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Topic 3: Quantative Chemistry By Bethan Poole (3.2 Use of amount of…
Topic 3: Quantative Chemistry
By Bethan Poole
3.1 Chemical measurement, conservation of mass ad the quantative interprtation of chemical equations
3.1.2 Relative Formula Mass
The Mr (RFM) of a compound is the sum of the Ar (RAM) of all the atoms in the numbers shown in the formula- No unit
RAM of atoms shown in Periodic Table
e.g What is the Relative Formula Mass of CO2? RFM=(12x1) + (16x2)= 44
Due to Conservation of mass - Relative Formula Mass of the reactants is equal to the Relative Formula Mass of products
3.1.3 Mass changes when a reactant or product is a gas
Some reactions appear to involve a change in mass
Non-closed system- Gases can enter or leave
e.g When Magnesium is burnt in the air to produce magnesium oxide , the mass of the solid increases (Manesium is combined with Oxygen and Oxygen has mass)
2Mg(S) + O2(G)-> 2MgO (S)
e.g Calcium Carbonate is heated , decomposes to form clacium oxide and carbon dioxide- Mass of solid decreases - one of products is gas gas- Escapes into the air. if mass of CO2 is included , total mass of all the reactants s equal to total mass of all products
CaCO3(S)-> CaO(S) + CO2(G)
3.1.1 Conservation of mass and balanced chemical equations
Conservation of mass
Chemical reation- Total mass of products is equal to total mass of rectants
Mass is conserved- No atoms lost or made
Chemical symbols must always be balanced to show this
With the same number of atoms of each element on each side of the equation
e.g When solid iron reacts with copper (II) sulftate solid copper and iron(II) sulfate solution is produced
Fe(S) + CuSO4 (AQ) -> Cu(S) +FeSO4 (AQ)
3.1.4 Chemical meaurement
Percentage uncertainty
Range of measurements/mean x 100
3.3 Yield and atom economy of chemical reations (Chemistry Only)
3.3.1 Percentage Yield
Yield
- The amout of product that a chemical reaction produces
Percentage Yield
- Comparison of what we really produce with what we could possible produce
It's not 100%
Some product is lost
Not all the chemicals react
Other chemical reactions occur
If reaction is reversible it may not go into completion
Equation
-
Percentage Yield= (Amount of Product Produced)/(Maximum Amount of Product Possible) X100
e.g Calculate how much calcium oxide can be produced from 50.0kg of calcium carbonate- caco3->caO + CO2- [40+12+(3x16)]->[40+16]+[12+(2x16)] = 100->56+44=100:56 This means 1kg of CaCO3 produces 0.56kg of CaO AND 50kg produces 28kg
THIS IS THEORETICAL YIELD
50kg of CaCO3 is expected to produce 28kg of CaO A company heats 50kg of calcium carbonate in a iln and obtais 22kg of CaO. Calculate percentage yield.Percentage Yield = 22/28 x100= 78.6%
3.3.2 Atom Economy
Amount of starting materials that end up as useful products
Calculating
1)
Identify the usful product
2)
Work out the Mr (RFM) of all products
3)
Use the formula to calculate atom economy
% Atom Economy = Mr useful products/ Total Mr of products x100
e.g- C2H5OH->
C2H4
+H2O (C2H4 is useful product) C2= Mr Values= (12x2)+(1x4)=28 H2O= Mr Values= (2x1)+16=18 **Percentage Atom Economy= 28/(28+18) x100 = 60.9%
Choosing Reaction Pathway
Comparing the atom economy of two competing reaction pathways is important
Although scientists have to consider- Cost of reactants percentage yield, rate of reaction, equilibrium position and usefulness of the by-products
3.5 Use of amount of substance in relation to volume of gases (Chemistry Only)
3.5.1 Use of amount of substance in relation to volume of gases (HT only)
Room temp. and pressure- one mole of any gas takes up a volume of 24dm3
Volume=Amount(Mol)X 24dm3
3.2 Use of amount of substance in relation to masses of pure substances
3.2.2 Amounts of substances in equations (HT only)
Masses of reactants can be calculated from balanced symbol equations
e.g Mg+2HCl MgCl2+H2 shows that one mole of magnesium reacts with two moles of hydrochloric acid to produce magnesium chloride and one mole of hyrogen gas
1) Write out balanced equation,2) Work out RFM (Mr) 3)Write it in Ratio
3.2.3 Using moles to balance equations (HT only)
Balancing equations- show the number of moles of each product and reactant- Can be used to calculate the mass of the reactants and products
Balance by- Calculate the masses of reactants and products by converting from moles to grams and converting the number of moles to simple ratios
3.2.4 Limiting Reactants (HT only)
Sometimes when two chemicals react together, one is completely used up during the reation called the
limiting reactant
Dictates when the reaction ends
Graph directly proportionate for the amount of product formed and the amount of limiting reaction Graph is - Straight, Positive gradient, Passes through origin (0,0)
3.2.5 Concentration of solutions
Measured in mol/dm3
Equation
concentration of a solution= Amount of substance(Mol)/Volume (dm3)
If 1.00 mole of solute is dissolved to form a solution that has a volume of 1.00dm3 the solution has a concentraion of 1.00mol/dm3
3.2.1 Moles (HT Only)
The mass of a mole of an atom- the Relative Atomic Mass in grams
One mole of atoms is equal to the
Avogadro constant
of particles
(6.02x10 23)
Equations-
Number of moles of an element= mass of element in grams/Relative Atomic Mass
and
Number of mass of a compound=Mass of compound in grams/Relative Formula Mass
e.g One mole of sodium cotains 6.02 x1023 atoms, The RAM of sodium is 23.0, One mole of sodium has a mass of 23.0g
3.4 Using concentration of solutions in mol/dm3 (Chemistry Only)
3.4.1 Using concentrations of solutions in mol/dm3 (HT only)
10cm3 of a solution of potassium hydroxide was titrated with a 0.010M solution of hydrochloric acid 13.35cm3 of acid was required for neutralization- calculate the concentraion of the potassium hydroxide solution
STEP 1- WRITE WHAT YOU KNOW
KOH (aq) Concentration=? Volume=10.0cm3
HCL (aq) Concentration= 0.10M Volume= 13.5cm3
STEP 2- WRITE A SYMBOL EQUATION TO IDENTIFY RATIOS
HCl(aq)+KOH (aq)-> KCl(aq)+H2O(l)
In the ratio of 1:1 so the moles are equal
STEP 3- CALCULATE MOLES FOR KNOWN SUBSTANCE
Moles= Vol x Conc
VOLUME HAS TO BE IN DM3
(divide by 1000)
(13.5/1000) x 0.1=0.0014 Moles
Moles of KOH= 0.0014
Concentration= mole/volume
Concentration of KOH= 0.0014/(10/1000)= 0014Mol/dm-3