 # 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.

% 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 $^{-1}$

$0.375$ 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 $C_ 2 H_ 6 .$ $\\$ $= 3 * 6.023 * 10^{ 23} = 18.069 *10 ^{23}$