Someone can help me please... -Calculate the moles of anhydrous (dry) KAl(SO 4 )

Question

Someone can help me please...

-Calculate the moles of anhydrous (dry) KAl(SO4)2 that were present in the sample.

-Calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2. Note: Report the ratio to the closest whole number.

-Write the empirical formula for the hydrated alum, based on your experimental results and answer to Question B. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(S04)2•4H20.

-Describe any visual differences between the hydrated alum and the dried, anhydrous form.

-How would the following errors affect the empirical formula for the compound? That is, will these errors cause the calculated number of moles of water in the hydrate to be artificially high or low?

-The student ran out of time and did not do the second heating.

-The student recorded the mass of the cup + sample incorrectly and started with 2.2 g of hydrated compound but used 2.0 g in the calculations.

-Cu(II) sulfate exists as a hydrate. It has many practical uses including use as a fungicide and pesticide. When mixed with chromium and arsenic it forms the wood preservative called CCA. CCA was used in pressure treated wood to protect wood from rotting due to insects and microbial agents. Because CCA-treated wood contains toxic heavy metals, its use has been discontinued for home use and children’s play sets.

Explanation / Answer

The number of moles could be found by the formula

            number of moles = n = weight in gram / molecular weight ...............1

            The weight in gram is found by the balance. The molecular weight of anhydrous KAl ( SO4)2 is calcualted as

                     molcular weight = 39.09 + 26.98 +2 (32.06+ 64)

                      molecular weight = 258.19amu

       By putting the data in equation 1 , the number of moles of anhydrous KAl(SO4)2 could be determined. If you want to calculate the one mole of KAl(SO4)2 then its

                                  1mole of KAl(SO4)2 = 258.19g

The emperical formula written will depends upon the ration between the hydrated water and potash alum.

it could be    KAl(SO4)2.10H2O

If students ran out the time and could not do the second heating then definatly it will effect becasue your alum will still be hydrated.

If you recorded 2.2g and used 2g in calculation it would affect your accuracy of calculation but you got to used the round off technique to avoid this mistake.

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