IB Chemistry S1.4: Counting particles by mass: The mole worked solutions

The named entity is the oxygen atom. Each ozone molecule contributes three: n(\mathrm O)=3(0.0450)=0.135\,\mathrm{mol}. Thus N(\mathrm O)=0.135(6.02\times10^{23})=8.13\times10^{22}. Option A counts ozone molecules instead.

Include both potassium atoms: 2(39.10)+1.01+30.97+4(16.00)=174.18. Relative formula mass is a ratio without units. Molar mass expresses grams per mole; its units are \mathrm{g\,mol^{-1}}.

For monatomic samples, N=(m/M)N_{\mathrm A}. In the count ratio, equal masses and N_{\mathrm A} cancel. Hence N_{\mathrm{He}}/N_{\mathrm{Xe}}=131.29/4.00=32.8. The lighter atoms give more entities per gram.

The whole formula contributes 2(35.45)+16.00=86.90 in relative mass. Oxygen contributes 16.00, so its mass percentage is 100(16.00/86.90)=18.4\%.

The empirical-unit molar mass is 2(12.01)+3(1.01)+14.01=41.06\,\mathrm{g\,mol^{-1}}. The multiplier is 123.18/41.06=3. Apply three to every subscript to obtain \mathrm{C_6H_9N_3}.

Solute amount is unchanged during the stated dilution. Therefore c_2=c_1V_1/V_2=0.168(2.00/32.0)=0.0105\,\mathrm{mol\,dm^{-3}}. The common volume units cancel; the denominator is the final volume.

At matching temperature and pressure, equal ideal-gas volumes have equal amounts. Helium particles are atoms, while the other two gases are molecular. Their atoms per particle are respectively one, two and five, so neither total atom count nor mass must agree.

The equation assigns one mole of aluminium chloride per mole of aluminium consumed. Its molar mass is 26.98+3(35.45)=133.33\,\mathrm{g\,mol^{-1}}. The maximum product mass is 0.0800(133.33)=10.6664\,\mathrm g. The recovery is 100(7.20/10.6664)=67.5\%.

(a) Use n=m/M: n=3.16/78.96=0.0400\,\mathrm{mol} [2]. (b) Use m=nM: m=0.0185(58.69)=1.09\,\mathrm g [2]. Each calculation has a relation point and a correct-result point; an unambiguous correct answer may receive both.

(a) The amounts are 1.81/180.95=0.01000276\,\mathrm{mol} Ta and 0.400/16.00=0.0250000\,\mathrm{mol} O [1]. Divide by the smaller amount to give 1:2.4993, compatible with 1:2.5; double both terms [1]. The formula is \mathrm{Ta_2O_5} [1]. (b) Individual Ta and O atoms carry different masses [1]. Dividing each measured mass by its elemental molar mass produces amounts that compare atom populations [1].

(a) The required amount is n=0.02500(0.2000)=0.005000\,\mathrm{mol} [1]. For \mathrm{K_2SO_4}, M=2(39.10)+32.07+4(16.00)=174.27\,\mathrm{g\,mol^{-1}}, giving m=0.005000(174.27)=0.8714\,\mathrm g [1]. (b) Transfer all solution and rinsings into the volumetric flask so the weighed solute is retained [1]. At room temperature, view the bottom meniscus at eye level and set it to the calibration mark; stopper and invert to homogenise [1].

(a) c=0.1000(18.40/20.00)=0.09200\,\mathrm{mol\,dm^{-3}} [1]. (b) Under the supplied model, two reading bounds give 0.05+0.05=0.10\,\mathrm{cm^3} [1]. The titre percentage uncertainty is 100(0.10/18.40)=0.5435\% [1]. Include the sample-volume percentage 100(0.03/20.00)=0.1500\%: the total is 0.30+0.5435+0.1500=0.9935\% [1]. (c) The absolute uncertainty is 0.09200(0.993478\ldots/100)=0.0009140, which rounds to 0.00091\,\mathrm{mol\,dm^{-3}} [1]. Report c=0.09200\pm0.00091\,\mathrm{mol\,dm^{-3}}. These are the stated additive bounds, not a general statistical prescription.

(a) Four unlabelled carbon positions each have two C/O single bonds, so they contribute eight hydrogens [1]. The two terminal OH groups contribute two more hydrogens and two oxygens: molecular formula \mathrm{C_4H_{10}O_2} [1]. (b) The common divisor of 4, 10 and 2 is 2 [1]. Dividing all three counts gives the simplest ratio \mathrm{C_2H_5O} [1]. It is an atom ratio, not a claim that a molecule contains half an oxygen atom.

