NCERT Solutions · Class 11 Chemistry Some Basic Concepts of Chemistry

Molar mass is the mass of one mole of a substance — add up the atomic masses of every atom in the formula, each counted the number of times it appears.Use the atomic masses (in u, numerically equal to g mol⁻¹ per atom): \[\text{H} = 1.008,\qquad \text{C} = 12.011,\qquad \text{O} = 16.00 \](i) H\(\displaystyle _2\)OThe formula has $\displaystyle 2$ H atoms and $\displaystyle 1$ O atom, so add \(\displaystyle 2\times(\text{mass of H})\) to \(\displaystyle 1\times(\text{mass of O})\): \[M(\text{H}_2\text{O}) = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.016\ \text{g mol}^{-1} \]The data $\displaystyle (1.008, 16.00)$ supports $\displaystyle 4$ significant figures, so round to: \[M(\text{H}_2\text{O}) = 18.02\ \text{g mol}^{-1} \](ii) CO\(\displaystyle _2\)The formula has $\displaystyle 1$ C atom and $\displaystyle 2$ O atoms: \[M(\text{CO}_2) = 1(12.011) + 2(16.00) = 12.011 + 32.00 = 44.011\ \text{g mol}^{-1} \]Rounding to $\displaystyle 4$ significant figures: \[M(\text{CO}_2) = 44.01\ \text{g mol}^{-1} \](iii) CH\(\displaystyle _4\)The formula has $\displaystyle 1$ C atom and $\displaystyle 4$ H atoms. This is the step people rush — count the subscript on H, not just "one carbon plus one hydrogen": \[M(\text{CH}_4) = 1(12.011) + 4(1.008) = 12.011 + 4.032 = 16.043\ \text{g mol}^{-1} \]Rounding to $\displaystyle 4$ significant figures: \[M(\text{CH}_4) = 16.04\ \text{g mol}^{-1} \]Answer: \(\displaystyle M(\text{H}_2\text{O}) = 18.02\ \text{g mol}^{-1}\), \(\displaystyle M(\text{CO}_2) = 44.01\ \text{g mol}^{-1}\), \(\displaystyle M(\text{CH}_4) = 16.04\ \text{g mol}^{-1}\)

Mass per cent of an element is that element's total mass in one mole of the compound, divided by the molar mass of the whole compound, times $\displaystyle 100$ — not the number of atoms, and not divided by a wrong denominator.Step $\displaystyle 1$: Find the molar mass of \(\displaystyle \text{Na}_2\text{SO}_4\).Using atomic masses \(\displaystyle \text{Na} = 23\ \text{u}\), \(\displaystyle \text{S} = 32\ \text{u}\), \(\displaystyle \text{O} = 16\ \text{u}\):\[M(\text{Na}_2\text{SO}_4) = 2(23) + 1(32) + 4(16) \] \[M(\text{Na}_2\text{SO}_4) = 46 + 32 + 64 = 142\ \text{g mol}^{-1} \]So $\displaystyle 1$ mole of \(\displaystyle \text{Na}_2\text{SO}_4\) has a mass of \(\displaystyle 142\ \text{g}\), and inside that mass sits \(\displaystyle 46\ \text{g}\) of Na, \(\displaystyle 32\ \text{g}\) of S, and \(\displaystyle 64\ \text{g}\) of O. Mass per cent compares each of these to the total \(\displaystyle 142\ \text{g}\) — this is where people slip: dividing by the number of atoms or by an individual atomic mass instead of the full molar mass gives a nonsense answer.Step $\displaystyle 2$: Mass per cent formula.\[\text{Mass \%\ of element} = \frac{\text{mass of that element in 1 mole of compound}}{\text{molar mass of compound}} \times 100 \]Step $\displaystyle 3$: Mass per cent of sodium (Na).\[\text{Mass \% Na} = \frac{46\ \text{g}}{142\ \text{g}} \times 100 = 32.394\ldots \% \]Rounding to two decimal places (the data — integer atomic masses — supports two decimal places here):\[\text{Mass \% Na} = 32.39\% \]Step $\displaystyle 4$: Mass per cent of sulphur (S).\[\text{Mass \% S} = \frac{32\ \text{g}}{142\ \text{g}} \times 100 = 22.535\ldots \% \] \[\text{Mass \% S} = 22.54\% \]Step $\displaystyle 5$: Mass per cent of oxygen (O).\[\text{Mass \% O} = \frac{64\ \text{g}}{142\ \text{g}} \times 100 = 45.070\ldots \% \] \[\text{Mass \% O} = 45.07\% \]Step $\displaystyle 6$: Check the three percentages add to $\displaystyle 100$%.\[32.39\% + 22.54\% + 45.07\% = 100.00\% \]This check confirms every atom in the formula was accounted for exactly once — a quick way to catch a dropped subscript or a wrong atomic mass.Answer: Mass per cent — Na = $\displaystyle 32.39$%, S = $\displaystyle 22.54$%, O = $\displaystyle 45.07$% (of \(\displaystyle \text{Na}_2\text{SO}_4\)).

