2.1 Masses of atoms and molecules
Relative Atomic mass, Ar = mean mass of one atom of the element/ mass of one atoms of C12 x 12
Relative Molecular mass, Mr = mean mass of molecule/ mass of one atom of C12 x 12
Atoms of the same element has the same atomic number.
Gaseous elements and elements in Group 7 are diatomic molecules - go around in pairs.
To calculate Mr add the Ar of all the atoms in its formula. Formulae of compunds may have brackets around them - you need to mulitply the Ar of each atom inside the bracket by the number outside the bracket. e.g. Mg(OH)2 = 24.3 = (2x16.0)=(2x1.0) = 58.3
Amount of a substance
Knowing the amount of each substance, unit is the mole, in a chemical reaction lets us know how much reactant is needed and how much useful prodcut is made.
The number of atoms in 12g of C12 is the avagardo constant, 6.022 x 10^23.
Avagardo principle - equal volumes of gases at the same pressure and concentration contain the same number of particles.
The mass of one mole of a substane is its Ar or Mr in g. For atoms: mass = Ar x n, For molecules: mass = Mrxn, where n is the amount of a substance in mol.
You can calculate the mas of ions if relative formula mass and mount of substance known. Electron has very small mass so ca be ingnored when calculating Mr. E,g: formula of sulfate ion in SO4 with 2- charge, Mr of ion = 32.1 = (4x16) = 96.1
2.04 The ideal gas equation
Boye's law: volume of a fixed mass of gas at constant temperature is inversly proportional to its pressure, V = 1/p, where p = pressure, v = volume and = is proportional.
Charle's Law: the volume of a fixed gas a constant pressure is proportional to its absoltue temperture, V= T, where v = volume, T = temp and = is propotional.
These equations put together gives the ideal gas equation: pV = nRT
- P - pressure in pascals, PA
- V is volume in cubic metres, m^3
- n is amount of gas measure in mols, mol
- R is the cas constant, 8.31 J/K/mol
- T is absolute temperture in kelvins, K.
2.05 - Using the ideal gas equation
temperature must be converted to C to K.
Pressure needs to be in Pa no kPa so you must convert it to Pa: 1kPa = 1000Pa, 1Pa =1/1000kpa
Volume is measure in cubic metres,m^3 not cubic decimeters, dm^3 or cubic centimeters, cm^3. 1m^3 = 1/1000 dm^3=100cm^3.
You can rearrange the equation pV=nRT to caclulate the volume, pressure and temperature of a gas.
2.06 Experiments to find Mr
To find the mass and number of moles of a substnace from a smaple you need to know the temp, pressure and volume, then use the ideal gas equation (n= pV/RT).
The Mr of a gas can be found by weighing a sample then measuring its volume, temp and pressure. Also works for volatile liquid - liquid that vapourises easily. The volume of a sample of a gas can be found using a gas syringe (glass syringe with low-friction plunger that moves out smoothly).
To get gases into the syringe: Some gases made from chem reactions or can be released from a pressurized cylinder but always important to find mass of gas released into gas syringe, done by measuring the mass before and after the gas enters. You can calculate the temp using a thermometer and the pressure using a barometer.
Volatile liquids are easily vapourised when heated, so syringe with rubber cap fitting used, put in oven so liquid will vaporuise when inside, second syringe silled iwth liquid which is injected through rubber cap into toher syringe, where vapourises. Gas expands and pushes plunger out - volume measured. Mass measured by weighing 2nd syringe before and fater injection.
2.07 - Molecular and Empirical formulae
Molecular formula - actual number of atoms of each element in 1 molecule of compound. Used in chemical calculations, to write equations & ionic compound - contains ions not compound.
Empirical formula - simplest whole number ratio of atoms of each element in compound e.g. C3H6 -> CH2. Often empirical and molecular formulae are the same. Can be calculate if you have the percentage composition by mass of each element in a compund.
Step 1: Get the mass of each element involved, if it in percentage, assume its percentage is its mass.
Step 2: Divide the mass of the element by its Ar, to give the number of moles of each element.
Step 3: Divide each number of moles by the smallest number
Step 4: Write out the empirical formula.
Molecular formula from empirical formula: need to know the Mr of compund to do this.
Step 1: Get the empirical formula and the Mr
Step 2: calculate the Mr of the empirical formula.
Step 3: Calculate the ratio of the two Mr values and mulitply the empricial formula by ratio.
2.07 - examples of calculating molecular and empir
Empirical formula: An oxide of sulfur conatains 50% by mass of sulfur and 50% by mass of oxygen.
