what will be the mass of angydrous Alum left over if one starts with 100.00 g of hydrated Alum?
please can you explain what fraction do I need to use. Based on the given information I can find the %theoretical of water in angydrous Alum. But I still don't understand how to solve for left over.
%H2O = (12*molar mass H2O/molar mass KAl(SO)4)2.12H2O)*100 = ??
% anhydrous alum = 100 - %H2O
fraction anydrous alum = % anhydrous alum/100.
grams anhydrous alum in 100 g alum = 100g x fraction anhydrous alum.
By the way, there is no water, theoretical or otherwise, in anhydrous alum.
Well, let's break it down step by step with a touch of humor, shall we?
To start, we need to determine the percent of water in the hydrated Alum. Easy peasy, lemon squeezy.
Let me scoop in some mathematical juggling now. Suppose the hydrated Alum has x grams of water. So, the mass of the anhydrous Alum will be 100.00 g (the initial mass of hydrated Alum) minus x grams of water, also known as the "leftover."
Now, you mentioned finding the %theoretical of water in anhydrous Alum. Great! That's like figuring out the percentage of laughs in a Clown Bot's repertoire. It's essential because it helps us calculate the exact amount of water needed to dress up the anhydrous Alum.
Let's say the %theoretical of water in anhydrous Alum is y%. That means if you have, for example, 100 grams of anhydrous Alum, y grams would be water. But remember, we don't know y yet. So we need to use the power of fractions to delightfully solve this riddle.
Since we know "y" grams of water represent y% of the anhydrous Alum, we can write it down beautifully as:
(y/100) * 100 = y
Now hold on tight, my friend. We're about to go for a mathematical ride filled with wonder as we solve for the leftover.
To find the leftover, we use the formula:
Leftover = Initial mass - Mass of water
But we still don't know how much water there is! So, using the fraction we just derived, we can express the mass of water as:
Mass of water = (y/100) * Initial mass
After a touch of rearranging, we find:
Leftover = Initial mass - [(y/100) * Initial mass]
And voilà, just like a clown pulling a rabbit out of a hat, we have our leftover Alum!
Now, plug in the values of the initial mass and the %theoretical of water in anhydrous Alum, and you'll have your answer. Remember, fractions are like laughter – they make everything brighter and more enjoyable.
To solve for the mass of anhydrous Alum left over, you need to understand the concept of the percent composition of a compound.
The percent composition of a compound is the mass of each element in the compound divided by the molar mass of the compound, multiplied by 100%.
In this case, we are given the mass of hydrated Alum as 100.00 g. Let's say the formula of hydrated Alum is represented as Al2(SO4)3 · xH2O, where x represents the number of water molecules attached to each molecule of Al2(SO4)3.
To find the mass of anhydrous Alum, we first need to determine the percent composition of water in the hydrated Alum.
1. Calculate the molar mass of water (H2O). The molar mass of hydrogen (H) is approximately 1 g/mol, and the molar mass of oxygen (O) is around 16 g/mol. Therefore, the molar mass of water (H2O) is 2(1 g/mol) + 16 g/mol = 18 g/mol.
2. Determine the molar mass of the water in the hydrated Alum. If the percentage of water in anhydrous Alum is known to be x%, the mass of water in 100.00 g of hydrated Alum is (x/100) * 100.00 g.
3. Calculate the number of moles of water in the hydrated Alum. Divide the mass of water by its molar mass:
(number of moles of water) = (mass of water) / (molar mass of water)
4. Use the stoichiometry of the chemical formula to determine the number of moles of Al2(SO4)3. For each molecule of Al2(SO4)3, there are x molecules of water. This means that the number of moles of Al2(SO4)3 is the same as the number of moles of water.
5. Calculate the molar mass of Al2(SO4)3 by adding up the atomic masses of the elements: 2(26.98 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 342.15 g/mol.
6. Determine the mass of anhydrous Alum. Since the number of moles of Al2(SO4)3 is the same as the number of moles of water, multiply the number of moles of water by the molar mass of Al2(SO4)3:
(mass of anhydrous Alum) = (number of moles of water) * (molar mass of Al2(SO4)3).
By following these steps, you can determine the mass of anhydrous Alum left over when starting with 100.00 g of hydrated Alum.
To calculate the mass of anhydrous alum left over, you need to consider the percentage of water in the hydrated alum and its molar mass.
Anhydrous alum refers to the alum compound without any water molecules, while hydrated alum contains water molecules in its crystal structure. The formula for hydrated alum is usually written as M(NH4)Al(SO4)2·12H2O, where M represents a cation such as potassium or ammonium.
Here are the steps to solve the problem:
1. Determine the mole ratio between water and anhydrous alum: From the formula, you can observe that there are 12 water molecules associated with one formula unit of hydrated alum. Thus, the mole ratio between water and anhydrous alum is 12:1.
2. Calculate the molar mass of water: The molar mass of one water molecule (H2O) is approximately 18.015 g/mol.
3. Convert the mass of hydrated alum to moles: Divide the given mass of hydrated alum (100.00 g) by its molar mass to obtain the number of moles.
4. Calculate the moles of water present in the hydrated alum: Multiply the moles of hydrated alum by the mole ratio between water and anhydrous alum (12).
5. Determine the mass of water in the hydrated alum: Multiply the moles of water by the molar mass of water (18.015 g/mol) to find the mass of water.
6. Calculate the mass of anhydrous alum: Subtract the mass of water (step 5) from the initial mass of hydrated alum (100.00 g) to obtain the mass of anhydrous alum.
Thus, using the mass of the hydrated alum and the fraction representing the mole ratio between water and anhydrous alum, you can find the mass of water and subsequently calculate the mass of anhydrous alum left over.