The empirical formula compares moles of each element, not their mass percentages — mass percent tells you how much stuff is there, moles tell you how many atoms are there, and only atom counts give a formula.Step $\displaystyle 1$: Treat the percentages as grams in a $\displaystyle 100$ g sample.Since the oxide is $\displaystyle 69.9$% iron and $\displaystyle 30.1$% dioxygen by mass, take $\displaystyle 100$ g of the compound. That sample contains\[\text{mass of Fe} = 69.9\ \text{g}, \qquad \text{mass of O} = 30.1\ \text{g} \]Step $\displaystyle 2$: Convert each mass to moles using \(\displaystyle n = \dfrac{m}{M} \), where \(\displaystyle n\) is moles, \(\displaystyle m\) is mass, and \(\displaystyle M\) is the molar mass of the element (Fe: $\displaystyle 55.85$ g/mol; O: $\displaystyle 16.00$ g/mol).\[n_{\text{Fe}} = \frac{69.9\ \text{g}}{55.85\ \text{g/mol}} = 1.2516\ \text{mol} \]\[n_{\text{O}} = \frac{30.1\ \text{g}}{16.00\ \text{g/mol}} = 1.8813\ \text{mol} \]This is the step people skip — you cannot compare $\displaystyle 69.9$ and $\displaystyle 30.1$ directly, because grams of iron and grams of oxygen are not the same "unit of stuff." Only after dividing by molar mass are the two numbers counting the same thing (moles of atoms), and only then can they be compared.Step $\displaystyle 3$: Find the simplest mole ratio by dividing every mole value by the smallest one.The smaller value here is \(\displaystyle n_{\text{Fe}} = 1.2516\ \text{mol}\), so divide both by it:\[\text{Fe}: \frac{1.2516}{1.2516} = 1.000 \]\[\text{O}: \frac{1.8813}{1.2516} = 1.503 \]Step $\displaystyle 4$: Convert the ratio to small whole numbers.A ratio of $\displaystyle 1$ : $\displaystyle 1.503$ is not yet whole numbers — $\displaystyle 1.503$ is close to \(\displaystyle 1\tfrac{1}{2} = \dfrac{3}{2}\). Multiplying both sides by $\displaystyle 2$ clears the fraction:\[\text{Fe}: 1 \times 2 = 2, \qquad \text{O}: 1.503 \times 2 = 3.006 \approx 3 \]So the whole-number mole ratio of Fe to O is \(\displaystyle 2 : 3\).Step $\displaystyle 5$: Write the empirical formula.Iron and oxygen combine in the ratio Fe : O = $\displaystyle 2$ : $\displaystyle 3$, so the empirical formula is \(\displaystyle \text{Fe}_2\text{O}_3 \).Answer: The empirical formula of the oxide is \(\displaystyle \mathrm{Fe_2O_3} \) (iron : oxygen mole ratio = $\displaystyle 2$ : $\displaystyle 3$).

