Quantitative Chemistry Topic 1 SL + HL

1.1 Mole concept and Avogadro’s Constant NA = Number of atoms in 12 gram 12C (def.) = 6.02*1023 units Mole = n = Amount of substance, with the unit mol

1 mol equals: 6.02*1023 Hydrogen atoms, H 6.02*1023 Hydrogen molecules, H2 6.02*1023 Water molecules, H20 6.02*1023 formula units of Sodium Chloride, NaCl If you have 1 mol of something, then you have 6.02*1023 units of that.

In 1 mol of H2-molecules there is 2 mol Hydrogen atoms 1 = 2 In 1 mol of H2O molecules there is 3 mol of atoms; 2 mol of H-atoms and 1 mol of O-atoms 1 = 3 =2 =1

1.2 Formulas A 12C-atom have by definition the mass =12 All other atoms or molecules masses relates to that mass. How much heavier or lighter they are. Masses of single atoms and single molecules, single formula units are therefore called: Relative atomic mass, Ar Relative molecular mass, Mr Relative formula mass, Mr Relative masses have no unit in IB (in other system the unit can be u or amu)

Relative mass => Mass of 1 mol Mass of 1 H-atom: Ar = 1.01 Mass of 1 mol H-atoms = 1.01g Atomic mass Mass of 1 H2-molecule: Ar = 2.016 Mass of 1 mol H2-molecule = 2.016 g Molecular mass Mass of NaCl = Ar = 58.44 Mass of 1 mol NaCl = 58.44 g Formula mass Different masses, same figures

Calculate the mass of one mole of a species from its formula. The term molar mass, M, will be used. Unit: g/mol If you know the formula of a compound, then you know the molar mass of the compound.

You find the masses of the atoms in the Periodic table The Molar mass of water , H2O M = 2*1+16 = 18g/mol The Molar mass of (NH4)2SO4 M= (14+4*1)2 + 32 + 16*4 = 132 g/mol The Molar mass of CuSO4*5H2O M= 63.5 + 32 + 16*4 + 5(2*1 + 16) = 249.5 g/mol

Relationship between the amount of substance in moles, mass and molar mass => The table Quantity Unit Mass, m g Molar mass, M g/mol (usually known) Mole, n mol Downwards: divide Upwards: multiply You can always do a unit analysis, to check your calculation

Examples You have 34 g of Ammonia. How many mole is that? Write the Formula of the compound. Write the table Fill in what you know and ? Calculate

1. Write the Formula of the compound. NH3

2. Write the table NH3 m g M g/mol n mol

3. Fill in what you know and ? NH3 m 34 g M 17 g/mol (from periodic table) n ? mol

4. Calculate NH3 Downwards m 34 g => divide M 17 g/mol n 34/17= 2 mol Answer: 2 mol

New example You have 0.50 mol of NaCl. How many gram is that?

1. Write formula 2. Write table NaCl m g M g/mol n mol

3. Fill in know and ? 4. Calculate NaCl m ? g M 58.5 g/mol (from periodic table) n 0.50 mol Upwards: multiply 0.5 * 58.5 = 29.25 g Answer: 29 g (significant figures)

Percent Composition of Compounds Percent composition of a compound is usually the Mass percent of the elements in a compound, e.g. H2O M = 18 g/mol Mass% H = 2/18 = 0.111 = 11.1% Mass% O = 16/18 = 0.888 = 88.8% (Very seldom is the % composition referred to number of atoms; e.g. H2O have 3 atoms % H= 2/3 = 67% and %O = 1/3 = 33%)

Empirical and Molecular formula Molecular formula: Shows the actual number of each atom/element in a compound, e.g. Ethane C2H6 Glucose C6H12O6 Empirical formula: Shows only the ratio of the elements in a compound, e.g. Ethane CH3 Glucose CH2O (Formulas of salts is empirical formulas)

Determining the Empirical formula You can determine the empirical formula from the percentage composition Use the table Assume that you have 100g of the compound Calculate number of moles Compare Mole-ration. The ratio give the formula

Example A compound is found to consist of: 70.58 % C, 5.93 % H, 23.49 % O (=100%) What’s the formula for this compound? Assume 100 g of the compound. Calculate number of moles

(divide with the lowest) C H O Mass% 70.58 5.93 23.49 (=100%) m 70.58 5.93 23.49 g (=100g) M 12.01 1.01 16.00 g/mol n 5.88 5.87 1.47 mol Check the ratio (divide with the lowest)

C H O 5.88 5.87 1.47 1.47 1.47 1.47 4 : 4 : 1 Empirical formula C4H4O

Molecular formula If you know the Empirical formula, C4H4O, and you know the Molar mass of the compound, 136 g/mol then you can calculate the Molecular formula. C4H4O M = 68 g/mol To Low C8H8O2 M = 136 g/mol Correct C12H12O3 M = 204 g/mol To High

