  is one where the substance retains its identity.   ◦ ◦ ◦ ◦ ◦ Examples of physical reactions: melting / freezing boiling / condensing subliming subdivision dissolving (solvation) Stoichiometry

  the atoms of one or more substances are rearranged to produce new substances with new properties. the atoms remain the same, but their arrangement changes:  

H 2 O



H 2 + O 2

   water is broken down into two substances, oxygen gas and hydrogen gas. The two new substances bear no resemblance to the water of which they are made.

Stoichiometry

a) b) c) d) a colour change gas production an energy change precipitate formation Stoichiometry

  refers to balancing chemical equations and the associated mathematics.

is possible because of the law of conservation

of mass.

Stoichiometry

Concise representations of chemical reactions Stoichiometry

CH

4 (

g

)

+ 2 O

2 (

g

)

CO

2 (

g

)

+ 2 H

2

O

(

g

) Stoichiometry

CH 4 (

g

)

+ 2

O 2 (

g

)

CO

2 (

g

)

+ 2 H

2

O

(

g

)

Reactants

appear on the left side of the equation.

Stoichiometry

CH

4 (

g

)

+ 2 O

2 (

g

) CO 2 (

g

)

+ 2

H 2 O (

g

)

Products

appear on the right side of the equation.

Stoichiometry

CH

4 (

g

)

+ 2 O

2 (

g

)

CO

2 (

g

)

+ 2 H

2

O

(

g

) The

states

of the reactants and products are written in parentheses to the right of each compound.

Stoichiometry

CH

4 (

g

)

+

2

O

2 (

g

)

CO

2 (

g

)

+

2

H

2

O

(

g

)

Coefficients

are inserted to balance the equation.

Stoichiometry

 Subscripts tell the number of atoms of each element in a molecule Stoichiometry

 Coefficients tell the number of molecules Stoichiometry

 Matter cannot be lost in any chemical reaction.

◦ all the atoms must equation.

balance

in the chemical ◦ ◦  same number and type on each side of the equation can only change the stoichiometric coefficients in front of chemical formulas.

Subscripts in a formula are

never

when balancing an equation.

changed Stoichiometry

 Balance the following equation: Na (s) + H 2 O (l)  H 2 (g) + NaOH (aq)  ◦ count the atoms of each kind on both sides of the arrow: The Na and O atoms are balanced (one Na and one O on each side) ◦ there are two H atoms on the left and three H atoms on the right.

Stoichiometry

 Balance the following equation: Na (s) + 2 H 2 O (l)  H 2 (g) + NaOH (aq)    the 2 increases the number of H atoms among the reactants.

O is now unbalanced. have 2 H 2 O on the left; can balance H by putting a coefficient 2 in front of NaOH on the right: Stoichiometry

 Balance the following equation: Na (s) + 2 H 2 O (l)  H 2 (g) + 2 NaOH (aq)    H is balanced O is balanced Na is now unbalanced, with one on the left but two on the right.

◦ put a coefficient 2 in front of the reactant: Stoichiometry

 Balance the following equation: 2 Na (s) + 2 H 2 O (l)    ◦ ◦ Check our work: ◦ 2 Na on each side 4 H on each side 2 O on each side Balanced !!

H 2 (g) + 2 NaOH (aq) Stoichiometry



CH 4 (g) + 2 O 2 (g)



CO 2 (g) + 2 H 2 O (g)

Stoichiometry



Pb(NO 3 ) 2 (aq) + 2 NaI (aq)



PbI 2 (aq) + 2 NaNO 3 (aq)

Stoichiometry

 ◦ ◦ ◦ ◦ Remember: when changing the coefficient of a compound, the amounts of all the atoms are changed.

do a recount of all the atoms every time you change a coefficient.

do a final check at the end.

fractions are not allowed. Multiply to reach whole numbers.

Stoichiometry

     (s) (l) (g) (aq) (ppt) solid liquid gas aqueous (meaning that the compound is dissolved in water.) precipitate (meaning that the reaction produces a solid which falls out of solution.)   ◦ Reaction conditions occasionally appear above or below the reaction arrow e.g., "Δ" is often used to indicate the addition of heat). Stoichiometry

 Complete 3.11 to 3.14

Stoichiometry

 Two or more substances react to form one product  Examples: N 2 ( g ) + 3 H 2 ( g ) C 3 H 6 ( g ) 2 Mg ( s ) + Br 2 ( l ) + O 2 ( g )    2 NH 3 ( g ) C 3 H 6 Br 2 ( l ) 2 MgO ( s ) Stoichiometry

Stoichiometry

 One substance breaks down into two or more substances  Examples: CaCO 3 ( s ) 2 KClO 3 ( s ) 2 NaN 3 ( s )    CaO ( s ) 2 KCl 2 Na + CO 2 ( g ) ( ( s ) s ) + O 2 ( g ) + 3 N 2 ( g ) Stoichiometry

