Quantitative aspects of chemical change

MOLE is the SI-unit for the

quantity of a substance

There are 6,02 x 1023

atoms in 12g Carbon-12

Avogadro’s constant (NA)

One mol of a substance is the amount of particles that are equal to the amount of atoms in 12g Carbon-12.

Calculating amount of mol:

n = 𝑚

𝑀mole

Mass (g)

Molar mass (g.mole-1)

Calculating amount of mol:

n = 𝑁

𝑁𝐴mole

Amount of particles

Avogadro’s constant6,02 x 1023 mole-1

One mole of any gas always has the same volume of 22,4dm3 at STP. (0oC and 101,3kPa)

n = 𝑉

𝑉𝑚mole

Volume of gas (dm3)

Molar gas volume22,4 dm3.mole-1

c = 𝑛

𝑉Concentrationmole.dm-3

mole

Volume of solution(dm3)

Concentration

n = 𝑚

𝑀

Conclusion

c = 𝑛

𝑉

Empirical formula is the simplest whole number ratio of atoms in a compound.

Empirical formula from percentages

A Scientist finds that a compound consists of 53,3% oxygen, 40% carbon and 6,6% hydrogen. Determine the empirical formula of this compound.

Start by assuming that you have 100g of the compound. Then 53,3g of the compound will be oxygen, 40g will be carbon and 6,6g will be hydrogen.

Now calculate the amount of mole of each element in the compound:

O: n = 𝑚

𝑀= 53,3

16= 3,33 mole

C: n = 𝑚

𝑀= 40

12= 3,33 mole

H: n = 𝑚

𝑀= 6,6

1= 6,66 mole

C : H : O3,33 : 6,66: 3,33

1 : 2 : 1

Empirical formula: CH2O

Empirical formula from percentages

A Scientist finds that a compound consists of 53,3% oxygen, 40% carbon and 6,6% hydrogen. Determine the empirical formula of this compound.

Now if they ask for the molecular formula if the molecular mass is 90 g.mole-1:

M(CH2O) = 30 g.mole-1:

Molecular formula:C3H6O3

Crystal water is the water that is bonded stoichiometrically in a crystal e.g. CuSO4·5H2O. (Copper sulphate pentahydrate)

A 3,34g monster of a hydrate with the formula SrS2O3.xH2O, contains 2,3g SrS2O3. Find the value of x.

M(SrS2O3 ) = (88)+(2 x 32)+(3x16)= 200 g.mole-1

For SrS2O3: n = 𝑚

𝑀= 2,3

200= 0,0115 mole

For H2O: n = 𝑚

𝑀= 3,34−2,3

18= 0,058 mole

SrS2O3 : H2O0,0115 : 0,058

1 : 5∴ x = 5.

Calculate the percentage of copper in:13.1 CuCl2

M(CuCl2) = (63,5)+(2 x 35,5)= 134,5 g.mole-1

% Cu = 63,5

134,5×

100

1= 47,21%

A standard solution is a solution with a known concentration.

15 Sweets cost R30

5 Sweets cost5

15× 30 = R10

Do you remember?

N2 + 3H2 2NH3

Mole ratios and balanced equations

1 mole N2 reacts with 3 mole H2 to produce 2 mole NH3

2 mole N2 reacts with 6 mole H2 to produce 4 mole NH3

0,5 mole N2 reacts with 1,5 mole H2 to produce 1 mole NH3

2KClO3 (s) 2 KCl (s) + 3 O2(g)

Calculate the mass of oxygen released when 29,4 g potassium chlorate is heated. The balanced equations is: 2KClO3 (s) 2 KCl (s) + 3 O2(g)

2 mole KClO3 produces 3 mol O2

n = 𝑚

𝑀=

29,4

122,5= 0,24 𝑚𝑜𝑙

M KClO3 = 122,5 𝑔.𝑚𝑜𝑙𝑒-1 M O2 = 32 𝑔.𝑚𝑜𝑙𝑒-1

0,24 mole KClO3 produces 0,36 mol O2

n = 𝑚

𝑀

0,36 = 𝑚

32

m = 11,52 𝑔

% Yield

% yield = 𝑟𝑒𝑎𝑙 𝑦𝑖𝑒𝑙𝑑

𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑×

100

1

The mass of oxygen released when 29,4 g potassium chlorate is heated is 11g. Calculate the percentage yield if the balanced equation is: 2KClO3 (s) 2 KCl (s) + 3 O2(g)

We did calculate that 11,52 𝑔 O2 is supposed to be produced

% yield = 11

11,52×

100

1= 95,49%

2SO2(g) + O2(g) 2SO3(g)

If 160g SO2 reacts with 32g O2, calculate the mass of SO3 that is produced

M(SO2) = 64 g.mol-1

n =𝑚

𝑀

=160

64

= 2.5 mole

Limiting reagents

M(SO3) = 80 g.mol-1

n =𝑚

𝑀

2 =𝑚

80

m = 160 g SO3

2.5 mole SO2 will react with 1,25 mole O2, but only 1 mole O2 is available. Thus 2 mole SO2 will react with 1 mole O2 to produce 2 mole SO3

M(O2) = 32 g.mol-1

n =𝑚

𝑀

=32

32

= 1 mole