Section 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Fundamental Chemical Laws

Law of Conservation of Mass

Law of conservation of mass: Matter is neither created nor destroyed during a reaction.
The mass of substances before a chemical change is always equal to the mass of substances after the change.
The mass of reactants is equal to the mass of products in any chemical reaction.

1 g hydrogen + 8 g oxygen >yields> 9 g water

2 g hydrogen + 16 g oxygen >yields> 18 g water

4 g hydrogen + 16 g oxygen >yields> 18 g water + 2 g hydrogen

Reactions occur in specific ratios

Chemical Equations

Symbols for Chemical Reactions

Symbol	Meaning
à	Reacts to yield. Points from reactants to products
+	Separates two or more reactants or two or more products
D (delta)	Heat added, used above the à
NR	No Reaction
(s)	solid material including precipitate
(l)	liquid
(g)	gas
(aq)	aqueous solution, i.e. dissolved in water

Chemical reactions involves one or more chemical reactants reacting to form one or more new chemical products

A chemical equation is used to describe a chemical reaction.

A chemical equation shows all the chemical reactants on the left and all the chemical products on the right. An arrow (à) separates the reactants from the products, pointing from reactants on the left to products on the right.

The physical state of the reactants and products can be indicated using an abbreviation in parenthesis after the chemical symbol. (optional, not required for a chemical equation)

A catalyst is a material that participates in a chemical reaction to speed up the reaction without being changed by the reaction. (A catalyst is shown above the reaction arrow)

The Meaning of a Chemical Equation

Shows the ratio of molecules

Shows the ratio of moles

Mass of reactants equals mass of products

Balancing Chemical Equations

To be in agreement with the conservation of mass, a chemical equation must be balanced. All elements must have equal amounts as both reactants and products. Equal amounts as shown by equal number of atoms.

A chemical equation is balanced by putting whole number coefficient in front of each chemical in order to balance the number of atoms between the reactants and products.

Example: Cobalt reacts with hydrochloric acid to yield cobalt chloride and hydrogen gas

Co(s) + HCl(aq) à CoCl₂(aq) + H₂(g) [not balanced, there are two hydrogen atoms and two chlorine atoms in the products but only one each in the reactants [Note: hydrogen gas is a diatomic molecule]

Co(s) + 2HCl(aq) à CoCl₂(aq) + H₂(g) [balanced]

Aluminum oxide reacts with chlorine gas to yield aluminum chloride and oxygen gas

Al₂O₃(s) + Cl₂(g) à AlCl₃(s) + O₂(g) [not balanced, two aluminum atoms in the reactant but one in the products; two chlorine atoms in reactants but three in products, and three oxygen atoms in the reactants and two in the products.]

2Al₂O₃(s) + 6Cl₂(g) à 4AlCl₃(s) + 3O₂(g) [balanced]

Guidelines for Balancing Chemical Equations

1. First verify that each chemical formula is correct with the correct subscripts.

2. Balance each element using whole number coefficient in front of each chemical (the number 1 is always implied and not written)

a) Begin to balance the equation using the most complex formula

b) Balance polyatomic ions as a single unit unless it decomposes [note: all atoms must balance]

c) Use whole number coefficients, A fraction can be used in an intermediate step in balancing the
equation, but must be removed by multiplying through the equation. [when multiplying through a
balanced equation containing a fraction, multiply all reactants and products equally]

d) Balance one element (or polyatomic ion) at a time, and repeat the process until all elements are
balanced, this may require several iterations.

3. After balancing the equation, verify that the equation is balanced by counting the atoms of each element in both the reactants and products. [count the atoms by multiplying the coefficient by the subscript for that element]

Example:

Under high pressure and temperature nitrogen gas can react with oxygen gas to yield dinitrogen pentaoxide

N₂(g) + O₂ (g) à N₂O₅(g)

Look at N₂O₅(g), it has 2 nitrogen and there are 2 nitrogen atoms as reactants [good]. However, there are 5 oxygen atoms in the products and two as reactants. The easiest way to balance this is to use the subscripts as coefficients for the other compound.

