If 1.50 g of H2C2O4.2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?  a. 0.34g b. 0.92g c. 1.07g d. 1.50g

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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Lets calculate the molecular weight of the hydrated chemical, H2C2O4.2H2O.

using the molecular masses of O=16, C=12 and H=1 gm/molar, we get

Molecular mass of the hydrated chemical = 2x1+2x12+4x16 + 2x2x1+2x16 = 126 gm/mole

and the molecular mass of water of hydration = 2x2x1 + 2x16 = 36 gm/mole

now for every 126 g hydrated chemical heated, 36 gm of water would be lost.

or, for 1.5 gm hydrated chemical, lost water = 36x1.5/126 gm = 0.429 gm

Thus, the amount of anhydrous chemical (H2C2O4) left = 1.5-0.429 gm = 1.071 gm 

Therefore, the correct answer is option C: 1.071 gm.

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