1.3 Chemical Equations A chemical reaction describes what is going on in a chemical reaction. Shows the Reactant(s) Shows the Product(s) Show the Ratio between the compounds (if correct balanced) Can show in which state the compounds are: (s) solid, (l) liquid, (g) gas, (aq) water solution (the most common)

Examples Reactants Products Propane + Oxygen  Carbon dioxide + water Word equation gives only the compounds. Equation must be balanced C3H8 + 5 O2  3 CO2 + 4 H2O Now we have a balanced equation = same number of all elements on both sides of the reaction arrow. The green numbers = subscripts; cannot be changed (a certain compound have a certain formula, if you change the formula you change the compound) The red numbers = coefficients; changes so that the reaction will be balanced (coefficients is valid only for a specific reaction)

What does an equation tell us? C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) The state of the compounds That you need 5 Oxygen molecules/Propane 1 Propane molecule will produce 3 Carbon dioxide molecules and 4 Water molecules Etc Or multiples thereof

Calculate mass etc in a reaction Examples If you start with 15 g of propane: Which mass of oxygen do you need and which mass of carbon dioxide and water do you get? Write the balanced reaction equation Write the table Fill in known and ?

C3H8 + 5 O2  3 CO2 + 4 H2O m 15g ? ? ? M 44g/mol 32g/mol 44g/mol 18g/mol n Calculate mol of propane (down =divide) Go to next compound on the “mol level”. Check the Ration between the compounds.

C3H8 + 5 O2  3 CO2 + 4 H2O m 15g ? ? ? M 44g/mol 32g/mol 44g/mol 18g/mol n 0.34mol 1:5 1.7mol 5:3 1.02mol 3:4 1.36mol 1. Go upwards = multiply to get the masses

C3H8 + 5 O2  3 CO2 + 4 H2O m 15g 54.5g 44.9g 24.5g M 44g/mol 32g/mol 44g/mol 18g/mol n 0.34mol 1:5 1.7mol 5:3 1.02mol 3:4 1.36mol

More examples Sodium reacts with chlorine and the product is sodium chloride. If you want to make 10 g of sodium chloride how many grams of sodium and chlorine do you need? Write the balanced equation Write the table Fill in known and ?

2 Na + Cl2  2 NaCl m ? ? 10g M 58.5g/mol n Mol of NaCl? Down = divide Go on “mol level” to other compounds. Check Ratio

2 Na + Cl2  2 NaCl m ? ? 10g M 23g/mol 71g/mol 58.5g/mol n 0.17mol 2:1 0.08mol 1:2 0.17mol 1. Up = multiply m 3.9g 6.1g 10g

More examples Ammonia, NH3, can be made from Nitrogen and Hydrogen. What mass of ammonia can you get from 168 g of Nitrogen and 24 g of Hydrogen? Write balanced equation Write table Fill in known and ? Calculate mol

N2 + 3 H2  2 NH3 m 168 g 24 g ? M 28 g/mol 2 g/mol 17 g/mol n 6 mol 12 mol Theoretical ratio between N2 and H2 => 1:3 Actual ratio between N2 and H2 => 1:2 We need 18mol of H2 if all N2 should react : Nitrogen is in excess. Hydrogen is limiting reactant, and decides amount of product that can be formed. m 168 g 24 g 136g M 28 g/mol 2 g/mol 17 g/mol n 12 mol 3:2 8mol

1.4 Theoretical and Percent yield The theoretical yield in the example was 136g. In real life nothing is perfect. Some of the reactants may not react or they react and become something else etc. Lets say that you only end up with 105g product. Percent yield = 105/136 = 77%

A gas have under standard condition the mol volume: 22.4 dm3/mol

Reacting gas volumes For a gas at a constant temperature and pressure, the volume is directly proportional to the number of moles of gas. The mol ration  The volume ratio

If you have 2 dm3 of Hydrogen gas, which volume of Oxygen do you need for complete reaction and which volume of water vapour will you get? (Constant: Pressure and Volume) 2 H2(g) + O2(g)  2 H2O(g) 2dm3 ? ? Mol Ratio 2 : 1 : 2 Volume Ratio => 2 H2(g) + O2(g)  2 H2O(g) 2dm3 1 dm3 2 dm3

Under other conditions The ideal gas equation pV=nRT p = Pressure; Pa V = Volume; m3 OBS! Usually in chemistry dm3 n = Mole; mol T = Temperatur; K OBS! K = 273 + oC R = Gas constant; 8.314 J/mol*K

