Worked solutions to student book questions Chapter 2...

Worked solutions to student book questions Chapter 2 Analysis by mass

Heinemann Chemistry 2 4th edition Enhanced Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. a Why was the soup sample in Worked Example 2.1 heated to 110°C? b Why was it necessary to weigh the sample four times?

A1. a The soup was heated above 100°C to evaporate water from the sample. b By repeatedly heating the sample until the mass remained unchanged, the analyst

could be sure that all the water had been removed.

Q2. Some laboratories use microwave ovens in place of conventional ovens to dry samples. What advantage could this have?

A2. Microwave ovens dry samples more rapidly than conventional ovens.

Q3. Soy sauce weighing 74.6 g was heated in an oven to constant mass. The final mass was 14.2 g. What percentage of water did the sauce contain?

A3. Mass water = 74.6 – 14.2 = 60.4 g

%Water = mass of watermass of sample

× 1

100 = 6.744.60 ×

1100 = 80.97 = 81.0%

(three significant figures)

Q4. Three brands of dog food were heated and dried to constant mass. The data recorded are shown in Table 2.2.

Table 2.2 Determination of water content of dog food

Dog food tested Mass of dog food sample (g) Final mass (g) Phydeaux Deluxe 19.8 3.9 K9 Budget 7.4 1.9 Fresh meat – buffalo mince 15.0 3.8 a Which brand contained the highest percentage of water? b Do you consider that water content is a good guide to the relative value of

different dog foods? What other factors might be important?



A4.

a Phydeaux Deluxe % water = 8.19

9.38.19 − × 1

100 = 80%

K9 Budget % water = 4.7

9.14.7 − × 1

100 = 74%

Fresh meat – buffalo mince % water = 0.15

8.30.15 − × 1

100 = 75%

b Water would not be a good guide to the nutritional value of the dog food. The amount of protein, carbohydrates, fats, vitamins and minerals need to be considered.

Q5. A student determined the water content of a sample of jam. The following measurements were obtained: Mass of evaporating dish: 20.22 g Mass of jam and evaporating dish

before heating: 30.95 g after heating: 27.22 g after more heating: 26.50 g after more heating: 26.49 g

What was the percentage, by mass, of water in the jam?

A5. Step 1 Find the mass of the moist jam by subtracting mass of evaporating dish. Mass of moist jam = 30.95 – 20.22 g = 10.73 g

Step 2 Find the mass of water. Mass of water = 30.95 – 26.49 g = 4.46 g

Step 3 Find the percentage of water in moist jam.

% (H2O) = 73.1046.4 × 100

= 41.57% Step 4 Express the answer with the correct number of significant figures. % (H2O) = 41.6%

Q6. Calculate the amount (in mole) of: a NaCl in 5.85 g of salt b Fe atoms in 112 g of iron c CO2 molecules in 2.2 g of carbon dioxide d Cl– ions in 13.4 g of nickel chloride (NiCl2) e O2– ions in 159.7 g of iron(III) oxide (Fe2O3)



A6. a Step 1 Write the formula to find amount.

n = massmolar

mass

Step 2 Substitute values and calculate. M(NaCl) = 58.442 g mol–1

n(NaCl) = 1mol g 58.442g 5.85

−

= 0.10009 mol Step 3 Express the answer with the correct number of significant figures. n(NaCl) = 0.100 mol

b Step 1 Write the formula to find amount.

n = massmolar

mass

Step 2 Substitute values and calculate. M(Fe) = 55.847 g mol–1

n(Fe) = 1mol g 55.847g 112

−

= 2.005479 mol

Step 3 Express the answer with the correct number of significant figures. n(Fe) = 2.01 mol

c Step 1 Write the formula to find amount.

n = massmolar

mass

Step 2 Substitute values and calculate. M(CO2) = 43.991 g mol–1

n(CO2) = 1mol g 43.991g 2.2

−

= 0.0500 mol

Step 3 Express the answer with the correct number of significant figures. n(CO2) = 0.050 mol



d Step 1 Write the formula to find amount.

n = massmolar

mass

Step 2 Substitute values and calculate. M(NiCl2) = 129.596 g mol–1

n(NiCl2) = 1mol g 129.596g 13.4

−

= 0.10339 mol

Step 3 Each mol of NiCl2 contains 2 mol of chloride ions. n(Cl–) = 0.10339 mol × 2 = 0.20679 mol Step 4 Express the answer with the correct number of significant figures. n(Cl–) = 0.207 mol = 2.07 × 10–1 mol

e Step 1 Write the formula to find amount.

n = massmolar

mass

Step 2 Substitute values and calculate. M(Fe2O3) = 159.69 g mol–1

n(Fe2O3) = 1mol g 159.69g 159.7

−

= 1.000 mol

Step 3 Each mol of Fe2O3 contains 3 mol of oxide ions. n(O2–) = 1.000 mol × 3 = 3.000 mol

Step 4 Express the answer with the correct number of significant figures. n(O2–) = 3.000 mol

Q7. Calculate the mass of: a 3.0 mol of oxygen molecules (O2) b 1.2 mol of aluminium chloride (AlCl3) c 2.0 mol of nitrogen atoms

A7. a Step 1 Write the formula to find mass. mass = amount × molar mass Step 2 Substitute values and calculate. M(O2) = 32 g mol–1 m(O2) = 3.0 mol × 32 g mol–1 = 96 g

Step 3 Express the answer with the correct number of significant figures. m(O2) = 96 g



b Step 1 Write the formula to find mass. mass = amount × molar mass

Step 2 Substitute values and calculate. M(AlCl3) = 133.34 g mol–1 m(AlCl3) = 1.2 mol × 133.34 g mol–1 = 160.00 g

Step 3 Express the answer with the correct number of significant figures. m(AlCl3) = 1.6 × 102 g

c Step 1 Write the formula to find mass. mass = amount × molar mass

Step 2 Substitute values and calculate. M(N) = 14.0 g mol–1 m(N) = 2.0 mol × 14.0 g mol–1 = 28 g

Step 3 Express the answer with the correct number of significant figures. m(N) = 28 g

Q8. A small oxygen cylinder carried by an ambulance has an internal volume of 1.42 L. What mass of oxygen is present at a pressure of 15 000 kPa and temperature of 15.0°C?

A8. Step 1 Write the formula for the mass of a gas. pV = nRT

n = RTpV

Mm =

RTpV

m = RT

pVM

Step 2 Convert pressure, temperature and volume into the appropriate units for use in the general gas equation.