Therefore:

N₂(g) + 5O₂ (g) à 2N₂O₅(g)

Now there are 10 oxygen atoms as both reactants and products. Look at nitrogen again. Now there are four nitrogen atoms as products and only 2 as reactants. Multiply the nitrogen reactants by 2 to give.

2N₂(g) + 5O₂ (g) à 2N₂O₅(g)

Check both oxygen and nitrogen again to verify that they are balanced.

Classifying Chemical Reactions

Introductory types of chemical reactions.

Combination reaction (synthesis reaction) is a reaction that combines two or more reactants into one product [it has two or more reactants and one product.]

Example: Fe(s) + S(s) à FeS(s)

Decomposition reaction: one compound is broken down into two or more compounds [one reactant and two or more products]

Example: Fe₂(CO₃)₃(s) à Fe₂O₃(s) + 3CO₂(g)

Combustion Reactions: Combustion reactions are rapid oxidation reactions, usually with the production of heat and light. These reactions consist of a compound reacting with oxygen to produce oxides. Combustion of hydrocarbons (fuels containing hydrogen and carbon) produce carbon dioxide and water.

Example: C₃H₈ + 5O₂ à 3CO₂ + 4H₂O
propane

The Mole

Avogadro's Number

Mole/Avogadro's number

To define a scale between the microscopic and macroscopic matter, it was decided to use the number of atoms that make up exactly 12 grams of carbon 12. This number is called Avogadro's number (N or N₀). This number was experimentally determined to contain 6.02 (6.022137) x 10²³ atoms in 12 grams of Carbon 12.

Mole = Avogadro's number of units

Similar to dozen

The amount of substance that contains Avogadro's number of species (atoms, molecules, or other) is called a mole (abbreviated as mol).

Molar Mass

molecular weight

The molecular weight (formula weight for ionic compounds) is the sum of the atomic masses of each atom in its formula. The unit will be amu for atomic scale masses and later g/mole for macroscopic amounts.

The mass of one mole of a substance is termed the molar mass (M. units of g/mol).

Examples of Molecular Weights.

O has molar mass of 16.00 g/mole

O₂ has the molar mass of (2 x 16.00) = 32.00 g/mole

H₂O has the molar mass of ((2 x 1.01) + 16.00) = 18.02 g/mole

Acetic acid, HC₂H₃O₂, has the molar mass of ((4 x 1.01) + (2 x 12.01) + (2 x 16.00)) = 60.06 g/mole

Percent Composition of Compounds

Percent composition of a compound is a list of the mass percent of each element in the compound. To calculate the percent composition, we need the relative mass of each element and mass of the compound. We can generate this from the chemical formula.

Example: What is the percent composition of methane, CH₄? The mass of carbon and hydrogen in one mole of methane is:

Mass Carbon = 1 x 12.01 amu x 1g/mol/amu x 1 mol = 12.01 g

Mass Hydrogen = 4 x 1.01 amu x 1g/mol/amu x 1 mol = 4.04 g

The molar mass of methane is:

Molar mass methane = 12.01g + 4.04g = 16.05g

The percent composition is:

%H = 4.04g/16.05g x 100% = 25.2%

%C = 12.01g/16.05g x 100% = 74.8%

Check [sum of percentages should be 100%]: 25.2% + 74.8% =100.0% [good]

Mass to Moles

The Molar Mass of a compound allows for the conversion between mass in grams of a compound and the number of moles of that compound.

The units of molar mass are g/mol, so we can convert between grams and moles.

Example:

If we have 35.0 grams of calcium sulfate, how many moles of calcium sulfate are present?

We need the molar mass, which we derive from the formula: CaSO₄

Molar mass is 136.1 g/mol CaSO₄

So we have mass and molar mass, how do we calculate number of moles?

The formula is: MM = mass/mol [just as the units show (g/mol)]

We can solve this for mol to give: mol = mass/MM

So: mol = 35.0 g/136.1 g/mol = 0.257 mol CaSO₄

Example:

If we have 0.576 mol of sodium sulfate, how many grams do we have?

