Topic 3: Quantative Chemistry By Bethan Poole

Topic 3: Quantative Chemistry By Bethan Poole

3.1 Chemical measurement, conservation of mass ad the quantative interprtation of chemical equations

3.3 Yield and atom economy of chemical reations (Chemistry Only)

3.5 Use of amount of substance in relation to volume of gases (Chemistry Only)

3.2 Use of amount of substance in relation to masses of pure substances

3.4 Using concentration of solutions in mol/dm3 (Chemistry Only)

3.1.2 Relative Formula Mass

3.1.3 Mass changes when a reactant or product is a gas

3.1.1 Conservation of mass and balanced chemical equations

3.1.4 Chemical meaurement

3.2.2 Amounts of substances in equations (HT only)

3.2.3 Using moles to balance equations (HT only)

3.2.4 Limiting Reactants (HT only)

3.2.5 Concentration of solutions

3.3.1 Percentage Yield

3.3.2 Atom Economy

3.4.1 Using concentrations of solutions in mol/dm3 (HT only)

3.5.1 Use of amount of substance in relation to volume of gases (HT only)

3.2.1 Moles (HT Only)

The mass of a mole of an atom- the Relative Atomic Mass in grams

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

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)

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

Percentage uncertainty

Range of measurements/mean x 100

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)

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

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)

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

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

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

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%

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

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

Room temp. and pressure- one mole of any gas takes up a volume of 24dm3

Volume=Amount(Mol)X 24dm3