Department of Engineering

Lyceum of the Philippines University

Cavite Campus

Matter: A Quick Review

Dimensional Analysis and Measurement

Modern Periodic Table and Electronic Structure of Atoms

Mole: Its Concepts and ApplicationsChemical Language: Reactions and

EquationsSolutions and Concentration Units

Gas Laws

Matter

Pure Substance

Element

monoatomic

polyatomic (molecule)

Compound ionic

molecule

Impure Substance (Mixture)

Homogeneous (i.e. solutions)

Heterogeneous (i.e. colloids and

suspensions)

approach used in problem solving

provides a systematic way of solving numerical problems in science and other disciplines as well as checking numerical solutions for possible error/s

units must be carried through all

calculations

correct use of

conversion factor/s to change one unit

to another

Given: Qty1 Find: Qty2

Solution:

factor/s

Conversion ×

quantity

Given = Unkown

( )1

2

12Qty

Qty × Qty = Qty

22Qty = Qty

Given: Qty1 Find: Qty2

Solution:

factor/s

Conversion ×

quantity

Given = Unkown

( )2

1

12Qty

Qty × Qty = Qty

( )

2

2

1

2Qty

Qty ≠ Qty

A. Base Quantities and Units

Quantity SI Metric English

length m

m, km, cm,

mm, m,

nm

mi, ft, yd, in

mass kgkg, Tg, mg,

cg, g, ngslug, oz

time ss, ms, hr,

min

s, hr, min,

day, wk,

mo, yr

A. Base Quantities and Units

Quantity SI Metric English

temperatur

eK C F

amount of

substancemol - -

electric

currentA - -

luminosity cd - -

B. Some Derived Quantities and Units

Quantity SI Metric English

area m2 m2, km2,

cm2, mm2

mi2, ft2, yd2,

in2

volume m3 km3, cm3,

mm3; mL

mi3, ft3, yd3,

in3; gal; qt

speed m/sm/min;

km/hr; cm/s

ft/s, mi/hr,

mi/min, in/s

density kg/m3

Mg/m3;

g/cm3;

g/mL

oz/qt;

oz/cm3

C. Some Conversion Factors

Please refer to QEE Review

2012 – Support Material, re:

Table 2

An aluminum foil is found to be 8.0 10-5

cm thick. What is it thickness in

micrometers?

m 10

μm 1 ×

cm 1

m 10 × cm 10 × 8.0 = thickness

6-

-2

5-

μm

μm 0.80or μm 10 × 8.0 = thickness-1

μm

A gas at 25 C exactly fills a container previously

to have a volume of 1.05 103 cm3. The container

plus the gas are weighed and found to have a mass

of 837.6 g. The container when emptied of all gas,

has a mass of 836.2 g. What is the density of the

gas at 25 C?

volume

mass = density

gas

gas

gas

( )

cm 10 × 1.05

g 836.2 - g 837.6 = density

33gas cm

g10 × 1.3 = density

3

3-

gas

most significant tool in ORGANIZING and SYSTEMATIC REMEM-BERING of chemical facts

list the important characteristics of all elements in increasing atomic no.

Isotopic Notation

13Al26.98

Atomic No. = # of p = # of e (neutral atom)

Mass No. = p + n

Isotopic

Notation

Atomic

No.

Mass

No.p n e

Electronic

structure

13Al2713 27 13 14 13

[10Ne] 3s2,

3p1

40Ar1818 40 18 22 18

[10Ne] 3s2,

3p6

13Al+3 13 27 13 14 10[2He] 2s2,

2p6

the mass, in grams, per mole of a substance (atom, ion, molecule)

how to find: just add the atomic weights of that substance (compound or molecules)

thus, in g/mol

MM= amu (monoatomic element)MM= FW (ions)MM= MW (molecules)

Molar Mass

atomic mass

unit/atomic weight

formula

mass/weight

Molecular

mass/weight

1 mol Fe weighs 55.85

g

MMFe = 55.85 g/mol

1 mol Fe2O3

weighs 159.70 g

MMFe2O3 =

159.70 g/mol

1 mol O2 weighs

32.00 g

MMO2 = 32.00

g/mol

1 mol O weighs 16.00 g

MMO = 16.00 g/mol

1 mol NaCl weighs

58.44 g

MMNaCl = 58.44

g/mol

1 mol H2O

weighs 18.02 g

MMwater =

18.02 g/mol

interconversion of mass of a substance, in grams, and the no. of particles of that substance (atoms, ions, or molecules)

the no. of moles of substance is central to stoichiometry

grams molesunit

particlesuse MM use

6.022 x 1023

Calculate the no. of moles of glucose,

C6H12O6 (MM = 180.0 g/mol), in 5.380 g of

this substance.

