Chapter 3, Problem 164AP

Interpretation Introduction

Interpretation:

The molar mass, percent composition, and empirical formula of ferrous fumarate and the mass of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ formed in combustion are to be calculated; and the reason as these tablets are not packed in unit-blister packaging is to be determined.

Concept Introduction:

The molar mass is calculated by adding the masses of each and every element multiplied by their number of atoms present (given in subscript). Its S.I. unit is $\text{g}/\text{mol}\text{.}$

To calculate the empirical formula, the use of combustion analysis is done. In this method, the number of moles of each element is calculated by using the given information, and then it is divided by the least number of moles of element present in the reaction. After that, it is converted into the smallest whole number ratio.

The molecular formula is a whole number, multiple of empirical formula, given by the expression, which is as follows:

$n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}.$

The percent composition of an element by its mass in a compound is calculated by the formula, given as follows:

$\text{percent by mass of an element}=\frac{n\times \text{mass of an element}}{\text{molecular mass of the compound}}\times 100\%$

where, n represents the elements and the number of atoms in the given molecule or compound.

Answer to Problem 164AP

Solution:

a)

Molar mass $=169.901\text{ g}$

$\begin{array}{l}\%\text{Fe}=32.872\%\\ \%\text{C}=28.346\%\\ \%\text{H}=1.1866\%\\ \%\text{O}=37.669\%\end{array}$

b)

${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$

c)

The mass of ${\text{CO}}_2$ is $1.039\text{ g}$ and the mass of ${\text{H}}_2\text{O}$ is $0.104\text{ g}$.

d)

The mass of iron present in the tablet is $21\text{ mg,}$ which is less than $30\text{mg}$. So, its packing in unit-blister is not important.

Explanation of Solution

a) The molar mass and percent composition of ferrous fumarate

The molar mass is calculated by adding the masses of each and every element multiplied by their number of atoms present (given in subscript). Its S.I. unit is $\text{g}/\text{mol}$.

So, the molar mass of ferrous fumarate is calculated as follows:

$\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{FeC}}_{\text{4}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}=55.85\text{ g/mol}+4\times 12.01\text{ g/mol}+2\times 1.008\text{ g/mol}+4\times 16.00\text{ g/mol}\\ =55.85\text{ g/mol}+48.04\text{ g/mol}+2.016\text{ g/mol}+64.00\text{ g/mol}\\ =169.90\text{ g/mol}\text{.}\end{array}$

The percent composition of an element in a compound is given by the expression, which is as follows:

$\text{percent by mass of an element}=\frac{n\times \text{mass of an element}}{\text{molecular mass of the compound}}\times 100\%.$

The mass percent of $\text{Fe}$ is calculated as follows:

$\begin{array}{c}\%\text{of Fe = }\frac{1\times 55.85\text{ g}}{169.90\text{ g}}\times 100\\ =\frac{\text{5585 }\cancel{\text{g}}}{310\cancel{\text{g}}}\\ =32.872\text{}\%.\end{array}$

The mass percent of $\text{C}$ is calculated as follows:

$\begin{array}{c}\%\text{of C = }\frac{4\times 12.04\text{ g}}{169.90\text{ g}}\times 100\\ =\frac{\text{4808 }\cancel{\text{g}}}{310\cancel{\text{g}}}\\ =28.346\%.\end{array}$

The mass percent of $\text{H}$ is calculated as follows:

$\begin{array}{c}\%\text{of H = }\frac{2\times 1.008\text{ g}}{169.90\text{ g}}\times 100\\ =\frac{\text{201}\text{.6 g}}{310\text{ g}}\\ =1.1866\text{}\%.\end{array}$

The mass percent of $\text{O}$ is calculated as follows:

$\begin{array}{c}\%\text{of O = }\frac{4\times 16.00\text{ g}}{169.90\text{ g}}\times 100\\ =\frac{\text{6400 }\cancel{\text{g}}}{310\cancel{\text{g}}}\\ =37.669\%.\end{array}$

b) Empirical formula of ferrous fumarate

The molecular formula is given by the expression, as follows:

$n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}}$

As, the molecular formula of ferrous fumarate is ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$.

The whole number, $n$ can be calculated by taking the greater common factor of the numbers which are in the subscript.

The greatest common factor of $1,4,2\text{ and }4$ is $1$ itself. So, $n$ is one.

