Question

A 3.530 g sample of an unknown hydrate of cobalt(II) bromide is heated until all the water of hydration is removed. The CoBr2 that remains has a mass of 2.362 g. (Include units in your answer.)

(a) How many moles of CoBr2 are in the sample?


(b) How many grams of water were lost in the dehydration?

(c) How many moles of water were lost?
.

(d) What is the value of "n" in the formula CoBr2 · n H2O? (1, 2, 3, 4...)


Please provide solution.

Answer 1

Question#47048 – Chemistry – Other

Question:

A 3.530 g sample of an unknown hydrate of cobalt(II) bromide is heated until all the water of hydration is removed. The CoBr₂ that remains has a mass of 2.362 g. (Include units in your answer.)

(a) How many moles of CoBr₂ are in the sample?

(b) How many grams of water were lost in the dehydration?

(c) How many moles of water were lost?

(d) What is the value of "n" in the formula CoBr₂ · n H₂O? (1, 2, 3, 4...)

Please provide solution.

Answer:

(a) The amount of moles of CoBr₂ can be estimated from the mass of CoBr₂ remained:

$$n(CoBr_2) = \frac{m(CoBr_2)}{M(CoBr_2)} = \frac{2.362\,g}{218.73\,g/mol} = 0.01080\,mol$$

(b) The change in mass of the sample is due to the released water vapor:

$$m(H_2O) = m_{sample} - m(CoBr_2) = 3.530\,g - 2.362\,g = 1.168\,g$$

(c) The amount of moles of water lost in dehydration:

$$n(H_2O) = \frac{m(H_2O)}{M(H_2O)} = \frac{1.168\,g}{18.02\,g/mol} = 0.06482\,mol$$

(d) The amount of moles of CoBr₂ relates to the amount of moles of water as 0.01080 : 0.06482 = 1 : 6.0002, which means that n = 6.

The formula of hydrate is CoBr₂ · 6H₂O.