of representing chemical

reaction• Reactants on the left side of equation.• Products on the right side.

Symbols Used in Writing Chemical EquationsSymbol Meaning yields, produces⇆ reversible reaction, equilibrium(s) solid phase(l) liquid phase(g) gas phase(aq) aqueous solution+ added to, and

Types of chemical reactions

• Combination– A + B C

• Decomposition– C A + B

• Combustion– Hydrocarbon (may

have -OH)+ O2 CO2 + H2O

– Produce flame

• Exchange/metathesis reactions– Single replacement

• A + BC B + AC

– Double replacement

• AB + CD AD + CB

Chemical Equation

• C2H5OH + 3O2 2CO2 + 3H2O

• Equation balanced

• 1 mole of ethanol reacts with 3 moles of oxygen

• to produce• 2 moles of carbon dioxide and 3

moles of water

Balancing equations-using coefficients

• Number placed before chemical formula in equation changes amount of substance

• Multiplier for formula– 2H2O means two molecules of water

• Two molecules of water consists of 4 hydrogen atoms and 2 oxygen atoms.

– 3H2O means three molecules of water, which stands

for 6 H atoms and 3 O atoms

• Absence of coefficient is understood to mean one

Calculating masses of reactants and products:

• Balance equation• Convert known mass of reactant or product

to moles of that substance• Use balanced equation to set up

appropriate mole ratios• Use appropriate mole ratios to calculate

number of moles of desired reactant or product

• Convert from moles back to grams if required

Example

• If 1.5 moles of C2H6 reacts, how many

moles of H2O will be formed?

• 2C2H6 + 7O2 4CO2 + 6H2O

1.50 mol C2H6 7 mol O2 32.0 g O2 = 168 g O2

2 mol C2H6 1 mol O2

Example

• If 160. grams of O2 reacts, how many

grams of CO2 will be formed?

160.g O2 1 mol O2 4 mol CO2 1 mol CO2= 126 g CO2

32.0 g O2 7 mol O2 44.0 g CO2

Example• A 12.00-gram sample of solid NaHCO3 is treated with an

excess of HCl and heated to remove the water. What is the change in the mass of the solid?

• NaHCO3 + HCl NaCl + CO2 + H2O• After the reaction, the solid that is present is NaCl. From

the equation, we know that 1 mole of NaHCO3 produces 1 mole of NaCl.

12.00 g NaHCO3 1 mol NaHCO3 1 mol NaCl 58.44 g NaCl = 8.348 g NaCl

84.01 g NaHCO3 1 mol NaHCO3 1 mol NaCl

• 8.348 g NaCl – 12.00 g NaHCO3 = -3.65 g-mass of solid decreases by 3.65 grams.

Example• A 3.75-gram sample of solid is either NaHCO3 or Na2CO3. When

treated w/excess HCl and heated, mass of solid increases by 0.38 gram. Identify original sample.

• We calculate the quantities of NaCl produced by 3.75 grams of NaHCO3 and Na2CO3 and we compare the differences in mass with that given in the problem statement.

3.75 g NaHCO3 1 mol NaHCO3 1 mol NaCl 58.44 g NaCl = 2.61 g NaCl 84.01 g NaHCO3 1 mol NaHCO3 1 mol NaCl

– 2.61 g NaCl – 3.75 g NaHCO3 = -1.14 g

3.75 g Na2CO3 1 mol Na2CO3 1 mol Na2CO3 1 mol NaCl = 4.13 g NaCl 106.0 g Na2CO3 2 mol NaCl 58.44 g NaCl

– 4.13 g NaCl – 3.75 G Na2CO3 = +0.38

• So the original sample is Na2CO3 since the mass increases by 0.38 gram

Homework:Read 3.6-3.8, pp. 102-112Q pp. 126-127, #76, 78, 82, 86, 90

(fun one)

Limiting Reagents (reactants)

• One reactant will be completely consumed (limiting reagent/limiting reactant) before other runs out– Reaction stops/no more product made– Determines, or limits, amount product

formed

• Reactant not completely consumed-excess reagent

Steps to determine limiting reagent:

1. Use equation to calculate stoichiometric mole ratio of reactants

2. Calculate mole ratio of reactants under experimental conditions given

3. Compare two mole ratios: If experimental mole ratio is larger than stoichiometric mole ratio, reactant in denominator is limiting reactant; if smaller, reactant in numerator is limiting reactant

4. In solving any problems involving calculations, always use limiting reactant

Given 11.95 grams of Fe(CrO2)2 and 6.52 grams of K2CO3 are reacted by the equation below, calculate

the mass of Fe2O3 produced, then all reactants, products, unreacted substances.

• Calculate number of moles of each and then compare using balanced equation

• Divide by coefficients

0.05338/4 = 0.013345

0.0472/8 = 0.0059

Therefore K2CO3 is limiting reagent

• To calculate how much is consumed, figure out yield using limiting reagent, then use that information to see how much of others used

• Subtract that from what is given

Theoretical YieldTheoretical Yieldamount of product predicted by calculations

• Calculations under ideal conditions• Under laboratory conditions, actual

yield usually less than theoretical yield• We define percent yield as actual yield

divided by theoretical yield multiplied by 100

• % yield = actual yield/theoretical yield x 100

• 50 g of silver nitrate is mixed with 50 g of hydrochloric acid in a water based solution. A white precipitate forms (silver chloride). The solution is filtered and the white precipitate collected and dried. The dried precipitate is measured to have a mass of 53.6 g. What is the

theoretical and percent yield?

Homework:Read 3.9, pp. 113-120Q pp. 127-128, #94 a/c, 96, 100Do one additional exercise and one challenge problem.Submit quizzes to me by email:http://www.cengage.com/chemistry/book_content/

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