  Rapid reactions that produce a flame Most often involve hydrocarbons reacting with oxygen in the air  Examples: CH 4 ( g ) + 2 O 2 ( g ) C 3 H 8 ( g ) + 5 O 2 ( g )   CO 2 ( g ) + 2 H 3 CO 2 ( g ) 2 O ( g ) + 4 H 2 O ( g ) Stoichiometry

 Complete Problems 3.15 to 3.20

Stoichiometry

   Sum of the atomic weights for the atoms in a chemical formula So, the formula weight of calcium chloride, CaCl 2 , would be Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu These are generally reported for ionic compounds Stoichiometry

  Sum of the atomic weights of the atoms in a

molecule

For the molecule ethane, C weight would be 2 H 6 , the molecular C: 2(12.01 g/mol) + H: 6(1.01 g/mol) 30.08 g/mol Stoichiometry

    formula and molecular weights are often called

molar mass.

The formula weight (in amu) is numerically equal to the molar mass (in g/mol).

Always report molar mass in g/mol.

Always round the atomic weights to 2 decimal places.

Stoichiometry

 Complete questions 3.21 and 3.22

Stoichiometry

Percentage composition

is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100.

% element = (number of atoms)(atomic weight) (molar mass of the compound) x 100 Stoichiometry

So the percentage of carbon in ethane (C 2 H 6 ) is… %C = (2)(12.01 g/mol) (2(12.01) + 6(1.01))g/mol = 24.02 g/mol x 100 30.08 g/mol = 79.85% Stoichiometry

• • the formula weight is 98.09 amu (do calculation) % H = 2(1.01 amu) x 100 = 2.06 % H 98.09 amu • % S = 32.07 amu x 100 = 32.69 % S 98.09 amu • % O = 4(16.00 amu) x 100 = 65.25 % O equal 100 % 98.09 amu the sum of the percent compositions will always Stoichiometry

  Complete questions 3.24 and 3.26

Calculate the % composition of each element in each compound.

Stoichiometry

  6.022 x 10 23 1 mole of 12 C has a mass of 12 g Stoichiometry

 ◦ ◦ By definition, these are the mass of 1 mol of a substance (i.e., g/mol) is the atomic weight on the periodic table The formula weight (in amu’s) will be the same number as the molar mass (in g/mol) Stoichiometry

Moles provide a bridge from the molecular scale to the real-world scale Stoichiometry

 ◦ ◦ ◦ Using the mass and molar mass to calculate the number of MOLES of a compound or element.

m n M

Mass amount of matter Molar mass grams moles grams per mole g mol g/mol number of = moles mass molar mass n = m M Stoichiometry

   If you have 20.2 grams of Potash, how many moles do you have?

Molar Mass KCl (calculated): 74.55 g/mol Mass KCl: 20.2 g  20.2 g = 0.271 moles KCl 74.55 g/mol if the units work out, you are correct.

Stoichiometry

    How many moles are present in 0.750 g of ammonium phosphate?

Ammonium phosphate: (NH 4 ) 3 PO 4 Molar mass: (3(14.01) + 12(1.01) + 30.98 + 4(16.00)) g/mol = 149.13 g/mol Number of moles = 0.750 g 149.13 g/mol = 5.03 x 10 -3 mol (NH 4 ) 3 PO 4 Stoichiometry

 ◦ ◦ What is the mass of 12.31 mol of NaOH ?

Step 1. Calculate molar mass     1 x 22.00 g/mol 1 x 16.00 g/mol 1 x 1.01 g/mol  40.00 g/mol Step 2. Calculate mass (12.31 mol)(40.00 g/mol) = 492.4 g NaOH Stoichiometry

    What is the mass of 2.1 x 10 -4 trioxide?

Mass = (2.1 x 10 -4 mol of diarsenic Diarsenic trioxide: As 2 O 3 Molar mass: (2(74.92) + 3(16.00)) g/mol = 197.84 g/mol mol)(197.84 g/mol) = 0.042 g As 2 O 3 Stoichiometry

How many moles of caffeine (C pound female? 8 H 10 N 4 O 2 ) is  Step 1. Molar Mass  8 x 12.01 g/mol     10 x 1.01 g/mol 4 x 14.01 g/mol 2 x 16.00 g/mol 194.22 g/mol  Step 2. Calculate Moles  12.00 g = 0.06179 mol caffeine  194.22 g/mol Stoichiometry

    We can also find the number of atoms of a substance using Avogadro’s Number: 6.022 x 10 23 atoms/mol Example: Calculate the number of atoms in 0.500 moles of silver?

0.500mol Ag x 6.022x10

23 atoms = 3.01 x 10

23

mol of Ag

atoms

Stoichiometry

  For molecules!

