Download Empirical Formula

Stoichiometry
Atomic Mass (atomic weight)
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Atoms are so small, it is difficult to discuss how much they weigh in grams
We use atomic mass units – an atomic mass unit (AMU) is one twelfth the mass
of the catbon-12 atom (the standard)
This standard gives us a basis for comparison
The decimal numbers on the periodic table are in amu’s
Example 1
 What is the atomic mass for hydrogen? _______For oxygen? _________
They are not whole numbers
 because they are based on the average mass of the common isotopes of an atom
 To calculate, multiply the % abundance of each isotope to the relative mass of
each and add together
Example 2 & 3
 There are 2 isotopes of lithium, one with a mass of 6.015 amu and an abundance
of 7.42%. The other isotope has a mass of 7.016 amu and an abundance of
92.58%. Calculate the atomic weight (average atomic mass) of lithium.
 There are 2 isotopes of nitrogen, one with an atomic mass of 14.0031 amu and
one with a mass of 15.0001 amu. What is the percent abundance of each?
The Mole (mol)
 A large number used for counting atoms
 It is the number of particles in exactly 12 grams of carbon 12.
 Equal to 6.022 x 1023 of anything, no matter what the substance
 6.022 x 1023 is called Avogadro’s number
1 mol = 6.022 x 1023 particles (atoms, molecules, formula units)
Example 4-5
 Convert 1.35 moles of carbon disulfide(CS2) into molecules?
 How many moles are in 3.59 x 1021 formula units of sodium nitrate, NaNO3
Molar Mass (molecular weight)
 Mass of 1 mole of a substance
1 mol = molar mass (g)
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For elements, it is the amount in grams numerically equal to the atomic mass
in amu’s
For compounds, add up the individual atomic masses of the elements that
make it up. Remember a subscript in a chemical formulas tells you how many
atoms of an element you have and this must be accounted for
Represented by M and has the units grams/mol
Example 6-10
 Calculate the mass of a carbon atom. _____________________
 Calculate the mass of 1 mole of carbon atoms. _________________
 Calculate the mass of 1 mole of lithium atoms. __________________
**** For any element atomic mass (AMU) = molar mass (grams).****
 What is the molar mass of sulfur dioxide?
 Calculate the molar mass of Al2O3
Conversions using Molar Mass
Examples 11-13
 How many moles of calcium carbonate are in a stick of chalk with a mass of
14.8 g.
 How many atoms are in 0.551 g of potassium ?
 How many H atoms are in 72.5 g of C3H8O ?
Mass Spectrometer
Percent Composition
 Percent that each element in a compound is composed of
 Calculate the molar mass of the compound. Then divide the total mass of each
element, by the molar mass(total mass of the compound), and multiply by 100.
 Check you answer to see if the total of the percent masses of all elements equals
100%
Example 14-15
 Calculate the mass percent of each element in C2H6O
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 Calculate the percent composition of water in the hydrate, CaSO4 · 2H2O
Empirical Formula
 The lowest whole number ratio of atoms in a chemical formula
 Ionic compounds formulas are ALWAYS empirical.
 From % composition, you can determine the empirical formula, based on mole
ratios
 Remember this rhyme: Percent to mass, mass to mole, divide by small, multiply
till whole (last step is not always mandatory)
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Steps to calculating empirical formula
1. If you are given a % composition, assume you have 100 grams and switch the
% unit to g. If you are given the actual number of grams of each element, use
those instead
2. Convert the grams of each element to moles of each element.
3. Divide each mole amount by the smallest mole amount.
4. If your answers are within .1 of whole number, round up or down and these
numbers become your subscripts.
5. If your answers are not within .1 of a whole number, you must find the
smallest number that you can multiply by all the mol amounts you have just
calculated to get them to be all whole numbers or within .1 of a whole
number.
Example 16-17
 A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and
5.34g O. Determine a formula for this substance.
 A compound has the following percent composition: 31.9%K, 29.0 % Cl and
39.1%O. What is the empirical formula?
Extra Practice if needed!
A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 %
H. What is the empirical formula of the substance?
Molecular Formula
 Empirical means the lowest ratio, molecular means the actual ratio of the
molecule that exists.
 To calculate, you need the molar mass of the molecular formula and the empirical
formula’s mass(you will probably have to calculate this
 Steps to calculate the molecular formula:
1. Divide the Molecular formula mass by the empirical formula mass to get a
whole number.
2. Multiply the whole number(factor) to the subscripts of the empirical formula
Example 18
A compound has an empirical formula of NO2. The colorless liquid, used in rocket
engines has a molar mass of 92.0 g/mole. What is the molecular formula of this
substance?
Calculating Empirical Formulas from Combustion Analysis
 Use the mass of CO2 to covert to moles of CO2 to moles of C to grams of C.
 Use the mass of H2O and convert to moles of H2O to mole H to g H
 Subtract the combined mass of C and H from the mass of the sample to get the
mass of oxygen. Convert it to moles.
 Simplify the mole ratios of C, H and O to get the empirical formula
Example 19: Calculate the empirical formula for the combustion of 11.5 g ethanol. 22.0
g CO2 and 13.5 g H2O are produced in the reaction.
Chemical Equations
 Sentences that describe what happens in a chemical reaction
 Reactants are the starting substances found on the left of an equation
 Products are the new substances created which are found on the right of an
equation
 Reactants
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Products
(yields)
 Law of Conservation of Mass: Matter can neither be created nor destroyed, just
changed in form.