The mole ratio in a balanced equation is a ratio of moles, not of grams — convert every mass to moles before you compare it against the equation.The reaction is combustion of carbon in dioxygen:\[\text{C(s)} + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \]This says $\displaystyle 1$ mole of carbon reacts with exactly $\displaystyle 1$ mole of dioxygen to give $\displaystyle 1$ mole of carbon dioxide — a $\displaystyle 1$ : $\displaystyle 1$ : $\displaystyle 1$ mole ratio.Molar masses, using \(\displaystyle \text{C} = 12\ \text{g mol}^{-1}\) and \(\displaystyle \text{O} = 16\ \text{g mol}^{-1}\):\[M(\text{O}_2) = 2 \times 16\ \text{g mol}^{-1} = 32\ \text{g mol}^{-1} \] \[M(\text{CO}_2) = 12 + 2\times16 = 44\ \text{g mol}^{-1} \](i) $\displaystyle 1$ mole of carbon burnt in airAir supplies far more dioxygen than $\displaystyle 1$ mole of carbon could ever need, so carbon is the reactant that runs out, and it alone fixes how much product forms. By the $\displaystyle 1$ : $\displaystyle 1$ mole ratio, $\displaystyle 1$ mol C gives $\displaystyle 1$ mol CO2.Converting moles to mass uses \(\displaystyle m = n \times M \), where \(\displaystyle m\) is mass, \(\displaystyle n\) is the amount in moles, and \(\displaystyle M\) is the molar mass:\[m(\text{CO}_2) = 1\ \text{mol} \times 44\ \text{g mol}^{-1} = 44\ \text{g} \](ii) $\displaystyle 1$ mole of carbon burnt in $\displaystyle 16$ g of dioxygenYou are given a mass of \(\displaystyle \mathrm{O_{2}}\) here, not moles, so convert it first using \(\displaystyle n = \dfrac{m}{M} \):\[n(\text{O}_2) = \frac{16\ \text{g}}{32\ \text{g mol}^{-1}} = 0.5\ \text{mol} \]This is the step that trips people up: you cannot set "$\displaystyle 1$ mole of C" directly against "$\displaystyle 16$ g of \(\displaystyle \mathrm{O_{2}}\)" — both sides of a mole ratio have to be in moles before you compare them.Now check which reactant runs out. The equation needs $\displaystyle 1$ mol \(\displaystyle \mathrm{O_{2}}\) for every $\displaystyle 1$ mol C, so the full $\displaystyle 1$ mol of carbon would need $\displaystyle 1$ mol \(\displaystyle \mathrm{O_{2}}\) — but only $\displaystyle 0.5$ mol \(\displaystyle \mathrm{O_{2}}\) is actually present. Dioxygen is used up first: it is the limiting reagent, and some carbon is left over unburnt.Because the mole ratio \(\displaystyle \mathrm{O_{2}}\) : \(\displaystyle \mathrm{CO_{2}}\) is $\displaystyle 1$ : $\displaystyle 1$, whatever \(\displaystyle \mathrm{O_{2}}\) reacts fixes the \(\displaystyle \mathrm{CO_{2}}\) formed:\[n(\text{CO}_2) = n(\text{O}_2)\ \text{reacted} = 0.5\ \text{mol} \] \[m(\text{CO}_2) = 0.5\ \text{mol} \times 44\ \text{g mol}^{-1} = 22\ \text{g} \](iii) $\displaystyle 2$ moles of carbon burnt in $\displaystyle 16$ g of dioxygenThe mass of dioxygen hasn't changed, so \(\displaystyle n(\text{O}_2) \) is still \(\displaystyle 0.5\ \text{mol}\) (same calculation as in (ii)). The equation still needs $\displaystyle 1$ mol \(\displaystyle \mathrm{O_{2}}\) per $\displaystyle 1$ mol C, so even $\displaystyle 2$ mol of carbon would demand $\displaystyle 2$ mol \(\displaystyle \mathrm{O_{2}}\) — far more than the $\displaystyle 0.5$ mol on hand. Dioxygen is again the limiting reagent, and it is exactly the same amount of it as in part (ii).Doubling the carbon does not double the \(\displaystyle \mathrm{CO_{2}}\) here, because carbon was never the reactant that ran out — the amount of product is set by whichever reactant is used up first, not by whichever one you have more of.So the \(\displaystyle \mathrm{CO_{2}}\) formed is again fixed by the $\displaystyle 0.5$ mol of \(\displaystyle \mathrm{O_{2}}\) that reacts:\[m(\text{CO}_2) = 0.5\ \text{mol} \times 44\ \text{g mol}^{-1} = 22\ \text{g} \]Answer: (i) $\displaystyle 44$ g \(\displaystyle \mathrm{CO_{2}}\); (ii) $\displaystyle 22$ g \(\displaystyle \mathrm{CO_{2}}\); (iii) $\displaystyle 22$ g \(\displaystyle \mathrm{CO_{2}}\) — dioxygen is the limiting reagent in both (ii) and (iii), so both give the same mass of \(\displaystyle \mathrm{CO_{2}}\) despite the extra carbon in (iii)