(a) The precipitate amount is 1.1466/143.32=0.008000279\,\mathrm{mol} [1]. Each AgCl contains one chloride, so [\mathrm{Cl^-}]=0.008000279/0.1000=0.08000\,\mathrm{mol\,dm^{-3}} [1]. (b) Each salt unit supplies two chlorides: n(\mathrm{MCl_2})=0.004000140\,\mathrm{mol} [1]. Thus M(\mathrm{MCl_2})=0.3808/0.004000140=95.1967\,\mathrm{g\,mol^{-1}} and A_{\mathrm r}(\mathrm M)=95.1967-2(35.45)=24.30 [1]. The relative atomic mass has no unit.

At common temperature and pressure, volumes track molecule amounts. Each parent molecule supplies 2(12.0/24.0)=1 N atom and 2(36.0/24.0)=3 F atoms. Thus the molecular formula is \mathrm{NF_3}. A doubled formula would predict different product volumes per parent molecule.

The residue contains 0.500(0.0120)=0.00600\,\mathrm{mol} HCl because the NaOH neutralisation ratio is one to one. Subtract it from the added acid: 0.0200-0.00600=0.0140\,\mathrm{mol} HCl reacted with ZnO. Two HCl are required per ZnO, so n(\mathrm{ZnO})=0.0140/2=0.00700\,\mathrm{mol}.

(a) Tin(II) loses two electrons; each iron(III) accepts one [1]. Balancing electrons gives \mathrm{Sn^{2+}+2Fe^{3+}\to Sn^{4+}+2Fe^{2+}} [1]. Both sides carry charge +8. (b) Subtract the anode reduction potential from the cathode reduction potential: E^{\ominus}_{\mathrm{cell}}=0.77-0.15=+0.62\,\mathrm V [1]. This positive value predicts forward thermodynamic feasibility under standard conditions [1]; it alone does not establish complete analytical conversion. Potentials are not multiplied by coefficients. (c) The added iron(III) amount is 0.1500(0.04000)=0.006000\,\mathrm{mol} [1]. Subtract the measured excess: 0.006000-0.001840=0.004160\,\mathrm{mol} consumed, corresponding to 0.002080\,\mathrm{mol} tin(II) [1]. Hence c=0.002080/0.02500=0.08320\,\mathrm{mol\,dm^{-3}} [1]. (d) Using the entire iron(III) charge would assign its unreacted excess to tin oxidation, raising the inferred tin concentration [1].

(a) n(\mathrm C)=n(\mathrm{CO_2})=1.056/44.01=0.02399455\,\mathrm{mol} [1]. Water contains two H atoms, giving n(\mathrm H)=2(0.4325/18.02)=0.04800222\,\mathrm{mol} [1]. (b) The original carbon and hydrogen masses are 0.2881745\,\mathrm g and 0.04848224\,\mathrm g. Oxygen is the remainder, 0.5287-0.2881745-0.04848224=0.1920433\,\mathrm g, or 0.01200270\,\mathrm{mol} [1]. The atom ratio is approximately 1.999:3.999:1, giving \mathrm{C_2H_4O} [1]. (c) Its empirical-unit mass is 44.06\,\mathrm{g\,mol^{-1}}; 132.2/44.06=3.00045, compatible with 3, so the molecule is \mathrm{C_6H_{12}O_3} [1]. (d) Less measured water lowers the inferred H amount without lowering the carbon amount, so H:C is too low [1].

(a) The base amount is 0.1000(0.03000)=0.003000\,\mathrm{mol}. The diprotic acid reacts with two hydroxides per molecule, so its aliquot amount is 0.001500\,\mathrm{mol} and the mixture concentration is 0.001500/0.02500=0.06000\,\mathrm{mol\,dm^{-3}} [1]. The total mixture amount is 0.06000(0.2500)=0.01500\,\mathrm{mol} [1]. The reference contributes 0.04000(0.1250)=0.005000\,\mathrm{mol}; subtracting this leaves 0.01000\,\mathrm{mol} from the weighed sample [1]. (b) M=0.9004/0.01000=90.04\,\mathrm{g\,mol^{-1}} [1]. The empirical-unit mass is 12.01+1.01+2(16.00)=45.02; the multiplier is 2, so the molecular formula is \mathrm{C_2H_2O_4} [1]. (c) Overshoot overestimates the mixture acid amount; subtracting the fixed reference amount also overestimates the weighed sample amount [1]. Since its weighed mass stays fixed, M=m/n is too low [1].