V = 1.42 L P = 15 000 kPa T = 15.0°C = (15.0 + 273) K = 288 K

Step 3 Calculate the mass of O2, using M(O2) = 32 g mol–1.

m(O2) = 28831.8

3242.100015×

××

= 284.797 g

Step 4 Express the answer with the correct number of significant figures. m(O2) = 285 g



Q9. Calculate the mass of the following gases: a 3.5 L of argon at SLC b 250 mL of ammonia (NH3) at STP

A9. a Step 1 Write the formula for the mass of a gas at SLC, where the molar

volume is 24.5 L mol–1.

n = 5.24

V

Mm =

5.24V

m = 5.24MV ×

Step 2 Calculate the m(Ar) where M(Ar) = 39.948 g mol–1.

m(Ar) = 1

1

molL5.24lmo g 948.39L5.3

−

−×

= 5.707 g Step 3 Express the answer with the correct number of significant figures. m(Ar) = 5.7 g

b Step 1 Write the formula for the mass of a gas at STP, where the molar volume is 22.4 L mol–1.

n = 4.22

V

Mm =

4.22V

m = 4.22MV ×

Step 2 Calculate the m(NH3) where M(NH3) = 17.0 g mol–1.

m(NH3) = 1

1

molL4.22lmo g 0.17L250.0

−

−×

= 0.1897 g

Step 3 Express the answer with the correct number of significant figures. m(NH3) = 0.190 g



Q10. Determine the percentage composition of the following compounds: a lead(IV) oxide (PbO2) b sodium carbonate (Na2CO3)

A10. a Step 1 Calculate the molar mass of lead(IV) oxide. M(PbO2) = 207.2 g mol–1 + (16.0 g mol–1 × 2) = 239.2 g mol–1

Step 2 Calculate the mass of lead in 1 mol (239.2 g) of PbO2. m(Pb) = 207.2 g

Step 3 Calculate the % of lead in PbO2.

% Pb = )(PbO

Pb)(

2Mm × 100

= g 239.2g 2.207 × 100

= 86.6220%

Step 4 Express the answer with the correct number of significant figures. % Pb = 86.6%

Step 5 Calculate the % of oxygen in PbO2. % O = 100 – % Pb = 100 – 86.6 = 13.4%



b Step 1 Calculate the molar mass of sodium carbonate. M(Na2CO3) = (22.9898 g mol–1 × 2) + 12.01115 g mol–1 +

(16.0 g mol–1 × 3) = 105.99 g mol–1

Step 2 Calculate the mass of sodium in 1 mol (105.99 g) of Na2CO3. m(Na) = (22.9898 × 2) g = 45.9796 g

Step 3 Calculate the % of sodium in Na2CO3.

% Na = )CO(Na

Na)(

32Mm × 100

= g 99.105g 9796.45 × 100

= 43.381%

Step 4 Express the answer with the correct number of significant figures. % Na = 43.4%

Step 5 Calculate the mass of carbon in 1 mol (105.99 g) of Na2CO3. m(C) = 12.01115 g

Step 6 Calculate the % carbon in Na2CO3.

% C = )CO(Na

C)(

32Mm × 100

= g 99.105g 01115.12 × 100

= 11.332 %

Step 7 Express the answer with the correct number of significant figures. % C = 11.3%

Step 8 Calculate the % oxygen in Na2CO3. % O = 100 – (% Na + % C) = 100 – (43.4 + 11.3) = 45.3%



Q11. A gaseous hydrocarbon that is used as a fuel for high-temperature cutting and welding of metals contains 92.3% carbon. a Determine its empirical formula. b If the molar mass of the hydrocarbon is 26 g mol–1, find its molecular formula.

A11. a Step 1 As this is a hydrocarbon it will contain carbon and hydrogen. Calculate the mass of hydrogen in a 100 g sample. m(H) = 100 – 92.3 g = 7.7 g

Step 2 Write the ratio by mass.

g 7.7 : g 92.3

H : C

Step 3 Calculate the ratio by amount (in moles).

(C)

g 92.3M

: (H)

g 7.7M

1mol g 01115.12g 92.3

− : 1mol g .007971g 7.7

− 7

7.6845 mol : 7.639 mol

Step 4 Divide by the smaller amount.

639.76845.7 :

639.7639.7

1.0059 : 1

Step 5 Round off to whole numbers. 1 : 1 Therefore, the empirical formula of the compound is CH.

b As the empirical formula is CH, the molecule must contain a whole number of CH units.

The molar mass of one of these units is (12.0 + 1.01) = 13.01 g mol–1.

The number of units in a molecule = unitoneofmassmolar

compound theof massmolar

= 1

1

mol g 13.10mol g 26

−

−

= 2 ∴The molecular formula of the compound is C2H2. (This is ethyne, commonly

called acetylene.)



Q12. When 1.66 g of tungsten (W) is heated in excess chlorine gas, 3.58 g of tungsten chloride is produced. Find the empirical formula of tungsten chloride.

A12.

Step 1 Calculate the mass of chlorine in this sample of tungsten chloride. m(Cl) = 3.58 – 1.66 g = 1.92 g

Step 2 Write the ratio by mass. W : Cl 1.66 g : 1.92 g


(W)

g 1.66M

: (Cl)

g 1.92M

1mol g 83.851g 1.66

− : 1mol g 5.453

g 1.92−

0.00903 mol : 0.0542 mol


00903.000903.0 :

00903.00542.0

Step 5 Round off to whole numbers 1 : 5.998 1 : 6 ∴The empirical formula of the compound is WCl6.



Q13. A sample of blue copper(II) sulfate crystals weighing 2.55 g is heated and decomposes to produce 1.63 g of anhydrous copper(II) sulfate. Show that the formula of the blue crystals is CuSO4.5H2O.

A13. Step 1 Calculate the amount of anhydrous copper sulfate.

n(CuSO4) = )CuSO()CuSO(

4

4Mm

= 5.159

63.1

= 0.0102 mol

Step 2 Calculate the amount of water in hydrated copper sulfate.

n(H2O) = 0.18

63.155.2 −

= 0.0511 mol

Step 3 Calculate the ratio of amount of anhydrous copper sulfate to amount of water.

O)(H

)(CuSO

2

4

nn =

0511.00102.0

= 5 ∴ The formula of the crystals is CuSO4.5H2O.



Q14. Magnesium reacts with hydrochloric acid according to the equation:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) If 10.0 g of magnesium reacts completely, calculate: a the mass of magnesium chloride that forms b the mass of hydrogen that forms.

A14. a Step 1 Write a balanced equation. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2 Calculate the amount of magnesium consumed.

n(Mg) = (Mg)(Mg)

Mm

= 1mol g 24.3g 10

−

= 0.4115 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of MgCl2 produced.

From the equation, 1 mol of MgCl2 is produced by 1 mol of Mg.

(Mg)

)(MgCl2

nn

= 11

n(MgCl2) = n(Mg) = 0.4115 mol

Step 4 Calculate the mass of magnesium chloride produced. M(MgCl2) = 95.211 g mol–1 m(MgCl2) = n(MgCl2) × M(MgCl2) = 0.4115 mol × 95.211 g mol–1 = 39.179 g

Step 5 Express the answer with the correct number of significant figures. m(MgCl2) = 39.2 g



b Step 1 Write a balanced equation. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2 Calculate the amount of magnesium consumed.

n(Mg) = (Mg)(Mg)

Mm

= 1mol g 24.3g 10

−

= 0.4115 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of hydrogen gas produced.

From the equation, 1 mol of H2(g) is produced by 1 mol Mg.

