Stoichiometry Introduction

Rx between Hydrogen and Oxygen can be described as:

Balanced equation:

Or

Or

Avogadros Number: (number of Molecules)

Or

Moles (amount of a substance containing avogadros number is a mole...so)

Converting Moles to Grams

Converting

Once you have the molar mass, you can easily convert from grams to moles,

and also from moles to grams.

Number of moles = (# of grams) / (molar mass)

Number of grams = (# of moles) * (molar mass)

Stoichiometry

Stoichiometry is the accounting, or math, behind chemistry. Given enough

information, one can use stoichiometry to calculate masses, moles, and

percents within a chemical equation

What You Should Expect

The most common stoichiometric problem will present you with a certain

amount of a reactant and then ask how much of a product can be formed.

Here is a generic chemical equation:

What You Should Expect

Here is a typically-worded problem:

Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

You will need to use molar ratios, molar masses, balancing and interpreting

equations, and conversions between grams and moles. If you struggled with

those in class, welcome to the club.

Go back and review them if you need to, because if you can't do that stuff, you

can't do stoichiometry.

Summary of molar ratios, molar masses, balancing and interpreting equations,

and conversions between grams and moles

Molar Ratio

A comparison of the number of moles of one substance in a chemical equation.

For example,

The ratio of sodium carbonate to potassium chloride to sodium chloride to

potassium carbonate is 1:2:2:1.

Molar mass, symbol M, is the mass of one mole of a substance in grams

(chemical element or chemical compound).

The base SI unit for mass is the kilogram but, for both practical and historical

reasons, molar masses are almost always quoted in grams per mole (g/mol or

g mol–1), especially in chemistry.

Balancing Chemical Equations

A chemical equation is said to be balanced when there are equal numbers of

each type of atom on each side.

Balancing is achieved by adjusting the stoichiometric coefficients – not the

subscripts in the chemical formulae.

Chemical Equations

A chemical equation describes a chemical reaction.

Example: CH4 + 2O2 CO2 + 2H2O

Each reactant and product is described by its chemical formula – reactants on

the left and products on the right.

A number in front of a chemical formula is a stoichiometric coefficient.

To calculate the percent (%) of each element in a compound

Example:

Calc the % composition by mass of each element in CaO (calcium oxide)

Example 1:

Calc the % composition by mass of each element in CaO (calcium oxide)

Example 2:

Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO4.5H2O)

An explanation:

Water of crystallisation is water chemically combined but not bonded to a host

molecule

Substances containing these crystals are said to be Hydrated those with no

crystals of this sort are said to be Anhydrous

Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO4.5H2O)

Solution:

Mr CuSO4.5H2O : H = 1; O = 16 ; S = 32 ; Cu = 63.5

Remember to count the number of atoms

*Just water

Empirical Formulas

Each Chemical Element is composed of atom

Each element is represented by a symbol

Some elements are present as small molecules (H2)

Chemical formulas tell us how many atoms are present in the element

Element Formula

Hydrogen H2

Chlorine Cl2

Nitrogen N2

Bromine Br2

Oxygen O2

Iodine I2

Fluorine F2

A compound is represented by a chemical formula e.g.

A molecule of water is represented by H2O

Called the Molecular Formula

Definition

The empirical formula of a compound indicates what elements are present in

the compound and the simplest whole number ratio in which the atoms of

these elements are present

Example 1

What is the empirical formula of Ethyne (C2H2)

Ratio is 2carbon:2hydrogen

Simplest WNR is 1:1

Therefore empirical formula is CH

Example 2

What is the empirical formula of Glucose(C6H12O6)

Ratio is 6 carbon:12 hydrogen: 6 Oxygen

Simplest WNR is 1:2:1

Therefore empirical formula is CH2O

Example 3

What is the empirical formula of Water (H2O)

Ratio is 2 hydrogen:1 Oxygen

Simplest WNR is 2:1

Therefore empirical formula is H2O

Using an empirical formula

Chemists needing to have a chemical analysed determine the empirical

formula using an elemental analyser then using the empirical formula they can

determine the molecular formula of the unknown chemical

Example of an Empirical formula from analytical data

On analysis a compound is found to contain 68.85% carbon ; 4.92% hydrogen ;

26.23% Oxygen What is the empirical formula ?

Answer: Imagine we have 100g of the compound then the ratios in grams is

68.85: 4.92: 26.23

We divide by the molar mass (periodic table) of each element

Carbon = 12

Hydrogen = 1

Oxygen = 16

68.85: 4.92: 26.23

We divide by the molar mass (Mr)

68.85÷12: 4.92÷1: 26.23÷16

This gives us a relative atomic mass (Ar) ratio

5.74:4.92:1.64

Divide each by the lowest Ar which in this case is 1.64

5.74:4.92:1.64

Divide these by the lowest Ar

5.74÷1.64 ; 4.92÷1.64; 1.64÷1.64

And we get the basis for an empirical formula for the unknown

3.5:3:1

The rule is it must be a whole number so!