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Equations should be balanced in order to satisfy the law of conservation of masssame number of each kind of atom on both sides of the equation must exist
__ CH4 + __ O2
 __ CO2 + __ H2O
1
4
2
C
H
O
1
2
3
Important Abbreviations : (s) solid
(l) liquid
(g) gas
(aq) aqueous
heat
∆
catalyst
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 Hints for balancing:
1. Never start with hydrogen or oxygen
2. Treat polyatomic ions as chunks not separate atoms
3. Do not put a coefficient of 1 down until the end when all atoms are balanced
4. If you have an odd number of atoms on one side and an even number of those
atoms of the other, place a 2 as a coefficient next to the chemical formula with
the odd numbers to make it even.
Examples 20-23
 __ Ca(OH) 2 + __ H3PO4  __ H2O + __ Ca3(PO4) 2
 __ Cr + __ S8  __ Cr2S 3
 __ KClO 3  ___ KCl + __ O 2
 Translate this and then balance: Solid iron(III) sulfide reacts with gaseous
hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas
Meaning of the coefficients in a Chemical Equation
 Used to describe a reaction in moles, and particles (molecules, formula units and
atoms) but not grams. For example: 2 H2 + O2  2 H2O would be interpreted as
2 molecules of hydrogen react with 1 molecule of water to produce 2 molecules of
water OR 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of
water!
 A Mole Ratio is the conversion factor between 2 different amounts in a balanced
chemical equation
2 KClO 3  2 KCl + 3 O 2 Possible Mole Ratios could be:
2 mol KClO 3 = 2 mol KCl
OR 2 mol KCl = 3 mol O 2 OR 2 KClO 3 = 3 mol O 2
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Stoichiometry: Given an amount of either reactant or product, you can determine
the other quantities in a reaction using dimensional analysis
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Steps to solve problems
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the
sought quantity
4. Convert moles of sought quantity into desired units
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Use conversion factors of [ molar mass (grams) = 1 mole] and the balanced
chemical equation mole ratio [? Mol A = ? Mol B]
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4 types of problems
1. mole –mole: 1 step process using a mole ratio to convert from moles of
substance A to moles of substance B
2. mole-mass: 2 step process using a mole ratio to convert from moles of
substance A to moles of substance B and then the molar mass
conversion factor to convert between moles of B to grams of B
3. mass-mole: 2 step process using the molar mass conversion factor to convert
from mass of A to moles of A and then the mole ratio to convert
from moles A to moles of B
4. mass-mass: 3 step process using the molar mass conversion factor to convert
from mass of A to moles of A, then a mole ratio used to convert
from moles A to moles of B, then a molar mass conversion factor
to convert between moles of B to mass of B
Example 24-26
 (Mole To Mol) How many moles of sulfur trioxide can be produced if 8.1
moles of oxygen react with sulfur. 2S + 3 O2  2 SO3
 (Mass To Mole) One way of producing oxygen O2 involves the
decomposition of potassium chlorate into potassium chloride and oxygen gas.
A 25.5 g sample of potassium chlorate is decomposed. How many moles of
O2 are produced?
2 KClO 3  2 KCl + 3 O 2
 (Mass to Mass)
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
Limiting Reagent
 Limiting reagent is the reactant that determines the amount of product formed ( it is
the one you run out of first)
 Excess Reagent The reactant that you don’t run out of.
 How to tell if the problem you are working is a limiting reagent: IF 2 AMOUNTS
OF REACTANT ARE GIVEN TO YOU IN THE PROBLEM!
 Steps to determine the LR(limiting reagent)
1. Pick A Product
2. Try ALL the reactants
3. The lowest answer will be the correct answer
4. The reactant that gives the lowest answer will be the limiting reactant
 Once you determine the LR, you should only start with it!
 To find the amount of Y and Z
A+B X+Y+Z
Let’s say B is the LR!
There is no need to use A to find Y and Z
It will give you the wrong answer – a lot of extra work for nothing
Example 27
10.0 g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride.
Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Shortcut to Finding Limiting Reagent
1. Take each reactant amount and convert to moles. If already in moles move to the next
step.
2. Divide each mole amount by the coefficient of that substance from the balanced
chemical equation.
3. Whichever is the smallest number will identify the limiting reagent (LR)
4. Use the original amount of the reactant from the problem to begin your conversion
process to determine either how much excess reagent you have left over or how much
product you will form.
Example 28:
2Al + Fe2O3  Al2O3 + 2Fe
In one process, 124 g of Al are reacted with 601 g of Fe2O3, calculate the mass of Al2O3
formed.
Finding Excess Reagent Practice
Example 29 : 10.0g of aluminum reacts with 35.0 grams of chlorine gas
2 Al + 3 Cl2  2 AlCl3
•We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are
produced. How much excess reagent in left over?
Percent Yield
 Percent Yield is the ratio in percent of amount of product you produced in the lab
compared to the amount of product that you should of produced if the reaction
went to completion and no problems arose in the lab
 Must always use the LR to determine the amount of product produced
 Theoretical yield- is the amount you would produce if everything went perfect;
use the balanced chemical equation for this one
 Actual Yield is the actual amount you make in lab under imperfect condition.
 % yield = Actual/Theoretical x 100% OR
 % yield = what you got/what you should of got x 100%
Example 30
 Aluminum burns in bromine producing aluminum bromide. In a lab, 6.0 g of
aluminum reacts with excess bromine. 50.3 grams of aluminum bromide are
produced. What are the three types of yield? 2 Al + 3 Br2  2 AlBr3