(Mg)

)(H2

nn =

11

n(H2) = n(Mg) = 0.4115 mol

Step 4 Calculate the mass of hydrogen produced. M(H2) = 2.0016 g mol–1 m(H2) = n(H2) × M(H2) = 0.4115 mol × 2.0016 g mol–1 = 0.08236584 g

Step 5 Express the answer with the correct number of significant figures. m(H2) = 0.0824 g



Q15. Iron metal is extracted in a blast furnace by a reaction between iron(III) oxide and carbon monoxide:

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g) To produce 1000 kg of iron, calculate: a the mass of iron(III) oxide required b the volume of carbon dioxide produced at SLC (25°C and 101.3 kPa)

A15. a Step 1 Write a balanced equation. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

Step 2 Calculate the amount of iron produced.

n(Fe) = (Fe)(Fe)

Mm

= 1mol g 55.8g 1000) kg (1000

−

×

= 17 921.15 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of iron(III) oxide required.

From the equation, 1 mol Fe2O3 produces 2 mol Fe.

(Fe)

)O(Fe 32

nn

= 21

n(Fe2O3) = 21 × n(Fe)

= 21 × 17 921.15 mol

= 8960.57 mol

Step 4 Calculate the mass of iron(III) oxide required. M(Fe2O3) = 159.694 g mol–1

m(Fe2O3) = n(Fe2O3) × M(Fe2O3)

= 8960.57 mol × 159.694 g mol–1

= 1 430 949.8 g

Step 5 Convert to kg.

mass in kg = 1000

g 949.8 430 1

= 1430.9 kg

Step 6 Express the answer with the correct number of significant figures. m(Fe2O3) = 1430 kg



b Step 1 Write a balanced equation. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

Step 2 Calculate the amount of iron produced.

n(Fe) = (Fe)

(Me)Mm

= 1mol g 55.8g 1000) kg (1000

−

×

= 17 921.15 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of carbon dioxide produced.

From the equation, 3 mol CO2 and 2 mol Fe are produced together.

(Fe))(CO2

nn =

23

n(CO2) = 23 × n(Fe)

= 23 × 17 921.15 mol

= 26 881.7 mol Step 4 Calculate the volume of carbon dioxide given that the molar volume of

a gas at SLC is 24.5 Lmol–1.

n(CO2) = mV

V

26 881.7 = 5.24

V

V = 26 881.7 × 24.5 = 658 602.15 L = 659 000 L (three significant figures)



Q16. A solution containing 10.0 g of silver nitrate is mixed with a solution containing 10.0 g of barium chloride. What mass of silver chloride precipitate is likely to be produced?

2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

A16.

Step 1 Write a balanced equation. 2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)

Step 2 To determine which reactant is in excess, calculate amount of each reactant divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate the amount of product formed. The other is the excess reactant.

Note: These calculations can only be used to determine the excess reactant. Continue the calculation, using original data.

)t(AgNOcoefficien

)AgNO(

3

3n =

2molg88.169g/0.10 1−

= 0.02943 mol

)t(BaClcoefficien

)BaCl(

2

2n = 1

molg24.208g/0.10 1−

= 0.0480 mol Hence AgNO3 is the limiting reactant.

Step 3 From the equation, 1 mol AgCl is produced from 1 mol AgNO3.

)AgNO(

)AgCl(

3nn =

11

n(AgCl) = n(AgNO3)

= 1molg88.169g0.10

−

= 0.05886 mol

Step 4 Calculate the mass. m(AgCl) = 0.05887 mol × 143 032 g mol–1 = 8.4372 g

Step 5 Express the answer with the correct number of significant figures. m(AgCl) = 8.44 g



Q17. A chemist determined the salt content of a sausage roll by precipitating chloride ions as silver chloride. If an 8.45 g sample of sausage roll yielded 0.636 g of precipitate, calculate the percentage of salt in the food. Assume that all the chloride is present as sodium chloride.

A17.

Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) → AgCl(s) Step 2 Calculate amount of AgCl.

n(AgCl) = 1molg32.143g636.0

−

= 0.004438 mol Step 3 From the equation, 1 mol of NaCl produces 1 mol of AgCl.

(AgCl)(NaCl)

nn =

11

n(NaCl) = 0.004438 mol Step 4 Calculate the mass of salt. m(NaCl) = 0.004438 × 58.44 = 0.2594 g Step 5 Convert to percentage.

% NaCl = 45.8

2594.0 × 100

= 3.0698% Step 6 Express the answer with the correct number of significant figures. % NaCl = 3.07%



Q18. An impure sample of iron(II) sulfate, weighing 1.545 g, was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was 0.315 g, calculate the percentage of iron in the sample.

A18. Step 1 Write an equation that is balanced for the appropriate element, Fe. 2FeSO4(s) + other reactants → Fe2O3(s) + other products

Step 2 Calculate amount of Fe2O3 precipitated.

n(Fe2O3) = 1molg694.159g315.0

−

= 0.001973 mol

Step 3 From the equation, 2 mol of FeSO4 is precipitated as 1 mol of Fe2O3.

)O(Fe)(FeSO

32

4

nn =

12

n(FeSO4) = 2 × 0.001973 mol = 0.003945 mol

Step 4 From the formula, 1 mol of Fe is present in 1 mol of FeSO4.

)(FeSO

(Fe)

4nn =

11

n(Fe) = 0.003945 mol

Step 5 Calculate the mass of Fe in the sample. m(Fe) = 0.003945 mol × 55.847 g mol–1 = 0.2203 g

Step 6 Convert to percentage.

% Fe = 545.12203.0 × 100

= 14.260%

Step 7 Express the answer with the correct number of significant figures. % Fe = 14.3%



Chapter review

Q19. Find the number of mol of: a Ca atoms in 60.0 g of calcium b NH3 molecules in 22 g of ammonia c H2O molecules in 20.0 g of CuSO4.5H2O d Cl– ions in 34 g of FeCl3

A19. a Step 1 Write formula for amount in moles.

n = Mm

Step 2 Calculate the amount of Ca.

n(Ca) = 1molg40.08g060.

