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C4: Quantitative chemistry, : - Coggle Diagram
C4: Quantitative chemistry
Relative formula mass
Ar, is an element's relative atomic mass (the bigger number)
Compounds have Mr which is just the relative atomic masses of all the atoms in the molecular formula.
E.g. Find the relative formula mass of MgCl2
Ar of Mg=24 & Ar of Cl=35.5
24+ (35.5 x 2) = 95
Mr of MgCl2=95
Harder questions
A mixture contains 20% iron ions by mass. What mass of iron chloride (FeCl2) would you need to provide iron ions for 50g of the mixture?
1) Find the mass of the iron in the mixture
The mixture contains 20% iron by mass, so in 50g there will be 50 x 20/100 = 10g of iron
2) Calculate the percentage mass of iron in iron chloride
Percentage mass of iron = 56/ 56 +(2 x 35.5) x100= 44.09...%
3) Calculate the mass of iron chloride that contains 10g of iron
Iron chloride contains 44.09...% of iron by mass, so there will be 10g of iron in 10/0.4409...= 23g
So you need 23g of iron chloride to provide the iron in 50g of the mixture
Percentage Mass of an element in a compound
Percentage mass of an element in a compound=
Ar x number of atoms in element/Mr of the compound x 100.
For example Ar of: sodium=23 & Ar of carbon=12 & Ar of oxygen=16
Mr for Na2CO3= (2x23) + 12 + (3x16) = 106
Percentage mass of sodium= 2x23= 46/106=43%
The Mole
Mole is the amount of a substance . One mole of any substance is just an amount that the substance contains 6.02 x 10^23 particles of.
The Avogadro constant: 6.02x10^23
Mass (g) = number of moles x Ar/Mr
Carbon has an Ar of 12. One mole of carbon weighs 12 g because the mass of one mole of the substance will have a mass in grams equal to the relative formula mass for that substance.
Example questions
If you have 48 g of Mg, how many moles of Mg do you have?
Mg Ar= 24
48/24 = 2 moles
What is the mass of 7.5 x 10^ -3 moles of aluminium sulfate?
Mr of Al2(SO4)3 = 342
(7.5 x 10^-3) x 342 = 2.6g
Balancing equations with moles
2g of hydrogen and 71g of chlorine react to make 73g of hydrogen chloride.
H2 + CL2 --> HCl
Ar:
Hydrogen= 1
Chlorine=35.5
Mr = 36.5
Divide the mass of each substance by its Mr to get moles
Hydrogen =2g/2 (Mr) = 1
Chlorine =71 g /71 (Mr)= 1
Hydrogen chloride =73g/36.5 = 2
Divide by the smallest number of moles which is 1 in this example.
Hydrogen= 1/1= 1
Chlorine= 1/1 =1
Hydrogen chloride= 2/1= 2
H2+Cl2 --> 2HCl
Balancing equations
H2 + Cl2 --> HCl (Not balanced)
H= 2 &Cl=2 on left side
H=1 & Cl= 1 on right side
To balance it on both sides there must be the same
H2 + Cl2 --> 2HCl (Now balanced)
H=2 Cl=2 on left side
H=2 & Cl = 2 on right side
Reacting masses
What masses of reactants and products are involved in the balanced symbol equation:
H2+Cl2 --> 2HCl
Mr H= 2 so 2g
Mr of Chlorine= 71 so 71g
Mr of HCl= 36.5 so 36.5 g
The ratio is 1:1:2
1 mole of H2 = 1x2=2
1 mole of Cl2 = 1x71=71g
2 moles of 2HCl = 2x 36.5= 73g
2.If you have a solution containing 100g of sodium hydroxide, what mass of chlorine gas do you need to convert it to bleach?
2NaOH + Cl2 --> NaOCl + NaCl + H2O
H=1
O=16
Na=23
Cl=35.5
NaOH=40g
Cl2= 71g
1 mole of sodium hydroxide has a mass of 40g. So 100 (mass)/40 (Mr)= 2.5 moles
The equation tells you that for every 2 moles of NaOH you need 1 mole of Cl to react with so 2.5/2=1.25 moles of Cl
1 mole of Cl has a mass of 71g
So you will need 1.25 (moles)x71 (Mr)= 88.75g of Cl to react with 100g of NaOH.
