Some Basic Concepts of Chemistry

Chemistry Class 11

NCERT

1   Calculate the molecular mass of the following:$\\$ (i)$H_2O$$\\$ (ii)$CO_2$$\\$ (iii)$CH_4$

Solution :

(i)$ H_ 2 O$$\\$ The molecular mass of water,$ H _2 O$$\\$ $= (2 *$ Atomic mass of hydrogen$) + (1 *$Atomic mass of oxygen)$\\$ $= [2(1.0084) + 1(16.00 u)]$$\\$ $= 2.016 u + 16.00 u$$\\$ $= 18.016$$\\$ $= 18.02 u$

(ii) $CO _2$$\\$ The molecular mass of carbon dioxide, $CO _2$$\\$ $= (1 *$Atomic mass of carbon$) + (2 * $Atomic mass of oxygen)$\\$ $= [1(12.011 u) + 2 (16.00 u)]$$\\$ $= 12.011 u + 32.00 u$$\\$ $= 44.01 u$$\\$ (iii) $CH _4$$\\$ The molecular mass of methane, $CH_ 4$$\\$ $= (1 *$ Atomic mass of carbon$) + (4 *$ Atomic mass of hydrogen)$\\$ $= [1(12.011 u) + 4 (1.008 u)]$$\\$ $= 12.011 u + 4.032 u$$\\$ $= 16.043 u$

2   Calculate the mass percent of different elements present in sodium sulphate $(Na_ 2 SO_ 4 )$ .

Solution :

The molecular formula of sodium sulphate is $Na_2 SO_ 4$$\\$ Molar mass of $Na _2 SO _4 =[(2 *23.0) + (32.066) + 4 (16.00)]$$\\$ $142.066 g$$\\$ Mass percent of an element =$\dfrac{\text{Mass ofthat element in thecompound}}{\text{Molar mass of thecompound}}*100$$\\$ $\therefore $ Mass percent of sodium:$\\$ $\dfrac{46.0g}{142.066g}*100$$\\$ $=32.379$$\\$ $=32.4\%$$\\$ Mass percent of sulphur:$\\$ $=\dfrac{32.066g}{142.066g}*100$$\\$ $=22.57$$\\$ $=22.6\%$$\\$ Mass percent of oxygen:$\\$ $=\dfrac{64.0g}{142.066g}*100$$\\$ $=45.049$$\\$ $=45.05\%$

3   Determine the empirical formula of an oxide of iron which has $69.9$% iron and $30.1$% dioxygen by mass.

Solution :

% of iron by mass $69.9$ % [Given]$\\$ % of oxygen by mass $30.1$ % [Given]$\\$ Relative moles of iron in iron oxide:$\\$ $\dfrac{\%\text{of iron by mass}}{\text{Atomic mass of iron}}$$\\$ $=\dfrac{69.9}{55.85}=1.25$ Relative moles of oxygen in iron oxide:$\\$ $=\dfrac{\%\text{of oxygen by mass}}{\text{Atomic mass or oxygen}}$$\\$ $=\dfrac{30.1}{16.00}$$\\$ $=1.88$$\\$ Simplest molar ratio of iron to oxygen:$\\$ $= 1.25 : 1.88$$\\$ $=1:1.5$$\\$ $2 : 3$$\\$ $\therefore $The empirical formula of the iron oxide is $Fe _2 O _3 $.

4   Calculate the amount of carbon dioxide that could be produced when$\\$ (i)$ 1$ mole of carbon is burnt in air.$\\$ (ii)$ 1$ mole of carbon is burnt in $16$ g of dioxygen.$\\$ (iii) $2$ moles of carbon are burnt in $16$ g of dioxygen.

Solution :

The balanced reaction of combustion of carbon can be written as:$\\$ $C+ O _2 \to C O _2$$\\$ (i) As per the balanced equation, $1$ mole of carbon burns in $1$ mole of dioxygen (air) to produce $1$ mole of carbon dioxide.$\\$ (ii) According to the question, only $16$ g dioxygen is available. Hence, it will react with $0.5$ mole of carbon to give $22$ g of carbon dioxide. Hence, it is a limiting reactant.$\\$ (iii) According to the question, only $16$ g dioxygen is available. It is a limiting reactant. Thus, $16$ g of dioxygen can combine with only $0.5$ mole of carbon to give $22$ g of carbon dioxide.

5   Calculate the mass of sodium acetate $(CH _3 COONa)$ required to make $500$ mL of $0.375$ molar aqueous solution. Molar mass of sodium acetate is $82.0245$ g mol $^{-1}$

Solution :

$0.375 $ M aqueous solution of sodium acetate$\\$ $\equiv 1000$ mL of solution containing $0.375$ moles of sodium acetate $\therefore $ Number of moles of sodium acetate in $500$ mL$\\$ $=\dfrac{0.375}{1000}*500$$\\$ $=0.1875$ mole$\\$ Molar mass of sodium acetate=$\\$ $82.0.245$ g mole$^{-1}$(Given)$\\$ $\therefore $ Required mass of sodium acetate $= (82.0245 $g mol$^{ -1} ) (0.1875$ mole)$\\$ $= 15.38 g$

6   Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41g mL ^-1$ and the mass percent of nitric acid in it being $69\%.$

Solution :