(a) Iron amount is 0.5585/55.85=0.01000\,\mathrm{mol} [1]. Oxygen removed from the dry oxide is 0.7985-0.5585=0.2400\,\mathrm g, or 0.01500\,\mathrm{mol} [1]. The Fe:O ratio is 1:1.5=2:3, so the dry formula is \mathrm{Fe_2O_3} [1]. (b) Dehydration removes 0.9787-0.7985=0.1802\,\mathrm g water, equal to 0.01000\,\mathrm{mol} [1]. There are 0.01000/2=0.005000\,\mathrm{mol} oxide empirical units, so x=0.01000/0.005000=2 [1]. (c) Retained water is included in the supposedly dry mass but absent from the final iron; subtracting the iron mass assigns that extra mass to oxygen, so the estimate is too high [1].

(a) Absorption removes 85.0-30.0=55.0\,\mathrm{cm^3} of carbon dioxide. Since \mathrm{2CO+O_2\to2CO_2}, the original CO volume is 55.0\,\mathrm{cm^3} [1]. The rest of the original fuel is hydrogen, 80.0-55.0=25.0\,\mathrm{cm^3} [1]. Both CO and hydrogen need half their gas volume of oxygen: 55.0/2+25.0/2=40.0\,\mathrm{cm^3}, consistent with 70.0-30.0=40.0\,\mathrm{cm^3} [1]. (b) From \mathrm{2H_2+O_2\to2H_2O}, water amount equals hydrogen amount: n=0.0250/22.7=0.00110132\,\mathrm{mol} [1]. Thus water mass is 0.00110132(18.02)=0.0198\,\mathrm g [1]. The liquid water volume is not inferred from a gas coefficient. (c) At fixed pressure a hotter gas has a larger volume for the same amount, so unequal temperatures would invalidate the mole-ratio interpretation of volume differences [1].

(a) Compare volume divided by coefficient: 25.0/1=25.0 and 60.0/3=20.0. Hydrogen limits [1]; ammonia volume is 2(20.0)=40.0\,\mathrm{cm^3} [1]. (b) Nitrogen left is 25.0-20.0=5.0\,\mathrm{cm^3}, giving total gas 5.0+40.0=45.0\,\mathrm{cm^3} [1]. At common temperature and pressure the mole fraction is the volume fraction, 40.0/45.0=0.889 [1]. (c) n(\mathrm{NH_3})=0.0400/22.7=0.00176211\,\mathrm{mol} [1]. With M=14.01+3(1.01)=17.04\,\mathrm{g\,mol^{-1}}, the mass is 0.00176211(17.04)=0.0300\,\mathrm g [1]. This stipulated model does not predict equilibrium conversion in an industrial reactor.

(a) n=2.022/101.11=0.0199980\,\mathrm{mol} [1]. (b) Include dissolved salt in the mass: m_{\mathrm{solution}}=52.022\,\mathrm g [1]. With \Delta T=-3.10\,\mathrm K, q_{\mathrm{solution}}=52.022(4.18)(-3.10)=-674.101\,\mathrm J [1]. (c) The dissolving process absorbs the heat lost by the solution: q_{\mathrm{process}}=+674.101\,\mathrm J [1]. Divide by amount and by 1000 to obtain \Delta H=+33.7084\,\mathrm{kJ\,mol^{-1}}, or +33.7\,\mathrm{kJ\,mol^{-1}} [1]. (d) Incoming heat reduces the observed temperature fall, so the inferred positive enthalpy is too small [1].

(a) Calcium chloride supplies 2(0.120)(0.0300)=0.00720\,\mathrm{mol} chloride and NaCl supplies 0.150(0.0200)=0.00300\,\mathrm{mol} [1]. Total chloride is 0.01020\,\mathrm{mol} in 0.0500\,\mathrm{dm^3}: concentration 0.204\,\mathrm{mol\,dm^{-3}} [1]. (b) The same chloride amount in 0.2500\,\mathrm{dm^3} gives 0.0408\,\mathrm{mol\,dm^{-3}} [1]. (c) The net equation is \mathrm{Ag^+(aq)+Cl^-(aq)\to AgCl(s)} [1]. Chloride limits because 0.01020<0.0120\,\mathrm{mol} at a 1:1 ratio [1]. The mass is 0.01020(143.32)=1.461864\,\mathrm g, or 1.46\,\mathrm g [1].