
A hydrate of magnesium iodide has a formula ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$. A $1.055{\text{g}}$ sample is heated to a constant weight of $0.695{\text{g}}$. What is the value of ${\text{x}}$?
Answer
586.2k+ views
Hint:Hydrate is a compound that contains water with a definite mass in the form of ${{\text{H}}_2}{\text{O}}$. Hydrates are often in the form of a crystal that can be heated and the water can be burnt off by turning it into steam. By measuring the compound before heating and after, the amount of water in the original hydrate can be determined and the formula discovered.
Complete step by step answer:
We use the word hydrate in our everyday lives. Heating of hydrates causes them to lose their crystalline structure. The substance that is left after the hydrate has lost its water. This is called anhydrate. Unknown hydrates are written with their base form; then an ${\text{x}}$ is placed before ${{\text{H}}_2}{\text{O}}$. ${\text{x}}$ is an unknown number which indicates the number of moles of water.
When ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ molecule is heated, some water molecules is removed.
The steps to find the value of ${\text{x}}$ are:
Determine the mass of the water that has left the compound.
This allows us to determine the mass of water that was in the hydrate and the mass of anhydrate. This is done by subtracting the mass of anhydrate from the mass of hydrate. This equals the mass of water.
Mass of ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ $ = 1.055{\text{g}}$
Mass of sample after heating $ = 0.695{\text{g}}$
Mass of water removed$ = $ Mass of ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ $ - $ Mass of sample after heating
Mass of water removed $ = 1.055{\text{g}} - 0.695{\text{g = 0}}{\text{.360g}}$
Convert the mass of water into moles
To convert, we divide the mass of water by molar mass of water to get moles of water.
Mass of water, ${{\text{m}}_{\text{w}}} = 0.360{\text{g}}$
Molar mass of water ${{\text{M}}_{\text{w}}} = 18{\text{g}}$
Number of moles of water ${{\text{n}}_{\text{w}}} = \dfrac{{{{\text{m}}_{\text{w}}}}}{{{{\text{M}}_{\text{w}}}}}$$ = \dfrac{{0.360{\text{g}}}}{{18{\text{gmo}}{{\text{l}}^{ - 1}}}}$$ = 0.02{\text{mol}}$
Hence the value of ${\text{x}}$ is $0.02{\text{mol}}$.
Note:
Molecular formula can be determined by the following steps:
-If given hydrate and anhydrate, subtract to get water.
-Convert mass of anhydrate and water to moles of each.
-Find the ratio of water to anhydrate in moles.
-Write the formula; ionic compound followed by ratio of water.
-Name the compound.
Complete step by step answer:
We use the word hydrate in our everyday lives. Heating of hydrates causes them to lose their crystalline structure. The substance that is left after the hydrate has lost its water. This is called anhydrate. Unknown hydrates are written with their base form; then an ${\text{x}}$ is placed before ${{\text{H}}_2}{\text{O}}$. ${\text{x}}$ is an unknown number which indicates the number of moles of water.
When ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ molecule is heated, some water molecules is removed.
The steps to find the value of ${\text{x}}$ are:
Determine the mass of the water that has left the compound.
This allows us to determine the mass of water that was in the hydrate and the mass of anhydrate. This is done by subtracting the mass of anhydrate from the mass of hydrate. This equals the mass of water.
Mass of ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ $ = 1.055{\text{g}}$
Mass of sample after heating $ = 0.695{\text{g}}$
Mass of water removed$ = $ Mass of ${\text{Mg}}{{\text{I}}_2}.{\text{x}}{{\text{H}}_2}{\text{O}}$ $ - $ Mass of sample after heating
Mass of water removed $ = 1.055{\text{g}} - 0.695{\text{g = 0}}{\text{.360g}}$
Convert the mass of water into moles
To convert, we divide the mass of water by molar mass of water to get moles of water.
Mass of water, ${{\text{m}}_{\text{w}}} = 0.360{\text{g}}$
Molar mass of water ${{\text{M}}_{\text{w}}} = 18{\text{g}}$
Number of moles of water ${{\text{n}}_{\text{w}}} = \dfrac{{{{\text{m}}_{\text{w}}}}}{{{{\text{M}}_{\text{w}}}}}$$ = \dfrac{{0.360{\text{g}}}}{{18{\text{gmo}}{{\text{l}}^{ - 1}}}}$$ = 0.02{\text{mol}}$
Hence the value of ${\text{x}}$ is $0.02{\text{mol}}$.
Note:
Molecular formula can be determined by the following steps:
-If given hydrate and anhydrate, subtract to get water.
-Convert mass of anhydrate and water to moles of each.
-Find the ratio of water to anhydrate in moles.
-Write the formula; ionic compound followed by ratio of water.
-Name the compound.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

