
How do I find the percentage composition of chlorine in $$CoC{l_2}$$?
Answer
556.5k+ views
Hint: The percentage composition for an element in a compound is the ratio of the amount of that element to the amount of the individual or each element which is present in the compound then multiplying it by 100. It helps in the chemical analysis of the compound. We need to measure the quantities in terms of grams of the elements here.
Complete step-by-step answer:
To find the percentage composition of the chlorine firstly we need to know the molecular mass of the compound given that is the cobalt chloride $$\dfrac{{70.906}}{{129.839}} \times 100 = 54.611\% $$ . So the molar mass of cobalt is 58.933195$$ gmo{l^{ - 1}}$$ and for chlorine is 35.453$$ gmo{l^{ - 1}}$$.
Now calculating the molar mass of cobalt chloride:
$$gmo{l^{ - 1}}$$= $$(1 \times 58.933195) + (2 \times 35.453) = 129.839gmo{l^{ - 1}}$$
Now we need to calculate the mass of the chlorine which is present in the compound of the cobalt chloride that is $$CoC{l_2}$$. The atoms of chlorine in cobalt chloride is 2. So now look at the following calculation for chlorine:
Chlorine= $$(2 \times 35.453) = 70.906gmo{l^{ - 1}}$$
The formula which is to be used for calculating the percentage composition is the following:
Percentage composition=$$\dfrac{{\text{mass of the element}}}{{\text{total molar mass of compound}}} \times 100$$
For the given compound the formula will be modified as the following:
Percentage composition= $$\dfrac{{\text{mass of chlorine}}}{{\text{total molar mass of}CoC{l_2}}} \times 100$$
Now substituting the values in the above formula:
Percentage combination= $$\dfrac{{70.906}}{{129.839}} \times 100 = 54.611\% $$
So the percentage composition of chlorine in cobalt chloride is 54.611%.
Note: This is an example of a colligative property- boiling point elevation, and it is not exclusive to water. It occurs any time when you add a non-volatile solute such as salt to a solvent such as water as addition of salt to pure water makes it impure hence resulting in an increase in boiling point of the solvent.
Complete step-by-step answer:
To find the percentage composition of the chlorine firstly we need to know the molecular mass of the compound given that is the cobalt chloride $$\dfrac{{70.906}}{{129.839}} \times 100 = 54.611\% $$ . So the molar mass of cobalt is 58.933195$$ gmo{l^{ - 1}}$$ and for chlorine is 35.453$$ gmo{l^{ - 1}}$$.
Now calculating the molar mass of cobalt chloride:
$$gmo{l^{ - 1}}$$= $$(1 \times 58.933195) + (2 \times 35.453) = 129.839gmo{l^{ - 1}}$$
Now we need to calculate the mass of the chlorine which is present in the compound of the cobalt chloride that is $$CoC{l_2}$$. The atoms of chlorine in cobalt chloride is 2. So now look at the following calculation for chlorine:
Chlorine= $$(2 \times 35.453) = 70.906gmo{l^{ - 1}}$$
The formula which is to be used for calculating the percentage composition is the following:
Percentage composition=$$\dfrac{{\text{mass of the element}}}{{\text{total molar mass of compound}}} \times 100$$
For the given compound the formula will be modified as the following:
Percentage composition= $$\dfrac{{\text{mass of chlorine}}}{{\text{total molar mass of}CoC{l_2}}} \times 100$$
Now substituting the values in the above formula:
Percentage combination= $$\dfrac{{70.906}}{{129.839}} \times 100 = 54.611\% $$
So the percentage composition of chlorine in cobalt chloride is 54.611%.
Note: This is an example of a colligative property- boiling point elevation, and it is not exclusive to water. It occurs any time when you add a non-volatile solute such as salt to a solvent such as water as addition of salt to pure water makes it impure hence resulting in an increase in boiling point of the solvent.
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