Law of constant proportion

Law of constant proportion

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Law of constant proportion or definite composition-

This law was stated by Louis Proust. According to this law,

‘ A pure chemical compound always contains same elements combined together in the same definite proportion by mass.’

Experimental Verification-

Carbon dioxide can be obtained by so many methods like,

(i) By respiration ( as byproduct )

(ii) By burning coal-

C + O₂ ——> CO₂

(iii) By heating lime stone-

CaCO₃ ———> CaO + CO₂

(iv) By heating Sodium bicarbonate-

2NaHCO₃ ——>Na₂CO₃ +H₂O + CO₂

(v) By the reaction of Dilute Hydrochloric acid & Calcium  carbonate-

2CaCO₃ + 2HCl ——–> CaCl₂ + H₂O + CO₂

It has been observed that CO2 obtained by every method contains,

C  :  O

12 : 32

3  :  8 (by weight)

Numerical problems –

problem 1)

(a) 1.08 gram of Cu- wire was allowed to react with HNO₃. The resulting solution was dried and ignited when 1.35 gram of CuO was obtained.

(b) In another experiment 2.3 gram of CuO was heated in presence of H₂, then 1.84 gram of Cu is obtained.

Show that the above data follows law of constant proportion .

Solution-

Cu           +      HNO₃    ——–>     Cu(NO₃)₂      ———–> CuO

1.08 gram                                                                                      1.35 gram

a)  1.35 gram  CuO contains Cu = 1.08 gram

100 gram CuO contains Cu = 1.08 X100/1.35 =80 gram

% of Cu in CuO =80%

% of O in CuO =(100-80) =20%

b) CuO      +      H₂ ——>   Cu         +     H₂O

2.3 gram                          1.84 gram

2.3 gram  CuO contains Cu = 1.84 gram

100 gram CuO contains Cu = 1.84 X100/2.3 =80 gram

% of Cu in CuO =80%

% of O in CuO =(100-80) =20%

% of Cu in both the cases is same .So data follows law of constant proportion.

problem 2-

1.375 gram CuO was reduced by H₂ & 1.098 gram Cu was obtained . In another experiment 1.178 gram of Cu was dissolved in HNO₃ & the resulting Cu – nitrate was converted into Cu by heating . The weight of CuO formed was 1.476 gram.

Show that  these results prove the law of constant proportion or composition.

Solution –

i)  CuO        +       H₂ —–> Cu             +       H₂O

1.375 gram                      1.098 gram

1.375  gram  CuO contains Cu = 1.098 gram

100 gram CuO contains Cu = 1.098 X100/1.375 = 79.8 gram = 80 gram

% of Cu in CuO =80%

ii)   Cu           +      HNO₃    ——–>     Cu(NO₃)₂      ———–> CuO

1.178 gram                                                                                      1.476 gram

1.476 gram  CuO contains Cu = 1.178 gram

100 gram CuO contains Cu = 1.178 X100/1.476 = 79.8 = 80 gram

% of Cu in CuO =80%

% of O in CuO =(100-80) =20%

% of Cu in both the cases is same .So data follows law of constant proportion.

Problem 3-

            Copper sulphate crystals contain 25.45 % Cu & 36.07 % H₂O. If the law of constant proportion is true then calculate the weight of Cu required to obtain 40 gram of crystalline copper sulphate ( CuSO₄.5H₂O)

Solution-

100 gram crystalline copper sulphate ( CuSO₄.5H₂O) contains Cu = 25.45 gram

40 gram crystalline copper sulphate ( CuSO₄.5H₂O) contains Cu = 25.45 X 40 /100=10.18 gram Ans.