Conservation of Mass
if the mass does change in a reaction it's down to two reasons:
if the mass increases, one of the reactants must be a gas that is found in the air and all the products are solid /liquid/aqueous.
Before the reaction, the gas is in the air but it's not contained in the reaction vessel
But when the gas reacts to form part of the product it becomes contained inside the reaction vessel so the total mass inside the reaction vessel increases
Or
If the mass decreases it's because one of the products is a gas. Before the reaction all the reactants are contained in the reaction vessel. If the vessel isn't enclosed then the gas will escape once it is formed, this will make the total mass decrease.
In a chemical reaction no atoms are destroyed or created. This means there are the same number and type of atoms on each side of the equation. No mass is lost or gained.
Limiting reactants
Reactions stop when one reactant is used up. Any other reactants are in excess. The reactant that is used up is called limiting reactant. The amount of product depends on the limiting reactant.
Examples
Calculate the mass of aluminium oxide formed when 135g of Aluminium is burned.
4AL+3O2 --> 2Al2O3
Calc Mr
Al=27
Al2O3= 102
Calc the number of moles of Al in 135g
135/27=5
Look at the ratio in the equation. 4 moles of Al react to produce 2 moles of Al2O3. So 5 moles of Al will react to form 2.5 moles of Al2O3.
Calc the mass of 2.5 moles of Al2O3.
2.5 x 102= 255g
Percentage yield
Percentage yield = actual mass of product produced/ maximum theoretical mass of product possible x 100
Very few reactions have a yield of 100%:
The raw materials may not be pure, some products are left behind in the apparatus, the reaction may have been reversible, the reaction may not have finished and reactions may give some unexpected products
The yield of a chemical reaction describes how much product is made.
Atom economy
It is important to maximise atom economy in industrial processes to conserve the Earth's resources and minimise pollution.
The atom economy of a reaction is a measure of the extent to which the atoms in starting materials end up in the desired products.
% Atom economy = Mr / Sum of Mr of reactants x 100
Concentrations
Mass (g) = conc (g/dm3) x volume (dm3)
The amount of a substance in a certain volume of a solution is called its concentration. The more solute there is the more concentrated it is. You can also increase concentration by evaporating the solvent.
cm3 to dm3 = /1000
Titrations
Titration is used to measure accurately what volumes of acid and alkali react together completely and to find an unknown concentration of a solution. The point at which the acid and alkali have reacted completely is called the end point.
Carrying out a titration
Measure a known volume of alkali into a conical flask using a volumetric pipette.
Now add a few drops of indicator (Litmus or methyl orange) to the solution in the flask and swirl.
Pour the acid you are going to use in a burette. The burette has markings on it to enable you to measure volumes accurately.
Record the reading of acid in the burette. Then open the tap to release a small amount of acid into the flask. Swirl the flask to make sure the solutions are mixed.
Keep repeating this until the indicator in the flask changes colour. Record the reading on the burette and work out the volume of acid that has been used. This volume is known as the titre.
Repeat the practical until you get concordant (precise) results and discard any anomolous results. Then calculate a mean value to give the most accurate answer possible
A pipette is used to measure out a fixed volume of solution.
A burette is used to measure the volume of the solution added. The solution you know accurately goes in the burette and the unkown one goes in the flask using a volumetric pipette
Titration Calculations
In a titration 20cm3 of 0.15 mol/dm3 HCl reacted with 25cm3 of NaOH. What was the concentration of the sodium hydroxide?
HCl+ NaOH -->. NaCl + H2O
The ratio between acid and alkali = 1:1
20cm3/1000= 0.02dm3
Volume of acid= 0.02dm3
Concentration of acid= 0.15
0.02x0.15= 0.03
moles of acid=0.003 so moles of alkali= 0.003
volume of alkali= 0.025
0.003/0.025= 0.12
concentration of alkali= 0.12 mol/dm3
Volumes of gases
24dm3 per mole is known as the molar gas volume.
number of moles of gas = volume of gas (dm3)/ 24dm3
Example
An air bag is inflated by 70g of Nitrogen. What volume would the nitrogen gas occupy?
Ar of N =14
Find number of moles
70g/ 28 (Mr) = 2.5 mol
2.Find volume
Volume= molesx24dm3
2.5x24=60dm3
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