Programme in mole calculation

Transcription

1 Programme in mole calculation Step 1 Start at the very beginning Atoms are very small indeed! If we draw a line 1 metre long, 6,000,000,000 (6 billion) atoms could be lined end to end. So a scientist cannot count atoms, ions or molecules directly. They are far too small and numerous. Instead a scientist counts particles by weighing. This is rather like in a bank. A bank cashier has not got the time to count every coin. She weighs the bags of coins on special scales which tell her how much they are worth. Hydrogen is the lightest atom, we say that hydrogen has a relative atomic mass of 1. In 1g of hydrogen atoms we say that we have 1 mole of hydrogen atoms. Experiments have shown that an atom of carbon weighs 12 times as much as an atom of hydrogen, so the relative atomic mass of carbon is 12. In 12g of carbon atoms we say that we have 1 mole of carbon atoms. Let's look at the following table which shows the relation between the mass of one mole element and the relative atomic mass. Element relative atomic mass mass of one mole element hydrogen 1 1g carbon 12 12g All answers throughout the programme should be collected to 3 sig. figures Class work 1: Use your Periodic Table to find out the mass of one mole of the following atoms. a. Helium b. Iron c. Potassium d. Sulphur e. Platinum f. Tin g. Silver h. Phosphorus. d. e. f. g. h. Actually one mole of atoms has 6.00x10 23 atoms. The scientific "magic number" is called Avogadro's number or Avogadro's constant (L).

2 P.2 Example 1: How many atoms are there in 6g of carbon? The relative atomic mass of carbon is 12. In 12g of carbon there are 6.00x10 23 atoms which equal one mole of carbon atoms. In 6g of carbon there is only half mole, 6 no. of moles of carbon = = Class work 2a: Use your periodic table to help you. How many atoms are there in a. 1.0g of helium b. 28.0g iron c. 13.0g potassium d. 6.4g sulphur e. 32.5g platinum f. 119g tin g. 27.0g silver h. 10.3g phosphorus? d. e. f. g. h. Class work 2b: How many moles of the following atoms? a. 3.0x10 23 hydrogen atoms a. b. 1.5x10 23 oxygen atoms b. c. 1.5x10 23 nitrogen atoms c. d. 3.0x10 23 carbon atoms. d. Put your answers in table form: Hydrogen oxygen nitrogen carbon number of atoms 3.0x x x x10 23 number of mole What do you discover?

3 P.3 Step 2 Masses into moles - moles into masses The following calculation will make use of the equation: actual mass of no. of moles of atoms = molar mass of the atoms the atoms Example 2: How many moles of atoms are there in 4g of calcium? [A r (Ca) = 40] By rearranging equation above, it is just as easy to find out the mass of a fraction of a mole. 4 no. of moles of calcium = = Example 3: What is the mass of 1/8 mole of copper? [A r (Cu) = 64] First equation above must be rearranged: mass = no. of moles x molar mass = 1/8 x 64 = 8 (g) (don't forget the unit) Classwork 3: 1. How many moles of atoms are there in a. 20g calcium b. 54g aluminium c. 11.2g iron d. 78g potassium e. 8g sulphur f. 2.4g magnesium? d. e. f. 2. What is the mass of a. 1/10 mole of sodium atoms b. 2 moles of silver atoms c. 1/3 mole of carbon atoms d. 8 moles of iron atoms e. 1/16 mole of magnesium atoms f. 1/4 mole of copper atoms? d. e. f.

4 P.4 Classwork 4: Here are the creep point questions. Avogadro's number = 6.00x How many atoms are there in a. 1 mole of carbon atoms b. 2 moles of oxygen atoms c. 1/2 mole of sulphur atoms d. 1/3 mole of lead atoms? e. What mass of magnesium contains 2.00x10 23 atoms? f. What mass of carbon contains 2.00x10 23 atoms? g. What mass of magnesium has five times as many atoms as 2g of carbon? h. What mass of potassium has the same number of atoms as 8g of magnesium? d. e. f. g. h. Step 3 Diatomic molecules Sometimes an element exists as molecules. Molecules are groups of atoms held together by chemical bonds. A hydrogen atom has a relative atomic mass of 1. A hydrogen gas molecule (H 2 ) has two hydrogen atoms, then the relative molecular mass (M r ) of the molecule is 2. To have one mole of molecules we should need to weigh out 2g of hydrogen gas. Let's look at the following table which shows the relation between the mass of one mole molecules and the relative molecular mass. molecule relative molecular mass mass of one mole molecules hydrogen (H 2 ) 2 2g nitrogen (N 2 ) 28 28g Referring to the equation on page 3, now we come to have a similar equation: actual mass of no. of moles of molecules = molar mass of the molecules the molecules