Molarity tells you moles of solute per litre of solution, and to get a mass you must first convert molarity into moles using the volume actually asked for.The formula connecting molarity to the amount of solute is\[\text{Molarity } (M) = \frac{\text{moles of solute}}{\text{volume of solution in litres}} \]Here \(\displaystyle M = 0.375\ \text{mol L}^{-1}\) and the volume of solution is \(\displaystyle 500\ \text{mL}\).Step $\displaystyle 1$ — convert the volume to litres. Concentration units are always in litres, not millilitres, so convert first: \[500\ \text{mL} = \frac{500}{1000}\ \text{L} = 0.500\ \text{L} \]Step $\displaystyle 2$ — find the moles of sodium acetate needed. Rearranging the molarity formula for moles of solute: \[\text{moles of solute} = M \times V_{\text{solution (L)}} \] \[\text{moles of CH}_3\text{COONa} = 0.375\ \text{mol L}^{-1} \times 0.500\ \text{L} = 0.1875\ \text{mol} \]Step $\displaystyle 3$ — convert moles to mass using the molar mass. Name the formula: mass is moles multiplied by molar mass, \[\text{mass} = n \times M_{\text{molar}} \] where \(\displaystyle n\) is the number of moles found above and \(\displaystyle M_{\text{molar}} = 82.0245\ \text{g mol}^{-1}\) is the mass of one mole of CH₃COONa (given).\[\text{mass of CH}_3\text{COONa} = 0.1875\ \text{mol} \times 82.0245\ \text{g mol}^{-1} \]Carrying out the multiplication in one go (no intermediate rounding): \[\text{mass} = 15.37959375\ \text{g} \]Rounding: the given data — $\displaystyle 500$ mL and $\displaystyle 0.375$ M — each carry $\displaystyle 3$ significant figures, so the answer should be reported to $\displaystyle 3$ significant figures, not to all the digits the calculator shows.\[\text{mass} \approx 15.4\ \text{g} \]A quick sanity check on the step people usually slip on: the volume used in \(\displaystyle M = n/V\) must be the volume of the solution you want to prepare ($\displaystyle 500$ mL total, after the solute is dissolved and made up to volume), not the volume of water — that is a molarity vs. molality confusion in disguise, and it doesn't come up here because the problem hands you the solution volume directly, but it's worth checking every time.Answer: $\displaystyle 15.4$ g of sodium acetate (CH₃COONa) is required to prepare $\displaystyle 500$ mL of $\displaystyle 0.375$ M solution.

Mass per cent tells you grams of solute per $\displaystyle 100$ g of solution, not per $\displaystyle 100$ g of solvent or per litre — you have to bring in the density to get to a volume.The plan: pick a convenient volume of solution, use density to find its mass, use the mass per cent to find the mass of \(\displaystyle \text{HNO}_3 \) in it, convert that mass to moles, and divide by the volume in litres.Step $\displaystyle 1$ — Take $\displaystyle 1$ L of solution as the sample.Density, \(\displaystyle \rho \), is mass per unit volume: \(\displaystyle \rho = \dfrac{\text{mass}}{\text{volume}} \), so mass \(\displaystyle = \rho \times \text{volume} \).