what is the percent composition by mass of water in a compound with a gram formula mass of 4905 g/mol
Do you hope to obtain a number out of this?
%H2O = (mass water/formula mass)*100 = ??
Without the mass of water or the formula of the compound, a number is not possible.
it only tells me that and that gypsum contains 2 moles of water for each 1 mole of calcium sulfate from CaSO4*2H20
That's all you need.
g H2O = 2 moles H2O = 2*molarmass H2O.
Then %H2O = (2*molar mass H2O/molar mass CaSO4.2H2O)*100 = ??
This looks simple enough; however, I don't believe CaSO4.2H2O has a molar mass even close to 4905 g/mol.
To determine the percent composition by mass of water in a compound, you first need to know the molar mass of water, which is 18 g/mol (1 g/mol for hydrogen and 16 g/mol for oxygen).
Given that the compound in question has a gram formula mass of 4905 g/mol, we need to find out how many moles of water are present in one mole of the compound.
To do this, we divide the gram formula mass of the compound by the molar mass of water:
4905 g/mol / 18 g/mol = 272.5 mol
This means that there are 272.5 moles of water in one mole of the compound.
Now, to calculate the percent composition by mass of water, we need to divide the mass of water by the total mass of the compound and multiply by 100:
Mass of water = molar mass of water x number of moles of water
= 18 g/mol x 272.5 mol
= 4905 g
Percent composition by mass of water = (mass of water / gram formula mass of the compound) x 100
= (4905 g / 4905 g/mol) x 100
= 100%
So, the percent composition by mass of water in the compound with a gram formula mass of 4905 g/mol is 100%.