−

= 1.4970 mol

Step 3 Express the answer with correct number of significant figures. n(Ca) = 1.50 mol

b Step 1 Calculate the amount, using appropriate formula.

n(NH3) = 1molg17g22

−

= 1.29 mol

Step 2 Express the answer with the correct number of significant figures. n(NH3) = 1.3 mol

c Step 1 Find molar mass of CuSO4.5H2O. M(CuSO4.5H2O) = 63.54 + 32.06 + (4 × 16.00) +

5((2 × 1.008) + 16.00) = 249.68 g mol–1

Step 2 Calculate the amount of CuSO4.5H2O.

n(CuSO4.5H2O) = 1molg68.249g0.02

−

= 0.0801 mol

Step 3 Since 1 mol of CuSO4.5H2O contains 5 mol of water molecules, find the amount of H2O.

n(H2O) = 5 × 0.0801 = 0.4005 mol

Step 4 Express answer with correct number of significant figures. n(H2O) = 0.401 mol



d Step 1 Calculate the amount of FeCl3.

n(FeCl3) = 1molg297.162g34

−

= 0.2095 mol

Step 2 As each mol of FeCl3 contains 3 mol Cl– ions, calculate the amount of Cl– ions.

n(Cl–) = 3 × 0.2095 mol = 0.628 mol

Step 3 Express the answer with the correct number of significant figures. n(Cl–) = 0.63 mol

Q20. Find the mass of: a 0.30 mol of zinc atoms b 0.16 mol of iron(III) oxide (Fe2O3) c 1.5 mol of ammonium phosphate ((NH4)3PO4)

A20. a Step 1 Use the formula for finding mass of substance. m = n × M

Step 2 Calculate the mass of Zn. m(Zn) = 0.30 mol × 65.38 g mol–1 = 19.61 g

Step 3 Express the answer with the correct number of significant figures. m(Zn) = 20 g

b Step 1 Calculate the mass, using the appropriate formula. m(Fe2O3) = 0.16 mol × 159.70 g mol–1

= 25.55 g Step 2 Express the answer with the correct number of significant figures. m(Fe2O3) = 26 g

c Step 1 Calculate the mass of (NH4)3PO4. m((NH4)3PO4) = 1.5 mol × 149.096 g mol–1 = 223.644 g

Step 2 Express the answer with the correct number of significant figures. m((NH4)3PO4) = 220 g = 2.2 × 102 g



Q21. A brand of toothpaste contains 0.22% by mass sodium fluoride (NaF). Calculate the mass of fluoride ions in a tube containing 120 g of the paste.

A21.

Step 1 Calculate the mass of NaF in the toothpaste tube. 0.22% by mass means 0.22 g NaF per 100 g toothpaste So in 120 g of toothpaste:

m(NaF) = 0.22 × 100120

= 0.264 g

Step 2 Write formula for amount in moles.

n = Mm

Step 3 Calculate molar mass. M(NaF) = 41.9 g mol–1

Step 4 Calculate the amount of NaF.

n(NaF) = 1molg9.41g264.0

−

= 0.006287 mol

Step 5 As 1 mol NaF contains 1 mol of F– ions, calculate the amount of F–. n(F–) = 0.006287 mol

Step 6 Calculate the mass of F– ions. m(F–) = 0.006287 mol × 19.00 g mol–1

= 0.1195 g

Step 7 Express the answer with the correct number of significant figures. m(F–) = 0.12 g



Q22. 6.00 g of helium gas was blown into a fairground balloon. On the day, the temperature was 28.0°C and the pressure inside the balloon was 103.4 kPa. Assuming it is infinitely elastic, to what volume did the balloon inflate?

A22. Step 1 Write the formula for the volume of a gas. pV = nRT

V = p

nRT

Step 2 Convert pressure and temperature into the appropriate units for use in the general gas equation.

P = 103.4 kPa T = 28.0°C = (28.0 + 273) K = 301 K

Step 3 Calculate the amount of He, using M(He) = 4.0 g mol–1.

n(He) = 1molg0.4g00.6

−

= 1.5 mol

Step 4 Calculate the volume of He.

V(He) = 4.103

30131.85.1 ××

= 36.286 L

Step 5 Express the answer with the correct number of significant figures. V(He) = 36.3 L



Q23. Calculate the volume of the following gases: a 1.50 mol of oxygen at STP b 28.0 g of nitrogen at STP c 17 g of sulfur dioxide at SLC d 1.2 × 1022 atoms of helium at SLC

A23. a Step 1 Write the formula for the volume of a gas at STP, where the molar

volume is 22.4 L mol–1.

n = 4.22

V

V = n × 22.4

Step 2 Calculate the V(O2). V(O2) = 1.50 mol × 22.4 L mol–1 = 33.6 L

Step 3 Express the answer with the correct number of significant figures. V(O2) = 33.6 L

b Step 1 Write the formula for the volume of a gas at STP, where the molar volume is 22.4 L mol–1.

n = 4.22

V

Mm =

4.22V

V = M

m 4.22×

Step 2 Calculate the V(N2) where M(N2) = 28 g mol–1.

V(N2) = 1

1

molg28lmo L 2.42g.028

−

−×

= 22.4 L

Step 3 Express the answer with the correct number of significant figures. V(N2) = 22.4 L



c Step 1 Write the formula for the volume of a gas at SLC, where the molar volume is 24.5 L mol–1.

n = 5.24

V

Mm =

5.24V

V = M

m 5.24×

Step 2 Calculate the V(SO2) where M(SO2) = 64 g mol–1.

V(SO2) = 1

1

molg64lmo L .542g0.17

−

−×

= 6.508 L

Step 3 Express the answer with the correct number of significant figures. V(SO2) = 6.5 L

d Step 1 Write the formula for the volume of a gas at SLC, where the molar volume is 24.5 L mol–1.

n = 5.24

V

A

particles ofnumber N

= 5.24

V

V = number of particles × A

24.5N

Step 2 Calculate the V(He).

V(He) = 23

122

10023.6lmo L 24.5 atoms 102.1

××× −

= 0.488 L

Step 3 Express the answer with the correct number of significant figures. V(He) = 0.49 L



Q24. A 1.22 g sample of pure gas extracted from the gases from a car exhaust occupied 991 mL at 24.0°C and 1.00 atmosphere pressure. a Calculate the amount of gas, in mol, present in the sample. b What is the molar mass of the gas? c Suggest the identity of the gas.

A24. a Step 1 Write the formula for the amount of a gas. PV = nRT

n = RTPV

Step 2 Convert pressure, temperature and volume into the appropriate units for use in the general gas equation.

P = 1.00 atm = (1.00 × 101.325) kPa = 101.325 kPa V = 991 mL = 0.991 L T = 24.0°C = (24.0 + 273) K = 297 K Step 3 Calculate the amount of gas.

n = 29731.8

991.0325.101××

= 0.0407 mol Step 4 Express the answer with the correct number of significant figures. n = 0.0407 mol

b Step 1 Write the formula for the molar mass of the gas.

M = nm

Step 2 Calculate the M of the gas.

M = mol0.0407g22.1

= 29.987 g mol–1

Step 3 Express the answer with the correct number of significant figures. M = 30.0 g mol–1

c Using the molar mass, the gas is NO.



Q25. Solutions of silver nitrate and potassium chromate react to produce a red precipitate of silver chromate:

2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq) If 0.778 g of precipitate is formed in a reaction, find: a the mass of potassium chromate that reacted b the mass of silver nitrate that reacted.

A25. a Step 1 Write a balanced equation. 2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)

Step 2 Calculate amount of precipitate, Ag2CrO4.

n(Ag2CrO4) = 1molg74.331g 778.0

−

= 0.002345 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of K2CrO4 required.

From the equation, 1 mol K2CrO4 produces 1 mol Ag2CrO4.

