Calculations in Chemistry IGCSE Chemistry Mole concept Relative

Calculations in Chemistry IGCSE Chemistry

�Mole concept

Relative atomic mass and Relative molecular mass �Why atomic mass or molecular mass has no unit? �Atoms and molecules are too small to weigh. �Relative atomic mass and Relative molecular mass are comparative values. �This shows how many times an atom or a molecule heavier than th of a Carbon atom.

The mole…. �Atomic mass or molecular mass with g after the number is called a mole. �Example: �Atomic mass of Oxygen is 16. � 16 g of oxygen(O) is 1 mole of oxygen atoms. �Molecular mass of Oxygen O 2 is 32. � 32 g of Oxygen (O 2) is 1 mole of oxygen molecules. �Molecular mass of water is 18. � 18 g of water is called 1 mole of water.

�Number �Mass mass of moles = (g) =Number of moles X Molar

Questions � 1. How many moles are there in 4 g of Sodium hydroxide? �Number of moles = = 0. 10 moles � 2. Find the number of moles present in 1. 28 g of oxygen molecules. �Number of moles = � � = 0. 04 moles

� 3. What is the mass of 0. 01 moles of sodium hydroxide? �Mass = 0. 01 X 40 g � = 0. 40 g � 4. How many grams 1. 5 moles of sulphuric acid weighs? �Mass = 1. 5 X 98 g � = 147 g

Try the following questions: � 1. What is the mass of 0. 02 moles of carbon dioxide gas? � 0. 88 g � 2. How many moles are present in 1 g of calcium carbonate, Ca. CO 3? � 0. 01 moles � 3. Find the number of moles present in 3 g of potassium hydroxide, KOH. � 0. 054 moles � 4. Calculate the mass of 100 moles of copper oxide, Cu. O � 8000 g � 5. 5. 2 g of sodium hydroxide, Na. OH or 3 g of potassium hydroxide, KOH contains more number of moles? Show your working clearly. � Na. OH = 0. 13 moles KOH = 0. 054 moles. So Na. OH

Avogadro constant �One mole of any substance contains 6. 02 X 10 atoms or molecules. � 6. 02 X 10 is a constant and is called Avogadro constant or Avogadro Number. �In other words, Avogadro Number is the number of atoms or number of molecules present in ONE mole of a substance. �Avogadro constant is used to find

Number of moles = Number of molecules Avogadro constant Number of molecules = Number of moles X Avogadro constant

Calculations �Remember always change mass or number of molecules into moles.

Examples: �How many molecules of Carbon dioxide are there in 0. 44 g of the gas? �Answer: �Number of moles = 0. 44 g/44 g � = 0. 01 moles �Number � of molecules = 0. 01 X 6. 02 X 10 = 6. 02 X 10

Calculate how many oxygen atoms are present in 1. 96 g of sulphuric acid? �Answer: �Moles of sulphuric acid = 1. 96 g/98 g � = 0. 02 moles �Number of molecules of sulphuric acid � = 0. 02 X 6. 02 X 10 � =1. 20 X 10 �Each sulphuric acid contains 4 oxygen atoms. �So number of oxygen atoms 0. 02 moles of H 2 SO 4 is

Mole ratio �When chemical reactions take place, the reactants and products are in a simple ratio (mole ratio) �For example, Zinc and hydrochloric acid react according to the following reaction: � Zinc gas + Hydrochloric acid Zinc chloride + Hydrogen � Zn � It + 2 HCl Zn. Cl 2 + H 2 means, zinc, Hydrochloric acid, Zinc chloride and Hydrogen moles are in the ratio 1: 2: 1: 1 � One mole of Zinc reacts with 2 moles of Hydrochloric acid to produce, One mole of Zinc chloride and 1 mole of Hydrogen gas. � The reactants and products are always in a ratio by moles and not by mass.

Find how many grams of copper oxide can be made from 3. 2 g of copper. �Answer: � 2 Cu + O 2 2 Cu. O �The balanced equation shows that two moles of copper reacts with 1 mole of oxygen to form 2 moles of copper(II) oxide. �So change 3. 2 g to moles �Moles of copper = 3. 2 g/64 g =0. 05 moles �Copper : Copper oxide is 1: 1 ratio. �So moles of copper(II) oxide = 0. 05 moles �Mass (g) = 0. 05 X 80 g � = 4 g

When Posphorus burns in oxygen, Phosporus(V)oxide is produced. In an experiment, 8 g of phsphorus(V)oxide is produced. Find the mass of Phosphorus reacted with oxygen. �Answer: �Equation: 4 P + 5 O 2 2 P 2 O 5 �Mole ratio is 4: 5: 2 �Moles of phosphorus(V)oxide= 8 g/142 g � = 0. 056 moles �Mole ratio Phosphorus oxide: phosphorus � 2: 4 �Moles of phosphorus burnt = 0. 056 X 2 � = 0. 112 �Mass of phosphorus = 0. 112 X 31 g � = 3. 472 g

Limiting reagent and excess �When there is a chemical reaction, the reactant which is used up completely is the limiting reagent. �The chemical which is left some unreacted at the end is called excess. �When the limiting reagent is finished, the reaction stops. �All mole ratios should be done using the limiting reagent.