6126

6126

6126glucoseOHC g 180.0

OHC mol 1 × OHC g .3805 = mol

6126glucoseOHC mol .029890 = mol

How many copper atoms, Cu (MM = 63.5

g/mol), are there in a traditional copper penny

weighing 3 g. Assume the penny to be 100%

copper.

Cu mol 1

atoms Cu 10 × 6.022 ×

Cu g 63.5

Cu mol 1 × Cu g 3 = atoms Cu

23

atoms Cu 10 × 3 = atoms Cu22

Consider the combustion of butane, C4H10, the

fuel in disposable cigarette lighters.

(l)22(g)2(g)10(l)4OH + CO → O + HC

(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →

2 mol

C4H10

13 mol

O2

8 mol

CO2

10 mol

H2O

2

2

104

2

1042CO mol 1

CO g 01.44 ×

HC mol 2

CO 8 × HC mol .341 = CO g

CO g 342 = CO g22

Consider the combustion of butane, C4H10, the fuel

in disposable cigarette lighters.

(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →

Determine the mass in grams of CO2 (MM = 44.01

g/mol) formed when 1.34 mol of C4H10 (MM = 58.14

g/mol) reacts.

104

104

2

104

2

2

2104HC mol 1

HC g 8.145 ×

O mol 13

HC mol 2 ×

CO g 32.00

O mol 1 × O g .006 = HC g

HC g .681 = HC g104104

Consider the combustion of butane, C4H10, the fuel

in disposable cigarette lighters.

(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →

Determine the mass in grams of C4H10 (MM = 58.14

g/mol) required to react with 6.00 g of O2 (MM = 32.00

g/mol).

SOLUTE >> the substance to be dissolved

SOLVENT >> the dissolving medium

Concentration units:

solution of V

mol = M Molarity, 1.

L in

solute

solvent of mass

mol = m Molality, 2.

kg in

solute

100 x solution of mass total

mass = percent mass 3.

solute

solute

mol/L 0.0444or M .04440 =Mglucose

A solution used for intravenous feeding contains 4.80 g

of glucose, C6H12O6 (MM = 180.16 g/mol) in 600.0 mL

of solution. What is the molar concentration of glucose?

solution of V

mol = M

L in

glucose

glucose

nsol'mL 1000

nsol'L 1× nsol'mL 0.006

OHC g 180.16

OHC mol 1×OHC g 4.80

= M6126

6126

6126

glucose

mol/kg 0.296or m .2960 =mglucose

A solution used for intravenous feeding contains 4.80 g

of glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g of

water. What is the molal concentration of glucose?

water of m

mol = m

kg in

glucose

glucose

water g 1000

water kg 1×water g 0.90

OHC g 180.16

OHC mol 1×OHC g 4.80

= m6126

6126

6126

glucose

mol/kg 0.296or m .2960 =mglucose

A solution used for intravenous feeding contains 4.80 g

of glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g of

water. What is the molal concentration of glucose?

water of m

mol = m

kg in

glucose

glucose

water g 1000

water kg 1×water g 0.90

OHC g 180.16

OHC mol 1×OHC g 4.80

= m6126

6126

6126

glucose

4%1 =mass)(by %NaCl

In a solution prepared by dissolving 24 g of NaCl in

152 g of water, determine the mass percent NaCl.

100 x solution of mass

mass = )/%(

NaCl

NaClg

g

100 x g) 152 + g (24

g 24 = )/%(

NaClg

g

A 1.13 molar solution of aqueous KOH (MM = 56.11

g/mol) has a density of 1.05 g/mL. calculate its

molality or molal concentration.

water of m

mol = m

kg in

KOH

KOH

] KOH mol 1

KOH g 56.11 × KOH mol 13.1[ - ]

nsol'mL 1

nsol' g 1.05 ×

nsol'L 1

nsol'mL 1000 × nsol'L 1[

KOH mol 13.1 = m

KOH

mol/kg 1.14or m .141 =mKOH