)CrOAg()CrOK(

42

42

nn =

11

n(K2CrO4) = 0.002345 mol

Step 4 Calculate the mass of K2CrO4. m(K2CrO4) = 0.002345 mol × 194.204 g mol–1 = 0.4554 g

Step 5 Express the answer with the correct number of significant figures. m(K2CrO4) = 0.455 g



b Step 1 Write a balanced equation. 2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)

Step 2 Calculate amount of precipitate, Ag2CrO4. n(Ag2CrO4) = m × M

= 1molg74.331g 778.0

−

= 0.002345 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of AgNO3 required.

From the equation, 2 mol AgNO3 produces 1 mol Ag2CrO4.

)CrOAg(

)AgNO(

42

3

nn

= 12

n(AgNO3) = 2 × 0.002345 mol = 0.00469 mol

Step 4 Calculate the mass. m(AgNO3) = 0.00469 mol × 169.88 g mol–1 = 0.7967 g

Step 5 Express the answer with the correct number of significant figures. m(AgNO3) = 0.797 g



Q26. Magnesium in distress flares burns in air according to the equation:

2Mg(s) + O2(g) → 2MgO(s) If 10.0 g of magnesium burns in air, calculate: a the mass of magnesium oxide produced b the mass of oxygen that reacts.

A26. a Step 1 Write a balanced equation 2Mg(s) + O2(g) → 2MgO(s)

Step 2 Calculate the amount of Mg.

n(Mg) = 1molg31.24g 0.10

−

= 0.4114 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of MgO produced.

From the equation, 1 mol MgO is produced by 1 mol Mg.

)Mg()MgO(

nn =

11

n(MgO) = 0.4114 mol

Step 4 Convert to mass. m(MgO) = n × M = 0.4114 mol × 40.31 g mol–1 = 16.58 g

Step 5 Express answer with correct number of significant figures. m(MgO) = 16.6 g



b Step 1 Write a balanced equation. 2Mg(s) + O2(g) → 2MgO(s)

Step 2 Calculate the amount of Mg.

n(Mg) = 1molg31.24g 0.10

−

= 0.4114 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of O2 used.

From the equation, 1 mol O2 reacts with 2 mol Mg.

)Mg()O( 2

nn

= 21

n(O2) = 24114.01× mol

= 0.2057 mol

Step 4 Convert to mass. m(O2) = 0.2057 mol × 32.0 g mol–1 = 6.5824 g

Step 5 Express answer with correct number of significant figures. m(O2) = 6.58 g



Q27. Lithium peroxide may be used as a portable oxygen source for astronauts. Calculate the volume of oxygen gas, measured at 25°C and pressure of 101.3 kPa, that is available from the reaction of 0.500 kg of lithium peroxide with carbon dioxide according to the equation:

2Li2O2(s) + 2CO2(g) → 2Li2CO3(s) + O2(g)

A27. Step 1 Write a balanced equation. 2Li2O2(s) + 2CO2(g) → 2Li2CO3(s) + O2(g) Step 2 Calculate the amount of Li2O2.

n(Li2O2) = 1

3

molg878.45g 10500.0−

×

= 10.898 mol Step 3 From the equation, 1 mol O2 is produced by 2 mol Li2O2.

)OLi(

)O(

22

2

nn =

21

n(O2) = 2

898.101× mol

= 5.449 mol Step 4 Convert pressure and temperature to the appropriate units for use in the

general gas equation. P = 101.3 kPa T = 25°C = (25 + 273) K = 298 K Step 5 Calculate the volume at the conditions given, using the general gas equation. pV = nRT

V = p

nRT

V(O2) = 3.101

29831.8449.5 ××

= 133.2 L Step 6 Express answer with correct number of significant figures. V(O2) = 133 L



Q28. A power station burns coal at 45.5 tonnes per hour (1 tonne = 106 g). Assuming the coal is pure carbon and that all the coal is oxidised completely to carbon dioxide gas on combustion, what volume of carbon dioxide is released to the atmosphere per hour when the atmospheric pressure is 758 mmHg and the temperature is 19.0°C?

A28.

Step 1 Write a balanced equation. C(s) + O2(g) → CO2(g)

Step 2 Calculate the amount of C.

n(C) = 1

6

molg12g 105.45

−

×

= 3.792 × 106 mol

Step 3 From the equation, 1 mol CO2 is produced from 1 mol C.

(C)

)(CO2

nn

= 11

n(CO2) = 3.792 × 106 mol

Step 4 Convert pressure and temperature to the appropriate units for use in the general gas equation.

P = 760758 × 101.325 kPa

= 101.058 kPa T = 19°C = (19 + 273) K = 292 K

Step 5 Calculate the volume at the conditions given, using the general gas equation. pV = nRT

V = p

nRT

V(CO2)= 058.101

2928.3110792.3 6 ×××

= 9.105 × 107 L

Step 6 Express answer with correct number of significant figures. V(CO2) = 9.10 × 107 L



Q29. Silicon steel is an alloy of the elements iron, carbon and silicon. An alloy sample was reacted with excess hydrochloric acid and the following reaction occurred:

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) The carbon and silicon in the alloy did not react with the acid. If an alloy sample with a mass of 0.160 g produced 62.0 mL of hydrogen gas, measured at SLC, calculate the: a amount of hydrogen evolved in the reaction b mass of iron that reacted to produce this amount of hydrogen c percentage of iron in the alloy.

A29. a Step 1 Write a balanced equation. Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

Step 2 Calculate the amount of H2(g) given the volume at SLC.

n(H2) = 1molL5.24L 062.0

−

= 2.5306 × 10–3 mol

Step 3 Express the answer with the correct number of significant figures. n(H2) = 2.53 × 10–3 mol

b Step 1 Write a balanced equation. Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

Step 2 From the equation, 1 mol Fe produces 1 mol H2.

)(H

(Fe)

2nn =

11

n(Fe) = n(H2) from part a = 2.53 × 10–3 mol

Step 3 Calculate the mass. m(Fe) = 2.53 × 10–3 mol × 55.8 g mol–1 = 0.14117 g

Step 4 Express the answer with the correct number of significant figures. m(Fe) = 0.141 g

c Step 1 Calculate the percentage of Fe in the alloy.

% (Fe) = )alloy(

(Fe)m

m × 100

= 160.0141.0 × 100

= 88.12%

Step 2 Express the answer with the correct number of significant figures. % (Fe) = 88.1%



Q30. Explain the meaning of the following terms: a relative atomic mass b relative molecular mass c mole d Avogadro’s number e molar mass f precipitate g gravimetric analysis h ionic equation

A30. a Relative atomic mass—the weighted mean of the relative masses of the isotopes

of an element on the 12C scale. b Relative molecular mass—the relative mass of a molecule on the 12C scale. c Mole—the amount of substance which contains the same number of specified

particles as there are atoms in exactly 12 g of 12C. d Avogadro’s number—the number of carbon atoms in exactly 12 g of 12C

(approximately 6.02 × 1023). e Molar mass—mass in grams of a mole of a substance. f Precipitate—solid which is formed during a chemical reaction that occurs in

solution. g Gravimetric analysis—analysis of a sample using measurement of mass. One step

in the procedure usually involves the formation of a precipitate. h Ionic equation—an equation for a reaction in which the ions that remain in

solution during the reaction (spectator ions) are omitted.



