Student copy of Stoichiometry sample calculations.pptx

IGCSE Chemistry Stoichiometry ppt

1. Stoichiometry

2. Stoichiometry “stochio” = Greek for element “metry” = measurement   Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.

3. SOI -Transformation of one quantitative relationship into another leads to balance and innovation in the real world. Key Concept- Relationships Related Concept- Balance , Transformation Global Context - Scientific and Technical Innovation

4. Prior Knowledge.

5. How is mass of an atom measured? One grain of sand contains millions of atoms, so atoms must be really small. Atoms are so small that their masses are not measured directly. Instead, all atoms are compared with the mass of carbon-12. The mass of an atom on this scale is called its relative atomic mass .

6. Relative atomic mass Mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12 Each atom has a different r.a.m. value.

7. Where are r.a.m. values found? The values of relative atomic mass (r.a.m.) are usually found in the periodic table. So you don’t have to work them out or remember them all! atomic number symbol relative atomic mass When looking up relative atomic mass in the periodic table, remember that it always the larger of the two numbers given.

8. Why isn’t r.a.m. always a whole number? Relative atomic mass (r.a.m.) is not always a whole number. For example, the r.a.m. of chlorine is 35.5 . The standard r.a.m. value of each element is actually the average relative atomic mass, which takes all the isotopes of each element into account. Chlorine has two isotopes: chlorine-35 (75%) and chlorine-37 (25%). average r.a.m. of chlorine = (35 x 75%) + (37 x 25%) = (35 x 0.75) + (37 x 0.25) = 26.25 + 9.25 = 35.5

9. Calculating average r.a.m. from isotopes To calculate the average r.a.m. of a mixture of isotopes, multiply the percentage of each isotope by its relative atomic mass and then add these together. Naturally-occurring bromine is composed of two isotopes: bromine-79 (50.5%) and bromine-81 (49.5%). What is the average r.a.m. of naturally-occurring bromine?

10. Solution average r.a.m. = (79 x 50.5%) + (81 x 49.5%) = (79 x 0.505) + (81 x 0.495) = 39.895 + 40.095 = 79.99 This figure can be rounded up.

11. What about the mass of compounds? Most substances are made of molecules, not individual atoms. Molecules are really small too, so can we work out their masses in the same kind of way? Of course! The mass of a molecule is called the relative formula mass . This is calculated by adding up the relative atomic masses of all the atoms in the molecule.

12. How is relative formula mass calculated? To find the relative formula mass of a compound, add up the relative atomic masses of all the atoms in its formula. Step 1 : Write down the formula of the molecule. Step 2 : Find the r.a.m. of each type of atom in the molecule. Step 3 : Multiply each r.a.m. by the number of atoms of that element and add these values together. What is the relative formula mass of water? Step 1: formula of water is H 2 O Step 2: r.a.m. values: hydrogen = 1, oxygen = 16 Step 3: relative formula mass = (2 x 1) + (1 x 16) = 18

13. Calculate the Relative formula mass for the following compounds. FeS CuSO 4 NH 4 Cl Al 2 (SO 4 ) 3

14. The Mole Concept

15. MOLE- the chemical counting unit It is difficult to find out the weight of a single atom or molecule- too small. A ‘counting unit’ has to be found- practical chemistry. The counting unit plays an important role when dealing with large no. of small objects. Eg 1. Banks weigh coins rather than count them Eg 2. no. of sweets from a jar can be found out from their mass

16. How can we find the no. of iron atoms in an iron block? Chemists count atoms and molecules by weighing them. The amount of the substance is taken as the relative formula mass of substance in grams. This is known as 1 mole of a substance (1 mol) . Each mole contains same no. of atoms, molecules or formula units. The no. per mole has been given by an Italian chemist named Amedeo Avogadro.

17. Avogadro ’ s Number Avogadro ’ s Number (symbol N ) is the number of atoms in 12.01 grams of carbon. Its numerical value is 6.02 × 10 23 . Therefore, a 12.01 g sample of carbon contains 6.02 × 10 23 carbon atoms.

18. The Mole The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro ’ s number of particles, that is 6.02 × 10 23 particles. 1 mol = Avogadro ’ s Number = 6.02 × 10 23 units We can use the mole relationship to convert between the number of particles and the mass of a substance . Eg: Mr of C = 12 therefore, mass of 1mole of C (molar mass) = 12g This mass (1 mole) contains 6.02 x 10 23 units of C atoms.

19. Molar Mass The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of iron is 55.85 amu. Therefore, the molar mass of iron is 55.85 g/mol. Since oxygen occurs naturally as a diatomic, O 2 , the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol.

20. Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of magnesium nitrate, Mg(NO 3 ) 2 ? The sum of the atomic masses is: 24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) = 24.31 + 2(62.01) = 148.33 amu The molar mass for Mg(NO 3 ) 2 is 148.33 g/mol.

21. Calculating no. of moles No. of moles = Given mass ____________ Molar mass

22. Solve How many moles are present in 60 g of Sodium hydroxide? Relative formula mass = Mr(NaOH) = 23 + 16+ 1= 40 Thus, Molar mass of NaOH = 40 g mol -1 Use the mole triangle for further calculations. No. of moles = 1.5

23. Mass-Mole Calculations What is the mass of 1.33 moles of titanium, Ti? We want grams, we have 1.33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti Formula: Mass = No. of moles x Molar mass = 1.33 x 47.88 = 63.68g

24. Percent Composition The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H 2 O is: 11% hydrogen and 89% oxygen All water contains 11% hydrogen and 89% oxygen by mass.

25. Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Lets practice using H 2 O. Assume you have 1 mole of the compound. One mole of H 2 O contains 2 mol of hydrogen and 1 of oxygen. molar mass of H 2 O = 18 g mol -1 2g Hydrogen + 16g Oxygen = 18g of H 2 O

26. Calculating Percent Composition Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water: 16 g Oxygen 18 g H 2 O × 100% = 89% 2 g Hydrogen 18 g H 2 O × 100% = 11%

27. Percent Composition Problem TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C 7 H 5 (NO 2 ) 3 . 7(12 g C) + 5(1 g H) + 3 (14 g N + 32 g O) = * g C 7 H 5 (NO 2 ) 3 *84 g C + 5 g H + 42 g N + 96 g O = 227 g C 7 H 5 (NO 2 ) 3 .

28. Percent Composition of TNT 84.07 g C 227.15 g TNT × 100% = 37.01% C 1.01 g H 227.15 g TNT H × 100% = 2.22% H 42.03 g N 227.15 g TNT × 100% = 18.50% N 96.00 g O 227.15 g TNT × 100% = 42.26% O

29. Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C 6 H 6 The empirical formula of benzene is CH. The molecular formula of octane is C 8 H 18 The empirical formula of octane is C 4 H 9 . The empirical formula might or might not be the same as the actual molecular formula! The empirical formula = molecular formula for IONIC COMPOUNDS - ALWAYS!

30. Steps 1.Start with the number of grams of each element, given in the problem. If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given. 2. Convert the mass of each element to moles using the molar mass from the periodic table . 3.Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number.

31. Calculating Empirical Formulas Use the following poem to remember the steps: Percent to mass Mass to moles Divide by small Multiply ‘til whole

32. Tabular presentation Sr. No Element symbol Percentage by mass Mass in 100 g Molar mass No. of Moles Simpl-est ratio Form-ula 1 A x 2 B 100 - x

33. EXAMPLE PROBLEM Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen.

34. Step One: Percent to Mass Let’s assume we have a 100. g sample of methyl acetate. This means that each element’s percent is also the number of grams of that element. 48.64% C = 48.64 g C 8.16% H = 8.16 g H 43.20% O = 43.20 g O

35. Step Two: Mass to Moles Convert each mass into moles using the molar mass of each element. 48.64 g C /12 g C = 4.053 mol C 8.16 g H / 1 g = 8. 16 mol H 43.20 g O x 16 g O = 2.700 mol O

36. Step Three: Divide by Small Oxygen accounts for the smallest number of moles in the formula, so divide each element by oxygen’s number of moles: 2.700 mol Carbon: 4.053 mol / 2.700 mol = 1.501 = 1.5 Hydrogen: 8.16 mol / 2.700 mol = 3.02 = 3 Oxygen: 2.700 mol / 2.700 mol = 1.000 = 1 Remember, we will want whole-number ratios

37. Your turn- Calculate the EF A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?

38. Molecular Formulas If the empirical formula is different from the molecular formula, the molecular formula will always be a simple multiple of the empirical formula. EX: The empirical formula for hydrogen peroxide is HO; the molecular formula is H 2 O 2 . In both formulas, the ratio of oxygen to hydrogen is 1:1

39. Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene. The actual molecular formula is some multiple of the empirical formula, (CH) n . Let that multiple be “ n ” . which is the “ multiplier ” Benzene has a molar mass of 78 g/mol. Find n to find the molecular formula. n= Mass of Molecular formula / Mass of Empirical formula = CH (CH) n 78 g/mol 13 g/mol n = 6 and the molecular formula is C 6 H 6 .

40. Calculating Molecular Formulas Step 1 – Calculate the empirical formula (if needed) Step 2 – GIVEN molecular mass /empirical formula molar mass = multiplier Step 3 – Multiply the empirical formula subscripts by the multiplier found in Step 2

41. EXAMPLE PROBLEM Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.