PowerPoint Presentation S.L.O.P - Paper 1 Unit 3 Quantitative Chemistry Number of elements in a formula Example: FeBr3 The number of elements is the same as the number of symbols. Each different...

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PowerPoint Presentation S.L.O.P - Paper 1 Unit 3 Quantitative Chemistry Number of elements in a formula Example: FeBr3 The number of elements is the same as the number of symbols. Each different symbol starts with a capital letter. FeBr3 So we have two different elements Fe = Iron, Br = Bromine Task 1: How many different elements are in each chemical formula? a) Fe2O3 b) CaCO3 c) H2 d) AgNO3 e) H2SO4 Number of atoms of each element in a formula Example: Na2CO3The small numbers on the bottom right of each symbol tell you how many atoms of that element you have. If there is no number to the right you have one atom of that element. Task 2: How many atoms of each element are in these formulae? Also give the total number of atoms: a) FeCl3 b) H2SO4 c) AgNO3 d) K2SO4 e) CH4 d) NH3 e) SiO2 Na2CO3 2 atoms of Na (sodium) 1 atom of C (carbon) 3 atoms of O (oxygen) = 6 atoms in total Periodic table on back! Name: Class: Revision 1.0: © L. Tull 2019 1 Use of brackets in formulae Example: Al2(SO4)3The number of atoms of each element inside the brackets is multiplied by the number on the bottom right outside the brackets. Task 3: How many atoms of each element are in these formulae? Also give the total number of atoms: a) Mg(NO3)2 b) Ca(OH)2 c) Fe(C2O4)3 d) Al2(CO3)3 Al2(SO4)3 2 atoms of Al (aluminium) 1 atom of S (sulfur) inside the brackets, however multiply by 3 as that’s the number outside: 1 x 3 = 3 atoms of S 4 atoms of O (oxygen) inside the brackets, however multiply by 3 as that’s the number outside: 4 x 3 = 12 atoms of O = 17 atoms in total Balancing equations Example: 4 Al + 3 O2→ 2 Al2O3 The number of atoms of each element either side of the arrow should be the same. __ Al + __ O2 → __ Al2O31. Draw a line down from the arrow in the unbalanced equation and list the symbols of all elements either side. Al = O = Al = O = __ Al + __ O2 → __ Al2O3 Al = O = Al = O = 2. Count the number of atoms of each element currently in the equation. 1 2 3 2 Continued on next page… 2 __ Al + __ O2 → __ Al2O3 Al = O = Al = O = 3. Work on one element (row) at a time, finding a number that both numbers fit into. e.g oxygen’s original 2 and 3 both fit into 6. Put the number you multiplied by into the correct gap in the equation. If at the end it didn’t need multiplying, just put a 1 in the gap or leave it blank. 1 2 3 2 Balancing equations continued… x 3 = 6 x 2 = 6 3 2 4. Note how when I multiplied Al2O3 by 2 to get 6 oxygens, I also multiplied the Al2 bit of it by 2. Track these changes as you go! __ Al + __ O2 → __ Al2O3 Al = O = Al = O = 1 2 3 2 x 3 = 6 x 2 = 6 3 2 x 2 = 4 __ Al + __ O2 → __ Al2O3 Al = O = Al = O = 1 2 3 2 x 3 = 6 x 2 = 6 3 2 x 2 = 4x 4 = 4 45. Now we can check that the numbers of atoms of each element are the same. Al = 4 Al = 4 O = 6 O = 6 The equation is balanced. Now apply what you’ve learned to the questions on the next page. 3 Task 4: Balance these equations: a) ___ Fe + ____ Cl2→ ___ FeCl3 b) ___ Mg + ___ HCl → ___ H2 + ___ MgCl2 c) ___ N2 + ___ O2 → ___ NO d) ___ N2 + ___ O2 → ___ NO2 e) ___ Fe2O3 + ____ C → ____ CO2 + ___ Fe f) ___ Na + ___ H2O → 2 NaOH + ___ H2 4 Calculating Relative Formula Mass (Mr) Example: FeCl3 FeCl3 1 atom of iron 1 x 56 = 56 Look for the elements’ symbols on the periodic table, you need to use the mass numbers (They’re the massive ones!). Note: Sometimes they give you the mass numbers of each element in the Q. 3 atoms of chlorine 3 x 35.5 = 106.5 56 + 106.5 = Mr = 162.5 g/mol You need to add up the masses of each atom of each element. Don’t forget the unit of “grams per mole” (g/mol). Example: Al2(SO4)3 Al2(SO4)3 This example has brackets, remember what we looked at earlier, multiple the number inside the brackets by the number on the outside to get the full number of atoms of that element. 2 atoms of Aluminium 2 x 27 = 54 3 atoms of Sulfur 3 x 32 = 96 12 atoms of Oxygen 12 x 16 = 192 54 + 96 + 192 = Mr = 342 g/mol Now apply what you’ve learned to the questions on the next page. 5 Task 5: Calculate the Relative Formula Mass (Mr) of each substance: Note: Use your copy of the periodic table to find the relative atomic mass: a) FeCl3 b) H2SO4 c) AgNO3 d) K2SO4 e) CH4 f) NH3 g) SiO2 h) Mg(NO3)2 i) Ca(OH)2 j) Fe(C2O4)3 k) Al2(CO3)3 6 Calculating Elemental Percentage Example: Al2O3Sometimes you will be asked a question like this:“What is the percentage by mass in Al2O3 , of aluminium? Step 1: Calculate Mr Step 2: Divide the total mass of the particular element by the Mr then x 100 Al2O3 2 atoms of Aluminium 2 x 27 = 54 3 atoms of Oxygen 3 x 16 = 48 54 + 48 = Mr = 102 g/mol Aluminium made up 54 out of the total of 102 so: 54 102 x 100 = 52.9 % Task 6: Calculate the percentage by mass for the specified element, round to 1 d.p: a) FeCl3 %Fe? b) H2SO4 %O? c) Ca(OH)2 %H? d) Al2(CO3)3 %C? 7 The Mole Chemical amounts are measured in moles, the symbol for the unit mole is mol. The mass of one mole of a substance is equal to its relative formula mass (Mr) in grams e.g. 1 mol of FeCl3 (Mr = 162.5 g/mol) = 162.5 g This is why the units of relative formula mass are “grams per mole” – It’s how many grams 1 mole would weigh! So two moles of a substance would be twice its Mr…. e.g. 2 mol of FeCl3 (Mr = 162.5 g/mol) = 325g This gives us the equation: Mass = Mr x mol Real world mass in grams (g) Relative formula mass in grams per mole (g/mol) How many moles of the substance you have (mol) This can be rearranged to give: mass Mr mol = mass mol Mr = The mass in this equation needs to be in grams, but modern papers are giving you masses in: • nanograms (ng) n x 10-9 • micrograms (μg) n x 10-6 • milligrams (mg) n x 10-3 kilograms (kg) n x 103 • tonnes (T/Mg) (megagrams) n x 106. Whatever number they give you (if not in grams), just multiply by the correct power as above and slot into the equation → remembering these is easier than remembering conversions! 8 The Mole examples….. Example 1: What is the mass of 3.5 moles of Methane (CH4)?, Mr = 16 g/mol Mass = Mr x mol Mass = 16 x 3.5 = 56 g Example 2: How many moles are in 348 g of aluminium oxide (Al2O3)? Mr = 102 g/mol Mol = mass / Mr Mol = 348 / 102 = 3.41 mol Example 3: What is the Mr of two moles of a compound that weighs 204 g? Mr = mass/mol Mol = 204 / 2 = 102 g/mol Therefore it must be aluminium oxide… Example 4: How many moles are in 5 kg of carbon dioxide (CO2)? Mr = 44 g/mol Mol = mass / Mr Mol = 5 x 103 / 44 = 113.6 mol Example 5: How many moles in 600 mg of paracetamol? (C8H9NO2) Mr = 151 g/mol Mol = mass / Mr Mol = 600 x 10-3 / 151 = 3.97 x 10-3 mol Make sure you know the powers of ten that each prefix stands for, this will save you time in your biology, physics and chemistry papers! 