\[\text{volume} = 1\ \text{L} = 1000\ \text{mL}, \qquad \rho = 1.41\ \text{g mL}^{-1} \]\[\text{mass of solution} = 1.41\ \text{g mL}^{-1} \times 1000\ \text{mL} = 1410\ \text{g} \]This $\displaystyle 1410$ g is the mass of the whole solution (acid + water) — the number people mistakenly plug into the moles formula is often the solution mass instead of the acid mass, or vice versa. Keep the two separate.Step $\displaystyle 2$ — Get the mass of \(\displaystyle \text{HNO}_3 \) from the mass per cent.Mass per cent of solute is defined as\[\text{mass \%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \]so mass of solute \(\displaystyle = \dfrac{\text{mass \%}}{100} \times \text{mass of solution} \). Here mass % of \(\displaystyle \text{HNO}_3 \) is $\displaystyle 69$, so $\displaystyle 69$ g of \(\displaystyle \text{HNO}_3 \) sit in every $\displaystyle 100$ g of solution:\[\text{mass of HNO}_3 = \frac{69}{100} \times 1410\ \text{g} = 972.9\ \text{g} \]Step $\displaystyle 3$ — Convert that mass to moles.Number of moles \(\displaystyle n = \dfrac{\text{mass}}{\text{molar mass}} \), where molar mass \(\displaystyle M \) is the mass of one mole of the substance.For \(\displaystyle \text{HNO}_3 \): \(\displaystyle M = 1(\text{H}) + 14(\text{N}) + 3\times16(\text{O}) = 1 + 14 + 48 = 63\ \text{g mol}^{-1} \).\[n(\text{HNO}_3) = \frac{972.9\ \text{g}}{63\ \text{g mol}^{-1}} = 15.44\ \text{mol} \]Step $\displaystyle 4$ — Convert to concentration.Molarity is moles of solute per litre of solution (not per litre of solvent — the water alone is never the denominator):\[\text{Molarity} = \frac{n(\text{HNO}_3)}{\text{volume of solution in L}} = \frac{15.44\ \text{mol}}{1\ \text{L}} = 15.44\ \text{mol L}^{-1} \]Because the volume chosen was exactly $\displaystyle 1$ L, the mole count from Step $\displaystyle 3$ already is the molarity — that's why picking $\displaystyle 1$ L as the sample size makes the arithmetic clean. The data ($\displaystyle 69$%, $\displaystyle 1.41$ g mL⁻¹) each carry about $\displaystyle 3$ significant figures, so the answer is kept to $\displaystyle 4$ significant figures at most, giving $\displaystyle 15.44$ mol L⁻¹.Answer: The concentration of nitric acid is $\displaystyle 15.44$ mol L⁻¹ ($\displaystyle 15.44$ M).

Copper sulphate's formula tells you the mole ratio directly — one mole of \(\displaystyle \text{CuSO}_4\) contains exactly one mole of Cu atoms.Step $\displaystyle 1$: Find the molar mass of \(\displaystyle \text{CuSO}_4\).Using atomic masses Cu = $\displaystyle 63.5$ u, S = $\displaystyle 32$ u, O = $\displaystyle 16$ u:\[M(\text{CuSO}_4) = 63.5 + 32 + 4(16) = 63.5 + 32 + 64 = 159.5\ \text{g mol}^{-1} \]Step $\displaystyle 2$: Convert the given mass of \(\displaystyle \text{CuSO}_4\) to moles.Moles is mass divided by molar mass, \(\displaystyle n = \dfrac{\text{given mass}}{\text{molar mass}} \):\[n(\text{CuSO}_4) = \frac{100\ \text{g}}{159.5\ \text{g mol}^{-1}} = 0.6270\ \text{mol} \]Step $\displaystyle 3$: Use the formula to move from moles of compound to moles of copper.This is the step people skip past too fast: the formula \(\displaystyle \text{CuSO}_4\) is a $\displaystyle 1$:$\displaystyle 1$:$\displaystyle 4$ ratio of Cu : S : O, so every mole of the compound hands you exactly one mole of Cu atoms — no factor of $\displaystyle 2$ or $\displaystyle 4$ belongs here, because there is only one Cu per formula unit.\[n(\text{Cu}) = n(\text{CuSO}_4) = 0.6270\ \text{mol} \]Step $\displaystyle 4$: Convert moles of Cu back to mass using Cu's own atomic mass.Mass is moles times molar mass, \(\displaystyle m = n \times M \), and here \(\displaystyle M(\text{Cu}) = 63.5\ \text{g mol}^{-1}\):\[m(\text{Cu}) = 0.6270\ \text{mol} \times 63.5\ \text{g mol}^{-1} = 39.81\ \text{g} \]Equivalently, in one line, the mass of copper is just the mass fraction of Cu in \(\displaystyle \text{CuSO}_4\) applied to $\displaystyle 100$ g:\[m(\text{Cu}) = 100\ \text{g} \times \frac{63.5}{159.5} = 39.81\ \text{g} \]The given mass ($\displaystyle 100$ g) has three significant figures, so the answer is reported to three significant figures.Answer: $\displaystyle 39.8$ g of copper can be obtained from $\displaystyle 100$ g of \(\displaystyle \text{CuSO}_4\).

