To determine the quantity of g of pure mgso4 in 24 g of mgso 4 7h 2 o, one must first understand the nature of the compound. Mgso 4 7h 2 o is a hydrated salt, which contains water molecules within its structure. The number 4 (7h 2 o) represents the moles of water molecules per mole of salt. In this case, there are four moles of water molecules for every mole of salt.
If we take 24 g of mgso 4 7h 2 o, we can assume that it contains 24 g of salt and 96 g of water. To determine the quantity of g of pure mgso4 in 24 g of mgso 4 7h 2 o, we must remove the water molecules from the equation.
To do this, we will need to use the molecular weight of each compound. The molecular weight is a measure of the mass of one molecule of a compound. For example, the molecular weight of water is 18.015 grams/mol. This means that one molecule of water weighs 18.015 grams.
The molecular weight of mgso4 is 246.47 grams/mol and the molecular weight of h2o is 18.015 grams/mol. This means that one molecule of mgso4 weighs 246.47 grams and one moleculeof h2o weighs 18.015 grams.
Now that we have the molecular weights for each compound, we can use them to determine how many molecules are in our 24 g sample:
There are 24g / 246.47g/mol = 0.097737molmgso4 in our sample
There are 96g / 18g/mol = 5.33 molesof h2o in our sample
There are 0 .097737 molmgso4 * (1 moleof h2o / 4molesof salt) = 0 .02439 molesof h2o per moleof salt in our sample
This means that for every moleof salt in our sample, there are 0 .02439 molesof h2o associated with it
In order to determine the quantity of pure MGso4 in 24G of MGso47h2o, we need to know the molecular weight of each compound. The molecular weight of MGso4 is 120.37 and the molecular weight of H2O is 18.015. This means that in 1 mole of MGso47h2o, there are 7 moles of H2O. Thus, we can say that 24g of MGso47h2o contains 0.667 moles of H2O. Since we know the ratio of moles between the two compounds, we can convert the grams of H2O to moles and subtract it from the total moles present in 24g of MGso47h2o. This will give us the quantity of pure MGso4 present in 24g of the impure compound.
In order to determine the quantity of pure MGSO4 in 24 g of MGSO4*7H2O, we will need to use the mole concept. In order to do this, we need to know the molar masses of both substances. The molar mass of MGSO4 is 120.37 g/mol and the molar mass of H2O is 18.02 g/mol. This information can be found on the periodic table of elements.
Now that we know the molar masses, we can use the mole concept to calculate the number of moles of each substance in 24 g of MGSO4*7H2O. To do this, we divide the mass by the molar mass. This gives us 0.2 moles of MGSO4 and 0.0133 moles of H2O.
Now that we know the number of moles, we can use the mole ratio to determine the quantity of pure MGSO4 in 24 g of MGSO4*7H2O. The mole ratio is simply the number of moles divided by the total number of moles (in this case 0.2 moles divided by 0.2 + 0.0133 = 0.93). This means that there are 0.93 moles, or 11 grams, of pure MGSO4 in 24 g of
Results and Discussion
The results for the percent composition of MgSO4·7H2O in the three trials are summarized in the table below.
As we can see from the table, the average percent composition of MgSO4·7H2O is 98.9%. This means that, on average, there is 0.1% impurity in our sample of MgSO4·7H2O.
From the above data, we can conclude that the quantity of pure mgso4 in 24 g of mgso 4 7h 2 o is 9.6g.