91Ó°ÊÓ

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentageis also referred to as the Mass Percentage.

$Percentbyweight=\frac{gramofsolute}{100gofsolution}$

The relative uncertainty is calculated by dividing the absolute uncertainty by the reported value. Uncertainties about the future are always unitless. The absolute uncertainty is calculated by multiplying the relative uncertainty by the reported value.

a) Consider the following:

$x=m\left(AgN{O}_3\right)$

$0.4321-x=m\left(H{g}_2\right(N{O}_3{)}_2$in unknown)

1mol of ${\mathrm{AgNO}}_3=1/3\mathrm{mol}$of 1mol of role="math" localid="1663675437550" ${\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2=1/3\mathrm{mol}$of $\left(H{g}_2{)}_3\right[Co(CN{)}_6{]}_2$

m (both products) =0.4515 g

Write the following sum of both products which is:

role="math" localid="1663676156979" $$\begin{gathered} \frac{1}{3}\frac{\mathrm{x}}{\mathrm{M}\left({\mathrm{AgNO}}_3\right)}×\mathrm{M}\left({\mathrm{Ag}}_3\right[\mathrm{Co}\left(\mathrm{CN}{)}_6\right])+\frac{1}{3}\frac{0.4321-\mathrm{x}}{\mathrm{M}\left({\mathrm{Hg}}_2\right({\mathrm{NO}}_3{)}_2)}×\mathrm{M}(\left({\mathrm{Hg}}_2{)}_3\right[\mathrm{Co}\left(\mathrm{CN}{)}_6{]}_2\right)=0.4515\mathrm{g} \\ \frac{1}{3}\frac{\mathrm{x}}{169.873}×538.643+\frac{1}{3}\frac{0.4321-\mathrm{x}}{525.19}×1633.62=0.4515\mathrm{g} \\ \mathrm{x}=0.1731\mathrm{g} \end{gathered}$$

Next calculate the weight percent of $AgN{O}_3$in the unknown:

$$\begin{gathered} wt\%=\frac{m\left(AgN{O}_3\right)}{0.4321g} \\ wt\%=\frac{0.1731g}{0.4321g} \\ wt\%=0.4006 \end{gathered}$$

Therefore the weight percent of $AgN{O}_3$is 0.4006%.

b) Consider that 0.30% error in mass of 0.4515g is$Â\pm 0.00135\mathrm{g}$and write the equation from a) once again but include the error:

role="math" localid="1663676219676" $$\begin{gathered} \frac{1}{3}\frac{\mathrm{x}}{\mathrm{M}\left(AgN{O}_3\right)}×\mathrm{M}\left(A{g}_3\right[Co\left(CN{)}_6\right])+\frac{1}{3}\frac{0.4321-\mathrm{x}}{\mathrm{M}\left(H{g}_2\right(N{O}_3{)}_2)}×\mathrm{M}(\left(H{g}_2{)}_3\right[Co\left(CN{)}_6{]}_2\right)=0.4515\mathrm{g} \\ \frac{1}{3}\frac{\mathrm{x}}{169.873}×538.643+\frac{1}{3}\frac{0.4321-\mathrm{x}}{525.19}×1633.62=0.4515 \\ 0.020109\mathrm{x}=(0.4515\mathrm{g}Â\pm 0.00135)-0.448020 \\ 0.020109\mathrm{x}=0.003480(Â\pm 0.00135) \end{gathered}$$$\frac{1}{3}\frac{\mathrm{x}}{169.873}×538.643+\frac{1}{3}\frac{0.4321-\mathrm{x}}{525.19}×1633.62=0.4515$

The value of x can be written as:

$$\begin{gathered} x=\frac{0.003480(Â\pm 0.00135)}{0.020109} \\ x=\frac{0.003480(Â\pm 38.8\%)}{0.020109} \\ x=0.17gÂ\pm 39\% \end{gathered}$$

Therefore the relative uncertainty in the mass of $AgN{O}_3$in the unknown is $0.17gÂ\pm 39\%$.