Question

64.36 g of an unknown cobalt (II) chloride hydrate was dehydrated in a crucible. The final mass of cobalt (II) chloride was 45.44.

a.) Using factor-label, calculate the mass of water lost and convert this into number of moles of water lost.

b.) Using factor-label, calculate the number of moles of CoCl₂​ remaining in the crucible.

c.) Determine the ratio of water moles lost to moles of CoCl_​2 ​remaining.

Answer 1

a) total mass of cobalt (II) chloride hydrate sample = 64.36 gm

final mass of cobal(II) chloride = 45.44 gm

mass of water lost = (total mass of cobalt (II) chloride hydrate sample) - (final mass of cobal(II) chloride)

mass of water lost = 64.36gm - 45.44gm = 18.92 gm

mass of water lost = 18.92 gm

molar mass of water = 18.01528 gm/mol

no. of mole = mass of compound / molar mass

no. of mole of water lost = 18.392 gm / 18.01528 gm mol^-1 = 1.05 mole

no. of mole of water lost = 1.05 mole

b) mass of CoCl₂ remain = 45.44 gm

molar mass of CoCl₂ = 129.839 g/mol

no. of mole = gm of compound / molar mass

no. of mole of CoCl₂ = 45.44 gm / 129.839 gm mol^-1= 0.34997 mole

no. of mole of CoCl₂ = 0.34997 mole

mole of water lost = 1.05 mole

mole of CoCl₂ remaining = 0.34997 mole

retio between mole of water lost to mole of CoCl₂ remaining = 1.05 mol : 0.34997 mol =

1.05 mol / 0.34997 mol = 3.00

retio of mole of water lost to mole of CoCl₂ remaining = 3 :​ 1