The empirical formula of the compound is given as follows:

$\begin{array}{c}\text{Empirical formula}=\frac{\text{ Molecularformula}}{n}\\ \text{=}\frac{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}{1}\\ ={\text{FeC}}_4{\text{H}}_2{\text{O}}_4.\end{array}$

c) Mass of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ would be produced by combustion of 1.00 g ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4.$

The balanced equation for the combustion of ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$ is given as follows:

${\text{2FeC}}_{\text{4}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+5O}}_{\text{2}}\to {\text{8CO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O+2Fe}\text{.}$

Since the molar ratio of ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$ to that of ${\text{CO}}_{\text{2}}$ is $2:8=1:4$.

So, the number of moles of ${\text{CO}}_2$ isgiven as follows:

$n_{{\text{CO}}_2}=\frac{4}{1}\times n_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}.$

The number of moles of ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$ isgiven by the expression, which is as follows:

$n_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}=\frac{m_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}}{M_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}}.$

Substitute $1.00g$ for $m_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}$ and $169.90\text{ g}/\text{mol}$ for $M_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}$ in the above equation.

$\begin{array}{c}{n}_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}=\frac{1.00\cancel{g}}{169.90\cancel{\text{g}}/\text{mol}}\\ =0.0058\text{ mol}\text{.}\end{array}$

Therefore, the number of moles of ${\text{CO}}_2$ is calculated as follows:

$\begin{array}{c}{n}_{{\text{CO}}_2}=\frac{4}{1}\times 0.0058\text{ mol}\\ \text{=0}\text{.0236 mol}\text{.}\end{array}$

The mass of ${\text{CO}}_2$ is given by the expression, which isas follows:

$n_{{\text{CO}}_2}\times M_{{\text{CO}}_2}=m_{{\text{CO}}_2}.$

Substitute $\text{0}\text{.0236 mol}$ for $n_{{\text{CO}}_2}$ and $44.01\text{ g}/\text{mol}$ for $M_{{\text{CO}}_2}$ in the above equation.

$\begin{array}{c}\text{0}\text{.0236 }\cancel{\text{mol}}\times 44.01\text{ g}/\cancel{\text{mol}}=m_{{\text{CO}}_2}\\ 1.039\text{ g}=m_{{\text{CO}}_2}.\end{array}$

Similarly, the molar ratio of ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$ to that of ${\text{H}}_2\text{O}$ is $2:2=1:1$.

So, the number of moles of ${\text{H}}_2\text{O}$ isgiven as follows:

$n_{{\text{H}}_2\text{O}}=n_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}.$

The number of moles of ${\text{H}}_2\text{O}$ isgiven by the expression, which is as follows:

$n_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}=\frac{m_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}}{M_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}}.$

Substitute $1.00g$ for $m_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}$ and $169.90\text{ g}/\text{mol}$ for $M_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}$ in the above equation.

$\begin{array}{c}{n}_{{\text{FeC}}_4{\text{H}}_2{\text{O}}_4}=\frac{1.00\cancel{g}}{169.90\cancel{\text{g}}/\text{mol}}\\ =0.0058\text{ mol}\text{.}\end{array}$

Therefore, the number of moles of ${\text{H}}_2\text{O}$ is $0.0058\text{ mol}\text{.}$

The mass of ${\text{H}}_2\text{O}$ is given by the expression, which is as follows:

$n_{{\text{H}}_2\text{O}}\times M_{{\text{H}}_2\text{O}}=m_{{\text{H}}_2\text{O}}.$

Substitute $0.0058\text{ mol}$ for $n_{{\text{H}}_2\text{Ol}}$ and $18.016\text{ g}/\text{mol}$ for $M_{{\text{H}}_2\text{O}}$ in the above equation.

$\begin{array}{c}0.0058\cancel{\text{mol}}\times 18.016\text{ g}/\cancel{\text{mol}}=m_{{\text{CO}}_2}\\ 0.104\text{ g}=m_{{\text{CO}}_2}.\end{array}$

Hence, the mass of ${\text{CO}}_2$ $=1.039\text{ g}$ and the mass of ${\text{H}}_2\text{O}$ $=0.104\text{ g}\text{.}$

d)It is not necessary for the tablets in the photograph to be in unit-dose blister packaging.

The mass of tablet is $65\text{ mg}$.

Iron in unit-blister packaging is more than $30\text{ mg}$.

The percentage of Fe in ${\text{FeC}}_4{\text{H}}_2{\text{O}}_4$ is $32.869\text{}\%$.

So, the mass of iron present in the tablet is calculated as follows:

$\begin{array}{c}\text{Mass}\text{of}\text{iron}=32.869\text{}\%\times 65\text{ mg}\\ =0.32869\times 65\text{ mg}\\ \approx \text{21 mg}\text{.}\end{array}$

It is clearly seen that this mass is less than 30 mg. Hence, it is not important to pack them in the unit–blister.