Quantitative aspects of chemical change sdfgsfgfgsgf Grade 10 Physical Science CAPS 2016
The mole concept
The mole concept Atoms are small chemists know this. But somewhere along the line they have to count them. The mole is the SI Unit for the measurement of the amount of matter. A mole is a large group of atoms or molecules, which can be described as a specific amount of matter, or the number of particles.
The mole concept Particles The particles of an element is called atoms. Na Na Na Na NaCl The particles of a covalent compound is called molecules. H 2 O O H H The particles of an ionic compound is called ions. O H H Na Cl Na Cl
The mole concept A mole is that quantity of matter that has the same number of particles as there are in 12g of carbon 12. How much is this number of particles? Avogardo s number Who is Avogardo? What is this number? 6.02 10 23 Some clever Italian scientist.
The mole concept How big is that? It s 602 000 000 000 000 000 000 000 That s nice.how BIG IS THAT?! Go and read on page 196
The mole concept This means that 1 mole of a substance contains 6.02 10 23 units Example: 1 mole of sodium means 6.02 10 23 sodium atoms 1 mole of H 2 O means 6.02 10 23 H 2 O molecules.
The mole concept So just like: A dozen = 12 A pair = 2 A case = 24 SO a Mole = 6.02 10 23
The mole concept So if I could put all of these particles onto a scale what would it s mass be? The mass of 1 mole of a substance is equal to it s atomic mass in grams. 1 mole Na = 23g 1 mole Mg = 24 g 1 mole O 2 = 32 g
The mole concept So if I could put all of these particles onto a scale what would it s mass be? The mass of 1 mole of a substance is equal to it s atomic mass in grams. 24 1 mole MgO = g 1 mole NaOH = 40 g 1 mole H 2 SO 4 = 98 g
The mole concept Molecular Mass Since particles are so small, it s difficult to use g or kg to calculate their masses. Hence we used the atomic mass unit (u)
The mole concept There s a difference Atomic mass Molar mass Formula mass Symbol: M Symbol: M Symbol: M Unit: g. mol 1 Unit: g. mol 1 Unit: g. mol Used 1 for an ionic Used for an Used for an compound. element compound Eg. M NaCl = 58.5g. mol 1 Eg. M Cu = 63.53g. mol 1 Eg. M H 2 O A= r MgSO 18g. mol 4 = 1 M Mg = 24g. mol 1 M HCl = 36.5g. A r KOH mol = 1 120g. mol 1 56g. mol 1 M Ca = M CO 2 = 40g. mol 1 44g. mol 1
The mole concept Relative Atomic mass Symbol: A r No unit Used for an element Eg.A r Cu = 63.53 A r Mg = 24 A r Ca = 40 There s a difference Relative Molar mass Relative Formula mass Symbol: M r No unit Used for a compound. Eg. M r H 2 O = 18 M r HCl = 36.5 M r MgSO 4 = M r CO 2 = 44 M r KOH = Symbol:M r No unit Used for an ionic compound. Eg.M r NaCl = 58.5 120 56
The mole concept Let s practice Substance Number of moles Mass in grams Particles Number of particles Carbon 1 12 Atoms 6.02 10 23 1 Cobalt 59 Atoms 6. 02 10 23 Carbon dioxide Sodium Chloride 1 1 44 58.5 molecule s ions 6. 02 10 23 2 6. 02 10 23
homework Exercise 20 pg. 198-199
The relationship between: mole mass molar mass n ( ) m (g) M (g.mol-1)
Relationship between mole, mass and molar mass The equation #1 mol ( ) mass (g) n = m M (g. mol 1 ) Molar mass TB. PG 229
Relationship between mole, mass and molar mass The equation #2 mol ( ) n = no. of particles N A Avogardo s number: 6. 02 10 23 TB. PG 229
Relationship between mole, mass and molar mass worksheet
Empirical formula This is the simplest ratio in which the elements of the compound bonds with each other.
H 2 O H H O This formula says that 1 water molecule has: 2 hydrogen atoms and 1 oxygen atom. 1 mole H 2 O contains 6.02 10 23 water molecules So that means there are 2 6.02 10 23 hydrogen atoms and 1 6.02 10 23 oxygen atoms.
Empirical formula worksheet
Homework Exercise 21 pg. 202-203
Percentage Composition Used to determine the composition of a substance according to a percentage division.
Example 1 Determine the percentage composition of NaHCO 3. Step 1 Find the formula mass of NaHCO 3. 84g. mol 1 Step 2 M Na M H M C M O = 23g. mol 1 = 1g. mol 1 = 12g. mol 1 = 16g. mo Find the atomic mass of each element in the compound.
