Mole Concept Problems

A comprehensive list of problems that I couldn’t solve at first glance

The ratio mass of oxygen and nitrogen of a particular gaseous mixture is 1:4. The ratio of number of their molecule is:

1. Given Ratio $\frac{w_{{\text{O}}_2}}{w_{{\text{N}}_2}}=\frac{1}{4}$
2. Core Concept Avogadro’s law states $N\propto n$, meaning the ratio of molecules equals the ratio of moles.
3. Execution $\frac{N_{{\text{O}}_2}}{N_{{\text{N}}_2}}=\frac{n_{{\text{O}}_2}}{n_{{\text{N}}_2}}=\frac{w_{{\text{O}}_2}}{M_w({\text{O}}_2)}\times \frac{M_w({\text{N}}_2)}{w_{{\text{N}}_2}}=\frac{1}{4}\times \frac{28}{32}=\frac{7}{32}\to 7:32$

An oxide (${\text{Fe}}_2{\text{O}}_n$) of iron contains 70% iron by mass. Calculate value of $n$.

1. Mass Percentage Formula The mass percentage of an element is given by $(\text{Total mass of element}/\text{Molar mass of compound})\times 100\%$. For ${\text{Fe}}_2{\text{O}}_n$, the total mass of iron is $2\times 56=112\text{ g/mol}$, and the total mass of oxygen is $16n\text{ g/mol}$.
2. Execution Set up the equation using the given percentage: $(112/(112+16n))\times 100=70$. Simplifying this yields $11200=70(112+16n)$, which reduces to $11200=7840+1120n$. Subtracting $7840$ from both sides gives $3360=1120n$, which directly solves to $n=3$.

Number of water molecules in the drop of water, if 1 mL of water has 20 drops and A is Avogadro’s number, is-

1. Mass of One Drop For liquid water, use the standard density of liquid water ($1\text{ g/mL}$) instead of the molar volume of a gas. Since 1 mL contains 20 drops, the volume of a single drop is $1/20=0.05\text{ mL}$, which corresponds directly to a mass of $0.05\text{ g}$.
2. Execution The molar mass of water ($H_{2}O$) is 18 g/mol. The total number of molecules is calculated by multiplying the moles of water by Avogadro’s number ($A$), yielding $\text{Molecules}=(\text{mass}/\text{molar mass})\times A=\frac{0.05A}{18}$.

The percentage composition of carbon by mole in methane is

1. Core Concept Methane (${\text{CH}}_4$) consists of 1 atom of carbon and 4 atoms of hydrogen per molecule, meaning $1\text{ mol}$ of methane contains $1\text{ mol}$ of carbon atoms and $4\text{ mol}$ of hydrogen atoms. Total atoms per mole of compound equals $1+4=5\text{ mol}$.
2. Execution The percentage composition by mole is calculated using mole fractions rather than mass fractions: $\%\text{ C by mole}=\frac{1}{5}\times 100\%=20\%$. Your scratchpad calculated the percentage composition by mass ($\frac{12}{16}\times 100=75\%$).

How much coulomb charge is present on 1g ion of $N^{3-}$ ?

1. Terminology Clarification The phrase “1g ion” is a classical chemical term that specifically means one mole of ions. Thus, the problem requires calculating the total magnitude of charge present on exactly $1\text{ mol}$ of nitride ($N^{3-}$) ions.
2. Execution Since each individual $N^{3-}$ ion carries a net charge of 3 elementary charges, $1\text{ mol}$ of these ions contains $3\text{ mol}$ of electronic charge. Multiplying this by Faraday’s constant ($96500\text{ C/mol}$) yields $\text{Total Charge}=3\times 96500\text{ C}=2.895\times 10^5\text{ C}$.

Advance Concepts