Stoichiometry and hydration of Mg nitrate

In summary, the problem involves determining the percentage of magnesium nitrate dihydrate in a mixture of magnesium nitrate and its hydrate. After heating the mixture, the mass decreased and the percentage of the hydrate can be calculated using the mass difference and molar masses. However, a mistake was made in converting the moles of water to moles of the hydrate, leading to incorrect results.
  • #1
Mitchtwitchita
190
0
Hey Guys, I'm having a little trouble with this problem:

A mixture of Mg(NO3)2 and its hydrate Mg(NO3)2*2H2O has a mass of 1.7242 g. After heating to drive off all the water, the mass is only 1.5447 g. What is the percentage of Mg(NO3)2*H2O in the original mixture? Report your answer to 4 significant figures.

Molar Masses: Mg(NO3)2=148.31 g/mol; Mg(NO3)2*2H2O = 184.34 g/mol.

So, 1.7242g - 1.5447g = 0.1795g

0.1795 g H2O x 1mol H2O/18.02g H2O = 0.009961mol H2O

0.009961 mol H2O x 1 mol Mg(NO3)2 / 2 mol H20 = 0.004981 mol Mg(NO3)2

0.004981 mol Mg(NO3)2 x 148.31g Mg(NO3)2/1 mol Mg(NO3)2 = 0.07387 g Mg(NO3)2

Therefore, 0.7387 g Mg(NO3)2 / 1.7242 g x 100% = 42.84%

Thus, % mass of Mg(NO3)2*2H20 = 57.16%

However, 0.004981 mol Mg(NO3)2*2H2O x 184.34g Mg(NO3)2*2H2O / 1 mol Mg(NO3)2*2H2O = 0.9181g Mg(NO3)2*2H2O

0.9181g Mg(NO3)2*(2)H2O / 1.7242g x 100% = 53.25%

Can anybody tell me where I'm going wrong?
 
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  • #2
Mitchtwitchita said:
Hey Guys, I'm having a little trouble with this problem:

A mixture of Mg(NO3)2 and its hydrate Mg(NO3)2*2H2O has a mass of 1.7242 g. After heating to drive off all the water, the mass is only 1.5447 g. What is the percentage of Mg(NO3)2*H2O in the original mixture? Report your answer to 4 significant figures.

Molar Masses: Mg(NO3)2=148.31 g/mol; Mg(NO3)2*2H2O = 184.34 g/mol.

So, 1.7242g - 1.5447g = 0.1795g

0.1795 g H2O x 1mol H2O/18.02g H2O = 0.009961mol H2O

0.009961 mol H2O x 1 mol Mg(NO3)2 / 2 mol H20 = 0.004981 mol Mg(NO3)2

0.004981 mol Mg(NO3)2 x 148.31g Mg(NO3)2/1 mol Mg(NO3)2 = 0.07387 g Mg(NO3)2

Therefore, 0.7387 g Mg(NO3)2 / 1.7242 g x 100% = 42.84%

Thus, % mass of Mg(NO3)2*2H20 = 57.16%

However, 0.004981 mol Mg(NO3)2*2H2O x 184.34g Mg(NO3)2*2H2O / 1 mol Mg(NO3)2*2H2O = 0.9181g Mg(NO3)2*2H2O

0.9181g Mg(NO3)2*(2)H2O / 1.7242g x 100% = 53.25%

Can anybody tell me where I'm going wrong?


Very close. Here is where you got off track...
0.009961 mol H2O x 1 mol Mg(NO3)2 / 2 mol H20 = 0.004981 mol Mg(NO3)2
This is actually the number of moles of the dihydrate of magnesium nitrate. Continue from here and you will get the correct answer.
 
  • #3


Hi there,

It seems like you have made a small error in your calculations. The correct percentage of Mg(NO3)2*2H2O in the original mixture is actually 53.25%, as you have calculated in your last step. This is because you have correctly determined that 0.004981 mol of Mg(NO3)2*2H2O is present in the original mixture, but then you multiplied it by the molar mass of Mg(NO3)2 instead of the molar mass of Mg(NO3)2*2H2O.

So the correct calculation would be: 0.004981 mol Mg(NO3)2*2H2O x 184.34g Mg(NO3)2*2H2O / 1 mol Mg(NO3)2*2H2O = 0.9181g Mg(NO3)2*2H2O

Then, 0.9181g Mg(NO3)2*2H2O / 1.7242g x 100% = 53.25%

Overall, your approach to the problem and use of stoichiometry is correct. Just be careful to use the correct molar mass for each compound in your calculations. Keep up the good work!
 

1. What is stoichiometry and why is it important in studying Mg nitrate hydration?

Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. In studying the hydration of Mg nitrate, stoichiometry is important because it helps us understand the exact amount of reactants and products involved in the reaction, as well as the ratio in which they combine.

2. What is the chemical formula for Mg nitrate and how does it relate to its hydration?

The chemical formula for Mg nitrate is Mg(NO3)2. This means that for every one magnesium ion (Mg2+) there are two nitrate ions (NO3-) present. In the process of hydration, water molecules surround and bond to these ions, resulting in hydrated Mg nitrate with the formula Mg(NO3)2•xH2O.

3. How does the hydration of Mg nitrate affect its physical properties?

The hydration of Mg nitrate increases its overall mass and changes its physical state from a solid to a hydrated solid or a solution. It also changes its melting and boiling points, as well as its solubility in water.

4. How can we determine the degree of hydration for Mg nitrate?

The degree of hydration for Mg nitrate can be determined by performing a gravimetric analysis, where the mass of the hydrated and anhydrous forms of the compound are compared. The difference in mass can be used to calculate the number of water molecules present in the hydrated form.

5. Why is the hydration of Mg nitrate an important factor in its use as a fertilizer?

Mg nitrate is commonly used as a fertilizer because it provides both magnesium and nitrogen, which are essential nutrients for plant growth. The hydration of Mg nitrate ensures that these nutrients are readily available for plants to absorb through their roots, making it an effective fertilizer for promoting healthy plant growth.

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