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Stoichiometry

Chapter 3 Chemical Stoichiometry

Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. 03_34 Detector plate

Ion-accelerating electric field Least massive ions Accelerated ion beam

Positive ions Sample Most massive ions

Electron beam Slits Magnetic field

Heating device to vaporize sample

Schematic diagram of a mass spectrometer. Average Elements occur in as mixtures of Carbon = 98.89% 12C 1.11% 13C <0.01% 14C

(0.9889)(12.0000 amu) = 11.87 amu (0.0111)(13.0034 amu) = 0.144 amu 12.01 amu The The number equal to the number of carbon in exactly 12 grams of pure 12C.

1 mole of anything = 6.022  1023 units of that thing Equal moles of substances have equal numbers of atoms, , ions, formula units, etc. Counting Atoms

Atoms are too small to be seen or counted individually. Atoms can only be counted by weighing them. • all jelly beans are not identical. • jelly beans have an average mass of 5 g. • How could 1000 jelly beans be counted? Jelly Beans & Mints Mints have an average mass of 15 g.

How would you count out 1000 mints?

Why do 1000 mints have a mass greater than 1000 jelly beans? Atomic Mass Unit

Atoms are so tiny that the gram is much too large to be practical. The mass of a single carbon is 1.99 x 10-23 g. The atomic mass unit (amu) is used for atoms and molecules. AMU’s and Grams

1 amu = 1.661 x 10 -24 g

Conversion Factors 1.661 x 10-24g/amu 6.022 x 1023amu/g The Mole

• One mole of rice grains is more than the number of grains of all rice grown since the beginning of time! • A mole of marshmallows would cover the U.S. to a depth of 600 miles! • A mole of hockey pucks would be equal in mass to the moon. The Mole

Substance Average Atomic Mass # Moles # Atoms (g) Na 22.99 1 6.022 x 1023 Cu 63.55 1 6.022 x 1023 S 32.06 1 6.022 x 1023 Al 26.98 1 6.022 x 1023 Measurements

dozen = 12 gross = 12 dozen = 144 ream = 500 mole = 6.022 x 1023 Avogadro’s number equals 6.022  1023 units Unit Cancellation

How many dozen eggs would 36 eggs be? (36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs

How many eggs in 5 dozen? (5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs Calculating Mass Using AMU’S 1 N atom = 14.01 amu

(23 N atoms)(14.01 amu/1N atom) = 322.2 amu Calculating Number of Atoms from Mass 1 O atom = 16.00 amu

(288 amu)(1 O atom/16.00 amu) = 18 atoms O AMU’s & Grams

1 atom C = 12.011 amu = 1.99 x 10-23g

1 mol C = 12.011 g

Use TI-83 or TI-83 Plus to store 6.022 x 1023 to A. Calculating Moles & Number of Atoms 1 mol Co = 58.93 g (5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms) = 8.30 x 10-4 mol Co

(8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co

Moles are the doorway grams <---> moles <---> atoms

A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.

CO2 = 44.01 grams per mole Calculating Mass from Moles

CaCO3 1 Ca = 1 (40.08 g) = 40.08 g 1 C = 1 (12.01 g) = 12.01 g 3 O = 3 (16.00 g) = 48.00 g 100.09 g

(4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3 Calculating Moles from Mass

Juglone 10 C = 10(12.01g) = 120.1 g 6 H = 6(1.008 g) = 6.048 g 3 O = 3(16.00 g) = 48.00 g 174.1 g (1.56 g juglone)(1 mol/174.1 g) = 0.00896 mol juglone Percent Composition

Mass percent of an element: mass of element in compound mass % =  100% mass of compound

For in iron (III) oxide, (Fe2O3) 111.69 mass % Fe =  100% = 69.94% 159.69 % Composition

. CuSO4 5 H2O 1 Cu = 1 (63.55 g) = 63.55 g 1 S = 1 (32.06 g) = 32.06 g 4 O = 4 (16.00 g) = 64.00 g

5 H2O = 5 (18.02 g) = 90.10 g 249.71 g % Composition (Continued) % Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu % S = 32.06 g/249.71 g (100 %) = 12.84 % S % O = 64.00 g/249.71 g (100 %) = 25.63 % O

% H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O

Check: Total percentages. Should be equal to 100 % plus or minus 0.01 %. Formulas

Ionic compounds -- empirical formula NaCl

CaCl2

Covalent compounds -- molecular formula -

C6H12O6

C2H6 Formula of a Compound Calculations Example for Problems 75-78 on page 125. A compound contains only carbon, , and nitrogen. When 0.1156 g of it is reacted

with , 0.1638 g of CO2 and 0.1676 g of water are collected.

Molar mass of CO2 is 44.01 g Molar mass of HOH is 18.02 g Formula of a Compound Calculations

(0.1638g CO2)(12.01g C/44.01g CO2) = 0.04470g C %C = (0.04470g/0.1156g)(100%) = 38.67% C

(0.1676 g HOH)(2.016 g H/18.02g HOH) = 0.01875g H %H = (0.01875g/0.1156g)(100%) = 16.22% H %H + %C + %N = 100% %N = 100% -38.67% -16.22% = 45.11% N Formula of a Compound Calculations (38.67g C)(1 mol/12.01g) = 3.220 mol C = 1 C 3.219 mol (16.22g H)(1 mol/1.008g) = 16.09 mol H = 5 H 3.219 mol (45.11g N)(1 mol/14.01g) = 3.219 mol N = 1 N 3.219 mol

Empirical Formula is CH5N. Formulas

molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH Empirical Formula Determination

1. Base calculation on 100 grams of compound. 2. Determine moles of each element in 100 grams of compound. 3. Divide each value of moles by the smallest of the values. 4. Multiply each number by an integer to obtain all whole numbers. Calculating Empirical Formulas 4.151 g Al & 3.692 g O (4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000

(3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500

1 Al (2) = 2 Al 1.5 O (2) = 3 O

Al2O3 Molecular Formulas

71.65 % Cl, 24.27 % C, & 4.07 % H (71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1 (24.27 g C)(1 mol/12.01g) = 2.021 mol/2.021 mol = 1 (4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2

(EM)x = (MM) (EF)x = (MF) (49.46)x = (98.96) (ClCH2)2 = Cl2C2H4 x = 2 Chemical Equations

Chemical change involves a reorganization of the atoms in one or more substances.