A percentage composition turns into a mole ratio only after you divide by atomic mass — dividing the raw mass percentages by each other gives a mass ratio, not the ratio of atoms.Since only percentages are given, assume a convenient sample size of $\displaystyle 100$ g of the oxide. Then the mass of each element in the sample equals its percentage in grams:mass of Fe \(\displaystyle = 69.9\ \text{g} \) mass of O \(\displaystyle = 30.1\ \text{g} \)Step $\displaystyle 1$ — convert mass to moles.Use \(\displaystyle n = \dfrac{m}{M} \), where \(\displaystyle n\) is the number of moles, \(\displaystyle m\) is the mass taken, and \(\displaystyle M\) is the atomic mass of the element (Fe \(\displaystyle = 55.85\ \text{g mol}^{-1} \), O \(\displaystyle = 16.00\ \text{g mol}^{-1} \)).\[n_{\text{Fe}} = \frac{69.9\ \text{g}}{55.85\ \text{g mol}^{-1}} = 1.252\ \text{mol} \]\[n_{\text{O}} = \frac{30.1\ \text{g}}{16.00\ \text{g mol}^{-1}} = 1.881\ \text{mol} \]Step $\displaystyle 2$ — find the simplest whole-number ratio.Divide both mole values by the smaller one (\(\displaystyle 1.252\ \text{mol}\), the Fe value) to bring the ratio down to its lowest terms:\[\text{Fe} : \frac{1.252}{1.252} = 1.000 \qquad\qquad \text{O} : \frac{1.881}{1.252} = 1.503 \]This is the step people rush: \(\displaystyle 1.503\) is not close enough to a whole number to round to \(\displaystyle 1\) or \(\displaystyle 2\) — it sits almost exactly at \(\displaystyle 1.5\). A ratio ending in \(\displaystyle .5\) is the signal to scale the whole ratio up by \(\displaystyle 2\) rather than round the single number on the spot:\[\text{Fe} : 1.000 \times 2 = 2.00 \qquad\qquad \text{O} : 1.503 \times 2 = 3.006 \approx 3 \]So the atoms combine in the ratio Fe : O \(\displaystyle = 2 : 3 \), giving the empirical formula \(\displaystyle \text{Fe}_2\text{O}_3 \).Step $\displaystyle 3$ — empirical formula vs. molecular formula.Going from an empirical formula to a molecular formula needs one more piece of data — the compound's molar mass — through \(\displaystyle n = \dfrac{\text{molecular formula mass}}{\text{empirical formula mass}}\). No molar mass is given here, and iron oxides (like other ionic/metal-oxide compounds) don't exist as discrete "molecules" with a multiple of the empirical unit anyway — the formula unit is the simplest whole-number ratio. So the empirical formula found above is also the formula reported for this oxide.As a check, the mass percentage of Fe predicted by \(\displaystyle \text{Fe}_2\text{O}_3 \) (molar mass \(\displaystyle 2(55.85) + 3(16.00) = 159.70\ \text{g mol}^{-1} \)) is\[\%\text{Fe} = \frac{2 \times 55.85}{159.70} \times 100 = 69.9\% \]which matches the given data.Answer: The oxide is \(\displaystyle \text{Fe}_2\text{O}_3 \) (iron(III) oxide), with Fe and O in the atom ratio $\displaystyle 2$ : 3.