Q31. 16.0 g of hydrogen sulfide is mixed with 20.0 g of sulfur dioxide and they react according to the equation:

2H2S(g) + SO2(g) → 2H2O(l) + 3S(s) a What mass of sulfur is produced? b What mass of reactant is left after the reaction?

A31. a Step 1 Write a balanced equation. 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)



)St(Hcoefficien

)SH(

2

2n = 2

molg0.34g/0.16 1−

= 0.23529 mol

)t(SOcoefficien

)SO(

2

2n = 1

molg0.64g/0.20 1−

= 0.3125 mol Hence H2S is the limiting reactant.

Step 3 From the equation, 3 mol of S is produced by 2 mol of H2S.

)S(H

)S(

2nn =

23

n(S) = 2

)SH(3 2n×

= 2

)mol g g/34.00.16(3 1−×

= 0.70588 mol

Step 4 Calculate the mass. m(S) = 0.70588 mol × 32.0 g mol–1 = 22.588 g

Step 5 Express the answer with the correct number of significant figures. m(S) = 22.6 g



b Step 1 As SO2 was in excess, calculate the amount of SO2 that reacted with all the H2S, using the mol ratio from the equation.

)SH()SO(

2

2

nn =

21

n(SO2)reacted = 2

molg0.34g/0.16 1−

= 0.23529 mol

Step 2 Calculate the amount of SO2 added initially.

n(SO2)initially = 1molg0.64g0.20

−

= 0.3125 mol

Step 3 Calculate the amount of SO2 in excess by subtracting the amount of SO2 that reacted from the amount of SO2 added initially.

n(SO2)excess = 0.3125 mol – 0.23529 mol = 0.07721 mol

Step 4 Calculate the mass of SO2 that was in excess. m(SO2) = 0.07721 g × 64.0 g mol–1 = 4.9412 g

Step 5 Express the answer with the correct number of significant figures. m(SO2) = 4.94 g



Q32. Calculate the volume of carbon dioxide gas produced, at SLC, when 5.00 g of calcium carbonate is added to a solution containing 5.00 g of nitric acid.

A32.

Step 1 Write a balanced equation. CaCO3(s) + 2HNO3(aq) → CO2(g) + H2O(l) + Ca(NO3)2(aq)



( )( )3

3

CaCOtcoefficienCaCOn =

1mol g 100.08

g 5.001−

= 0.04996 mol

( )( )3

3

HNOtcoefficienHNOn =

2mol g 63

g 5.001−

= 0.03968 mol Hence HNO3 is the limiting reactant.

Step 3 From the equation, 1 mol of CO2 is produced by 2 mol of HNO3.

( )( )3

2

HNOCO

nn =

21

n(CO2) = ( )2

mol g 63g 5.001 1−×

= 0.03968 mol

Step 4 Calculate the volume of CO2 at SLC when the molar volume is 24.5 L mol–1.

n(CO2) = ( )1

2

mol L 24.5CO

−

V

V(CO2) = n(CO2) × 24.5 L mol–1 = 0.03968 mol × 24.5 L mol–1 = 0.9722 L

Step 5 Express the answer with the correct number of significant figures. V(CO2) = 0.972 L



Q33. The following compounds are used in fertilisers as a source of nitrogen. Calculate the percentage of nitrogen, by mass, in: a ammonia (NH3) b ammonium nitrate (NH4NO3) c urea (CO(NH2)2).

A33. a Step 1 Calculate the molar mass of NH3. M(NH3) = 17.0 g mol–1

Step 2 From the formula, 1 mol of N is present in 1 mol of NH3. Calculate the mass of N in 1 mol (17.0 g) of NH3. m(N) = 14.0 g

Step 3 Calculate the % of nitrogen in NH3.

% N = 3)NH(

)N(M

m × 100

= 17.014.0

× 100

= 82.35%

Step 4 Express the answer with the correct number of significant figures. % N = 82.4%

b Step 1 Calculate the mass of 1 mol of NH4NO3. M(NH4NO3) = 80.2 g mol–1

Step 2 From the formula, 2 mol of N are present in 1 mol of NH4NO3. Calculate the mass of N in 1 mol (80.2 g) of NH4NO3. m(N) = 2 × 14.0 g = 28.0 g

Step 3 Calculate the % of nitrogen in NH4NO3.

% N = 80.228.0 × 100

= 34.91%

Step 4 Express the answer with the correct number of significant figures. % N = 35.0%



c Step 1 Calculate the molar mass of CO(NH2)2. M(CO(NH2)2) = 60.0 g mol–1

Step 2 From the formula, 2 mol of N are present in 1 mol of CO(NH2)2. Calculate the mass of nitrogen in 1 mol (60.0 g) of CO(NH2)2. m(N) = 2 × 14.0 g = 28.0 g

Step 3 Calculate the % of nitrogen in CO(NH2)2.

% N = 60.028.0 × 100

= 46.67 %

Step 4 Express the answer with the correct number of significant figures. % N = 46.7 %

Q34. Find the empirical formula of: a a compound that contains 65.2% scandium and 34.8% oxygen by mass b an oxide of copper that contains 89% copper by mass c a polymer used to make plastic drain pipes, which contains 38.4% carbon, 4.84%

hydrogen and 56.7% chlorine by mass.

A34. a Step 1 Write the ratio by mass. Sc : O 65.2 g : 34.8 g


(Sc)

g 65.2M

: (O)

g 34.8M

1molg956.44g2.65

− : 1molg0.16g8.34

−

1.450 mol : 2.175 mol


1.4501.450 :

1.4502.175

1 : 1.5

Step 5 Express as integers by multiplying by 2. 2 : 3 ∴ The empirical formula of the compound is Sc2O3.



b Step 1 As this is an oxide of copper, calculate the percentage oxygen in the compound.

% O = (100 – 89)% = 11%

Step 2 Write the ratio by mass. Cu : O 89 g : 11 g


(Cu)

g 89M

: (O)g 11

M

1molg54.63g89

− : 1molg0.16g11

−

1.401 mol : 0.6875 mol


0.68751.401 :

0.68750.6875

2.038 : 1

Step 5 Round off to whole numbers. 2 : 1 ∴ The empirical formula of the compound is Cu2O.

c Step 1 Write the ratio by mass. C : H : Cl 38.4 g : 4.84 g : 56.7 g


(Cu)

g 38.4M

: (H)

g 4.84M

: (Cl)

g 56.7M

1molg0.12g4.38

− : 1molg00.1g84.4

− : 1molg5.35g7.56

−

3.2 mol : 4.84 mol : 1.597 mol


597.1

g 3.2 : 597.1

4.84 : 597.1597.1

2.00 : 3.04 : 1

Step 4 Round off to whole numbers. 2 : 3 : 1 ∴ The empirical formula of the compound is C2H3Cl.



Q35. Gypsum is hydrated calcium sulfate (CaSO4•xH2O). A residue of 5.65 g of anhydrous calcium sulfate is obtained by heating 7.15 g of gypsum. Determine the empirical formula of gypsum.

