
How does the law of definite proportions apply to hydrates?
Answer
448.8k+ views
Hint:To apply the law of definite proportions on hydrates we should know what is the law of definite proportions and what hydrates are. The law of definite proportions, as the name indicates is a law that states about the definite ratio of masses of elements in a compound. Hydrates are the water containing salts. We know the masses ratio of elements of water. We arbitrarily choose a hydrated salt and we will check the ratio of masses of elements of water in hydrates.
Complete answer:
The elements combine to form compounds. The compounds have elements in a fixed and constant ratio by mass and the ratio does not depend upon the preparation method of the source of the compounds.
For example, consider water, we can take water from anywhere. The ratio of oxygen and hydrogen in water by mass will always be $16:2$. So, the chemical formula of water will be ${{\text{H}}_{\text{2}}}{\text{O}}$. The ratio of oxygen and hydrogen will not change.
Hydrates are the salts that contain water. Hydrates contain water in a fixed amount. For example, Magnesium sulphate forms heptahydrate. The formula is ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$. The copper sulphate forms pentahydrate. The formula is ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$.
Now, let's consider the copper sulphate pentahydrate, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$. It contains five water molecules.
On heating water removes and we get copper (II) sulphate anhydrous.
${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\,\mathop \to \limits^\Delta {\text{CuS}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{5}}{{\text{H}}_{\text{2}}}{\text{O}}$
We get five water molecules from the heating of copper sulphate pentahydrate. The ratio of oxygen and hydrogen in each water molecule by mass is $16:2$.
The molar mass of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ is $159.6$ g/mol.
The molar mass of water is g/mol.
So, the molar mass of five water molecules is,
$18 \times 5\, = \,90$
So, the molar mass of five water molecules is $90$g.
The ratio of copper sulphate ${\text{CuS}}{{\text{O}}_{\text{4}}}$ and seven water molecules is,
$\dfrac{{{\text{CuS}}{{\text{O}}_{\text{4}}}}}{{{\text{7}}{{\text{H}}_{\text{2}}}{\text{O}}}}\, = \,\dfrac{{ \approx 2}}{1}$
So, in copper sulphate pentahydrate the ratio of copper sulphate and water is also in definite ratio almost$1:2$.
Therefore, the law of definite proportions tells about the definite ratio of salt and water in hydrates.
Note:Hydrated salt contains salt and water. The chemical formula of hydrated salt formula contains a formula of salt followed by the dot after that and a coefficient showing the number of water molecules followed by the chemical formula of water. The Law of definite proportion is used for the stoichiometric calculation. If we are calculating the ratio of oxygen and hydrogen in any reaction, if we get oxygen and hydrogen in $16:2$ by mass we can say the compound will be water. If the ratio comes to $32:2$ we can ensure the compound is not water. The compound will have the formula HO so, it will be hydrogen peroxide ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ .
Complete answer:
The elements combine to form compounds. The compounds have elements in a fixed and constant ratio by mass and the ratio does not depend upon the preparation method of the source of the compounds.
For example, consider water, we can take water from anywhere. The ratio of oxygen and hydrogen in water by mass will always be $16:2$. So, the chemical formula of water will be ${{\text{H}}_{\text{2}}}{\text{O}}$. The ratio of oxygen and hydrogen will not change.
Hydrates are the salts that contain water. Hydrates contain water in a fixed amount. For example, Magnesium sulphate forms heptahydrate. The formula is ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$. The copper sulphate forms pentahydrate. The formula is ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$.
Now, let's consider the copper sulphate pentahydrate, ${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}$. It contains five water molecules.
On heating water removes and we get copper (II) sulphate anhydrous.
${\text{CuS}}{{\text{O}}_{\text{4}}}{\text{.5}}{{\text{H}}_{\text{2}}}{\text{O}}\,\mathop \to \limits^\Delta {\text{CuS}}{{\text{O}}_{\text{4}}}\,{\text{ + }}\,{\text{5}}{{\text{H}}_{\text{2}}}{\text{O}}$
We get five water molecules from the heating of copper sulphate pentahydrate. The ratio of oxygen and hydrogen in each water molecule by mass is $16:2$.
The molar mass of ${\text{CuS}}{{\text{O}}_{\text{4}}}$ is $159.6$ g/mol.
The molar mass of water is g/mol.
So, the molar mass of five water molecules is,
$18 \times 5\, = \,90$
So, the molar mass of five water molecules is $90$g.
The ratio of copper sulphate ${\text{CuS}}{{\text{O}}_{\text{4}}}$ and seven water molecules is,
$\dfrac{{{\text{CuS}}{{\text{O}}_{\text{4}}}}}{{{\text{7}}{{\text{H}}_{\text{2}}}{\text{O}}}}\, = \,\dfrac{{ \approx 2}}{1}$
So, in copper sulphate pentahydrate the ratio of copper sulphate and water is also in definite ratio almost$1:2$.
Therefore, the law of definite proportions tells about the definite ratio of salt and water in hydrates.
Note:Hydrated salt contains salt and water. The chemical formula of hydrated salt formula contains a formula of salt followed by the dot after that and a coefficient showing the number of water molecules followed by the chemical formula of water. The Law of definite proportion is used for the stoichiometric calculation. If we are calculating the ratio of oxygen and hydrogen in any reaction, if we get oxygen and hydrogen in $16:2$ by mass we can say the compound will be water. If the ratio comes to $32:2$ we can ensure the compound is not water. The compound will have the formula HO so, it will be hydrogen peroxide ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ .
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