9 Task 7: Using the mole equation: a) Write the mole equation in its basic form: b) Write the mole equation rearranged to make mol the subject: c) Write the mole equation rearranged the make Mr the subject: d) Calculate the number of moles in: i) 50g of Calcium Mr (Ca) = 40g/mol ii) 100 g of Zinc Mr (Zn) = 65g/mol iii) 300 tonnes of Iron chloride, Mr (FeCl3) = 162.5 g/mol e) Calculate the mass of: i) 0.5 moles of Cobalt, Mr (Co) = 59 g/mol ii) 0.75 moles of Methane, Mr (CH4) = 16 g/mol iii) 12 moles of Zinc chloride, Mr (ZnCl2) = 136 g/mol 10 Task 8: Moles challenge – For levels 5 and higher: How many moles of each substance are there? You will first need to calculate Mr in order to use the equation mol = mass / Mr a) 150g of Bromine – Br2 b) 540g of calcium chloride – CaCl2 c) 460g of Iron (II) hydroxide – Fe(OH)2 What is the mass of each substance? You will first need to calculate Mr in order to calculate mass using the equation mass = Mr x mol d) 3 moles of Vanadium, V e) 0.75 moles of Lead oxide Pb2O f) 4 moles of Aluminium chloride AlCl3 Calculate the Mr of the unknowns and attempt to identify them by writing a formula that matches the Mr (there are multiple solutions for each). g) 4 moles of X, has a mass of 68 g, what is the Mr of X and suggest its identify by writing a formula that matches the Mr you calculated. h) 5.4 moles of Y has a mass of 162g, Calculate the Mr of Y and suggest its identity by writing a formula that matches the Mr you calculated. 11 Avogadro’s Constant The real world masses of atoms doesn’t match our metric system. However the mass = Mr x mol equation makes it so that 1 mole of a substance weights exactly its Mr in grams. This is because the mole is based on a number called Avogadro’s constant which translates the mass of atoms to the scale of our
Answered 1 days AfterJan 12, 2021

Answer To: PowerPoint Presentation S.L.O.P - Paper 1 Unit 3 Quantitative Chemistry Number of elements in a...

Poulami answered on Jan 14 2021
143 Votes
Task 1
a) Fe2O3: 2 elements
b) CaCO3: 3 elements
c) H2: 1 element
d) AgNO3: 3 elements
e) H2SO4: 3 elements

Task 2
a) FeCl3: 4 atoms (1 Fe, 3 Cl)
b) H2SO4: 7 atoms (2 H, 1 S, 4 O)
c) AgNO3: 5 atoms (1 Ag, 1 N, 3 O)
d) K2SO4: 7 atoms (2 K, 1 S, 4 O)
e) CH4: 5 atoms (1 C, 4 H)
f) NH3: 4 atoms (1 N, 3 H)
g) SiO3: 4 atoms (1 Si, 3 O)
Task 3
a) Mg(NO3)2: 1 Mg, 2 N, 6 O (total: 9 elements)
b) Ca(OH)2: 1 Ca, 2 O, 2 H (total: 5 elements)
c) Fe(C2O4)3: 1 Fe, 6 C, 12 O (total: 19 elements)
d) Al2(CO3)3: 2 Al, 3 C, 9 O (total 14 elements)
Task 4
a) 2 Fe + 3Cl2 = 2FeCl3
b) Mg + 2HCl = H2 + MgCl2
c) N2 + O2 = 2NO
d) N2 + 2O2 = 2NO2
e) 2Fe2O3+ 3C = 3CO2+ 4Fe
f) 2Na + 2H2O = 2NaOH + H2
Task 5
a) FeCl3 = 162.204 g/mol
b) H2SO4 = 98.07848 g/mol
c) AgNO3= 169.9 g/mol
d) K2SO4= 174.259 g/mol
e) CH4= 16.043 g/mol
f) NH3= 17.0 g/mol
g) SiO2= 28 g/mol
h) Mg(NO3)= 148.32 g/mol
i) Ca(OH)2= 74.093 g/mol
j) Fe(C2O4)3= 437.20 g/mol
k) Al2(CO3)3= 233.99 g/mol
Task 6
a) FeCl3 = 34.429% Fe
b) H2SO4 = 65.254% O
c) Ca(OH)2 = 2.7208% H
d) Al2(CO3)3 = 15.399% C
Task 7
a)
· Molar mass of a Substance =...
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