Ch 3 stoichiometry calculations with chemical formulas
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Ch. 3 Stoichiometry: Calculations with Chemical Formulas. Law of Conservation of Mass. Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged. Stoichiometry. The quantitative nature of chemical formulas and chemical reactions. Reactants.

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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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Ch 3 stoichiometry calculations with chemical formulas

Ch. 3 Stoichiometry: Calculations with Chemical Formulas


Law of conservation of mass

Law of Conservation of Mass

  • Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.


Stoichiometry

Stoichiometry

  • The quantitative nature of chemical formulas and chemical reactions


Reactants

Reactants

  • The chemical formulas on the left of the arrow that represent the starting substances

    2H2 + O2 2H2O

Reactants


Products

Products

  • The substances that are produced in the reaction and appear to the right of the arrow

    2H2 + O2 2H2O

Products


Ch 3 stoichiometry calculations with chemical formulas

  • Because atoms are neither created nor destroyed in any reaction a chemical equation must have the same number of atoms of each element on either side of the arrow


Balancing chemical equations

Balancing Chemical Equations

CH4 + O2 CO2 + H2O

C=1 C=1

H=4 H=2

O=2 O=3


Balancing chemical equations1

Balancing Chemical Equations

CH4 + O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 X 2 =4

O=2 O=3


Balancing chemical equations2

Balancing Chemical Equations

CH4 + O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 x 2 = 4

O=2 O= 2 + 2x1 = 4


Balancing chemical equations3

Balancing Chemical Equations

CH4 + 2O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 x 2 = 4

O=2 x 2 = 4 O= 2 + 2x1 = 4


Combustion reactions

Combustion Reactions

  • Rapid reactions that produce a flame.

  • Most combustion reactions involve O2 as a reactant

  • Form CO2 and H2O as products


Combustion reactions1

Combustion Reactions

C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)

C= 3 C=1 X 3 = 3

H=8 H=2 X 4 = 8

O= 2 X 5 = 10 O=(2 X3)+(1X4)=10


Combination reactions synthesis

Combination Reactions (synthesis)

  • 2 or more substances react to form 1 product.


Combination reactions synthesis1

Combination Reactions (synthesis)

2Mg(s) + O2(g) 2MgO(s)

Mg=1 x 2=2 Mg= 1 x 2=2

O= 2 O=1 x 2 = 2


Decomposition reaction

Decomposition Reaction

  • 1 substance undergoes a reaction to produce 2 or more substances


Decomposition reaction1

Decomposition Reaction

CaCO3 (s)  CaO (s) + CO2(g)

Ca=1 Ca=1

C=1 C=1

O=3 O=1+2=3


3 methods of measuring

3 Methods of Measuring

  • Counting

  • Mass

  • Volume


Example 1

Example 1

  • If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?


Example 11

Example 1

  • Count: 1 dozen apples = 12 apples

  • Mass: 1 dozen apples = 2.0 kg apples

  • Volume: 1 dozen apples = 0.20 bushels

    apples

    Conversion Factors:

  • 1 dozen2.0 k.g1 dozen

    12 apples 1 dozen 0.20 bushels


Example 12

Example 1

  • 0.50 bushel x 1 dozen x 2.0 kg =

    0.20 bushel 1 dozen

    = 5.0 kg


Avogadro s number

Avogadro’s Number

  • Named after the Italian scientist Amedo Avogadro di Quaregna

  • 6.02 x 10 23


Mole mol

Mole (mol)

  • 1 mol = 6.02 x 10 23 representative particles

  • Representative particles: atoms, molecules ions, or formula units (ionic compound)


Mole mol1

Mole (mol)

  • Moles= representative x 1 mol

    particles 6.02 x 10 23


Example 2 atoms mol

Example 2 (atoms  mol)

  • How many moles is 2.80 x 10 24 atoms of silicon?


Example 2

Example 2

  • 2.80 x 10 24 atoms Si x 1 mol Si

    6.02 x 10 23 atoms Si

    = 4.65 mol Si


Example 3 mol molecule

Example 3 (mol  molecule)

  • How many molecules of water is 0.360 moles?