The formula is Na₂SO₄

And Molar Mass is: 142.0 g/mol Na₂SO₄

Solve the formula for molar mass [MM = mass/mol] for mass to get: mass = MM(mol)

Solve for mass: 142.0 g/mol (0.576 mol) = 81.8 g Na₂SO₄

Particles to Mass

Avogadro's number along with the molar mass can be used to correlate mass with the number of particles or to determine the mass of a particle.

This is a two step process; Avogadro's number is used to convert between number of particles and moles.

And the molar mass is used to convert between moles and mass (grams).

Example:
How much mass does 3.15 x 10²³ atoms of zinc, Zn, have?

(3.15 x 10²³ atoms / 6.02 x 10²³ particles/mole) x 65.39 g/mole
= 0.523 mol Zn x 65.39 g/mol
= 34.2 grams Zn

Example:
What is the number of atoms in 1.75 g calcium, Ca?

(1.75 g Ca/ 40.08 g/mole) x 6.02 x 10²³ particles/mole = 2.63 x 10²² atoms
0.0437 moles x 6.02 x 10²³ particles/mole = 2.63 x 10²² atoms

Mass of an Atom or Molecule

What is mass of a single atom of copper, Cu?

(1 atom / 6.02 x 10²³ particles/mole) x 63.55 g/mole = 1.06 x 10^-22 g

What is mass of a single molecule of methanol, CH₃OH? [12.01 + 16.00 + (4 x 1.01) = 32.05 g/mole]

(1 molecule / 6.02 x 10²³ particles/mole) x 32.05 g/mole = 5.32 x 10^-23 g

Moles: The Basics

To Calculate Moles

· moles = # atoms x (1 mole/6.02 x 10²³ atoms)

· moles = # grams x (1 mol/MM(grams))

· moles = M(mol/Liter) x V (Liters) Valid only for solutions

To Calculate from Moles

· # atoms = moles x (6.02 x 10²³ atoms/1 mole)

· # grams = moles x (MM (grams)/1 mole)

· M(mol/Liter) = (moles/V(Liters)) Valid only for solutions

· V (Liters) = moles x (1/M)(Liters/mole) Valid only for solutions

Mass (¸ MM) (* MM) Mass

Molarity & Volume (M·V) mol A (B/A) mol B (¸ M) V

(¸ V) M

P, V, & T (PV/RT) (PV = nRT) P or V

V @ STP (¸ 22.4 L/mol) (* 22.4 L/mol) V @ STP

Empirical Formula

Empirical Formula is the lowest whole number ratio of atoms of each element in the formula of a molecule or ions in an ionic compound. This formula can be, but doesn't have to be, smaller than the molecular formula or the formula unit of an ionic compound. The empirical formula is usually an experimentally calculated formula and is not generally used after the actual molecular formula is determined. The molecular formula represents the composition that is actually found in nature. For example hydrogen is always found as a diatomic molecule, H₂.

Since the ratio of atoms is the same as the ratio of moles, we can use the ratio of moles to calculated the empirical formula.

We can convert mass into moles, so we can calculate the empirical composition from a mass analysis of the compound or from the mass percent composition.

Example:

0.500g of tin reacted with nitric acid to give 0.635 g of tin oxide, what is the empirical formula of tin oxide?

Mass tin = 0.500g

Mass oxygen = 0.635g - 0.500g = 0.135g

Moles tin = 0.500g/118.71g/mol = 0.00421 mol

Moles oxygen = 0.135g/16.00g/mol = 0.00844 mol

Empirical formula = Sn_0.00421O_0.00844 = Sn₁O₂ = SnO₂

[divide both mole numbers by smaller value]

Empirical Formulas from Percent Composition

Example:

The mass percent of manganese fluoride is:

59.1% Mn and 40.9% F; what is the empirical formula?

Assume a total of 100 g and convert percent into grams;

Mass Mn = 59.1 g

Mass F = 40.9 g

Moles Mn = 59.1g/54.94g/mol = 1.08 mol

Moles F = 40.9g/19.00g/mol = 2.15 mol

Ratio'd to smallest number:

Moles Mn = 1.08 mol/1.08 = 1.00 moles

Moles F = 2.15 mol/1.08 = 1.99 = 2 moles

Formula = MnF₂

Molecular Formula

The molecular formula represents the composition that is actually found in nature. For example hydrogen is always found as a diatomic molecule, H₂.

