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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