Example 6.0 g Carbon burns in Oxygen. Give the volume of formed Carbon dioxide. Temperature = 400K Pressure = 1.0 kPa =100 000Pa Balanced Equation Table Fill in known and ? Calculate

C + O2  CO2 m 6. 0g M 12g/mol n 0. 50 mol  1 : 1  0 C + O2  CO2 m 6.0g M 12g/mol n 0.50 mol  1 : 1  0.50 mol Use the gas law: pV = nRT V = nRT/p = 0.50*8.314*400/100000 = 0.017m3 V= 17 dm3

Solute + Solvent  Solution 1.5 Solutions Solute + Solvent  Solution = Substances = the liquid You often have to dissolve chemicals to make them react

The composition of Solutions can be given in many ways, e.g. Mass percent = Mass of substance/Mass of solution Volume percent = volume of solute/ totale Vol   Mol fraction = Xa = na/(na+nb)  Molality = moles of solute/kg of solvent Gram/dm3 (IB sullabus) Molarity = moles of solute/ dm3 of solution (IB sullabus) Concentration in mol/dm3 is often represented by square brackets around the substance under consideration, eg [HCl]

Examples You dissolve 59.5 g of Potassium Bromide in water to 750 cm3. What’s the concentration/molarity? Write compound formula Write the table; add volume and concentration Fill in known and ?

KBr m 59,5 g M 119 g/mol n V 0.75 dm3 c ? OBS! Don’t mix up the V in solutions and the V in gas equation. Gas equations relation to the table is the number of moles; n

KBr Down = divide m 59,5 g M 119 g/mol n 0. 500 mol V 0. 75 dm3 c 0 KBr Down = divide m 59,5 g M 119 g/mol n 0.500 mol V 0.75 dm3 c 0.667 mol/dm3 [KBr] = 0.667 M (Capital M as a unit is often used instead of mol/dm3, don’t mix up with M for Molar mass) Square brackets = concentration of….

More Examples How many grams of NaCl is it in 200 cm3 of a solution with the concentration of 0.50 mol/dm3? Write compound formula Write the table; add volume and concentration Fill in known and ?

NaCl m. Up = multiply M 58. 5 g/mol n V 0. 200 dm3 C 0 NaCl m ? Up = multiply M 58.5 g/mol n V 0.200 dm3 C 0.50 mol/dm3 ___________________________________________________ m 5.9 g n 0.10 mol

More examples The concentration of chloride ions, if you dissolve 10 g AlCl3 in 0.50 dm3 in water? Write equation Write the table Fill in known and ?

AlCl3 + aq  Al3+ (aq) + 3 Cl- (aq) Salts dissociate in water solutions: AlCl3 + aq  Al3+ (aq) + 3 Cl- (aq) m 10g M 133.5 g/mol n  1 : 3  V c ?

AlCl3 + aq  Al3+ + 3 Cl- m 10g OBS! [Cl-] = 0.45 mol/dm3 M 133.5 g/mol n 0.075 mol  1 : 3  0.225 mol V 0.5 dm3 0.50 dm3 c 0.15 mol/dm3 0.45 mol/dm3 OBS! [Cl-] = 0.45 mol/dm3 [AlCl3] = [Al3+] = 0.15 mol/dm3

More Examples You have 0.25 dm3 of NaOH solution with a concentration of 0.50 M. Which volume of a 0.10 M Sulphuric acid do you need to neutralise the NaOH solution? The mass of the formed Sodium sulphate? Write balanced equation Write the table Fill in known and ?

2 NaOH + H2SO4  Na2SO4 + 2 H2O M ? M 142 g/MOL n 2:1 1:1 V 0.25 dm3 ? C 0.50mol/dm3 0.10 mol/dm3

up =multiply down =divide 2 NaOH + H2SO4  Na2SO4 + 2 H2O M 8 up =multiply down =divide 2 NaOH + H2SO4  Na2SO4 + 2 H2O M 8.9g M 142 g/mol n 0.125mol 2:1 0.0625mol 1:1 0.0625mol V 0.25 dm3 0.625 dm3 C 0.50mol/dm3 0.10 mol/dm3

Dilution You have 0.75 dm3 of 0.3 M HCl. You add 0.25 dm3 of water. The new concentration? Write equation Write the table Fill in known and ?

HCl(aq) before  HCl(aq) after m M n 0. 225 mol 1:1 0. 225 mol V 0 HCl(aq) before  HCl(aq) after m M n 0.225 mol 1:1 0.225 mol V 0.75 dm3 1.00 dm3 c 0.3 mol/dm3 0.225 mol/dm3 ( New volume = V = 0.75 + 0.25 = 1.00 dm3)