Some information about the compound, such as molar mass, is necessary to convert the empirical formula into the molecular formula

Example:

Aspirin has an empirical formula of C₉H₈O₄ and a molar mass of 180 g/mol, what is the molecular formula?

The empirical mass is 9 x 12 + 8 x 1 + 4 x 16 = 108 + 8 + 64 = 180 g

The ratio is 180/180 = 1.

So the molecular formula is C₉H₈O₄

Example 2:

Ethylene glycol has a molar mass of 62 g/mole. If the percent composition is 38.7% C, 9.74% H, and 51.6% O, what is the empirical and molecular formulas?
Assume 100 grams and convert to moles
38.7 g C /12.01 g/mole C = 3.22 moles
9.74 g H / 1.01 g/mole H = 9.64 moles
51.6 g O /16.00 g/mole O = 3.22 moles
divide by smallest number
3.22 moles/3.22 moles = 1 mole C
9.64 moles/3.22 moles = 2.99 moles H = 3 moles H
3.22 moles/3.22 moles = 1 mole O

Empirical formula is CH₃O

Empirical mass is 12.01 +3(1.01) + 16.00 = 31.04 g/mole
Molecular/empirical ratio = 62/31 = 2
Molecular formula is C₂H₆O₂

The Meaning of a Chemical Equation

Shows the ratio of molecules

Shows the ratio of moles

Mass of reactants equals mass of products

Mole/Stoichiometric Ratio

Stoichiometry - is the ratio of reaction of reactants and products

Just as two molecules of hydrogen react with one molecule of oxygen to form two molecules of water, two moles of hydrogen react with two moles of oxygen to form two moles of water. Avogadro's number is the scale factor between the microscopic and the macroscopic. The same stoichiometric ratios carry over from the molecule to the mole.

Stoichiometric Calculations: Amounts of Reactants and Products

Stoichiometry - is the ratio of reaction of reactants and products

In balancing a chemical equation, we make sure that the number of atoms of each element are the same on both sides of the reaction by adding coefficients in front of the compounds

This provides the ratio that molecules of these reactants react and the ratio the products are formed, and the ratio that the products form relative to the reactants. These ratios will not change for this reaction.

However, the coefficients are valid for more than the ratio of molecules. The ratios are valid for the number of moles of each compound in the reaction.

For gases, the ratios are valid for the number of liters of gases.

2H₂ (g) + O₂ (g) à 2H₂O (g)

In this reaction 1 molecule oxygen reacts with 2 molecules of hydrogen to form 2 molecules of water.

Also, 1 mole oxygen reacts with 2 moles of hydrogen to form 2 moles of water.

And, 1 liter oxygen gas reacts with 2 liters of hydrogen gas to form 2 liters of water gas.

But this is only the ratio that these compounds react. They can react in any quantities that maintain this ratio.

So:

0.5 mole oxygen reacts with 1 mole of hydrogen to form 1 mole of water.

2 molecules oxygen reacts with 4 molecules of hydrogen to form 4 molecules of water.

3 moles oxygen reacts with 6 moles of hydrogen to form 6 moles of water.

10 liters oxygen gas reacts with 20 liters of hydrogen gas to form 20 liters of water gas.

These ratios can be used to create unit factors to convert between reactants and products.

1 = (1 mole oxygen/2 moles of hydrogen)
1 liter oxygen gas/2 liters of water gas
2 molecules of hydrogen/2 molecules of water = 1 molecules of hydrogen/1 molecules of water
2 liters of hydrogen gas/1 liter oxygen gas

Using the Molar Mass of the reactants and products, the mass of the reactants and products can be related.

Stoichiometry allows you to relate, or convert between moles, mass, and gas volume of the various reactants and products.

The conversions can be between;
moles - moles
moles - mass
moles - volume
mass - mass
mass - volume
volume - volume

Calculations Involving a Limiting Reactant

Stoichiometric Quantities: Quantities of reactants in the stoichiometric ratios.

