topic 1: quantitative chemistry ib chemistry sl. states of matter (p. 2-4) determined by the kinetic...

TOPIC 1: QUANTITATIVE CHEMISTRY

IB Chemistry SL

STATES OF MATTER (P. 2-4)

Determined by the kinetic energy of the particles

Kinetic Molecular theory states…All matter consists of particles in motionAs the temperature increases, the movement of the particles increase

STATES OF MATTERThree common states of matter existSolid –particles closely packed in fixed positions; forces between particles strong enough to restrict movement; have fixed shapeLiquid –particles closely packed but not in fixed position; forces between particles weak enough to allow movement but fix the volume; no fixed shapeGas –forces between particles negligible; particles free to move around in space provided; no fixed shape or volume

KINETIC MOLECULAR THEORY When temperature increases enough for the particles to have sufficient energy to overcome the intermolecular forces, a change of state occurs

IN YOUR

NOTEBOOK:

Draw a diagram of

the molecules for

using arrows to

represent

movement for

SOLID, LIQUID, and

GAS.

THINK ABOUT IT (THEN SHARE!) 1. Explain why the temperature of a boiling liquid does not increase despite energy being constantly applied.

2. Deduce which would be more painful, scalding your skin with water vapour or boiling water.

3. Explain why you might feel cold and shiver when you get out of the water at the beach on a very hot, windy day.

Matter: any substance that

occupies space and has mass

Mixture: a combination of two or more pure

substances that retain their individual properties

Pure substance: has a definite and constant

composition

Homogeneous solution:

has both uniform

composition and

properties throughout,

eg salt water, metal alloys

Heterogeneous solution:

has non-uniform

composition and varying

properties, eg salad dressing, paint, garden

soil

Element: made up of atoms that each have the same atomic

number, eg Pb, Hg, Br

Compound: made up of a

combination of atoms or ions in a

fixed ratio and having different properties from the constituent

elements, eg H2O, CO2, NaCl

p. 4-6

THE SEPARATION OF MIXTURES PRE-LAB

Materials:

Hot plates

Electronic balance

Test tubes

Beakers

Conical flasks

Filter Funnel

Filter paper

Distillation apparatus

Bunsen burners

Ring Stands

•This experiment will be conducted in groups of two.

•Please turn in one lab per group.

•We will meet in the Chemistry Lab tomorrow!

Complete the pre-lab portion of the investigation.

What else would you like? – Different apparatus? Alternative soluble and insoluble solids?

SugarSaltCoffee

ShellsSand

Pepper

CHEMICAL EQUATIONS (P. 7-10)

Shows the changes during a reaction as reactants are rearranged into productsProvides the ratio of reactants (what starts the reaction) and products (what is made)Provides information on the state of the substances: (s), (g), (l), (aq)The ratio, in the form of coefficients, illustrates the conservation of matter–matter is neither created nor destroyed, simply rearranged

Generally-•Balance what occurs only once (or smallest amount) on each side first.•Balance H and O last.•Worry about charges when we get to the chapter on Redox.

TRY IT!

Zinc metal reacts with hydrochloric acid to form the salt zinc chloride. Hydrogen gas is evolved.

Hydrogen gas and oxygen react together to form water.

At a high temperature, calcium carbonate decomposes into calcium oxide and carbon dioxide.

TYPES OF REACTIONS

Combination (synthesis): combination of two or more substances to produce a single productEx: C(s) + O2(g) CO2(g)

Decomposition: a single reactant being broken down into two or more productsEx: CaCO3(s) CaO(s) + CO2(g)

TYPES OF REACTIONS

Single replacement: one element replaces another in a compoundEx: Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

Double replacement: occurs between ions in solution to form insoluble substancesand weak or non-electrolytes (AKA metathesis)Ex: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Combustion: a reaction between a hydrocarbon and oxygenEx: 2C4H10 + 13 O2 8CO2 + 10H2O

HOMEWORK

Read and take notes on the section titled “The Atom Economy” on page 11 in your textbook.

Complete the activity at the end of the section (a-c) in your notebook. (I will check tomorrow when you come in.)

CHEMICAL EQUATIONS Write the reaction equation for the combustion of methane.

Balance the reaction.

Nitrogen (N2) and oxygen (O2) react in the cylinders of car engines to form nitrogen monoxide. Give a balanced equation for this reaction.

CHEMICAL EQUATIONS After nitrogen monoxide escapes into the environment, it reacts with oxygen to produce nitrogen dioxide. Give a balanced equation for this reaction.

Nitrogen dioxide can further react with oxygen and water to produce nitric acid, one of the ingredients in acid rain. Give a balanced equation for this reaction.

