Quantitative Chemistry in Chemical Reactions
Quantitative Chemistry
Relative Formula Mass (M
r
)
Total of all the relative
atomic masses of all the
atoms in the compound
% mass of an element
in compound
=
A
r
x number of atoms
of that element
M
r
of the compound
x100
e.g. % by mass of chloring in MgCl
2
1.
Ar of Mg is 24, there are 2 atoms 2 x 24 = 48
2.
Mr of compound (see opposite) – MgCl
2
= 95
Calculation = (48/95) x 100 =
50.1%
1 Mole
= 6.02 x 10
23
particles
(atoms, molecules, ions)
= A
r
or M
r
in grams
e.g.
A
r
Mg is 24
and
1 mole Mg
is 24g
M
r
MgCl2 is 95
and
1 mole
is 95g
Total mass of reactants
=
Total mass of products
H2O would
therefore
be 36g
Mass may seem
to increase if
oxygen from the
air reacts
Mass may seem
to decrease if a
gas escapes into
atmosphere
1. Calculate the number of moles in
34g of LiOH. (Li – 7 O – 16 H – 1)
M
r
LiOH = (1 x7) + (1 x 16) + (1 x 1)
= 7 + 16 + 1 = 24
Number of moles = 34/24 = 1.4
mol
2. Calculate the mass of water in
4.5mol (H – 1, O – 16)
Mr water H
2
O = (2 x 1) + (1 x 16) = 18
4.5 = mass / 18
4.5 x 18 = mass
= 81
g
Learn and
rearrange
Working out the balanced symbol equations
from the masses
1.
Calculate number of moles by mass/A
r
2.
Divide number of moles of each substance
by the smallest number of moles
3.
If any are not whole numbers – multiply all
by the same amount to become whole
4.
Write the balanced equation
8.1g ZnO reacts completely with 0.6g of carbon to
form 2.2g of carbon dioxide and 6.5g zinc
1.
M
r
ZnO = (65 +16) = 81 M
r
C = 12
M
r
CO
2
= 12 +(2 x 16) = 44 M
r
Zn 65
ZnO = 8.1 / 81 = 0.1mol C =0.6 / 12 = 0.05mol
CO2 = 2.2 / 44 = 0.05 mol Zn = 6.5 / 65 = 0.1mol
2. Divide by the smallest 0.05
ZnO = 0.1/0.05 = 2 C = 0.05/0.05 = 1
CO
2
= 0.05/0.05 = 1 Zn = 0.1 / 0.05 = 2
3. Equation
2
ZnO + C
→ CO2 +
2
Zn
Limiting Reactants
In a reaction to produce sodium sulfide, Na
2
S, 9.2 g of sodium is reacted with 8.0 g of sulfur.
Which reactant is in excess and which is limiting?
Step 1: Write a balanced equation:
2
Na + S
→ Na
2
S (2 : 1 mols of Na react with S)
Step 2: Calculate the relative formula masses (M
r
): Na = 23, S=32 Na
2
S = (2 x 23) + 32 = 78
Step 3: Calculate the number of moles
Na = 9.2/23 = 0.4 mols S = 8.0/32 = 0.25 mols
but 1.6 : 1 mols of Na reacted with S
Step 4: Identify the reactant in excess / limiting
•
Na must be limited because there is less than 2mols (needs to be double of S but it is
not
•
S is in excess because it is more than half the number of moles of Na
Look at p127 for an example to show
how to calculate the mass of a
reactant
Concentrations of solution
Concentration
(g/dm
3
)
=
Mass of solutes (g)
Volume of solvent (dm
3
)
Exam tip!
Change volumes to dm
3
1dm
3
= 1000cm
3
What mass of sodium chloride would be
needed to form a 3.2g/dm
3
in a 45cm
3
solution
1.
Change volume to dm
3
45/1000 = 0.045dm
3
2. Calculate mass
3.2 = mass / 0.045
3.2 x 0.045 = mass
= 0.144g
6 mark calculation questions
Make sure
you can do
this!
How many moles of NaOH are
in a 25cm
3
solution with a
concentration of 56 g/dm
3
?
1.
Change volume to dm
3
25/1000 = 0.025dm
3
2. Calculate mass 56 = mass / 0.025
56 x 0.025 = mass
= 1.4
Use the moles, mass and M
r
equation
3. Calculate M
r
NaOH = (1x23)+(1x16)+ (1x1)
= 23 + 16 + 1 = 40
4. Number of moles = mass / M
r
= 1.4 / 40
Number of moles = 0.035 mols
Step 1 and 2 use
this equation
Step 3 and 4 use
this equation
Explore the fundamentals of quantitative chemistry, from calculating percentages by mass to determining limiting reactants. Learn how to work out balanced symbol equations and solve concentration questions with ease. Enhance your knowledge of moles, molecular masses, and more in chemical reactions.
