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CHE 183 GENERAL CHEMSTRYProperties of Matter Matter: Anything that has mass and occupies space(volume)

Composition: Parts or components of matter ex. Composition of water H2O, 11.9% H and 88.81% O

Properties: Distinguishing features physical and chemical properties

ATOM

atomelementsmoleculecompounds CONTAINING

8 SUBSTANCEELEMENTS Na, H, K,..COMPOUNDS NaCl, H2OONE TYPE OF ATOMTWO OR MORE ATOMSCHEMICALLY COMBINED2 or more substances mixed togethernot chemically combined

. MIXTURESHeterogeneous Mixture not uniform throughouthave obvious phasesOil+WaterHomogeneous Mixture : uniform throughoutSalt+Water. MIXTURES

11PHYSICAL AND CHEMICAL CHANGESPhysical Change: changes that occur with no changein the chemical composition.Example: Ice melting. (You still have H2O)Chemical Change: different chemical substances are formed.Example: paper burning.

10-6 METER = 1 MKRON

10-2METER =1CENTMETER

10-3 METER = 1 MLMETER

10 +3 METER = 1 KLOMETER

Non SI Units

To convert C to F F= (9/5)xC+32 To convert F to CC= (5/9) x(F-32)Example: An automobile engine coolant has antifreeze protection to a temperature of -22C0.Will this coolant offer protection to temperatures as low as -150F ?Answer:First we must convert to F to C C0= (5/9) x(F-32)C0 = (5/9)x (-15-32)C = - 26 C0 protection to a temperature of -22C0 so -26 < -22C0 No protection at this temperature

or g/cm3Density is the ratio of mass to volume

Shape is rectangular prism

Percent as a Conversion Factor:Example: A seawater sample contains % 3.5g of sodium Chloride(NaCl)This means:

3.5 g NaCl OR 100g seawater 100g seawater 3.5 g NaCl Example:

How many kilograms of ethanol are present in 25 L of a solution that is %90 gasoline -%10 ethanol by mass?The density of the solution is 0.71 g/mlAnswer:The asked : How many kg of ethanol= XThe Given: 25 L of a solution solution: %90 gasoline-%10ethanol density of the solution: 0.71 g/ml?

M= d x V M solution = 0.71 g x (25 L sol.x1000ml ) ml 1 L

X ethanol (kg) = 17.8 kg sol. X 10 eth. = 1,78 kg 100 sol. =1.8kg ethanol M solution = 17.750 g = 17.750 g x 1 kg = 17,8 kg 10000g

ex: 3.63 cm 13.129 cm + 123.1 cm139.859 cm = 139.9 cm55 so

H2 + 1/2 O2 H2O2 g + 16 g = 18gExample:7.12 g sample of magnesium is heated with 1.80 g of bromine .All the bromine is used up and 2.07g of magnesium bromide is the only product.What mass of magnesium remains unreacted??AnswerMass of Before Reaction = Mass of After Reaction7.12g Mg + 1.80g Br = 2,07 MgBr2 + Rest MgRest Mg = 6,85g

Example:A 0,100 g sample of magnesium when combined with oxygen yields 0,166 g magnesiumoxide. We wish to make exactly 2,00 g of magnesium oxide.What masses of magnesium and oxygen must we combine to do this??? Answer 0,100 g Mg = X g Mg 0,166 g MgO 2,00 g MgO

X =1,20g Mg

Magnesum + Oxygen Magnesiumoxide1,20g + Y = 2,00g Y = 0,80 g O2

Example Write a symbol for the species with 47 protons ,61 neutrons ,and 47 electron.

x

Answer: x47+61= 108 4747-47=0

Atomic mass: The average mass of atoms

10 x 0,19 + 11 x 0,81 = 10,811 amu1 amu = atomic mass unit

1 amu = 1 proton = 1 neutron41

1 mole S = S atoms = 32 g

1 mole S S atoms

And 32 g S 1 mol SExample:How many atoms of gold are present in 5.07x10-3 mol Au (gold)?Answer:

atoms5.07x10-3 mol Au1 mole Aux=3,05x1021 Au atomWhat is the mass of 2,35x1024 atoms of Cu (Copper) ?? (1mol Cu = 63.54g) Example:Answer2,35x1024 Cu atom x 1 mole Cu x 63.54g Cu 6,02x1023atom 1mol Cu = 248g CuExample How many lead-206 atoms are present in 8,27x10-3 mol Pb ?

Lead- 206 naturel abundance percent: 24,1 % Answer8,27x10-3 mol Pb x 6,02 x1023 Pb atom = 49,785x 1020 1 mol Pb =4,98x1021 Pb atom

4,98x1021 Pb atom x 24,1 (Lead-206) = 1,2x 1021 100 Pb atom

1 mol K :39,10 g

Example:

Rhenium-187 is a radioactive isotope that can be used to determine the age of meteorites. A 0,100 mg sample of Re contains 2,02x 1017 atoms of 187Re. What is the percent abundance of Rhenium-187 in the sample??

1 mol Re :186,2 g0,100 mg Re x 1 g x 1mol Re =5,37 x 10-7 mol Re 1000 mg 186,2 g

5,37 x 10-7 mol Re x 6.02x 1023 atom = 3,23 x 1017 Re atom 1 mol Re

2,02x 1017 (187Re atom) = x 3,23 x 1017 Re atom 100

X = % 62.5 187Re atom

THE CHEMCAL FORMULA OF A COMPAUND NDCATES

The element present

The relative numbers of atoms of each element in compound

An empirical formula is the simplest formula for a compound; it shows the types of atoms present and their relative numbers

Acetic acidFormaldehyde Glucose ALL HAVE THE SAME EMPIRICAL FORMULA

A molecular formula is based on an actual molecule of a compound. the molecular formula is a multiple of the empirical formula.

Molecular formula of Acetic acidA structural formula shows the order in which atoms are bonded together in a molecule and by what types of bonds

Acetic acid

66Mg: 24,305 Cl:35,453 g66

Au :196,97 g