The empirical formula is 7:6:2 or C7H6O2

Another Example

A compound containing only C (52.17%); H (13.04%); O (34.79%)

You are told The Mr is 92

What is the empirical formula?

What is the Molecular formula?

Remember C = 12, O = 16 and H = 1

Solution

To answer this problem we need to under stand...if we have 100 grams of the

compound then Carbon will make up 52.17 grams; Hydrogen, 13.04g and

Oxygen, 34.79g

Convert each of these masses to moles by dividing by the relative atomic mass

to get a mole ratio

If we then divide by the smallest mole ratio we can get a simplest whole

number mole ratio

Remember

Sometimes when we divide by the smallest mole ratio we do not get a whole

number and it is necessary to multiply by a factor to get a whole number

Compound contains C (65.11%);H (8.83%);O (26.06%)

Find the empirical formula

Empirical formulas from combination data

Example

0.72 g of Magnesium is heated in excess oxygen, 1.2g of Magnesium Oxide is

formed what is the empirical formula of Magnesium Oxide

Mass of Magnesium Oxide formed = 1.2g

Mass of Oxygen consumed = 1.2 – 0.72 = 0.48g

Moles of Magnesium atoms consumed = 0.72÷24 = 0.03

Moles of Oxygen atoms consumed = 0.48÷16 = 0.03

Ratio of Magnesium to Oxygen atoms = 0.03 : 0.03 = 1:1

Empirical Formula = MgO

Empirical formulas from decomposition data

5.8g of an oxide of Iron was heated with carbon and 4.2g of iron was formed

What is the empirical formula of the oxide ?

Mass of Iron in the compound = 4.2g

Mass of Oxygen in the compound = 5.8 - 4.2g = 1.6

Moles of iron atoms in the compound = 4.2÷ 56 = 0.075

Moles of oxygen atoms in the compound = 1.6 ÷ 16 = 0.1

Ratio of Iron atoms to Oxygen atoms in the compound = 0.075 : 0.1

Whole numbers = 3:4

Empirical formula = Fe3O4

Molecular Formulas

Empirical formulas are used only for ionic compounds

Covalent compounds are more complex (molecules) and so we use molecular

formulae

Definition

The Molecular formula of a compound indicates the actual number of atoms

of an each element present in the molecule of the compound

To find the molecular formula of a compound we need to know the empirical

formula and the relative molecular mass

Example

Empirical formula for benzene is CH & Mr = 78. what’s the molecular formula?

Answer

Formula mass = 12+1 = 13

Mr = 78

Number of CH units in a benzene molecule is = 78/13 = 6

Molecular formula is = C6H6

Another Example

Mr of propene = 42. The solution contains 85.7% C ; 14.3% H by mass.

Molecular Formula ?

Answer

Carbon = 85.7 ÷ 12 = 7.14

Hydrogen = 14.3 ÷ 1 = 14.3

Simplest ratio = 1:2

Therefore empirical formula = CH2

Therefore empirical formula = CH2

Formula Mass of CH2 = 14

Relative Molecular Mass of propene = 42

Number of CH2 units in a propene molecule = 42 / 14 = 3

Molecular formula of Propene = C3H6

% composition by Mass

If the empirical formula of a compound is known the % by mass of each

element present can be calculated

To calculate the percent (%) of each element in a compound

Example:

Calc the % composition by mass of each element in CaO (calcium oxide)

Example

Calc the % composition by mass of Nitrogen in ([NH4]2SO4)

Example

Calc. the % water of crystallisation in Copper II Sulfate crystals (CuSO4.5H2O)

An explanation:

