Stepanovite is mineral that is a naturally-occurring example of a class of compo

Question

Stepanovite is mineral that is a naturally-occurring example of a class of compounds called "metal organic frameworks (MOFs)". These have potential applications for the storage of hydrogen gas as fuel, removal of CO_2 from industrial emissions, the separation of gases (purifying O_2directly from the air), and the fabrication of photovoltaic cells. Natural stepanovite occurs as a hydrate, like CuSO_4 middot 5 H_2O or Kal(SO_4)_2 middot 12 H_2O The elemental analysis of the anhydrous portion of stepanovite is as follows: b) A 1.000 g sample of the hydrate form of stepanovite contains 1.178 times 10^21 molecules (formula units, actually). What is the molar mass of the hydrate form of stepanovite? c) Using the molar mass of the hydrate and anhydrous forms of stepanovite, calculate the number of waters of hydration present in this compound, and write out the formula for the hydrate.

Explanation / Answer

33. b) Given: 1.000 g of hydrate form of stepanovite contains 1.178*1021 molecules.

First calculate the moles: 1mol = 6.022*1023 molecules

So, 1.178*1021 molecules / 6.022*1023 molecules = 0.1956*10-2 moles = 0.001956 moles of hydrate

Now, molar mass of hydrate = 1.000g / 0.001956 moles = 511.25 g/mol

c) Now, calculate the mass of anhydrous form from the mass percent composition of elements.

First calculate the moles for each element: Assume 100g of compound is present.

Sodium(Na) = 6.261g/23g/mol = 0.2722 mol

magnesium(Mg) = 6.619g/24g/mol = 0.2758 mol

Iron(Fe) = 15.21g/56g/mol = 0.2716 mol

carbon(C) = 19.63g/12g/mol = 1.6358 mol

oxygen(O) = 52.29g/16g/mol = 3.2681 mol

now divide by smallest number to get the ratios of elements.

and we get the emperical formula as: NaMgFeC6O12

calculate the mass = 23+24+56+(6*12)+(12*16) = 367 g/mol

now, calculate the water lost = 511.25g(hydrate)-367g(anhydrous form) = 144.25 g

from this mass calculate the moles of water lost = 144.25 g/ 18g/mol = 8 moles of water

Now, determine the whole number ratio of anhydrous form to water

moles of anhydrous form = 5.72 moles

so, there is whole number ratio of 1:1

thus, formula of hydrate = NaMgFeC6O12.H2O

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