5 P.5 Example 4: How many moles of nitrogen molecules are present in 7g of nitrogen gas? First note that the question did not ask about nitrogen atoms. It concerns nitrogen molecules, N 2. Each molecule contains two atoms of nitrogen. Since each atom has a relative atomic mass of 14, then the relative molecular mass of nitrogen molecule is 2 x 14 = 28. Example 5: What is the mass of 10 1 mole of oxygen gas? [Ar (O) = 16] Oxygen is a diatomic gas also. It has the relative molecular mass of 2 x 16 = 32. Now rearrange the basic equation on page 4: mass = number of mole x relative molecular mass = 10 1 x 32 = 3.2 (g) Classwork 5: 1. How many moles of a. chlorine molecules are present in 7.1g of chlorine gas, b. chlorine atoms are present in 7.1g of chlorine gas, c. oxygen molecules are present in 64g of oxygen gas? d. oxygen atoms are present in 64g of oxygen gas? d. 2. What is the mass of a. 1/8 mole of oxygen gas b. 1/4 mole of bromine gas c. 2 moles of chlorine gas d mole of iodine? d.

6 P.6 Step 4 Compounds Some compounds are also made up of molecules. But a compound contains atoms of different elements joined together by chemical bonds. The formula for water is H 2 O. This means that a molecule of water contains two atoms of hydrogen and one atom of oxygen. To work out the mass of one mole of a compound we use the same idea as in Step 3. The relative atomic masses of all the atoms in the compound are added together. Let's look at the following table which shows the relation between the mass of one mole compound and the relative formula (molecular) mass. molecule relative (formula) molecular mass mass of one mole compound water (H 2 O) 1x2+16 = 18 18g ammonia (NH 3 ) 14+1x3 = 17 17g Example 6: What is the mass of one mole of carbon dioxide? [A r (C) = 12, A r (O) = 16] The carbon dioxide molecule contains 1 carbon atom and 2 oxygen atoms. The relative molecular mass is equal to x2 = 44. The mass of one mole of carbon dioxide is 44g. Example 7: What is the relative formula mass of magnesium nitrate Mg(NO 3 ) 2? [A r (Mg) = 24 A r (N) = 14, A r (O) = 16] Each mole of magnesium nitrate contains 1 mole magnesium ions, 2 moles of nitrogen and 6 moles of oxygen. So the relative formula mass is x14 + 6x16 = 148. Classwork 6: Find the relative formula mass of a. copper(ii) oxide b. sulphur trioxide c. copper(ii) sulphide d. copper(ii) carbonate e. zinc nitrate f. ammonium carbonate. d. e. f.

7 P.7 Referring to the equation on pages 3 and 5, now we have another similar equation: Example 8: actual mass of the substance no. of moles of any substance = molar mass of the formula How many moles are contained in 72g of water? The relative molecular mass of water is 18, so 72 no. of moles of water = = 4 18 Example 9: How many moles of hydrogen atoms and oxygen atoms are contained in 72g of water? Since each mole of water contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms, 4 moles of water contain 8 moles of hydrogen atoms and 4 moles of oxygen atoms. Classwork 7: Find the number of moles contained in a. 40g copper(ii) sulphate b. 282g of zinc nitrate c. 60g sulphur trioxide d. 32g ammonium carbonate. d. Classwork 8: Find the mass of a. 0.5 mole copper(ii) carbonate a. b. 1/4 mole copper(ii) oxide b. c. 1/10 mole of sulphur trioxide c. d mole ammonium carbonate. d.