A35. Step 1 Calculate the mass of water present in the sample. m(H2O) = m(CaSO4•xH2O) – m(CaSO4) = 7.15 g – 5.65 g = 1.50 g

Step 2 Write the ratio by mass. CaSO4 : H2O 5.65 g : 1.50 g


1molg08.136g65.5

− : 1molg0.18g50.1

−

0.0415 mol : 0.08333 mol


0415.0

0.0415 : 0415.0

0.08333

1 : 2.008

Step 5 Round off to whole numbers. 1 : 2 ∴ The empirical formula of the compound is CaSO4.2H2O.



Q36. A 2.203 g sample of an organic compound, CxHyOz, was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidised to 3.23 g of carbon dioxide. a Find the empirical formula of the compound. b Another sample was analysed in a mass spectrometer. The mass spectrum

produced showed that the molar mass of the compound was 60.0 g mol–1. What is its molecular formula?

A36.

a n(H) = 2n(H2O) = 1832.1 = 0.15 mol

m(H) = 0.15 × 1 = 0.15 g

n(C) = n(CO2) = 4423.3 = 0.073 mol

m(C) = 0.073 × 12 = 0.88g m(O) = 2.203 – m(C) – m(H) = 2.203 – 0.88 – 0.15 = 1.173g

C : H : O Mass 0.88 : 0.15 : 1.17

No mole 1288.0 :

115.0 :

1617.1

= 0.073 : 0.15 : 0.073 = 1 : 2 : 1

Empirical formula is CH2O

b mass of empirical formula = 12+1 ×2 + 16 = 30 molecular mass is 60, hence molecular formula must be double empirical formula, i.e. C2H4O2

Q37. What mass of barium chloride (BaCl2) will remain after a 15.0 g sample of the hydrated salt BaCl2•2H2O is heated to drive off all of the water?

A37. Step 1 Find the percentage of BaCl2 in the sample.

% BaCl2 = O)H2BaCl(

)(BaCl

22

2

⋅MM × 100

= 34.24434.208 × 100

= 85.266%

Step 2 Calculate the mass of the sample which is BaCl2. m(BaCl2) = 85.266% of 15.0 g = 12.79 g

Step 3 Express the answer with correct number of significant figures. m(BaCl2) = 12.8 g



Q38. A student is given solutions of lead(II) nitrate, copper(II) chloride and barium hydroxide. a Using Table 2.5 on page 23 of the student book, name the precipitates that could

be made by mixing together pairs of solutions. b Write full and ionic equations for each of the reactions.

A38. a lead(II) chloride, lead(II) hydroxide, copper(II) hydroxide b Pb(NO3)2(aq) + CuCl2(aq) → PbCl2(s) + Cu(NO3)2(aq)

Pb2+(aq) + 2Cl–(aq) → PbCl2(s) Pb(NO3)2(aq) + Ba(OH)2(aq) → Pb(OH)2(s) + Ba(NO3)2(aq) Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s) CuCl2(aq) + Ba(OH)2(aq) → Cu(OH)2(s) + BaCl2(aq) Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

Q39.

Design a flowchart to show how the salt content of a savoury spread could be determined by gravimetric analysis.

A39. 1 Weigh a sample of the savoury spread. ↓ 2 Mix the sample with water to dissolve Cl– ions. ↓ 3 Filter the mixture. ↓ 4 Add excess silver nitrate solution to precipitate silver chloride. ↓ 5 Filter the precipitate and wash with water. ↓ 6 Dry the precipitate in an oven at 110°C. ↓ 7 Weigh the precipitate. ↓ 8 Repeat steps 6 and 7 until a constant mass is obtained.



Q40. The iodide ions in a solution containing 0.300 g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed?

A40.

Step 1 Write a balanced equation. Ag+(aq) + I–(aq) → AgI(s)

Step 2 Calculate amount of NaI.

n(NaI) = 1molg89.149g300.0

−

= 0.002001 mol

Step 3 Use the ratio of amounts of substances to calculate the amount of AgBr produced.

From the equation, 1 mol AgI is produced by 1 mol NaI.

(NaI)(AgI)

nn =

11

n(AgI) = 0.002001 mol

Step 4 Calculate the mass. m(AgI) = 0.002001 g × 234.77 g mol–1 = 0.4698 g

Step 5 Express the answer with the correct number of significant figures. m(AgI) = 0.470 g

Q41. A precipitate of Fe2O3, of mass 1.43 g, was obtained by treating a 1.5 L sample of bore water. What was the concentration of iron, in mol L–1, in the water?

A41. Step 1 Calculate amount of Fe2O3.

n(Fe2O3) = 1molg7.159g43.1

−

= 0.008954 mol

Step 2 From the formula, 2 mol Fe are obtained from 1 mol Fe2O3. n(Fe) = 2 × 0.008954 = 0.017908 mol

Step 3 Calculate the concentration in mol L–1.

c(Fe) = L5.1

mol 017908.0

= 0.01193 mol L–1

Step 4 Express the answer with the correct number of significant figures. c(Fe) = 0.012 mol L–1



Q42. The chlorine in a 0.63 g sample of a chlorinated pesticide, DDT (C14H9Cl5), is precipitated as silver chloride. What mass of silver chloride is formed?

A42.

Step 1 Write an appropriately balanced equation. C14H9Cl5(aq) + 5Ag+(aq) → 5AgCl(s) + other products

Step 2 Calculate the amount of DDT (C14H9Cl5).

n(C14H9Cl5) = 1molg462.354g63.0

−

= 0.001777 mol

Step 3 From the equation, 5 mol of AgCl is precipitated from 1 mol of C14H9Cl5. n(AgCl) = 5 × 0.001777 mol = 0.008886 mol

Step 4 Calculate the mass of AgCl. m(AgCl) = 0.008886 × 143.32 = 1.2698 g

Step 5 Express the answer with the correct number of significant figures. m(AgCl) = 1.3 g



Q43. A 0.693 g sample of a silver alloy used to make cutlery is dissolved completely in nitric acid. Excess sodium chloride solution is added to produce a precipitate of silver chloride. The precipitate is filtered, dried and found to weigh 0.169 g. a Find the percentage of silver in the alloy. b If the precipitate was not completely dry when weighed, what effect would this

have on the answer for part a?

A43. a Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) → AgCl(s)

Step 2 Calculate amount of AgCl precipitated.

n(AgCl) = 1molg32.143g169.0

−

= 0.001179 mol

Step 3 From the equation, 1 mol of Ag+ from the alloy is precipitated as 1 mol of AgCl.

(AgCl)

)(Agnn +

= 11

n(Ag+) = 0.001179 mol

Step 4 Calculate the mass of silver in the alloy. m(Ag) = 0.001179 mol × 107.87 g mol–1 = 0.1272 g

Step 5 Convert to percentage.

% Ag = 693.0

1272.0 × 100

= 18.355%

Step 6 Express the answer with the correct number of significant figures. % Ag = 18.4%

b The result would be too high.



Q44. A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride was added to precipitate barium sulfate. What was the total mass of barium sulfate produced?

A44.

