© boardworks ltd 2003 quantitative chemistry. © boardworks ltd 2003 the formula of molecular...

© Boardworks Ltd 2003

QuantitativeChemistry


The formula of molecular compounds

Water

Carbon dioxide

MethaneFormulaName

C HH

HH

CO O

H

HO

• Molecular compounds have formulae that show the type and number of atoms that they are made up from.

CH4

CO2

H2O


The formula of ionic compounds• Ionic compounds are giant

structures.• There can be any number of

ions in an ionic crystal - butbut always a definite ratio of ions.

Name Ratio FormulaSodium chloride 1:1

Magnesium chloride 1:2

Aluminium chloride 1:3

Aluminium Oxide 2:3

+ -+-

+-

-+ +

+-+

-

- --+

++ -

+-+-

-+ +

Sodium chloride

A 1:1 ratio

NaCl

AlCl3

Al2O3

MgCl2


Ions with groups of atoms• Some ions are single atoms with a charge.

• Other ions consist of groups of atoms that remain intact throughout most chemical reactions.

• E.g. Nitrate and sulphate ions commonly occur in many chemical reactions.

Chloride ClCl--

nitride NN3-3-

Sulphide SS2-2-

Cl-

N3-

S2-

nitrate

NONO33--

Sulphate

SOSO442-2-

NO O-

O

S

O

O-O-

O


Use of brackets in formulae• Ions like nitrate and sulphate remain unchanged

throughout many reactions. • Because of this we tend to think of the sulphate ion as a

“group” rather than a “collection of individual” sulphur and oxygen atoms.

• This affects how we write formulae containing them. Aluminium sulphate contains two Al ions and three sulphate ions.

• We write it as AlAl22(SO(SO44))3 3 Not AlAl22SS33OO1212

• Similar rules apply to ions such as nitrate NO3-,

hydroxide OH-, etc.


Use the information to write out the formula for the compound.

1) Calcium bromide (One calcium ion, two bromide ions)

2) Ethane(Two carbon atoms, six hydrogen atoms)

3) Sodium oxide(Two sodium ions, one oxygen ion)

4) Magnesium hydroxide(One magnesium ion, two hydroxide ions)

5) Calcium nitrate(One calcium ion, two nitrate ions)

CaBr2

C2H6

Na2O

Mg(OH)2

Ca(NO3)2

Activity


The Masses of chemicals


Atomic Mass of elements• The atoms of each element have a different mass.• Carbon is given a relative atomic mass (RAM) of 12Carbon is given a relative atomic mass (RAM) of 12..• The RAM of other atoms compares them with carbon.• Eg. Hydrogen has a mass of only one twelfth that of carbon

and so has a RAM of 1.• Below are the RAMs of some other elements.

Element Symbol Times as heavy as carbon R.A.M

Helium He one thirdBeryllium Be three quarters

Molybdenum Mo EightKrypton Kr SevenOxygen O One and one thirdSilver Ag Nine

Calcium Ca Three and one third

4

12

96

84

16

108

40


Emperical Formula• When a new compound is discovered we have to

deduce its formula.• This always involves getting data about the

masses of elements that are combined together.• What we have to do is work back from this data to

calculate the number of atoms of each element and then calculate the ratio.

• In order to do this we divide the mass of each atom by its atomic mass.

• The calculation is best done in 5 stages:


Formula from Composition by mass.


• A chloride of silicon was found to have the following % composition by mass: Silicon 16.5%: Chlorine 83.5%

(Atomic. Mass Si=28: Cl=35.5)

SubstanceSubstance Silicon ChlorideSilicon Chloride

1. Elements SiSi Cl

2. Mass of each element (g per 100g)

3. Mass / Atomic Mass

4. Ratio

5. Formula

16.516.5 83.583.5

16.5/28 =0.5916.5/28 =0.59 83.5/35.5 =2.3583.5/35.5 =2.35

ClCl÷Si = (2.35 ÷ 0.59) = (3.98) ÷Si = (2.35 ÷ 0.59) = (3.98) Ratio of Ratio of ClCl:Si =4:1:Si =4:1

SiSiClCl44

Divide biggest by smallest

Activity


• We found 3.2g of copper reacted with 0.8g of oxygen. What is the formula of the oxide of copper that was formed? (At. Mass Cu=64: O=16)

SubstanceSubstance Copper oxideCopper oxide

1. Elements CuCu OO

2. Mass of each element (g)


4. Ratio

5. Formula

3.23.2 0.80.8

3.2/64 =0.053.2/64 =0.05 0.8/16 =0.050.8/16 =0.05

1:11:1

CuOCuO


• We found 5.5g of manganese reacted with 3.2g of oxygen. What is the formula of the oxide of manganese formed? (Atomic. Mass Mn=55: O=16)

SubstanceSubstance Manganese oxideManganese oxide

1. Elements MnMn OO

2. Mass of each element (g)


4. Ratio

5. Formula

5.55.5 3.23.2

5.5/55 =0.105.5/55 =0.10 3.2/16 =0.203.2/16 =0.20

1:21:2

MnOMnO22


• Calculate the formula of the compounds formed when the following masses of elements react completely:

(Atomic. Mass Si=28: Cl=35.5)

Element 1Element 1 Element 2Element 2 Atomic MassesAtomic Masses FormulaFormula

Fe = 5.6g Cl=106.5g Fe=56 Cl=35.5

K = 0.78g Br=1.6g K=39: Br=80

P=1.55g Cl=8.8g P=31: Cl=35.5

C=0.6g H=0.2g C=12: H=1

Mg=4.8g O=3.2g Mg=24: O=16

FeClFeCl33

KBrKBr

PClPCl55

CHCH44

MgOMgO

Activity


EMPIRICAL FORMULAEMPIRICAL FORMULA• The empirical formulaempirical formula represents the smallest

ratio of atoms present in a compound.