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Presentation Transcript
Quantitative Chemistry Ar x number of atoms of that element Relative Formula Mass (Mr) = % mass of an element in compound x100 Mr of the compound e.g. % by mass of chloring in MgCl2 1. Ar of Mg is 24, there are 2 atoms 2 x 24 = 48 2. Mr of compound (see opposite) MgCl2 = 95 Total of all the relative atomic masses of all the atoms in the compound Calculation = (48/95) x 100 = 50.1% 1 Mole = 6.02 x 1023 particles (atoms, molecules, ions) Mass may seem to increase if oxygen from the air reacts = Ar or Mr in grams Mass may seem to decrease if a gas escapes into atmosphere e.g. Ar Mg is 24 and 1 mole Mg is 24g Mr MgCl2 is 95 and 1 mole is 95g H2O would therefore be 36g Total mass of reactants = Total mass of products
Working out the balanced symbol equations from the masses Learn and rearrange 1. Calculate number of moles by mass/Ar 2. Divide number of moles of each substance by the smallest number of moles 3. If any are not whole numbers multiply all by the same amount to become whole 4. Write the balanced equation 1. Calculate the number of moles in 34g of LiOH. (Li 7 O 16 H 1) Mr LiOH = (1 x7) + (1 x 16) + (1 x 1) = 7 + 16 + 1 = 24 8.1g ZnO reacts completely with 0.6g of carbon to form 2.2g of carbon dioxide and 6.5g zinc Number of moles = 34/24 = 1.4 mol 1. Mr ZnO = (65 +16) = 81 Mr C = 12 Mr CO2= 12 +(2 x 16) = 44 Mr Zn 65 2. Calculate the mass of water in 4.5mol (H 1, O 16) ZnO = 8.1 / 81 = 0.1mol C =0.6 / 12 = 0.05mol CO2 = 2.2 / 44 = 0.05 mol Zn = 6.5 / 65 = 0.1mol Mr water H2O = (2 x 1) + (1 x 16) = 18 2. Divide by the smallest 0.05 ZnO = 0.1/0.05 = 2 C = 0.05/0.05 = 1 CO2 = 0.05/0.05 = 1 Zn = 0.1 / 0.05 = 2 4.5 = mass / 18 4.5 x 18 = mass = 81g 3. Equation 2 ZnO + C CO2 + 2 Zn
Limiting Reactants In a reaction to produce sodium sulfide, Na2S, 9.2 g of sodium is reacted with 8.0 g of sulfur. Which reactant is in excess and which is limiting? Step 1: Write a balanced equation: 2Na + S Na2S (2 : 1 mols of Na react with S) Step 2: Calculate the relative formula masses (Mr): Na = 23, S=32 Na2S = (2 x 23) + 32 = 78 Step 3: Calculate the number of moles Na = 9.2/23 = 0.4 mols S = 8.0/32 = 0.25 mols but 1.6 : 1 mols of Na reacted with S Step 4: Identify the reactant in excess / limiting Na must be limited because there is less than 2mols (needs to be double of S but it is not S is in excess because it is more than half the number of moles of Na Look at p127 for an example to show how to calculate the mass of a reactant
Concentrations of solution 6 mark calculation questions Make sure you can do this! How many moles of NaOH are in a 25cm3 solution with a concentration of 56 g/dm3? Mass of solutes (g) Concentration (g/dm3) = Volume of solvent (dm3) Exam tip! Change volumes to dm3 1dm3 = 1000cm3 Step 1 and 2 use this equation 1. Change volume to dm3 25/1000 = 0.025dm3 2. Calculate mass 56 = mass / 0.025 56 x 0.025 = mass What mass of sodium chloride would be needed to form a 3.2g/dm3 in a 45cm3 solution = 1.4 1. Change volume to dm3 Step 3 and 4 use this equation Use the moles, mass and Mr equation 3. Calculate Mr NaOH = (1x23)+(1x16)+ (1x1) = 23 + 16 + 1 = 40 4. Number of moles = mass / Mr = 1.4 / 40 45/1000 = 0.045dm3 2. Calculate mass 3.2 = mass / 0.045 3.2 x 0.045 = mass = 0.144g Number of moles = 0.035 mols