The Na2SO4 and Al2(SO4)3 react independently with the BaCl2. Treat them as separate reactions and write two balanced equations. Add the masses of BaSO4 from each to find the total mass. Step 1 For the Na2SO4 solution, write a balanced equation. BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

Step 2 Calculate amount of Na2SO4.

n(Na2SO4) = 1molg04.142g500.0

−

= 0.003520 mol

Step 3 From the equation, 1 mol of BaSO4 is precipitated from 1 mol of Na2SO4.

)SO(Na

)(BaSO

42

4

nn =

11

n(BaSO4) = 0.003520 mol

Step 4 Calculate mass of BaSO4 from this reaction with Na2SO4. m(BaSO4) = 0.003520 mol × 233.4 g mol–1 = 0.821568 g

Step 5 For the Al2(SO4)3 solution, write a balanced equation. 3BaCl2(aq) + Al2(SO4)3(aq) → 3BaSO4(s) + 2AlCl3(aq)

Step 6 Calculate amount of Al2(SO4)3.

n(Al2(SO4)3) = 1molg14.342g500.0

−

= 0.001461 mol

Step 7 From the equation, 3 mol of BaSO4 is precipitated from 1 mol of Al2(SO4)3.

))(SO(Al

)(BaSO

342

4

nn =

13

n(BaSO4) = 3 × 0.001461 mol = 0.004383 mol

Step 8 Calculate mass of BaSO4 from this reaction with Al2(SO4)3. m(BaSO4) = 0.004383 mol × 233.4 g mol–1 = 1.0229 g

Step 9 Calculate the total mass of BaSO4 precipitated from both reactions. m(BaSO4) = (0.821568 + 1.0229) g = 1.84456 g

Step 10 Express the answer with the correct number of significant figures. m(BaSO4) = 1.85 g



Q45. Water pollution can result from the phosphates added to washing powders to improve the stability of their suds. The phosphorus in a 2.0 g sample of washing powder is precipitated as Mg2P2O7. The precipitate weighs 0.085 g. a What is the percentage, by mass, of phosphorus in the washing powder? b Suppose you were in charge of an advertising campaign to promote the washing

powder. Would you advertise the percentage of phosphorus or phosphate in the product? Explain.

A45. a Step 1 Find the percentage of P in Mg2P2O7.

% P = 56.22297.302× × 100

= 27.83%

Step 2 Using the % P in Mg2P2O7, calculate the mass of P in precipitate. m(P) = 27.83% of 0.085 g

= 100

83.27 × 0.085 g

= 0.02366 g

Step 3 Calculate the percentage P in 2.0 g of washing powder.

% P = g2.0

g0.02366 × 100

= 1.183%

Step 4 Express the answer with the correct number of significant figures. % P = 1.2% b You would need to consider the fact that the percentage of phosphate in the

washing powder is greater than the percentage of the phosphorus.



Q46. A 2.10 g sample of a commercial antacid powder is treated with excess hydrochloric acid. The volume of carbon dioxide evolved is 430 mL, measured at 21.0°C and 109.6 kPa pressure. If magnesium carbonate is the active ingredient in the antacid, calculate the percentage of magnesium carbonate in the sample.

A46.

Step 1 Write a balanced equation. MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)

Step 2 Convert pressure, volume and temperature to appropriate units for use in the general gas equation.

P = 109.6 kPa V = 430 mL = 0.430 L T = 21°C = (21 + 273) K = 294 K

Step 3 Calculate the amount of CO2(g).

n(CO2) = RTpV

= 29431.8

430.06.109××

= 0.01928 mol

Step 4 From the equation, 1 mol MgCO3 produces 1 mol CO2.

)CO(

)MgCO(

2

3

nn

= 11

n(MgCO3) = 0.01928 mol

Step 5 Calculate the mass. m(MgCO3) = 0.01928 mol × 84.3 g mol–1 = 1.6261 g

Step 6 Calculate the % MgCO3.

% MgCO3 = )sample()MgCO( 3

mm

× 100

= 10.2

6261.1 × 100

= 77.435%

Step 7 Express the answer with the correct number of significant figures. % MgCO3 = 77.4%



Q47. When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of phosphorus is produced. The molar mass of the oxide is 284 g mol–1. a Determine the empirical formula of the phosphorus oxide. b Determine the molecular formula of the phosphorus oxide.

A47. a Step 1 Write the ratio by mass. P : O 0.100 g : (0.228 – 0.100) g 0.100 g : 0.128 g


1molg974.30g100.0

− : 1molg0.16g128.0

−

0.003229 mol : 0.008 mol


008.0

003229.0 : 008.0008.0

0.4036 : 1

Step 4 Express as integers by multiplying by 5. 2 : 5 ∴ The empirical formula of the compound is P2O5. b As the empirical formula is P2O5, the molecule must contain a whole number of

P2O5 units. The molar mass of one of these units is ((2 × 30.974) + (5 × 16)) =

141.948 g mol–1. The number of units in a molecule = molar mass of the compound/molar mass of

one unit

= 1

1

molg948.141mol g284

−

−

= 2 ∴ The molecular formula of the compound is P4O10.



Q48. Excessive salt intake in the diet can cause high blood pressure and heart disease. The salt content of a 14.96 g sample of powdered chicken soup was determined by dissolving it in water to make a volume of 250.0 mL. A 20.00 mL volume of this stock solution was pipetted into a conical flask and excess silver nitrate was added. The silver chloride precipitate that formed was then filtered, washed and dried. Its mass was 0.246 g. a Write an ionic equation for the formation of the silver chloride precipitate. b Calculate the amount, in mol, of silver chloride that was produced.

Assume all the chloride in the powdered soup came from sodium chloride (common salt).

c Determine the amount of sodium chloride in the 20.00 mL volume of stock solution in the conical flask.

d Calculate the amount of sodium chloride in 250.0 mL of the stock solution. e What mass of sodium chloride was in the sample?

A48.

a Ag+(aq) + Cl–(aq) → AgCl(s) b Step 1 Write a balanced equation. Ag+(aq) + Cl–(aq) → AgCl(s)

Step 2 Calculate amount of AgCl produced. Note: Round off to the appropriate number of significant figures for

this part of the answer, but keep all the digits running in your calculator for the next parts of the question. Do this for all parts.

n(AgCl) = 1molg32.143g246.0

−

= 0.001716 mol = 0.00172 mol

c From the equation for the reaction in the 20.00 mL of stock solution, 1 mol of Cl– ions produces 1 mol of AgCl.

(AgCl)

)(Clnn −

= 11

n(Cl–) in 20.00 mL = 0.001716 mol = 0.00172 mol

d Calculate the amount of salt in the 250.0 mL of stock solution.

n(Cl–) in 250.00 mL = 0.001716 × 20.00250.0 mol

= 0.021455 mol = 0.0215 mol

e Calculate the mass of salt in 250.0 mL. m(NaCl) in 250.00 mL = 0.021455 mol × 58.44 g mol–1 = 1.25386 g = 1.25 g

Worked solutions to student book questions Chapter 2...

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