Stoichiometry is one where the substance retains its identity. Examples of physical reactions: ◦ melting / freezing ◦ boiling / condensing ◦ subliming

Chapter 3Stoichiometry:

Calculations with Chemical Formulas and Equations

Stoichiometry

is one where the substance retains its identity.

Examples of physical reactions:

◦ melting / freezing◦ boiling / condensing ◦ subliming◦ subdivision◦ dissolving (solvation)

Physical Change

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:

H2O H2 + O2

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.

Chemical Change

Stoichiometry

a) a colour changeb) gas productionc) an energy changed) precipitate formation

How to Recognize a Chemical Change

Stoichiometry

refers to balancing chemical equations and the associated mathematics.

is possible because of the law of conservation of mass.

Stoichiometry

Stoichiometry

Concise representations of chemical reactions

Chemical Equations

Stoichiometry

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Anatomy of a Chemical Equation

Stoichiometry

Reactants appear on the left side of the

equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Stoichiometry

Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Stoichiometry

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

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Stoichiometry

Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Stoichiometry

Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each

element in a molecule

Stoichiometry

Subscripts and Coefficients Give Different Information Coefficients tell the number of molecules

Stoichiometry

Matter cannot be lost in any chemical reaction.◦all the atoms must balance in the

chemical equation. 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

changed when balancing an equation.

Balancing Chemical Equations

Stoichiometry

Balancing Chemical Equations Balance the following equation:

Na(s) + H2O(l) H2 (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


Na(s) + 2 H2O(l) H2 (g) + NaOH(aq)

the 2 increases the number of H atoms among the reactants.

O is now unbalanced. have 2 H2O on the left; can balance H by

putting a coefficient 2 in front of NaOH on the right:

Stoichiometry


Na(s) + 2 H2O(l) H2 (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


2 Na(s) + 2 H2O(l) H2 (g) + 2 NaOH(aq)

Check our work:◦ 2 Na on each side◦ 4 H on each side◦ 2 O on each side

Balanced !!

Stoichiometry

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

CH4 (g) + O2 (g) CO2 (g) + H2O

(g)

Stoichiometry

Pb(NO3)2 (aq) + 2 NaI (aq)

PbI2 (aq) + 2 NaNO3 (aq)

Pb(NO3)2(aq) + NaI(aq) PbI2(aq) + NaNO3(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) - solid (l) - liquid (g) - gas (aq) - aqueous (meaning that the compound

is dissolved in water.) (ppt) - 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).

States of Reactants and Products

Stoichiometry


Complete 3.11 to 3.14

Reaction Types

Stoichiometry

Examples:N2 (g) + 3 H2 (g) 2 NH3 (g)

C3H6 (g) + Br2 (l) C3H6Br2 (l)

2 Mg (s) + O2 (g) 2 MgO (s)

Two or more substances react to form one product

Combination Reactions

Stoichiometry

2 Mg (s) + O2 (g) 2 MgO (s)

Stoichiometry

Examples:CaCO3 (s) CaO (s) + CO2 (g)

2 KClO3 (s) 2 KCl (s) + O2 (g)

2 NaN3 (s) 2 Na (s) + 3 N2 (g)

One substance breaks down into two or more substances

Decomposition Reactions

Stoichiometry

Examples:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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

Rapid reactions that produce a flame

Most often involve hydrocarbons reacting with oxygen in the air

Combustion Reactions

Stoichiometry

Complete Problems 3.15 to 3.20

Balancing Chemical Reactions

Formula Weights

Stoichiometry

Sum of the atomic weights for the atoms in a chemical formula

So, the formula weight of calcium chloride, CaCl2, would be

Ca: 1(40.08 amu) + Cl: 2(35.45 amu)

110.98 amu These are generally reported for ionic

compounds

Formula Weight (FW)

Stoichiometry

Sum of the atomic weights of the atoms in a molecule

For the molecule ethane, C2H6, the molecular weight would be

Molecular Weight (MW)

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.

Molecular & Formula Weights

Stoichiometry

Complete questions 3.21 and 3.22

Formula & Molecular Weights

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.

Percent Composition

% element =(number of atoms)(atomic weight)

(molar mass of the compound)x 100

Stoichiometry

So the percentage of carbon in ethane (C2H6) is…

Percent Composition

%C = (2)(12.01 g/mol)(2(12.01) + 6(1.01))g/mol

24.02 g/mol30.08 g/mol

= x 100

= 79.85%

Stoichiometry

• the formula weight is 98.09 amu (do calculation)

• % H = 2(1.01 amu) x 100 = 2.06 % H98.09 amu

• % S = 32.07 amu x 100 = 32.69 % S 98.09 amu

• % O = 4(16.00 amu) x 100 = 65.25 % O 98.09 amu

the sum of the percent compositions will always equal 100 %

% Composition of H2SO4

Stoichiometry

Complete questions 3.24 and 3.26 Calculate the % composition of each

element in each compound.