Mass percent of nitric acid in the sample =$ 69\% $ [Given]$\\$ Thus, $100 g$ of nitric acid contains $69 g$ of nitric acid by mass.$\\$ Molar mass of nitric acid $(HNO _3 )$ $\\$ $=\{1+14+3(16)\}g mol^{-1}$ $\\$ $= 1 + 14 + 18$ $\\$ $= 63 g mol ^{-1}$ $\\$ $\therefore $ Number of moles in $69 g$ of $HNO _3$ $\\$ $=\dfrac{69 g}{63 g mol^{-1}}$ $\\$ $= 1.095$ mol $\\$ Volume of $100g$ of nitric acid solution$\\$ $=\dfrac{\text{Mass of solution}}{\text{density of solution}}$ $\\$ $=\dfrac{100g}{1.41 gmL^{-1}}$ $\\$ $= 70.92 mL = 70.92 * 10 ^{- 3} L$ $\\$ Concentration of nitric acid $\\$ $\dfrac{1.095 mole}{70.92*10^{-3}L}$ $\\$ $= 15.44 mol / L$ $\\$ $\therefore $Concentration of nitric acid =$ 15.44 mol/L$

7   How much copper can be obtained from $100 g$ of copper sulphate $(CuSO_ 4 )?$

Solution :

$1$ mole of $CuSO_ 4$ contains $1$ mole of copper.$\\$ Molar mass of $CuSO _4 = (63.5) + (32.00) + 4(16.00)$ $\\$ $= 63.5 + 32.00 + 64.00$ $\\$ $= 159.5 g$ $\\$ $159.5 g$ of $CuSO _4$ contains $63.5 g$ of copper.$\\$ $\Rightarrow 100g $ of $ CuSO_4 $ will contain $ \dfrac{63.5*100g}{159.5}$ of copper $\\$ $\therefore $ Amount of copper that can be obtained from $100g CuSO_ 4 =\dfrac{63.5*100}{159.5}$ $\\$ $=39.81 g $

8   Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are $69.9$ and $30.1$ respectively. Given that the molar mass of the oxide is $159.69g \ mol ^{- 1}$

Solution :

Mass percent of iron (Fe) =$ 69.9\% $(Given)$\\$ Mass percent of oxygen (O) = $30.1\%$ (Given)$\\$ Number of moles of iron present in the oxide =$\dfrac{69.90}{55.85}$ $\\$ $= 1.25$ $\\$ Number of moles of oxygen present in the oxide =$\dfrac{30.1}{16.0}$ $\\$ $=1.88$ $\\$ Ratio of iron to oxygen in the oxide,$\\$ $= 1.25 : 1.88$ $\\$ $=\dfrac{1.25}{1.25}:\dfrac{1.88}{1.25}$ $\\$ $= 1 : 1.5$ $\\$ $= 2 : 3 $ $\\$ $\therefore $ The empirical formula of the oxide is $Fe _2 O_ 3 .$ $\\$ Empirical formula mass of $Fe _2 O_ 3 = [2(55.85) + 3(16.00)] g$ $\\$ Molar mass of $Fe _2 O_ 3 = 159.69 g$ $\\$ $\therefore n=\dfrac{\text{Molar mass}}{\text{Empiricical formula mass}}$ $\\$ $=\dfrac{159.69 g}{159.7 g}$ $\\$ $=0.999$ $\\$ $=1$(approx)$\\$ Molecular formula of a compound is obtained by multiplying the empirical formula with $n.$ $\\$ Thus, the empirical formula of the given oxide is $Fe_ 2 O_ 3 $ and $n$ is $1.$$\\$ Hence, the molecular formula of the oxide is $ Fe _2 O_ 3 .$

9   Calculate the atomic mass (average) of chlorine using the following data:$\\$ $\begin{array}{|c|c|}\hline & \text{% Natural Abundance} & \text{Molar mass}\\ \hline ^{35}Cl & 75.77 & 34.9689\\ \hline ^{37}Cl & 24.23 & 36.9659 \\ \hline \end{array}$

Solution :

The average atomic mass of chlorine$\\$ =[(Fractional abundance of $^{35}Cl$)(Molar mass of $^{35}Cl$)+(Fractional abundance of $^{37}Cl$)(Molar mass of $^{37}Cl$)]$\\$ $=[\{(\dfrac{75.77}{100})(34.9689 u) \}+\{(\dfrac{24.23}{100})(36.9659 u)\}]$ $\\$ $=26.4959 + 8.9568$$\\$ $= 35.4527 u $$\\$ $\therefore $ The average atomic mass of chlorine =$ 35.4527 u$

10   In three moles of ethane $(C_ 2 H_ 6 )$, calculate the following:$\\$ (i) Number of moles of carbon atoms.$\\$ (ii) Number of moles of hydrogen atoms.$\\$ (iii)Number of molecules of ethane.

Solution :

(i) $1$ mole of $C _2 H _6 $ contains $2$ moles of carbon atoms.$\\$ $\therefore $ Number of moles of carbon atoms in $3$ moles of $C _2 H_ 6 .$ $\\$ $= 2 * 3 = 6$ $\\$ (ii)$ 1$ mole of $ C_ 2 H_ 6$ contains $6$ moles of hydrogen atoms.$\\$ $\therefore $ Number of moles of carbon atoms in $3$ moles of $C_ 2 H_ 6 .$ $\\$ $= 3 * 6 = 18$ $\\$ (iii) $1$ mole of $C _2 H _6 $ contains $6.023 * 10^{ 23}$ molecules of ethane.$\\$ $\therefore $ Number of molecules in $3$ moles of $C_ 2 H_ 6 .$ $\\$ $= 3 * 6.023 * 10^{ 23} = 18.069 *10 ^{23}$