Water of crystallisation is water chemically combined but not bonded to a host

molecule. Substances containing these crystals are said to be Hydrated those

with no crystals of this sort are said to be Anhydrous

Structural Formulas

Molecular formulas of covalent compounds give the chemist a lot of detail

about the compound and the molecules in it

The structural formula shows the arrangement of the atoms within a molecule

of the compound

IF THE STRUCTURAL FORMULA IS KNOWN THEN THE EMPIRICAL AND

MOLECULAR FORMULAS ARE EASY

Example

Structural formula of ethene is

To get the Molecular formula, count the atoms C2H4

To get the empirical formula just assess the simplest ratio – CH2

Chemical Equations

Tells us what substances are in and what is produced during a chemical

reaction

The correct formula for each of the reactants and the products must be used

and the equation must be balanced

To balance an equation the formulas cannot be altered just multiplied

Example

Carbon reacts with oxygen to form carbon monoxide

Balance the equation from this unbalanced one

There is one carbon atom on the left

There is one carbon atom on the right

There are two oxygen atoms on the left

There is one oxygen atom on the right

The equation is unbalanced

We must check the equation and multiply the atoms to ensure that they

balance out

There are two carbon atoms on the left

There are two carbon atoms on the right

There are two oxygen atoms on the left

There are two oxygen atoms on the right

The equation is balanced

Example

Unbalanced equation

There is one carbon atom on the left

There is one carbon atom on the right

There are four hydrogen atoms on the left

There are two hydrogen atoms on the right

There are two oxygen atoms on the left

There are three oxygen atoms on the right

The equation is unbalanced

Multiply the atoms until balanced

The equation is short 2 oxygen atoms on the left and 2 hydrogen atoms on the

right

Calculations based on balanced equations

A balanced equation gives the relative amounts of each of the reactants

consumed and each of the products produced in a chemical reaction

If the amount of any one of the reactants consumed is known then it is

possible to calculate the amounts of the other reactants and products

If the amount of methane consumed is known we can calculate the amounts of

carbon dioxide and water produced. We can also calculate the amount of

oxygen consumed

The reverse is also true ...if we know the amount of a product then we can

calculate the others

example

Methane burns in air

If 2.5 moles of methane are reacted fully with oxygen calculate the number of

moles of carbon dioxide and oxygen produced and the number of moles of

oxygen consumed

We know that 2.5 moles of methane were consumed

Calculating Masses of Reactants or products from Balanced Chemical

Equations

Example

Magnesium reacts with Oxygen to produce MgO

The first step is to change the given quantity (gas or volume of gas at s.t.p.)to

moles

To achieve this divide the given mass by the molar mass

If we burn 9g of Mg in excess O2 (enough to react all the Mg) what mass of

MgO will be formed

Solution

First write down the equation and circle what we want to know

Which is the same as :

Another way of achieving the same thing

9 g of Mg = 9 ÷ 24 = 0.375

From the equation

USE YOUR MOLE MAP

Calculations of volumes of gaseous reactants or products from

balanced chemical equations

example

Propane burns in air according to the equation

What volume of O2 (measured at s.t.p.) is needed for complete combustion of

11g of propane

Answer:

11g of C3H8 = 11/44 moles = 0.25

Another type of calculation

A solution of NaOH is reacted with enough H2SO4 Solution to easily neutralise

it. The equation for the reaction is

On evaporation of the water 284g of Sodium Sulfate are obtained. Calculate :

A. The number of moles of H2SO4 acid consumed

B. The number of water molecules formed

C. The mass of the NaOH used to make up the solution

Answer: A. The number of moles of H2SO4 acid consumed

284g NaSO4 = 284 ÷ 142 moles = 2

From the equation

Answer: B. The number of water molecules formed

B. Avagadros number = 6 x 1023

From the equation

Answer: c. The mass of the NaOH used to make up the solution

c. Relative Molecular mass (Mr)

From the equation

Calculations involving excess of one reactant

Sometimes when a chemical reaction occurs there is excess of one of the

reactants. If the balanced equation is known and the initial quantities in the

reaction are known then it is possible to identify the chemical that is in excess.

The substance that is not present in excess is called the limiting reactant as it is

the amount of this substance that will dictate how much product will be

produced

Example

Zinc reacts with sulphuric acid according to the equation

A 250 cm3 aqueous solution containing 9.8g sulfuric acid is added to 13g Zinc

(a) Show that zinc is present in excess

(b) Calcuate the mass of Zinc Sulfate formed

(c) Calculate the volume of hydrogen gas (measured at s.t.p) formed

Answer: a : Show that zinc is present in excess

Moles of zinc present initially = 13 ÷ 65 = 0.2

Moles of Sulfuric acid present initially = 9.8 ÷ 98 = 0.1

Answer: b : Calcuate the mass of Zinc Sulfate formed

Sulfuric acid is the limiting reactant

Answer: c : Calculate the volume of hydrogen gas (measured at s.t.p) formed

Percentage Yields

When a real reaction occurs the amounts of product isolated are often less

than those calculated.

This may be due to the reaction been reversible, or to some of the products

reacting further to form yet other products or to loss of product during

purification

Percentage yield accounts for these factors

Percentage Yields

Example

10.2 g of ethanol (C2H5OH)were heated with aluminium Oxide and 1.7g ethene

(C2H4) were formed

Calculate the percentage yield of ethene

10.2 g of C2H5OH = 10.2 ÷ 46 moles = 0.22