A representation of a :

C2H5OH + 3O2 → 2CO2 + 3H2O reactants products Physical States

• solid (s) • liquid (l) • (g) • aqueous (aq) Important Equation Symbols

• yields ------> • heat ------>

cat. • catalyst ------> light • light ------>

H2SO4 • catalyst ------> elect. • electricity ------> Chemical Equations Quantitative Significance

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s) This equation means

4 Al atoms + 3 O2 molecules ---give--->

2 molecules of Al2O3

4 moles of Al + 3 moles of O2 ---give--->

2 moles of Al2O3 Balancing Equation Prerequisites Student must have memorized: • 44 chemical symbols • Table 2.3 on page 61 in text • Table 2.4 on page 62 • Table 2.5 on page 66 • Table 2.6 on page 67 • Type I, II, III, and acid nomenclature • Count HOFBrINCl Four Steps in Balancing Equations

1. Get the facts down. 2. Check for diatomic molecules (subscripts). 3. Balance charges on compounds containing a metal, ammonium compounds, and acids (subscripts). 4. Balance the number of atoms (coefficients). a. Balance most complicated first. b. Balance other elements. c. Balance hydrogen next to last. d. Balance oxygen last. Balancing Equations Caution The identities (formulas) of the compounds must never be changed in balancing a chemical equation! Only coefficients can be used to balance the equation- subscripts will not change! Chemical Equation

C2H5OH + 3O2 → 2CO2 + 3H2O The equation is balanced.

1 mole of reacts with 3 moles of oxygen to produce 2 moles of and 3 moles of water Balancing Equations

When solid ammonium dichromate decomposes, it produce solid chromium(III) oxide, nitrogen gas, and water vapor.  (NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + HOH(g)

 (NH4)2Cr2O7(s) ----> Cr2O3(s) + N2(g) + 4HOH(g) Calculating Masses of Reactants and Products

1. Balance the equation. 2. Convert mass to moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to grams, if necessary. Gram to Mole & Gram to Gram

__Al(s) + __I2(s) ---> __AlI3(s)

2Al(s) + 3I2(s) ---> 2AlI3(s)

How many moles and how many grams of aluminum iodide can be produce from 35.0 g of aluminum? Gram to Mole & Gram to Gram

2Al(s) + 3I2(s) ---> 2AlI3(s)

(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al) = 1.30 mol AlI3

(35.0 g Al) (1 mol/26.98 g)(2 mol AlI3/2 mol Al)(407.68 g/1 mol) = 529 g AlI3 Limiting Reactant

The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. Solving a Stoichiometry Problem

1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams. Limiting Reactant Problem

If 56.0 g of Li reacts with 56.0 g of N2, how many grams of Li3N can be produced?

__Li(s) + __N2(g) ---> __Li3N(s)

6 Li(s) + N2(g) ---> 2 Li3N(s)

(56.0 g Li) (1 mol/6.94g)(1 mol N2/6 mol Li) (28.0 g/1 mol) = 37.7 g N2

Since there were 56.0 g of N2 and only 37.7 g used, N2 is the excess and Li is the Limiting Reactant. Limiting Reactant Problem

6 Li(s) + N2(g) ---> 2 Li3N(s)

(56.0 g Li)(1 mol/6.94g)(2 mol LiN3/6 mol Li) (34.8 g/1 mol) = 93.6 g Li3N

How many grams of nitrogen are left?

56.0g N2 given - 37.7 g used = 18.3 g excessN2 %

Values calculated using stoichiometry are always theoretical yields!

Values determined experimentally in the laboratory are actual yields! Limiting Reactant & % Yield

If 68.5 kg of CO(g) is reacted with 8.60 kg of H2(g), what is the theoretical yield of that can be produced?

__H2(g) + __CO(g) ---> __CH3OH(l)

2 H2(g) + CO(g) ---> CH3OH(l)

(68.5 kg CO)(1 mol/28.0 g)(2 mol H2/1 mol CO)

(2.02 g/1mol) = 9.88 kg H2 Limiting Reactant & % Yield

2 H2(g) + CO(g) ---> CH3OH(l)

Since only 8.60 kg of H2 were provided, the H2 is the limiting reactant, and the CO is in excess.

(8.60 kg H2)(1000 g/1 kg)(1 mol/2.02 g)(1 mol CH3OH/2 4 mol H2)(32.0 g/1 mol) = 6.85 x 10 g CH3OH Limiting Reactant & % Yield

2 H2(g) + CO(g) ---> CH3OH(l)

If in the laboratory only 3.57 x 104 g of

CH3OH is produced, what is the % yield? actual _ yield %Yield = (100%) theoretical _ yield 3.57x104 g %Yield = (100%) 6.85x104 g % Yield = 52.1 %