MEASUREMENT AND UNITS - SYSTEME INTERNATIONAL (SI)

Mass – kg1000 g = 1 kg1000 mg = 1 g

Time – s60 sec = 1 min60 min = 1 hr

Temperature – KBased on thermal motionAbsolute zero (all thermal motion ceases)

K = ˚C + 273

Volume –m3(cubic metre)1 L = 1 dm3= 1000 mL

Pressure –Pa (pascal)1 atmosphere = 1.01 x 105Pa = 101 kPa

1.1.1 APPLY THE MOLE CONCEPT TO SUBSTANCES

AMOUNTS OF SUBSTANCES

All matter composed of varying types of substancesAtoms – building blocks of matter; extremely small; exist as varieties in universe (elements)

Compounds – substances composed of atoms joined together in whole-number ratios

Because of size, atoms and compounds cannot be counted directlyCounted based on mass and using mole concept

1.1.1 APPLY THE MOLE CONCEPT TO SUBSTANCES

THE MOLE•Latin for heap; defined as number of carbon atoms in a 12.01 g sample

•1 mole = average atomic mass

• 1 mole = 6.02 x 1023 particles (Avogadro’s number)

•The mole is just a way to define a quantity

•Similar to the word ‘dozen’1.1.1 APPLY THE MOLE CONCEPT TO

SUBSTANCES

THE MOLE 1 mole of an element is equal to relative atomic mass (Ar) of that element –this mass found on the periodic tableMolar Mass = Expressed in units g mol-1

1 mole of a compound/molecule is equal to the relative mass (Mr) of that compound/moleculeMr= sum of all Ar of the elements making up compound

Subscripts in compound indicate amount of atoms

Ar/Mr and mass measured allow for determination of number of moles present

MOLAR MASSMolar mass is equal to mass (g) of 1 mol of the substance

Determine molar mass by summing masses of all component atoms

Find the molar mass of ammonia.Find the molar mass of water.What is the mass of 1 mole of methane?What is the mass of 1 mole of calcium carbonate?

1.1.2 DETERMINE THE NUMBER OF PARTICLES AND THE AMOUNT OF SUBSTANCE (IN MOLES)

MOLE CONVERSIONS

Number of moles (n) = mass (m)/ molar mass (M)

Number of particles (N) = number of moles(n) x Avogadros’s constant(L)

•Calculate the number of moles in 4.00 g of NaOH

•What is the mass of 2.50 moles of water?

•How many molecules are in 2.50 moles of water?

CLARIFICATI0N OF UNITS

Relative molecular mass (Mr) of CO2 = 44

Molar Mass of CO2 = 44 g/mol

Molecular Mass of CO2 = 44 a.m.u

PRACTICE 1. What is the mass of 2 moles of Lithium?

How many atoms are in 2 moles of Li atoms?

2. What is the mass of 4.5 moles of CaCO3?

MORE PRACTICE

1. 6g of hydrogen reacts with fluorine to produce hydrogen fluoride. What mass of HF fluoride is produced?

2. 0.1 moles of nitrogen reacts with hydrogen to make ammonia. What mass of ammonia is produced?

EMPIRICAL AND MOLECULAR FORMULAS•Empirical formula –lowest whole-number ratio; does not necessarily show the actual number of atoms involved

•Molecular formula – shows the number of each atom in a molecule

CALCULATING EMPIRICAL FORMULASIf experimental data is given1. Determine mass of elements involved (if percent composition data is given, assume a 100g sample)2. Convert mass to moles ( moles = mass/molar mass)3. Divide all molar amounts by the smallest molar amount; the numbers found are the amount of each atom in the empirical formula

If necessary force the ratio into whole numbers (multiply everything by 2, 3, etc.)

EMPIRICAL FORMULAS PRACTICE

Smog is common in cities throughout the world. One component of smog is PAN (peroxyacylnitrate) which consists of 20.2 % C, 11.4 % N, 65.9 % O, and 2.50 % H by mass. Determine the empirical formula of PAN, showing your work.

MOLECULAR FORMULA The actual number of atoms in the molecule

1. Determine the empirical formula

2. Divide the Mr of the molecule (given) by the Mr of the empirical; the number is the factor that all the subscripts of the empirical must be multiplied by

MOLECULAR FORMULA PRACTICE

Upon analysis, a sample of an acid with a molar mass of 194.13 g mol-1 was found to contain 0.25 g of hydrogen, 8.0 g of sulfur, and 16.0 g of oxygen. Determine the empirical and molecular formula.

Which is both an empirical and a molecular formula?

A. C5H12

B. C5H10

C. C4H8

D. C4H10

(Total 1 mark)

MASS RELATIONSHIPS IN CHEMICAL REACTIONS (STOICHIOMETRY) (P. 20-22)

Balanced chemical equations allow for the determination of theoretical yields of products and required amounts of reactants needed

Convert known masses to moles (if needed)

Use the molar ratio (from the balanced chemical equation) to convert to the unknown

The combustion of hydrocarbon fuels is an environmental concern as it add to the carbon dioxide levels in the atmosphere. Calculate the mass of CO2 produced when 100 g of propane is burned according to the equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

LIMITING REAGENTS If amounts of all reactants are given, determine the limiting reagent (produces the lowest amount of product) and the reagent in excess

Calculate theoretical yield for one product starting from each reactant; limiting reagent will produce the lowest amount.

If 14 g of each reactant is present, which is in excess? How many moles of product is made?

N2 + 3H2 2NH3

ASSIGNMENT (DUE TODAY) In the textbook:

Complete the Quick Questions (#1-3) on page 22.