Example 3

Example 3

  • 0.360 mol H2O x 6.02 x 10 23 molecules H2O1 mol H2O

    =2.17 molecules H2O


The mass of a mole of an element

The Mass of a Mole of an Element

  • The atomic mass of an element expressed in grams = 1 mol of that element = molar mass

Molar mass S

Molar mass Hg

Molar mass C

Molar mass Fe


Ch 3 stoichiometry calculations with chemical formulas

6.02 x 10 23 atoms S

6.02 x 10 23 atoms Hg

6.02 x 10 23 atoms C

6.02 x 10 23 atoms Fe


Example 4 mol gram

Example 4 (mol  gram)

  • If you have 4.5 mols of sodium, how much does it weigh?


Example 4

Example 4

  • .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na

    1 mol Na


Example 5 grams atoms

Example 5 (grams  atoms)

  • If you have 34.3 g of Iron, how many atoms are present?


Example 5

Example 5

  • 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms

    55.8 g Fe 1 mol Fe

    =3.70 x 10 23 atoms Fe


The mass of a mole of a compound

The Mass of a Mole of a Compound

  • To find the mass of a mole of a compound you must know the formula of the compound

  • H2O  H= 1 g x 2

    O= 16 g

    18 g = 1 mole = 6.02 x 10 23

    molecules


Example 6 gram mol

Example 6 (gram  mol)

  • What is the mass of 1 mole of sodium hydrogen carbonate?


Example 6

Example 6

  • Sodium Hydrogen Carbonate = NaHCO3

  • Na=23 g

  • H=1 g

  • C=12 g

  • O=16 g x3

  • 84 g NaHCO3 = 1 mol NaHCO3


Mole volume relationship

Mole-Volume Relationship

  • Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical


Avogadro s hypothesis

Avogadro’s Hypothesis

  • States that equal volumes of gases at the same temperature and pressure contain the same number of particles

  • Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible


Standard temperature and pressure stp

Standard Temperature and Pressure (STP)

  • Volume of a gas changes depending on temperature and pressure

  • STP= 0oC (273 K)

    101.3 kPa (1 atm)


Standard temperature and pressure stp1

Standard Temperature and Pressure (STP)

  • At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume


Conversion factors

Conversion Factors

  • AT STP

  • 1 mol gas22.4 L gas

    22.4 L gas 1 mol gas


Example 7

Example 7

  • At STP, what volume does 1.25 mol He occupy?


Example 71

Example 7

  • 1.25 mol He x 22.4 L He = 28.0 L He

    1 mol He


Example 8

Example 8

  • If a tank contains 100. L of O2 gas, how many moles are present?


Example 81

Example 8

  • 100. L O2 X 1 mol O2 = 4.46 mol O2

    22.4 L O2


Calculating molar mass from density

Calculating Molar Mass from Density

  • The density of a gas at STP is measured in g/L

  • This value can be sued to determine the molar mass of gas present


Example 9

Example 9

  • A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.


Example 91

Example 9

  • 1 mol gas x 22.4 L gas X 3.58 g gas =

    1 mol gas 1 L gas

    Molar Mass= 80.2 g


Percent composition

Percent Composition

  • The relative amounts of the elements in a compound

  • These percentages must equal 100


Percent composition1

Percent Composition

  • %element = mass of element x 100

    mass of compound


Example 10

Example 10

  • Find the mass percentage of each element present in Al2 (CO3)3


Example 101

Example 10

  • Al2(CO3)3

  • Al= 27 g x 2 = 54 g / 234 g x 100=23%

  • C= 12 g x 3 = 36 g/ 234 g x 100= 15%

  • O = 16 g x 9 = 144 g / 234 g x 100=62%

    234 gAl2(CO3)3


Empirical formula

The simplest whole number ratio of atoms in a compound

The formula obtained from percentage composition

Ex CH , CH4, H2O, C3H8

NOT C2H4, or C6H12O6 these could be simplified

Empirical Formula


Example 111

Example 11

  • Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical fromula.