Percent Composition

Moles

Stoichiometry

Avogadro’s Number

6.022 x 1023

1 mole of 12C 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)

Molar Mass

Stoichiometry

Using Moles

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

Stoichiometry

Molar Calculations Using the mass and molar mass to calculate

the number of MOLES of a compound or element.◦ m Mass grams

g◦ n amount of matter moles mol◦ M Molar mass grams per mole

g/mol

number of = mass moles molar mass

n = mM

Stoichiometry

Sample Calculation

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: (NH4)3PO4

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 = 5.03 x 10-3 mol 149.13 g/mol (NH4)3PO4

Example

Stoichiometry

Example 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 mol of diarsenic trioxide?

Diarsenic trioxide: As2O3

Molar mass: (2(74.92) + 3(16.00)) g/mol = 197.84

g/mol Mass = (2.1 x 10-4 mol)(197.84 g/mol)

= 0.042 g As2O3

Example

Stoichiometry

ExampleHow many moles of caffeine (C8H10N4O2 ) is required to reach the 12.0 g lethal dose in a 140 pound female? 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

Calculating Atoms…

We can also find the number of atoms of a substance using Avogadro’s Number:

6.022 x 1023 atoms/mol Example: Calculate the number of atoms in 0.500 moles

of silver?0.500mol Ag x 6.022x1023 atoms = 3.01 x

1023 atoms mol of Ag

Stoichiometry

The Same Goes…

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 1023 molecules = 3.00 x 1024 molecules

mol of CH4

Stoichiometry

How many molecules are present in 0.045 mol of H2O?

Number of atoms = (0.045 mol)(6.022 x 1023

particles/mol) = 2.7 x 1022 molecules H2O

Example

Stoichiometry

How many moles are present in 8.75 x 1026 molecules of Vitamin C?

Moles = 8.75 x 1026 molecules = 1450 mol 6.022 x 1023 particles/molVitamin C

Example

Stoichiometry

What is the mass of 2.1 x 1022 molecules of N2O?

Step 1. Convert molecules to moles:◦ Moles = 2.1 x 1022 molecules = 0.0349

mol N2O 6.022 x 1023 particles/mol

Step 2. Convert moles to mass:◦ Mass = (0.0349 mol)((2(14.01) + 16.00)

g/mol) = 1.54 g N2O

Example

Stoichiometry

How many molecules are present in 4.65 t of CO2?

Step 1: Convert mass to moles◦ Moles = (4.65 t)(1000 kg/t)(1000 g/kg) = 1.06 x

105 (12.01 + 2(16.00)) g/mol mol CO2

Step 2: Convert moles to molecules◦ Molecules = (1.06 x 105 mol)(6.022 x 1023

particles/mol) = 6.36 x 1028 molecules CO2

Example

Stoichiometry

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

The Mole

Finding Empirical Formulas

Stoichiometry

Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition

Stoichiometry

1. Assume we start with 100 g of sample.2. The mass percent then translates as the

number of grams of each element in 100 g of sample.

3. From these masses, calculate the number of moles (use atomic weight from Periodic Table)

4. The lowest number of moles becomes the divisor for the others. (gives a mole ratio greater than 1)

5. Adjust mole ratios so all numbers are whole (1, 2, etc)

6. The lowest whole-number ratio of moles is the empirical formula.

Finding empirical formula from mass percent data:

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: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

Stoichiometry


Calculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol

Stoichiometry


These are the subscripts for the empirical formula:

C7H7NO2

Stoichiometry

Complete questions 3.43 to 3.46

Empirical Formulas

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.

Molecular Formulas

Stoichiometry


Stoichiometry

Combustion Analysis

Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this◦ C is determined from the mass of CO2 produced

◦ H is determined from the mass of H2O produced◦ O is determined by difference after the C and H have

been determined

Stoichiometry

Elemental Analyses

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

Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometry

Stoichiometric CalculationsFrom 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


C6H12O6 + 6 O2 6 CO2 + 6 H2O

Stoichiometry

Work on questions 3.57 to 3.66 Copy examples done in class.


Limiting Reactants

Stoichiometry

How Many Cookies Can I Make?

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

How Many Cookies Can I Make?

In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

Stoichiometry

Limiting Reactants

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

Stoichiometry

Limiting Reactants

The limiting reactant is the reactant you’ll run out of first (in this case, the H2)

Stoichiometry

Limiting Reactants

In the example below, the O2 would be the excess reagent

Stoichiometry


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

Theoretical Yield

Stoichiometry

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

Percent Yield

Actual YieldTheoretical YieldPercent Yield = x 100

Stoichiometry


Documents

Stoichiometry is one where the substance retains its identity. Examples of physical reactions: ◦ melting / freezing ◦ boiling / condensing ◦ subliming