ch. 3 stoichiometry: calculations with chemical formulas
TRANSCRIPT
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
• The chemical formulas on the left of the arrow that represent the starting substances
2H2 + O2 2H2O
Reactants
Products
• The substances that are produced in the reaction and appear to the right of the arrow
2H2 + O2 2H2O
Products
• 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
CH4 + O2 CO2 + H2O
C=1 C=1
H=4 H=2
O=2 O=3
Balancing Chemical Equations
CH4 + O2 CO2 + 2H2O
C=1 C=1
H=4 H=2 X 2 =4
O=2 O=3
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 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
• Rapid reactions that produce a flame.
• Most combustion reactions involve O2 as a reactant
• Form CO2 and H2O as products
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)
• 2 or more substances react to form 1 product.
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
• 1 substance undergoes a reaction to produce 2 or more substances
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
• Counting
• Mass
• Volume
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 1
• Count: 1 dozen apples = 12 apples• Mass: 1 dozen apples = 2.0 kg apples• Volume: 1 dozen apples = 0.20 bushels applesConversion Factors:• 1 dozen 2.0 k.g 1 dozen
12 apples 1 dozen 0.20 bushels
Example 1
• 0.50 bushel x 1 dozen x 2.0 kg =
0.20 bushel 1 dozen
= 5.0 kg
Avogadro’s Number
• Named after the Italian scientist Amedo Avogadro di Quaregna
• 6.02 x 10 23
Mole (mol)
• 1 mol = 6.02 x 10 23 representative particles
• Representative particles: atoms, molecules ions, or formula units (ionic compound)
Mole (mol)
• Moles= representative x 1 mol
particles 6.02 x 10 23
Example 2 (atoms mol)
• How many moles is 2.80 x 10 24 atoms of silicon?
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)
• How many molecules of water is 0.360 moles?
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 atomic mass of an element expressed
in grams = 1 mol of that element = molar mass
Molar mass S
Molar mass C
Molar mass Hg
Molar mass Fe
6.02 x 10 23 atoms S
6.02 x 10 23 atoms C
6.02 x 10 23 atoms Hg
6.02 x 10 23 atoms Fe
Example 4 (mol gram)
• If you have 4.5 mols of sodium, how much does it weigh?
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)
• If you have 34.3 g of Iron, how many atoms are present?
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
• 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)
• What is the mass of 1 mole of sodium hydrogen carbonate?
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
• Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical
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)
• Volume of a gas changes depending on temperature and pressure
• STP= 0oC (273 K)
101.3 kPa (1 atm)
Standard Temperature and Pressure (STP)
• At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume
Conversion Factors
• AT STP
• 1 mol gas 22.4 L gas
22.4 L gas 1 mol gas
Example 7
• At STP, what volume does 1.25 mol He occupy?
Example 7
• 1.25 mol He x 22.4 L He = 28.0 L He
1 mol He
Example 8
• If a tank contains 100. L of O2 gas, how many moles are present?
Example 8
• 100. L O2 X 1 mol O2 = 4.46 mol O2
22.4 L O2
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
• A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.
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
• The relative amounts of the elements in a compound
• These percentages must equal 100
Percent Composition
• %element = mass of element x 100
mass of compound
Example 10
• Find the mass percentage of each element present in Al2 (CO3)3
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 g Al2(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
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 11
Assume that you have 100 grams of the compound therefore:
Hg = 73.9 % 73.9 g
Cl= 26.1% 26.1 g
Example 11
Step 2: Change grams of your compound to moles
Hg = 73.9 g x 1 mol =0.368 mol Hg 200.6gCl= 26.1 g x 1 mol = 0.735 mol Cl 35.5 g
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 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 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
• 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 12
• The empirical formula of ascorbic acid is C3H4O3. The molecular weight of ascorbic acid is 176 amu. Determine the molecular formula.
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 12
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 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
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)
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
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 (gg)
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 (gg)
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)
• 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)
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 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
• The quantity of the product that is calculated to form when all of the limiting reactant reacts.
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 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 = Actual Yield X 100
Theoretical 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 (% 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)