Example 112

Example 11

Assume that you have 100 grams of the compound therefore:

Hg = 73.9 %  73.9 g

Cl= 26.1%  26.1 g


Example 113

Example 11

Step 2: Change grams of your compound to moles

Hg = 73.9 g x 1 mol =0.368 mol Hg

200.6g

Cl= 26.1 g x 1 mol = 0.735 mol Cl

35.5 g


Example 114

Example 11

Step 3: Find the lowest number of moles present

Hg = 73.9 g x 1 mol =0.368 mol Hg

200.6g

Cl= 26.1 g x 1 mol = 0.735 mol Cl

35.5 g

0.368 < 0.735


Example 115

Example 11

Step 4: Divide by the lowest number of moles to obtain whole numbers

Hg = 73.9 g x 1 mol = 0.368 mol = 1

200.6g 0.368 mol

Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2

35.5 g 0.368 mol


Example 116

Example 11

Step 5: Put the whole numbers into the empirical formula

Hg = 73.9 g x 1 mol = 0.368 mol = 1

200.6g 0.368 mol

Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2

35.5 g 0.368 mol

HgCl2


Molecular formulas

Molecular Formulas

  • The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula

  • We can obtain the molecular formula from the empirical formula IF we know the molecular weight of the compound.


Example 121

Example 12

  • The empirical formula of ascorbic acid is C3H4O3. The molecular weight of ascorbic acid is 176 amu. Determine the molecular formula.


Example 122

Example 12

Step 1: First determine the molecular weight of the empirical formula

C3H4O3

C= 12 amu x 3

H= 1 amu x 4

O= 16 amu x 3

88 amu


Example 123

Example 12

Step 2: Divide the molecular weight of the molecular formula by the molecular weight of the empirical formula

C3H4O3176 amu = 2

C= 12 amu x 3 88 amu

H= 1 amu x 4

O= 16 amu x 3

88 amu


Example 124

Example 12

Step 3: Multiply the empirical formula by the number calculated in step 2

176 amu = 2

88 amu

(C3H4O3) x 2 = C6H8O6


Quantitative information from a balanced equation

Quantitative Information from a Balanced Equation

2 H2 (g) + O2 (g)  2 H2O (l)

2 molecules 1 molecule2 molecules

Or since we can’t count out 2 molecules

2 mol 1 mol 2 moles

The coefficients in a chemical reaction can be interpreted as either the relative number of molecules (formula units) involved in the reaction OR the relative number of moles


Example 13 mol mol

Example 13 (mol mol)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many moles of O2 do you need to react with 5 moles of C4H10?


Example 13

Example 13

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many moles of O2 do you need to react with 5 moles of C4H10?

5 mol C4H10 x 13 mol O2 = 32.5 mol O2

2 mol C4H10


Example 14 g g

Example 14 (gg)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many grams of O2 do you need to react with 50.0 g of C4H10?


Example 14 g g1

Example 14 (gg)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many grams of O2 do you need to react with 50.0 g of C4H10?

50.0 g C4H10 x 1 mol C4H10 x 13 mol O2 x 32 g O2=179 g O2

58 g C4H10 2 mol C4H10 1 mol O2


Limiting reactants reagents

Limiting Reactants (Reagents)

  • The reactant that is completely consumed

  • It determines or limits the amount of product that forms

  • The other reactant(s) are called excess reagents


Example 14 limiting reactants

Example 14 (limiting reactants)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

If you have 25.0 g O2 and 25.0 g of C4H10, what is the limiting reactant?


Example 14 limiting reactants1

Example 14 (limiting reactants)

  • 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

    25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2

    58 g C4H10 2 mol C4H10

    25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

    32 g O2 13 mol O2

    0.481 mol < 1.72 mol C4H10 is the limiting reactant


Theoretical yield

Theoretical Yield

  • The quantity of the product that is calculated to form when all of the limiting reactant reacts.


Example 15 theoretical yield

Example 15 (Theoretical Yield)

  • 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

    25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2

    58 g C4H10 2 mol C4H10

    25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

    32 g O2 13 mol O2

    0.481 mol < 1.72 mol C4H10 is the limiting reactant Calculate the theoretical yeild


Example 15 theoretical yield1

Example 15 (Theoretical Yield)

25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

32 g O2 13 mol O2

0.481 mol CO2 X 44 g CO2= 21.2 g CO2

1 mol CO2

If all of the limiting reactant (25.0 g C4H10) reacts than 21.2 g of CO2 will form.


Percent yield

Percent Yield

Percent Yield = Actual Yield X 100

Theoretical yield


Example 15 yield

Example 15 (% Yield)

  • A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?


Example 15 yield1

Example 15 (% Yield)

A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?

87.9 g / 105 g x 100 = 83.7%

(actual) (theoretical)


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