A 2.500 gram sample of hydrate of calcium sulfate loses 0.532 grams of water when heated. Determine the mass percent of water in the hydrate and the formula of the hydrate.
It lost 0.532 g H2O out of a 2.500 g sample.
%H2O = (mass water lost/mass sample) x 100 = approximately 22 but you need to go through it to do it more exacting than that.
Now take 100 g sample. That will give you
22% H2O (or the value you recalculate)
78% CaSO4 (found by 100%-% water)
22/molar mass H2O = moles H2O
78/molar mass CaSO4 = moles CaSO4.
Now find the ratio of the two numbers in small whole numbers. The easy way to do this is to divide the small number by itself give 1.000 for that. Then divide the other number by the same small number. Round to whole numbers and you have the formula written this way.
CaSO4.xH2O where x is what you find.
20.92 %
To find the mass percent of water in the hydrate, we need to calculate the ratio of the mass of water lost to the initial mass of the hydrate, and then multiply by 100%.
1. Calculate the mass of the water lost:
Mass of water lost = 0.532 grams
2. Calculate the initial mass of the hydrate:
Initial mass of hydrate = 2.500 grams
3. Calculate the mass percent of water in the hydrate:
Mass percent of water = (Mass of water lost / Initial mass of hydrate) * 100%
= (0.532 grams / 2.500 grams) * 100%
= 21.28%
Therefore, the mass percent of water in the hydrate is 21.28%.
Now, let's determine the formula of the hydrate.
1. Convert the masses of water and anhydrous salt into moles:
Moles of water lost = Mass of water lost / Molar mass of water
= 0.532 grams / 18.01528 g/mol
= 0.0296 mol
Moles of anhydrous salt = Initial mass of hydrate / Molar mass of anhydrous salt
= 2.500 grams / (40.078 g/mol + 32.066 g/mol + (4 x 16.00 g/mol))
= 0.0296 mol
2. Find the mole ratio of water to anhydrous salt:
Moles of water / Moles of anhydrous salt = 1 / 1
3. Determine the formula of the hydrate:
The formula of the hydrate is CaSO4 . H2O,
where the dot (.) indicates that water of hydration is present.
Therefore, the formula of the hydrate is CaSO4 . H2O.
To determine the mass percent of water in the hydrate, we need to calculate the amount of water in the sample and divide it by the initial mass of the sample.
The initial mass of the sample is 2.500 grams. The mass of water lost is 0.532 grams.
Therefore, the mass of water in the hydrate is 0.532 grams.
To calculate the mass percent, we can use the formula:
Mass percent = (Mass of water / Initial mass) x 100
Mass percent = (0.532 g / 2.500 g) x 100 = 21.28%
The mass percent of water in the hydrate is 21.28%.
Now, to determine the formula of the hydrate, we need to find the ratio of the moles of water to moles of anhydrous compound in the sample.
The molar mass of water (H2O) is 18.015 g/mol. The molar mass of calcium sulfate (CaSO4) is 136.144 g/mol.
First, let's calculate the moles of water:
Moles of water = mass of water / molar mass of water
Moles of water = 0.532 g / 18.015 g/mol = 0.0295 mol
Next, let's calculate the moles of anhydrous compound (CaSO4):
Moles of anhydrous compound = (initial mass - mass of water) / molar mass of CaSO4
Moles of anhydrous compound = (2.500 g - 0.532 g) / 136.144 g/mol = 0.0160 mol
Now, we need to find the ratio of moles of water to moles of anhydrous compound:
Ratio = Moles of water / Moles of anhydrous compound
Ratio = 0.0295 mol / 0.0160 mol = 1.84
The ratio of moles of water to moles of anhydrous compound is 1.84.
To determine the formula of the hydrate, we need to determine the simplest whole number ratio between the moles of water and the moles of anhydrous compound. The simplest ratio can be approximated as follows:
Multiply the ratio by 10 to eliminate decimal places: 1.84 x 10 = 18.4
Round the ratio to the nearest whole number: 18
The formula of the hydrate is CaSO4·18H2O.