Step 3 Calculate!!!! mass of element (g. mol 1 ) formula mass (g. mol 1 ) 100 = Na = 27.38% H = 1.19% C = 14.29% O = 57.14%
Homework EXERCISE 22 PG. 205-206
Step 1 Find the formula mass of NaHCO 3. Step 2 Find the atomic mass of each element in the compound. Step 3 mass of element (g. mol 1 ) formula mass (g. mol 1 ) 100 =
Compound Composition Used to determine the amount of mol water from crystallization.
Example 1 Weigh out 2.5g of CuSO 4. nh 2 O crystals. Dry the crystals by heating them gently in a crucible. This is done to remove the water. Weigh the crystals again.
Example 1 At start: 2.5g of CuSO 4. nh 2 O After drying: 1.6g of CuSO 4 So, in 2.5g of CuSO 4. nh 2 O 1.6g was CuSO 4 and 0.9g was H 2 O
Example 1 Just like Empirical Formula n CuSO4 = m M = 1.6 159.5 = 0.01mol n H2 O = m M = 0.9 18 = 0.05mol
Example 1 Ratio CuSO 4 : H 2 O 0.01: 0.05 1: 5
Example 1 SO!!!! 1 CuSO 4. 5H 2 O crystals.
Example 2 Weigh out 6.25g of MgSO 4. nh 2 O crystals. Dry the crystals by heating them gently in a crucible. This is done to remove the water. Weigh the crystals again.
Example 2 At start: 6.25g of MgSO 4. nh 2 O After drying: 4g of MgSO 4 So, in 6.25g of MgSO 4. nh 2 O 4g was MgSO 4 and 2.25g was H 2 O
Example 1 Just like Empirical Formula n MgSO4 = m M = 4 120 = 0.03mol n H2 O = m M = 2.25 18 = 0.13mol
Example 2 Ratio MgSO 4 H 2 O 0.03: 0.13 1: 4
Example 2 SO!!!! 1 MgSO 4. 4H 2 O crystals.
Workbook Exercise Write down and complete the following example in your workbook. Determine the number of moles of water of crystallization in 5.4g of AlCl3. nh2o.
Molar gas volume
1 mole of a gas occupies a volume of 22.4 dm 3 at STP. O 2 O 2 What is STP??? O 2 O 2 O 2 O 2 22. 4dm 3 Standard temperature and pressure o or 273K 101.3kPa
Molar gas volume The equation # 1 mol ( ) volume (dm 3 ) n = V V m (22. 4dm 3 ) Molar gas volume TB. PG 229
Molar gas volume The equation #2 mol ( ) mass (g) n = m M (g. mol 1 ) Molar mass TB. PG 229
Molar gas volume The equation #3 mol ( ) n = no. of particles N A Avogardo s number: 6. 02 10 23 TB. PG 229
Example 1 Calculate the mass of 5.6dm 3 of CO 2 at STP. What are you given? What do you want to find?
1.1 3.1 5.1 2.1 Homework 4.1 EXERCISE 23 pg. 207-209
Concentration This is the quantity of a dissolved solute in mole per dm 3 of solution.
Concentration The equation # 1 mol concentration ( ) (mol. dm 3 ) USE IF YOU ARE GIVEN MOLE c = n V (dm 3 ) Volume
Concentration The equation # 1 mass concentration ( g ) (mol. dm 3 ) USE IF YOU ARE GIVEN MASS (g. mol 1 ) Molar mass c = m MV (dm 3 ) Volume
Oh! But converting is so VERY vital 1000 1000 1000 mm 3 cm 3 dm 3 m 3 1000 1000 1000
AND 1cm 3 = 1ml 1dm 3 = 1 litre
Example 1 Calculate the concentration of a solution that contains 0.2 mol copper (II) sulphate in 250cm 3 water. What are you given? What do you want to find?
2.1 Homework EXERCISE 24 pg. 211-212
Stoichiometric calculations
A chemical reaction occurs when substances (reactants) react chemically and produce new substances (products). We represent such a reaction with a chemical formula.
2H 2 + O 2 2H 2 O + 2 mole hydrogen reacts with 1 mole of oxygen to form 2 moles of water. 4 hydrogen atoms react with 2 oxygen atoms to form 2 water molecules. 2 hydrogen molecules react with 1 oxygen molecule to form 2 water molecules. 4g of H 2 reacts with 32g of O 2 to form 36g of H 2 O
C + O 2 CO 2 + 1 mole carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide 1 carbon atom react with 2 oxygen atoms to form 1 carbon dioxide molecule 12g of C reacts with 32g of O 2 to form 44g of CO 2
Calculations
Example 1 mass-mass calculation Calculate the mass of oxygen obtained when 14.7g of potassium chlorate decomposes completely to form potassium chloride.
Example 2 mass-volume calculation What mass of potassium chlorate must be heated to release 90dm 3 of oxygen at STP?
HOMEWOR K EXERCISE 25 PG. 215-217