Question

Propagation of error. A mixture containing only silver nitrate and mercurous nitrate was dissolved in water and treated with excess sodium cobalticyanide, ${\mathbf{Na}}_{\mathbf{3}}\mathbf{\left[}\mathbf{Co}\mathbf{\right(}\mathbf{CN}{\mathbf{)}}_{\mathbf{6}}\mathbf{]}$ to precipitate both cobalticyanide salts:

$$\begin{gathered} AgN{O}_{3}FM169.873 \\ A{g}_3[Co{(CN)}_6]FM538.643 \\ H{g}_2{(N{O}_3)}_{2}FM525.19 \\ {\left(H{g}_2\right)}_3[Co{\left(CN\right)}_6{]}_{2}FM1633.62 \end{gathered}$$

(a) The unknown weighed 0.4321g and the product weighed 0.4515g. Find wt%${\mathrm{AgNO}}_3$ in the unknown. Caution: Keep all the digits in your calculator or else serious rounding errors may occur. Do not round off until the end.

(b) Even a skilled analyst is not likely to have less than a 0.3% error in isolating the precipitate. Suppose that thege is negligible error in all quantities, except the mass of product, which has an uncertainty of0.30%. Calculate the relative uncertainty in the mass of${\mathrm{AgNO}}_3$ in the unknown.

Short answer

(a)The weight percent of ${\mathrm{AgNO}}_3$in the unknown is 0.4006%.

(b)The relative uncertainty in the mass of ${\mathrm{AgNO}}_3$in the unknown is $0.17\mathrm{g}\pm 39\%$.

Step-by-step solution

Determining the weight percent and relative uncertainity.

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.

Determining the sum of both products.

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

Calculating the weight percent of AgNO3

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%.

Calculating the relative uncertainty.

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

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

The value of x can be written as:

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

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