Chapter 3 Form 4_lis(07)

CHAPTER 3:

CHEMICAL FORMULAE AND EQUATIONS

Miss Stephenie Nilus

ObjectiveThe relationship between the number of moles and mass

The relationship between the number of moles and number of particles

The relationship between the number of moles and volume of gas

Empirical and molecular formulae

The relationship between the number of moles and…

MASS

Number of moles = Mass ÷

Molar mass

NUMBER OF PARTICLES

Number of moles =

Number of particles ÷

NA

VOLUME OF GAS

Number of moles = Volume of gas ÷ Molar volume

of gas

Review

Definitions

I. Relative atomic mass (Ar)

II. Relative molecular mass (Mr)

Q & A1) Find the relative molecular mass of the

following.a)Br2

b)N2

c)NO2

d)CoCl2.6H20

e)PbSO4

f)CuSO4.5H2Oa)160b)28c)46d)238e)303f)250

2. A compound has the formula NaXO3. The relative molecular mass of the compound is 151. Find the relative atomic mass of X.

Answer: 80

3. Calculate how many more times 3 bromine atoms are heavier then 1 aluminium atom.

Answer: 8.89

The relationship between the number of moles and number of particles

÷ NA

× NA

Nu. of particles Mole

* NA = 6.02 × 1023

1 mole CO2 contains 6x1023 CO2 atoms.So, Nu. 0f particles = 0.25 x 6x1023 = 1.5 x1023 molecules

1 mole Mg2+ contains 6x1023 Magnesium ions.So, Nu. 0f particles = 0.5 x 6x1023 = 3 x1023 ions

1 mole Pb contains 6x1023 lead atoms.So, Nu. 0f particles = 3.5 x 6x1023 = 2.1 x1024 atoms

1. Calculate the number of particles in; (NA=6x1023)a)3.5 moles lead, Pb

b)0.5 mole magnesium(II) ions, Mg2+

c)0.25 mole carbon dioxide, CO2

Q &A

2. Calculate the number of moles of the following substances;

a) 1.5x1024 atoms of silvers, Ag =2.5 molesb) 3x1022 oxide ions, O2- =0.05 molesc) 4.5x1023 nitrogen molecules, N2 = 0.75moles

3. Find the number of atoms in;a) 1.5 moles of Cl2 = 1.8x1024 atoms

b) 0.75 moles of NO2 = 1.35x1024 atoms

Q &A

1. Calculate the number of particles in 8.5g of ammonia, NH3.

2. Calculate the mass of 1.2 × 1022 zinc atoms.

3. Calculate the mass of 3 × 1023 ethanol molecules, C2H5OH.

The relationship between the number of moles and mass

I. Mole– One mole of the atom of any substance has the

same number of particles which is 6.02 × 1023 particles (NA).

• Generally,

Mass (g)÷ Ar/Mr

× Ar/MrMole

÷ NA

× NA

Nu. of particles

Q & A

1. Calculate the mass of the following;a) 0.3 mole atom chlorine, Cl = 10.65gb) 1/8 mole of CO2 = 5.5gc) 3.2 moles ammonia gas, NH3 =54.4g

2. Find the number of moles in a) 31.5g Copper, Cu = 0.49 moleb) 45g ethane gas, C2H6 = 1.5 mole

3. Calculate the number of particles ina) 20g methane, CH4 = 7.5x1023 moleculesb) 372.6g lead, Pb = 1.08x1024 atomsc) 4.68g water, H20 =1.56x1023 molecules

4. Find the mass of the followinga) 2x1024 atoms of oxygen, O = 52.8gb) 1.2x1022 molecules of nitrogen dioxide,

NO2 = 0.92gc) 1.56x1023 atoms of silver, Ag = 28.08g

The relationship between the number of moles and volume of gas

Mass (g)

× Ar/Mr

÷ Ar/Mr

÷ NA

× NAMoleNu. of

particles

Volume of gas

× 22.4dm3

or 24dm3

÷ 22.4dm3 or 24dm3

s.t.p = 22.4dm3

r.t.p = 24dm3

Q & A

1. Calculate the volume occupied by the following gases at s.t.p

a) 3.5 moles of ammonia gas, NH3 = 78.4dm3

b) 2.8g carbon monoxide, CO =2.24 dm3

c) 1.5x1023 molecules of oxygen, O2 = 5.6dm3

2. Calculate the volume occupied by 1.6g of sulphur dioxide at room conditions. = 0.6dm3

3. Calculate the number of moles of 720cm3 of ammonia, NH3 at room temperature.

= 0.03mole

4. Calculate the number of molecules in 0.48dm3 of oxygen at s.t.p =1.26 molecules

5. Calculate the mass of 1.2dm3 of methane, CH4 at s.t.p = 0.857g

Empirical and molecular formulae

• Empirical formula – the simplest ratio of the atoms that combine to form a compound.

• Simplest ratio of a compound is also the simplest mole ratio of the elements.

• Ex : H2O (Water)

• Molecular formula – The actual numbers of the atoms that combine to form a compound.

• Ex: Empirical formula for ethane compound (CH3)n = 30 n = 2

Molecular formula = C2H6

• Ex: Find the empirical formula of given information

• Solution

Element M O

Mass (gram) 2.4 1.6

Relative atomic mass 48 16

Element M OMass 2.4g 1.6gNumber of moles

2.4 ÷ 48 = 0.05mole

1.6 ÷ 16=0.1 mole

Simplest ratio 0.05 ÷ 0.05=1 mole

0.1 ÷ 0.05=2mole

Thus, empirical formula of compound is MO2

1. 2.5g of X combined with 4g of Y to form compound with the formula XY2. If the relative atomic mass of Y is 80, determine the relative atomic mass of X.

2. Boric acid is used to prepare prawns and fish. Chemical analysis of the compound shows that it contain 4.8%hydrogen, 7.7% boron and the rest is oxygen. Determine the empirical formula of boric acid. ( RAM :H=1, B=11, O=16)

3. A gaseous hydrocarbon X contains 85.7% of carbon. Determine the empirical formula of X. (RAM: H=1, C=12)

Q &A

4. A metal X combines with 4.32g oxygen to form 13.68g of a metallic oxide of X. Find the empirical formula of the metallic oxide. [RAM of X=52]

5. A compound CxHyOz contains 40% carbon and 53.3% oxygen. If the relative molecular mass of the compound is 180, find its;a) empirical formulab)molecular formula

6. A metallic oxide, XO2 contains 63.22% of the metal X by mass. Calculate the relative atomic mass of the metal X.

7. A metallic oxide, M2O3 is formed when 5.4g of the metal M combines with m gram of oxygen. Find the value of m. [RAM: O=16, M=27]

ConclusionThe relationship of the number of moles

Mass (g)

× Ar/Mr

÷ Ar/Mr

÷ NA

× NAMoleNu. of

particles

× 22.4dm3

or 24dm3

÷ 22.4dm3 or 24dm3

Volume of gas s.t.p = 22.4dm3

r.t.p = 24dm3

* NA = 6.02 × 1023

ConclusionEmpirical and molecular formulae

Step 1 •Write the mass or percentage of each element

Step 2 •Calculate the number of moles

Step 3 •Divide each number by smallest number to obtain simplest ratio

Step 4 •Write the empirical formula

Thank you