Step 1: Assume mass was 100g, so 50g of Oxygen and 50g of Sulfur
Step 2: Sulfur - 50/ 32.1 = 1.558; Oxygen - 50/16.0 = 3.125
Step 3: Divide masses by 1.558 -> Sulfur - 1.558/1.558 = 1; Oxygen - 3.125/1.558 = 2
Step 4: work out empirical formula -> SO2.
Molecular formulae from empirical formulae: Empirical formula of butane is C2H5 and the mr of butane is 58.
Step 1: Mr of empirical formula; (2x12)+(5x1) = 29.
Step 2: Calculate ratio of Mr values; 58/ 29 = 2
Step 3: Mulitply empricial formula by 2; C4H10
2.08 - Balancing equations
No atoms are lost or made during reactions, just re-arranged. Chemical equations show formulae of different substances in reaction & need to be balaned to show relative amounts.
Reactant - chemical that react together; products - chemical made in reation
Need to ensure there are equal numbers of atoms on each side of the equation - balanced:
Step 1: Chose 1 element in equation & make equal to number of that element on other side.
Step 2: Count each atom to check equation now balanced, adjusting elements that aren't
Step 3: repeat step 2 until equation balanced.
It is acceptable to have fractions in balanced equations which helps if equation difficult to balance. Having a fraction in an equation dosen't mean there is a fraction of a mole, just that that fraction of the mole reacts with the moles of the other molecules.
Sometimes you don't need to count all the atoms e.g. Al + H2SO4 -> Al2(SO4)3 +H2. SO4 appears on both sides, so it is easier to count it as a block, so there are 3 SO4's present on the right so just add three to H2SO4, then use the steps above to balance:
2Al + 3H2SO4 -> Al2(SO4)3 +3H2.
2.09 - Ionic equations
Ionic equations - ions that react togeter ignoring the ones that don't take part in the reaction and are useful to describe neutralization, precipitation and displacement reactions.
Spectator ions - ions that do not take part in a reaction but 'watch' the reaction happening without reacting. E.g. the reaction between hydrochloric acid ad sodium hydrooxide:
HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l). This can be written out as ions, then the specator ions can be cancelled out H+(aq) + Cl-(aq) + Na+(aq) +OH-(aq) -> Na+(aq) + Cl-(aq) + H2O(l) to give: H+(aq) + OH-(aq) -> H2O(l)
Precipitation reactions - useful for identifying ions in solutions. E.g. Silver nitrate solution used to identify halide ions in solution. Coloured silver compounds form in reaction which are insoluble in water so form cloudy percipitate.
2.10 - reacting masses
Law of conservation of mass - mass of products is equal to mass of reactants. Allows to calculate mass of product that could be made in reaction, or reactants needed to produce product, mass calculations.
Step 1: Write a balanced equation
Step 2: Write the Mr values underneath
Step 3: Multiply the big numbers in the equations
Step 4: Convert to grams
Step 5: Divide all masses by largest mass
Step 6: Mulitply by amount of product needed
Using moles to calculate reacting masses:
Step 1: Write balanced equation
Step 2: Underline two substnaces mentioned in question
Step 3: Write down under underlined substances the information known about it
Step 4: You will know two things about 1, use this info to calculate the number of moles
Step 5: Use balanced equation to calculate number of moles of other substance
Step 6: Use this to calculate the mass of this substance.
2.11 - percantage yield
Yield of a chemical reaction is the amount of product formed. Can calculate theoretical yield (max poss amount of product from reactant) using reacting mass calcs. Unlikely to get theoretical yield. Usualkly actual yield < theoretical yield as: reactants may be impure; reaction may not go to completion; some of product my be left in container; may be difficult to purify product.
In lab this may be anoying as have to use more reactants, in industry, wasteful and expensive.
Percantage yield - how close to the theoretical yield actual yield is. higher % yeild, closer to actual yield. % yield = actual yield/theoretical yield x 100.
The Haber process - process that makes ammonia from nitrogen and hydrogen. Ammonia used widely for fertilizers, without haber process, hard to make enough food for everyone. Hard to get high % yield, reaction is reversible and does not go to completion.
Limiting reactant - the reactant that is used up in a reaction which causes another reactant to not be used up, to be in excess. Due to reactants in mixed proportions. Limiting reactant completly used up in reaction si determines theoretical yield.
2.12 - Atom economy
Atom economy - proportion of reactnats that are transfered into useful products. Reactions with a high atom economy are efficient and produce less waste -> important for enviroment and sustainable development. Efficient processes are sustainable.
% atom economy = mass of desired product/ total mass of reactants x 100.
Possible to predict atom economy using Mr of reactants and products:
% atom economy= relative mass of desiered products/ relative mass of reactants x 100. To do this you need to kow the balanced equation for the reaction and Ar or Mr of products.
Sometimes relative masses of reactants are easier to calculate than that of products.
Atom economy can be used to compare different ways of desired products.