Limiting Reactant (Reagent)

The stoichiometry of the chemical equations determine the ratios that the chemicals react.

What happens when the ratio of reactants present is not equal to the stoichiometric ratio?

The reactants will not react completely. However, at least one reactant will be used up. And at least one reactant will have excess left over.

The reactant that is used up, is the limiting reactant, it limits the extent of the reaction.

Pt/825°C
4NH₃(g) + 5O₂(g) à 4NO(g) + 6H₂O(g)

In this reaction, 4 moles of ammonia reacts with 5 moles of oxygen;
or 1.25 moles of oxygen reacts for every 1 mole of ammonia.
Alternatively, 0.8 moles of ammonia reacts with 1 mole of oxygen.

If we have equal molar amounts of both reactants (1 mole each), they cannot both react completely. Since less ammonia is used for each amount of oxygen, then oxygen will be used up first. Oxygen is the limiting reactant.

Actual: 1 O₂/NH₃ Theoretical: 1.25 O₂/NH₃ Since actual O₂ is less than theoretical, relative to NH₃, then O2 limits the reaction.

Actual: 1 NH₃/O₂ Theoretical: 0.8 NH₃/O₂ Since actual NH₃ is greater than theoretical, relative to O₂, then NH₃ does not limit the reaction and O₂ does limit the reaction.

Theoretical yield

The maximum amount of product that could be derived from the available reactants is called the theoretical yield.

You would get the theoretical yield of products if all available reactants went into the products. i.e. when the limiting reactant is used up.

Actual Yield

The amount of product produced is called the actual yield.

The actual yield can vary from the theoretical yield for various reasons, including;
Not all reactant was consumed; equilibrium conditions, not well mixed, reaction interrupted,
Not all product accounted for; product splattered, stuck on container walls, still in solution;
Product not pure; still wet, has unreacted reactants or co-products in it, filter paper.

Percent Yield

The percent yield is a measure of how close to the theoretical yield was actually achieved.

Percent yield = actual yield/theoretical yield x 100%

The percent yield can be greater than or less than 100%. Ideally it is close to 100%.

Molarity

An expression of concentration

Molar Concentration, also Molarity, symbol (M), is defined as

Molarity = moles of solute/ liters of solution

M = n/V or n = MV

Molarity is often more useful since chemicals react on a mole stoichiometry basis and not a weight basis.

Writing Unit Factors

The unit factor from molarity is moles/liter or it can also be written moles/1,000 mL

Solving Molarity Problems

The concept of molarity allows you to associate moles of solute with volume of solution. Using molar mass also allows a connection of mass of solute with volume of solution.

How many grams of NaCl is required to make 1.000 L of 1 M solution?

1.000 L x 1 mole/L x 58.44 g/mole = 58.44 grams NaCl

149.1 grams of KCl was added to enough water to make 1.000 L of solution. What is the Molarity?

(149.1 g KCl/74.55 g/mol)/ 1.000L = 2.000 moles/L = 2.000 M KCl

What volume of 8.00 M KCl solution is required to make 500 mL of 2.00 M KCl solution?

(2.00 M x 0.500 L) = (8.00 M x ? L)

(2.00 M x 0.500 L)/8.00 M = 0.125 L = 125 mL

Concentration of Ions

When an ionic compound dissolves, the Molarity represents the concentration of the compound in solution. It does not necessarily represent the concentration of ions in solution. the concentration of ions will depend on the number of ions produced by the compound.

Dilution

Since concentration (molarity) times volume gives moles [MV = n], this can be used to calculate dilution volumes using

M₁V₁ = n = M₂V₂

Stoichiometry Involving Concentration

Titration is commonly used to determine the concentration of an unknown compound. A solution of know concentration is reacted with a solution of unknown concentration, with the amount added is controlled until the reaction is just complete. This provides enough information to calculate the concentration of the unknown.

M₁V₁ (n₂/n₁) = M₂V₂

or

(M₁V₁/n₁) = (M₂V₂/n₂)