THEORETICAL AND EXPERIMENTAL YIELD (P. 23-24)

Comparison of the experimental yield to the theoretical yield

Why is the actual yield of a reaction almost always smaller than the theoretical yield?

Calcium carbonate decomposes on heating as shown below.

CaCO3 CaO + CO2

When 50g of calcium carbonate are decomposed, 7g of calcium oxide are formed. What is the percentage yield of calcium oxide?

Sodium thiosulfate may be produced by boiling solid sulfur and an excess of Na2SO3, determine the theoretical yield of the product. If only 50.0 g of product is collected, determine the percent yield. Your starting amount of S8 is 15.50 g.

S8(s) + 8Na2SO3(aq) 8Na2S2O3(aq)

GASEOUS VOLUMES (P. 24-30)

Molar ratio from a balanced chemical equation provides information on volume of gases that are used/produced; coupling this with the fact that 1 mole of any gas is always the same volume at a given temperature pressure allows for easy calculations when gases are reacting.

STANDARD TEMPERATURE AND PRESSURE•Standard conditions of temperature and pressure (STP) are 273 K (0°C) and 101.3 kPa (1 atm) pressure.

•The molar gas volume is 22.4 dm3 (L).

•Room Temperature and Pressure (RTP) = 298 K (25°C) and 1 atm.

•Molar volume at RTP is 24 dm3.

MOLAR VOLUME Moles(n) = Volume (V) / molar volume (Vmolar)

1. Calculate the moles of chlorine in 44.8 cm3 of the gas at STP

2. Calculate the volume occupied by 4.40 g of carbon dioxide at STP.

3. What volume of hydrogen gas is produced when 0.056 g of lithium reacts with water? Assume the volume is measured at STP.

IDEAL GAS LAW Gas volumes are not only affected by the number of moles present but also by pressure and temperature

This leads to the development of the Ideal Gas Law – how gases behave in ideal situations

PV = nRT

Gases only deviate from this law at high pressures and low temperature/volumes due to increasing IMFs

P = pressure (kPa)V = volume (dm3 or L)n = moles (mol)R = 8.31J K-1 mol -1

T = temperature (K) - MUST BE IN KELVIN!!! (if given data in °C, convert by adding 273)

NOTES - BOOK PAGES 24-30

Kinetic theory of gases

Definitions:

Ideal gas

Real gas

STP

Molar volume

Formulas and graphical relationships (if possible):

Boyle’s law

Charles’ law

Gay-Lussac’s law

Combined gas law

Ideal gas law

Equation

Units for variables

Gas constant

Worked example

SOLVE THESE AT THE END OF YOUR NOTES(SHOW YOUR WORK)

1. Calculate the molar mass of a gas if it is found that 15.0g of the gas occupies 3.00 x 104 mL at 581°C and 50.0 kPa. The actual molar mass is found to be different by 10%, account for this.

2. Gas in a 2.0 dm3 box is at a pressure of 100kPa and 200 K temperature. If the gas is heated to 300K, what is the new pressure?

3. A gas has it’s pressure doubled and volume tripled. What is the effect on it’s temperature (in K)?

SOLVE PROBLEMS USING THE IDEAL GAS EQUATION PV=NRT

Calculate the molar mass of a gas if it is found that 15.0g of the gas occupies 3.00 x 104 mL at 581°C and 50.0 kPa. The actual molar mass is found to be different by 10%, account for this.

PV = nRT ma

ssMolar mass

PV = mass x R x T Molar

mass

Mass x R x T= MmassPV

15.0 x 8.31 x 85450.0 x 30.0L =Mmass

Mmass = 71.1 g/mol (3 SF)

How do you account for the 10% difference?

SOLVE PROBLEMS BETWEEN T,P, AND V FOR A FIXED MASS OF AN IDEAL GAS

Gas in a 2.0 dm3 box is at a pressure of 100kPa and 200 K temperature. If the gas is heated to 300K, what is the new pressure?

P1V1__________ T1

=P2V2__________

T2

SOLUTIONS (P. 31-33)

•Solute – the less abundant component in a solution

•Solvent – the more abundant component in a solution

•Solution – mixtures of two components

SOLVENT

SOLUTESOLUTION

CONCENTRATION OF SOLUTIONS

Concentration is the composition of a solution. As more and more solute dissolves in the solvent, the solution becomes more and more concentrated. When the solvent cannot dissolve any more solute, it is saturated.

Ex: [HCl] = 1 mol dm-3 or 1M

𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑀 )= 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒(𝑚𝑜𝑙)𝑣𝑜𝑙𝑢𝑚𝑒𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛(𝑑𝑚3)

Units: mol/dm3 or mol dm-3

Determine the concentration (in mol dm-3) of the following solutions:

A) 250 cm3 containing 45.0 g of MgSO4

B) 0.25 dm3 containing 0.210 g of NaHCO3

Determine the mass of solute required to make the following solutions:

A) 250.0 mL of 0.1 M AgNO3

B) 0.500 L of 0.40 M Mg(NO3)2