• The molecular formulamolecular formula gives the total number of atoms of each element present in one molecule of a compound.

The empirical formula is the simplest formula The empirical formula is the simplest formula

and the molecular formula is the and the molecular formula is the ““truetrue”” formula.formula.


Empirical Formulas

The empirical formula for a compound is the formula with the smallest whole-number mole ratio of the elements.

– The empirical formula might or might not be the same as the actual molecular formula!

– The empirical formula = molecular formula for IONIC COMPOUNDS - ALWAYS!


Molecular Formulas

• If the empirical formula is different from the molecular formula, the molecular formula will always be a simple multiple of the empirical formula.– EX: The empirical formula for hydrogen peroxide

is HO; the molecular formula is H2O2.– In both formulas, the ratio of oxygen to hydrogen

is 1:1


Calculating Molecular Formulas

• Step 1 – Calculate the empirical formula (if needed)

• Step 2 – GIVEN molecular mass (experimental)/empirical formula molar mass = multiplier

• Step 3 – Multiply the empirical formula subscripts by the multiplier found in Step 2


EXAMPLE PROBLEM

• Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.


Step One: Find Empirical Formula

• 40.68 % C = 40.68 g C x 1mol C = 3.390 mol C 12.0 g C• 5.08 % H = 5.08 g H x 1 mol H = 5.03 mol H 1.01 g H• 54.24% O = 54.24 g O x 1 mol O = 3.390 mol O 16.0 g OC: H:O3.390 mol: 5.03 mol: 3.390 mol3.390 mol 3.390 mol 3.390 mol1 : 1.48: 1 1 : 1.5 : 1, multiply by 2 C2H3O2


Step Two: Divide Molar Masses

• Molar mass empirical formula = (2 x 12.0 g/mol) + (3 x 1.01 g/mol) + (2 x 16.0 g/mol) = 59.0 g/mol

• Given molar mass = 118.1 g/mol

• Multiplier = 118.1 g/mol = 2 59.0 g/mol


Step Three: Use Multiplier

• Empirical Formula = C2H3O2

• x 2 from step two

• Molecular formula = C4H6O4


Percentage Composition• It is sometimes useful to know how much of a compound

is made up of some particular element. • This is called the percentage composition by mass.

% Z = (Number of atoms of Z) x (atomic Mass of Z)Formula Mass of the compound

0

20

40

60

80

%

Carbon OxygenE.g. % of oxygen in carbon dioxide (Atomic Masses: C=12. O=16)

Formula = Number oxygen atoms

= Atomic Mass of O = 16 Formula Mass CO2 =

% oxygen =

CO2

2

12 +(2x16)=442 x 16 / 44 = 72.7%


Formula Atoms of O

Mass of O

Formula Mass

%age Oxygen

MgO 1

K2O 1

NaOH 1

SO2 2

• Calculate the percentage of oxygen in the compounds shown below

32+(2x16)=64

32

23+16+1=4016

(2x39)+16 =94

16

24+16=4016 16x100/40=40%

16x100/94=17%

16x100/40=40%

32x100/64=50%

% Z = (Number of atoms of Z) x (atomic Mass of Z)Formula Mass of the compound

Activity


• Nitrogen is a vital ingredient of fertiliser that is needed for healthy leaf growth.

• But which of the two fertilisers ammonium nitrate or urea contains most nitrogen?

• To answer this we need to calculate what percentage of nitrogen is in each compound

Activity


Formula Atoms of N

Mass of N

Formula Mass %age Nitrogen

NH4NO3 2 28

CON2H4 2 28

• Formulae: Ammonium Nitrate NH4NO3: Urea CON2H4

28x100 /80 = 35%

28x100 /60 = 46.7%

14+(1x4)+14+(3x16)=80

12+16+(2x14+(4x1)=

60

And so, in terms of % nitrogen urea is a better fertiliser than ammonium nitrate

01020304050

1st Qtr

Amm.Nitrate UreaAtomic masses H=1: C=12: N=14: O=16

Activity


What is the formula of a compound containing 1.4g nitrogen and 3.2g of oxygen? (N=14 O=16)

1. N2O

2. NO3. NO2

4. N2O3


What is the formula of a compound containing 6.5g zinc and 1.6g oxygen?(Zn=65 O=16)

1. ZnO2. Zn2O3

3. ZnO2

4. Zn2O


What is the formula of a compound formed between Cr3+ ions and O2- ions?

1. CrO2. Cr2O3

3. CrO2

4. Cr3O2


What is the percentage nitrogen in ammonium sulphate (NH4)2SO4?

1. 21%2. 42%3. 63%4. 84%

© boardworks ltd 2003 quantitative chemistry. © boardworks ltd 2003 the formula of molecular...

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