�Example 1 � 3 Mg + Al 2 O 3 3 Mg. O + 2 Al �If 2 moles of Magnesium is used with 1 moles of aluminium oxide, which is the limiting agent? Which chemical is in excess? How many moles in excess? �Mole ratio of Mg and Al 2 O 3 is 3: 1 �It means for 2 moles of Mg, 0. 67 moles of Al 2 O 3 is enough. But we have 1 mole of Al 2 O 3. So Al 2 O 3 is excess and Mg is the limiting reagent. �Moles of Al 2 O 3 (excess) left after reaction is

If mass is given, change all masses into moles first. Example 2 2 Cu + O 2 2 Cu. O In a reaction, 1. 92 g of copper is oxidised with 0. 64 g of oxygen. (a) Find which chemical is the limiting agent (b) How many g of the other chemical (excess) remaining at the end? (c) What mass of Copper (II)oxide produced? Answer Change mass in to moles Number of moles of copper = 1. 92 g/64 g = 0. 03 moles Number of moles of oxygen = 0. 64 g/32 g = 0. 02 moles Cu : O 2 mole ratio is 2: 1 It means, for 0. 03 moles of copper only 0. 015 moles of oxygen is enough. But we have 0. 02 moles. Copper is the limiting reagent and oxygen is excess.

�Excess oxygen = 0. 020 – 0. 015 = 0. 005 moles � Mass of excess oxygen = 0. 005 X 32 g = 0. 16 g �Moles of copper oxide (make mole ratio with limiting reagent) 2: 2 That is 0. 03 moles �Mass of copper oxide = 0. 03 X 80 g = 2. 40 g �Now try out the following: � 2 Na + S Na 2 S �If 0. 46 g of sodium and 0. 30 g of sulphur reacted, �(a) find the limiting reagent and excess reagent. �Sulphur �(b) find the mass of excess reagent left. � 0. 0012 moles Na which is 0. 028 g �(c) find what mass of sodium sulfide formed.

Molar volume �Gases have volume. Volume of a gas depends on two factors : �Temperature and Pressure �Volume decreases when pressure is increased. �Volume increases when the temperature is increased.

Molar Volume �One mole of any gas at Room Temperature and Pressure has a volume of 24 dm^3 (24000 cm^3) �Room temperature is considered as 20 degree C and atmospheric pressure is 1 atmosphere (760 mm of mercury) �It means 1 mole of oxygen gas is 24000 cm 3 � 3 moles of oxygen gas is 3 X 24000 cm 3 at RTP. � 24 dm 3(1 mole) is called molar volume of gases

�Moles X molar volume = volume of a gas �Moles = volume of a gas/molar volume Remember to convert volume or mass to moles first. Example 1 What is the volume of 1. 1 g of carbon dioxide gas at RTP? First convert mass in to moles Moles of CO 2 = 1. 1 g/44 g = 0. 025 moles Volume = moles X molar volume = 0. 025 X 24000 cm 3 = 600 cm 3

�Example 2 �Find out the mass of 8000 cm 3 of sulphur dioxide gas, SO 2 at RTP. �First change the volume into moles. �Moles = Volume/molar volume � 8000 cm 3/24000 cm 3 � = 0. 33 moles �Now change the moles in to mass �Mass = moles X molar mass � 0. 33 X 64 g = 21. 12 g

% Yield �We can calculate and predict how much of a product is formed during a reaction using the mole ratio. �How're in a real experiment, the predicted (calculated)amount may be different that we get in actual experiment. �This can be due to: � 1. Some reactants may be left unreacted. � 2. Change of substances in to other forms such as gases � 3. Some reactants may be lost to surrounding by overflowing, splashing etc � 4. Experimental errors and low quality measuring devices.

� %yield = Actual yield X 100 Theoretical yield � Example 1: � In an experiment, 20 g of Calcium carbonate when heated gave 10 g of calcium oxide is formed. Find the percentage yield of calcium oxide. � Ca. CO 3 Ca. O + CO 2 � %yield = Actual yield X 100 Theoretical yield � To work out the above problem, we need the actual yield and theoretical yield(Calculated yield). � Actual yield is given as 10 g of Calcium oxide. � Theoretical yield should be calculated by using mole ratio method.

�Ca. CO 3 Ca. O + CO 2 �We can find the mass of calcium oxide formed from the mass of calcium carbonate given. �But first convert mass in to moles. �Moles of Ca. CO 3 = 20 g/100 g � = 0. 20 moles �Mole ratio Ca. CO 3: Ca. O is 1: 1 �So, moles of Ca. O is also 0. 20 �Mass of Ca. O = 0. 20 X 56 g � = 11. 20 g �Now insert the values �% yield = Actual yield X 100% Theoretical yield That is 10 g X 100% = 89. 28%

�Example 2 � 2 Na + 2 H 2 O 2 Na. OH + H 2 � 115 g of sodium was reacted with water which produced 4. 2 g of hydrogen gas. What is the % yield of hydrogen gas? Answer �Moles of sodium = 5. 00 moles �Moles of H 2 = 2. 50 moles (2: 1) �Mass of H 2 = 5. 00 g �% yield = 84%

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