Avogadro’s number represents the number of particles, whether

atoms

or

molecules

.  Eg. Calculate the number of molecules in 4.99mol of methane. 4.99 mol x 6.022 x 10 23 molecules = 3.00 x 10 24 mol of CH 4 molecules Stoichiometry

 How many molecules are present in 0.045 mol of H 2 O?

 Number of atoms = (0.045 mol)(6.022 x 10 = 2.7 x 10 22 23 particles/mol) molecules H 2 O Stoichiometry

 How many moles are present in 8.75 x 10 26 molecules of Vitamin C?

 Moles = 8.75 x 10 26 6.022 x 10 23 molecules particles/mol = 1450 mol Vitamin C Stoichiometry

 What is the mass of 2.1 x 10 22 molecules of N 2 O?

  ◦ Step 1. Convert molecules to moles: Moles = 2.1 x 10 22 6.022 x 10 23 molecules particles/mol = 0.0349 mol N 2 O ◦ Step 2. Convert moles to mass: Mass = (0.0349 mol)((2(14.01) + 16.00) g/mol) = 1.54 g N 2 O Stoichiometry

 How many molecules are present in 4.65 t of CO 2 ?

 ◦ Step 1: Convert mass to moles Moles = (4.65 t)(1000 kg/t)(1000 g/kg) = 1.06 x 10 5 (12.01 + 2(16.00)) g/mol mol CO 2  ◦ Step 2: Convert moles to molecules Molecules = (1.06 x 10 5 = 6.36 x 10 28 mol)(6.022 x 10 23 molecules CO 2 particles/mol) Stoichiometry



Complete questions 3.32 to 3.42, even (leave 3.32 for last)

Stoichiometry

One can calculate the empirical formula from the percent composition Stoichiometry

1.

2.

3.

4.

5.

6.

Assume we start with 100 g of sample.

The mass percent then translates as the number of grams of each element in 100 g of sample.

From these masses, calculate the number of moles (use atomic weight from Periodic Table) The lowest number of moles becomes the divisor for the others. (gives a mole ratio greater than 1) Adjust mole ratios so all numbers are whole (1, 2, etc) The lowest whole-number ratio of moles is the empirical formula.

Stoichiometry

The compound

para

-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Assumptions: 1. These are % by mass; in 100g of the compound, the percentages represent masses in grams.

2. Mass divided by molar mass gives moles of each atom.

Stoichiometry

Assuming 100.00 g of

para

-aminobenzoic acid, each percentage converts into a mass: C: H: N: O: 61.31 g x 5.14 g x 10.21 g x 1 mol 12.01 g 1 mol 1.01 g 1 mol 14.01 g 23.33 g x 1 mol 16.00 g = 5.105 mol C = 5.09 mol H = 0.7288 mol N = 1.456 mol O Stoichiometry

Calculate the mole ratio by dividing by the smallest number of moles: C: 5.105 mol 0.7288 mol = 7.005  7 H: 5.09 mol 0.7288 mol = 6.984  7 N: O: 0.7288 mol 0.7288 mol = 1.000

1.458 mol 0.7288 mol = 2.001  2 Stoichiometry

These are the subscripts for the empirical formula: C 7 H 7 NO 2 Stoichiometry



Complete questions 3.43 to 3.46

Stoichiometry

  Once we know the composition of a compound the next step is to determine the molar mass. ◦ ◦ From these two pieces of information the actual molecular formula can be determined: take the ratio of the actual molar mass divided by the molar mass determined by the empirical formula multiply the atoms in the empirical formula by the results of the ratio.

Stoichiometry

 Complete questions 3.47 to 3.50

Stoichiometry

 Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this ◦ C is determined from the mass of CO 2 produced ◦ ◦ H is determined from the mass of H 2 O produced O is determined by difference after the C and H have been determined Stoichiometry

Compounds containing other elements are analyzed using methods analogous to those used for C, H and O Stoichiometry

 Complete questions 3.51 & 3.52

Stoichiometry

The coefficients in the balanced equation give the ratio of

moles

of reactants and products Stoichiometry

From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) Stoichiometry

C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O Stoichiometry

  Work on questions 3.57 to 3.66

Copy examples done in class.

Stoichiometry

  You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat) Stoichiometry

 In this example the sugar would be the

limiting reactant

can make , because it will limit the amount of cookies you Stoichiometry

The limiting reactant is the reactant present in the smallest stoichiometric amount Stoichiometry

 The limiting reactant is the reactant you’ll run out of first (in this case, the H 2 ) Stoichiometry

In the example below, the O 2

excess reagent

would be the Stoichiometry

 Complete questions 3.69 to 3.76

Stoichiometry

  ◦ The theoretical yield is the amount of product that can be made In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures Stoichiometry

A comparison of the amount actually obtained to the amount it was possible to make

Percent Yield = Actual Yield Theoretical Yield x 100

Stoichiometry

 Complete questions 3.77 to 3.80

Stoichiometry