The average atomic mass of an element is a weighted average — each isotope's mass counts in proportion to how much of it actually exists in nature, not a plain average of the two masses.Formula: for an element with isotopes of molar mass \(\displaystyle M_1, M_2, \ldots\) present in percentage abundance \(\displaystyle a_1, a_2, \ldots\),\[\text{Average atomic mass} = \frac{a_1 M_1 + a_2 M_2 + \cdots}{100} \]where \(\displaystyle a_i\) is the % natural abundance of isotope \(\displaystyle i\) and \(\displaystyle M_i\) is its molar mass (dividing by $\displaystyle 100$ turns the percentages into fractions that add to $\displaystyle 1$).For chlorine there are two isotopes: The step people skip is checking that the abundances add to $\displaystyle 100$%: \(\displaystyle 75.77 + 24.23 = 100.00\), so no third isotope is missing and the formula above applies directly.Substituting:\[\text{Average atomic mass} = \frac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100} \]Work out each product separately.\[75.77 \times 34.9689 = 2649.60\ (\text{to 2 d.p.}) \]\[24.23 \times 36.9659 = 895.68\ (\text{to 2 d.p.}) \]Add the two contributions (each isotope's mass weighted by how much of it is present):\[2649.60 + 895.68 = 3545.28 \]Now divide by $\displaystyle 100$ to undo the percentage scaling:\[\text{Average atomic mass} = \frac{3545.28}{100} = 35.4528\ \text{u} \]The input data (abundances and masses) are given to $\displaystyle 4$ significant figures, so the result should be rounded to the same precision, not carried out to six digits just because the calculator shows them.\[\text{Average atomic mass} \approx 35.45\ \text{u} \]This is very close to the periodic-table value of chlorine's atomic mass ($\displaystyle 35.45$ u), which makes sense because that tabulated value is itself this same natural-abundance-weighted average.Answer: $\displaystyle 35.45$ u

Every molecule of \(\displaystyle \text{C}_2\text{H}_6 \) is a fixed ratio of atoms — the subscripts tell you how many atoms of each element are in ONE molecule, and that ratio scales directly with moles.The formula \(\displaystyle \text{C}_2\text{H}_6 \) means one molecule of ethane contains $\displaystyle 2$ carbon atoms and $\displaystyle 6$ hydrogen atoms. Since a mole is just a fixed count of particles (Avogadro's number of them), this $\displaystyle 2$ : $\displaystyle 6$ ratio holds equally at the mole level: one mole of \(\displaystyle \text{C}_2\text{H}_6 \) contains $\displaystyle 2$ mol of C atoms and $\displaystyle 6$ mol of H atoms.(i) Moles of carbon atoms\[\text{moles of C} = \text{moles of } \text{C}_2\text{H}_6 \times \left(\dfrac{2 \text{ mol C}}{1 \text{ mol } \text{C}_2\text{H}_6}\right) \]Substituting $\displaystyle 3$ mol of ethane:\[\text{moles of C} = 3 \text{ mol } \text{C}_2\text{H}_6 \times \dfrac{2 \text{ mol C}}{1 \text{ mol } \text{C}_2\text{H}_6} = 6 \text{ mol C} \](ii) Moles of hydrogen atoms\[\text{moles of H} = \text{moles of } \text{C}_2\text{H}_6 \times \left(\dfrac{6 \text{ mol H}}{1 \text{ mol } \text{C}_2\text{H}_6}\right) \]Substituting $\displaystyle 3$ mol of ethane:\[\text{moles of H} = 3 \text{ mol } \text{C}_2\text{H}_6 \times \dfrac{6 \text{ mol H}}{1 \text{ mol } \text{C}_2\text{H}_6} = 18 \text{ mol H} \](iii) Number of molecules of ethaneThis step is the one people slip on: multiplying by Avogadro's number converts moles of a substance into a NUMBER of that substance's particles — here, whole ethane molecules, not atoms. Use Avogadro's number, \(\displaystyle N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \), which is the number of particles in exactly one mole:\[\text{number of molecules} = \text{moles of } \text{C}_2\text{H}_6 \times N_A \]Substituting:\[\text{number of molecules} = 3 \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} \]\[= 18.066 \times 10^{23} = 1.8066 \times 10^{24} \]The given data ($\displaystyle 3$ mol) has one significant figure in the multiplier but Avogadro's number itself carries $\displaystyle 4$ significant figures, so the product is reported to $\displaystyle 4$ significant figures:\[\text{number of molecules} \approx 1.807 \times 10^{24} \text{ molecules} \]Answer: $\displaystyle 6$ mol of carbon atoms, $\displaystyle 18$ mol of hydrogen atoms, and \(\displaystyle 1.807 \times 10^{24} \) molecules of ethane.