Calculations in Chemistry- part 2
Molar volume
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What is the mass of:
600cm^3 of Ammonia gas NH3 at RTP?
0.43g
1000mL of Methane CH4 gas at RTP?
0.67g
4800 cm^3 Oxygen gas O2 at RTP?
6.40g
10000 cm^3 of Nitrogen dioxide NO2 gas at RTP?
19.17g
Molar volume
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What is the volume of:
0.003 moles of hydrogen H2 at RTP?
72cm^3
8g of oxygen O2 gas at RTP?
6000 cm^3
120g of sulphur dioxide SO2 gas at RTP?
45000 cm^3
5.6 g of Nitrogen N2 gas at RTP?
4800 cm^3
Reacting volumes
• Gases react according to the mole ratio in a
balanced equation.
• Volume of a gas is proportional to the moles.
• 2H2(g) + O2(g)  2H2O(g)
• In the above reaction, hydrogen, oxygen and
water vapour are in the ratio 2:1:2.
• If 100 cm^3 of hydrogen were completely
reacted, 50cm^3 oxygen are used and
100cm^3 water vapour are produced.
• Nitrogen and Hydrogen react according to the
equation.
• N2(g) + 3H2(g)  2NH3(g)
• If 150cm^3 of ammonia are formed, what will
be the volume of nitrogen and hydrogen
reacted?
• N2:NH3 = 1:2 So nitrogen reacted is 75cm^3
• N2: H2 = 1:3 So hydrogen reacted is 225cm^3
• In the following reaction 200 cm^3 of Sulphur dioxide
gas was mixed with 175cm^3 oxygen.
• 2SO2(g) + O2(g)  2SO3(g)
• (a) Balance the equation
• (b) Which gas will be completely used up?
• SO2 (limiting agent)
• (c) What is the volume of SO3 produced?
• 200cm^3
• (d) Name gases present when the reaction finished?
• SO3 and remaining O2
• (e) What is the final volume of the gaseous mixture?
• 200cm^3 (SO3)+ 75cm^3 (O2) = 275cm^3
• Methane burns in air according to the following
equation:
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
• In an experiment, 600cm^3 of methane were burnt in
1500cm^3 of O2.
• (a) Which chemical is the limiting reagent?
• Methane is the limiting agent
• What is the volume of CO2 and H2O are produced?
• CO2 = 600cm^3
H2O = 2 X 600 = 1200cm^3
• What volume of unreacted gas is left?
• Unreacted gas is Oxygen = 1500 – 1200 = 300cm^3
• What is the final volume at the end of the experiment?
• Final volume contains CO2, H2O and excess O2
• 600+ 1200+ 300 = 2100cm^3
% Yield
IGCSE Chemistry
• A student was asked the following calculation:
• Copper metal is made from copper(II) oxide by
heating it with carbon powder. The equation is as
follows:
• 2CuO + C  2Cu + CO2
• 8.0g of copper (II) oxide was heated with 4.0g of
carbon powder.
• (a) How many moles of copper(II) oxide was used?
• (b) How many moles of carbon were mixed?
• (c) What is the limiting reagent?
• (d) What mass of copper is made in this calculation?
• (e) How many g of the other reagent is also left at
the end of the reaction?
• (f) How many moles of CO2 were produced?
• (g) What is the volume of this CO2 at room
temperature and pressure?
The student made calculations and
made the following results.
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(a) moles of copper(II) oxide = 0.1
(b) moles of carbon mixed = 0.33
(c) Limiting agent is Copper(II) oxide
(d) mass of copper made = 6.4g
(e) g of other reagent(excess) left = 3.36g
(f) Moles of CO2 produced = 0.05
(g) Volume of CO2 produced = 0.05 x 24000
= 1200 cm3
• The teacher asked the student about the expected amount of
copper.
• The student read the previous calculations and said 6.4g
• The teacher then asked the student to conduct an experiment
using 8.0g of copper(II) oxide and 4.0g of carbon.
• The student conducted the experiment using the following set
up
• What is left in the test tube when the experiment is over?
• What the student might have seen with lime water? Why?
• What can be the reason the test tube is slanted slightly down?
• The student separated copper produced from
remaining carbon and weighed. She found that
the weight is smaller than what she calculated.
That is 4.8g
• Why you think the amount of copper is smaller
than what she calculated?
• What you call the amount of product calculated
and the amount of product really produced in
an experiment?
• If the formula to calculate % yield is:
find out the percentage yield of copper.
Try this question now:
• Zinc displaces copper from copper(II) sulphate by
displacement reaction according to the following
equation:
• CuSO4 + Zn  ZnSO4 + Cu
• In a reaction 3.2g of copper(II) sulphate was
mixed with 3.0g of Zinc.
• (a) How many moles of copper(II) sulphate are
present in 3.2g?
• 0.02 moles
• (b) How many moles of Zinc is present in 3.0g?
• 0.046 moles
• What chemical is the limiting agent?
• Copper(II) sulphate is the limiting agent
• How many moles of the other chemical is left
at the end of the reaction?
• 0.046 – 0.020 = 0.026 moles of Zinc
• What mass is it?
• 0.026 x 65g = 1.69g
• What mass of copper is produced? (theoretical
yield)
• 0.02 x 64g = 1.28 g
• In an experiment, 1.10g of copper was
produced (Experimental/actual yield). Find out
the % yield of copper
FORMULA
IGCSE Chemistry
Formula
• Two important types of formulae in Chemistry
are:
• 1. Empirical formula
This shows which elements are present in a
compound and the ratio of various atoms in one
molecule. E.g Empirical formula of Benzene is CH
• 2. Molecular formula
Molecular formula shows actual number of
various atoms in a molecule.
E.g Molecular formula of Benzene is C6H6
• When you multiply Empirical formula with a
number, you get the molecular formula.
• In the example of benzene, when you multiply
Empirical formula with 6, you get the molecular
formula
• (CH) X 6  C6H6
• You can calculate the number if you know the
empirical formula and molecular mass.
• First find the empirical formula mass by adding
various atomic masses.
• Then divide the molecular mass with empirical
formula mass, you get the number.
Molecular formula
• Molecular mass will be given in the question.
• In the example of Benzene, molecular mass is
given as 78 and empirical formula is CH
• Find Empirical formula mass first = 12 + 1 = 13
• Divide molecular mass with empirical formula
mass to get number (n)
• 78/13 = 6
• Now multiply Empirical formula with 6 you get
molecular formula
• (CH) X 6 = C6H6
How to find the empirical formula?
• To find the empirical formula, you need the mass
(g) or (%) of different element.
• Step 1:convert mass (%) in to moles by dividing
with molar mass of the element.
• Step 2:Find the mole ratio by dividing each
moles calculated in step one with the smallest
mole.
• Step 3: Round off the moles to the nearest whole
number and write the empirical formula.
Example
• A hydrocarbon contains 92.3% C and 7.7% H.
Its molecular mass is 78. find the empirical
formula and molecular formula of the
compound.
• Empirical formula
Empirical formula = CH
• Now find out n
• 78/13 = 6
• Molecular formula is (CH) X 6
• C6H6
Another example
• Determine the empirical formula of the
compound containing 37.5% C, 12.5% H, and
50.0% O by weight.
• Deduce the molecular formula of the
compound if the molecular mass is 64.
• Empirical formula mass = 12 + 4 + 16 = 32
• n = molecular mass/empirical formula mass
64/32 = 2
Molecular formula is (CH4O) X 2
C2H8O2
END OF PART 2