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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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



Balancing chemical equations
Balancing Chemical Equations reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

CH4 + O2 CO2 + H2O

C=1 C=1

H=4 H=2

O=2 O=3


Balancing chemical equations1
Balancing Chemical Equations reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • Rapid reactions that produce a flame.

  • Most combustion reactions involve O2 as a reactant

  • Form CO2 and H2O as products


Combustion reactions1
Combustion Reactions reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 2 or more substances react to form 1 product.


Combination reactions synthesis1
Combination Reactions (synthesis) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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


Decomposition reaction1
Decomposition Reaction reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • Counting

  • Mass

  • Volume


Example 1
Example 1 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 0.50 bushel x 1 dozen x 2.0 kg =

    0.20 bushel 1 dozen

    = 5.0 kg


Avogadro s number
Avogadro’s Number reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • Named after the Italian scientist Amedo Avogadro di Quaregna

  • 6.02 x 10 23


Mole mol
Mole (mol) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 1 mol = 6.02 x 10 23 representative particles

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


Mole mol1
Mole (mol) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • Moles= representative x 1 mol

    particles 6.02 x 10 23


Example 2 atoms mol
Example 2 (atoms reaction a chemical equation must have the same number of atoms of each element on either side of the arrow mol)

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


Example 2
Example 2 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow molecule)

  • How many molecules of water is 0.360 moles?


Example 3
Example 3 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 0.360 mol H2O x 6.02 x 10 23 molecules H2O 1 mol H2O

    =2.17 molecules H2O


The mass of a mole of an element
The Mass of a Mole of an Element reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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


6.02 x 10 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow23 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow gram)

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


Example 4
Example 4 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • .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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow atoms)

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


Example 5
Example 5 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow mol)

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


Example 6
Example 6 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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


Conversion factors
Conversion Factors reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • AT STP

  • 1 mol gas22.4 L gas

    22.4 L gas 1 mol gas


Example 7
Example 7 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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


Example 71
Example 7 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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

    1 mol He


Example 8
Example 8 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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


Example 81
Example 8 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • The relative amounts of the elements in a compound

  • These percentages must equal 100


Percent composition1
Percent Composition reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • %element = mass of element x 100

    mass of compound


Example 10
Example 10 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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


Example 101
Example 10 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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

C3H4O3 176 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

2 H2 (g) + O2 (g)  2 H2O (l)

2 molecules 1 molecule 2 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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 reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

Percent Yield = Actual Yield X 100

Theoretical yield


Example 15 yield
Example 15 (% Yield) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

  • 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) reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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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