Question

To find the Ce4+ content in a solid sample, 4.3718 g of the solid sample were dissolved and treated with excess iodate to precipitate the Ce4+ as Ce(IO3)4. The precipitate was collected, washed well, dried, and ignited to produce 0.1848 g of CeO2 (FM 172.114). What was the weight percentage of Ce (AM 140.116) in the original sample?

Answer 1

Answer:

$\%Ce=3.44\%$

Explanation:

Hello,

At first, we compute the percent of $Ce$ into the $CeO_2$ by using their respective molar masses as shown below:

$\%Ce=\frac{1*140.116}{172.114}*100\%=81.4\%$

Next, we compute the grams of $Ce$ into the 0.1848-g sample of $CeO_2$ as follows:

$m_{Ce}=0.814*0.1848g=0.15gCe$

Finally, the weight percentage of $Ce$ is computed as:

$\%Ce=\frac{0.15gCe}{4.3718g}*100\%\\\%Ce=3.44\%$

Best regards.

Answer 2

Answer:

3.43 %

Explanation:

We need  to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.

0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2

0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce

.00107 mol Ce x 140.116 g Ce/ mol  =  0.150 g Ce

0.150 g Ce/ 4.3718 g sample  x 100 = 3.43 %