8 P.8 Revision exercise 1 Take Avogadro's constant = 6.0x g hydrogen gas contain mole(s) of hydrogen molecules, mole(s) of hydrogen atoms, hydrogen molecules and hydrogen atoms x10 23 particles of CO 2 is equivalent to mole(s) of CO 2, g of CO Find the relative molecular mass of a. sulphuric acid, H 2 SO 4 b. ammonium sulphate, (NH 4 ) 2 SO 4 c. hydrated copper(ii) sulphate, CuSO 4 5H 2 O. 4. What is the mass of a. 2 moles of magnesium atoms b. 1/2 mole of copper atoms c mole of calcium atoms? 5. How many atoms are there in a. 1 mole of carbon atoms b. 0.5 mole of oxygen atoms c. 1/3 mole of sulphur atoms? 6. How many atoms are there in a. 39.0g of potassium b. 6.35g of copper c. 60.0g of calcium? 7. What is the mass of a. 4 moles of water b. 0.5 mole of sodium thiosulphate, Na 2 S 2 O 3 c mole of ammonium dichromate, (NH 4 ) 2 Cr 2 O 7?

9 20g sugar 10g sugar P.9 Step 5 Molarity a unit of concentration of solution Concentration is a measure of how much solute dissolved in a unit volume of the solution. For example, if you dissolve 10g sugar in water so that the volume of solution is 10cm 3, then the concentration of the sugar solution is 10g in 10cm 3 or 1g/cm 3. Then the unit of concentration is g/cm 3, kg/cm 3, kg/m 3 etc. It is very convenient to use figure to compare the concentrations of different solutions. For example, two solutions Molarity is one kind of units of concentration the unit is mol/dm 3. no.of moles of solute molarity = volume of solution in dm3 molarity = (M) = 10.0M no. of moles of NaCl = mass of NaCl formula mass of NaCl 58.5 = 1 no. of moles of NaCl = 58.5 Class work 9 : 1. Calculate the concentration of the following solutions. a g sodium chloride in 250cm 3 solution no. of moles of sodium chloride =

10 1. In order to prepare the following solutions, how many grams of the solution is required? P.10 b g copper(ii) sulphate in 0.5dm 3 solution. c. 44.8g sulphur dioxide in 700cm 3 solution. Class work cm 3, 0.5M copper(ii) chloride solution no. of moles of copper(ii) chloride required = mass of copper(ii) chloride required = 250cm 3, 0.45M sodium hydroxide solution 2dm3, 1.5M potassium nitrate solution 2. What is the volume of the solutions in order to make a. 0.1M solution by dissolving 13.45g copper(ii) chloride? no. of moles of copper(ii) chloride dissolved = volume of the solution (dm 3 ) = b M solution by dissolving 1.825g hydrogen chloride? no. of moles of hydrogen chloride dissolved = volume of the solution (dm 3 ) =

11 P.11 Answer of the class work Class work 1 a. 4.00g b. 56.0g c. 39.0g d. 32.0g e. 195g f. 119g g. 108g h. 31.0g Class work 2a a. 1.5x10 23 b. 3.0x10 23 c. 2.0x10 23 d. 1.2x10 23 e. 1.0x10 23 f. 6.0x10 23 g. 1.5x10 23 h. 2.0x10 23 Class work 2b a. 0.5 b c d. 0.5 Hydrogen oxygen nitrogen carbon number of atoms 3.0x x x x10 23 number of mole Class work 3 1. a b c d e f a. 2.30g b. 216g c. 4.00g d. 448g e. 1.50g f. 15.3g Class work 4 a. 6x10 23 b. 12x10 23 c. 3x10 23 d. 2x10 23 e. 8.00g f. 4.00g g. 20.0g h. 13.0g Class work 5 1. a b c d a. 4.00g b. 40.0g c. 142g d. 63.5g Class work 6 a b c d. 124 e. 189 f Class work 7 a b c d

12 P.12 Class work 8 a. 61.8g b. 19.9g c. 8.00g d. 24.0g Revision exercise , 1.50, 4x10 23, 9x , a b. 132 c a. 48.0g b. 31.8g c. 30.0g 5. a. 6x10 23 b. 3x10 23 c. 2x a. 6x10 23 b. 6x10 22 c. 9x a. 72.0g b. 79.0g c. 63.0g class work 9 1. a. 0.25, 1.0M b. 0.6M c. 1.0M class work a. 0.1, 13.45g b. 4.5g c g 2. a. 0.1, 1.0dm 3 b. 0.05, 2.0dm 3